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Christian7e ffirs.tex V1 - 08/16/2013 2:53 P.M. Page i
ANALYTICAL CHEMISTRY SEVENTH EDITION Gary D. Christian
University of Washington
Purnendu K. (Sandy) Dasgupta
University of Texas at Arlington
Kevin A. Schug
University of Texas at Arlington
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To Nikola from Gary—for your interests in science. You have a bright future,wherever your interests and talents take you Philip W. West from Sandy—wherever you are Phil, sipping your martini with 1 ppm vermouth, you know how it was: For he said, I will give you, A shelter from the storm. . . . Dad from Kevin—well its not hardcore P. Chem., but it is still quite useful. Thanks for your love, support, and guidance through the years
VP & Publisher: Editorial Assistant: Senior Marketing Manager: Designer: Associate Production Manager:
Petra Recter Ashley Gayle/Katherine Bull Kristine Ruff Kenji Ngieng Joyce Poh
This book was set in 10.5 Times Roman by Laserwords Private Limited and printed and bound by Courier Kendallville. The cover was printed by Courier Kendallville. This book is printed on acid free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2014, 2004 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201)748-6011, fax (201)748-6008, website http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return mailing label are available at www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. Library of Congress Cataloging-in-Publication Data Christian, Gary D., author. Analytical chemistry. -- Seventh edition / Gary D. Christian, University of Washington, Purnendu K. (Sandy) Dasgupta, University of Texas at Arlington, Kevin A. Schug, University of Texas at Arlington. pages cm Includes index. ISBN 978-0-470-88757-8 (hardback : alk. paper) 1. Chemistry, Analytic--Quantitative--Textbooks. I. Dasgupta, Purnendu, author. II. Schug, Kevin, author. III. Title. QD101.2.C57 2014 543--dc23 2013019926 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1
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Contents Chapter 1 Analytical Objectives, or: What Analytical Chemists Do
1
1.1 What Is Analytical Science?, 2 1.2 Qualitative and Quantitative Analysis: What Does Each Tell Us?, 3 1.3 Getting Started: The Analytical Process, 6 1.4 Validation of a Method—You Have to Prove It Works!, 15 1.5 Analyze Versus Determine—They Are Different, 16 1.6 Some Useful Websites, 16
Chapter 2 Basic Tools and Operations of Analytical Chemistry
20
2.1 The Laboratory Notebook—Your Critical Record, 20 2.2 Laboratory Materials and Reagents, 23 2.3 The Analytical Balance—The Indispensible Tool, 23 2.4 Volumetric Glassware—Also Indispensible, 30 2.5 Preparation of Standard Base Solutions, 42 2.6 Preparation of Standard Acid Solutions, 42 2.7 Other Apparatus—Handling and Treating Samples, 43 2.8 Igniting Precipitates—Gravimetric Analysis, 48 2.9 Obtaining the Sample—Is It Solid, Liquid, or Gas?, 49 2.10 Operations of Drying and Preparing a Solution of the Analyte, 51 2.11 Laboratory Safety, 57
Chapter 3 Statistics and Data Handling in Analytical Chemistry
62
3.1 Accuracy and Precision: There Is a Difference, 62 3.2 Determinate Errors—They Are Systematic, 63 3.3 Indeterminate Errors—They Are Random, 64 3.4 Significant Figures: How Many Numbers Do You Need?, 65 3.5 Rounding Off, 71 3.6 Ways of Expressing Accuracy, 71 3.7 Standard Deviation—The Most Important Statistic, 72 3.8 Propagation of Errors—Not Just Additive, 75 3.9 Significant Figures and Propagation of Error, 81 3.10 Control Charts, 83 3.11 The Confidence Limit—How Sure Are You?, 84 3.12 Tests of Significance—Is There a Difference?, 86 3.13 Rejection of a Result: The Q Test, 95 3.14 Statistics for Small Data Sets, 98 3.15 Linear Least Squares—How to Plot the Right Straight Line, 99 3.16 Correlation Coefficient and Coefficient of Determination, 104 3.17 Detection Limits—There Is No Such Thing as Zero, 105 3.18 Statistics of Sampling—How Many Samples, How Large?, 107 3.19 Powering a Study: Power Analysis, 110 3.20 Use of Spreadsheets in Analytical Chemistry, 112 3.21 Using Spreadsheets for Plotting Calibration Curves, 117 iii
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3.22 Slope, Intercept, and Coefficient of Determination, 118 3.23 LINEST for Additional Statistics, 119 3.24 Statistics Software Packages, 120
Chapter 4 Good Laboratory Practice: Quality Assurance and Method Validation 132 4.1 What Is Good Laboratory Practice?, 133 4.2 Validation of Analytical Methods, 134 4.3 Quality Assurance—Does the Method Still Work?, 143 4.4 Laboratory Accreditation, 144 4.5 Electronic Records and Electronic Signatures: 21 CFR, Part 11, 145 4.6 Some Official Organizations, 146
Chapter 5 Stoichiometric Calculations: The Workhorse of the Analyst
149
5.1 Review of the Fundamentals, 149 5.2 How Do We Express Concentrations of Solutions?, 152 5.3 Expressions of Analytical Results—So Many Ways, 159 5.4 Volumetric Analysis: How Do We Make Stoichiometric Calculations?, 166 5.5 Volumetric Calculations—Let’s Use Molarity, 169 5.6 Titer—How to Make Rapid Routine Calculations, 179 5.7 Weight Relationships—You Need These for Gravimetric Calculations, 180
Chapter 6 General Concepts of Chemical Equilibrium 6.1 Chemical Reactions: The Rate Concept, 188 6.2 Types of Equilibria, 190 6.3 Gibbs Free Energy and the Equilibrium Constant, 191 6.4 Le Chˆatelier’s Principle, 192
6.5 Temperature Effects on Equilibrium Constants, 192 6.6 Pressure Effects on Equilibria, 192 6.7 Concentration Effects on Equilibria, 193 6.8 Catalysts, 193 6.9 Completeness of Reactions, 193 6.10 Equilibrium Constants for Dissociating or Combining Species—Weak Electrolytes and Precipitates, 194 6.11 Calculations Using Equilibrium Constants—Composition at Equilibrium?, 195 6.12 The Common Ion Effect—Shifting the Equilibrium, 203 6.13 Systematic Approach to Equilibrium Calculations—How to Solve Any Equilibrium Problem, 204 6.14 Some Hints for Applying the Systematic Approach for Equilibrium Calculations, 208 6.15 Heterogeneous Equilibria—Solids Don’t Count, 211 6.16 Activity and Activity Coefficients— Concentration Is Not the Whole Story, 211 6.17 The Diverse Ion Effect: The Thermodynamic Equilibrium Constant and Activity Coefficients, 217
Chapter 7 Acid–Base Equilibria
188
222
7.1 The Early History of Acid—Base Concepts, 222 7.2 Acid–Base Theories—Not All Are Created Equal, 223 7.3 Acid–Base Equilibria in Water, 225 7.4 The pH Scale, 227 7.5 pH at Elevated Temperatures: Blood pH, 231 7.6 Weak Acids and Bases—What Is the pH?, 232 7.7 Salts of Weak Acids and Bases—They Aren’t Neutral, 234 7.8 Buffers—Keeping the pH Constant (or Nearly So), 238 7.9 Polyprotic Acids and Their Salts, 245 7.10 Ladder Diagrams, 247 7.11 Fractions of Dissociating Species at a Given pH: α Values—How Much of Each Species?, 248 7.12 Salts of Polyprotic Acids—Acid, Base, or Both?, 255
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7.13 Physiological Buffers—They Keep You Alive, 261 7.14 Buffers for Biological and Clinical Measurements, 263 7.15 Diverse Ion Effect on Acids and Bases: c Ka and c Kb —Salts Change the pH, 266 7.16 log C—pH Diagrams, 266 7.17 Exact pH Calculators, 269
Chapter 8 Acid–Base Titrations
9.5 Other Uses of Complexes, 336 9.6 Cumulative Formation Constants β and Concentrations of Specific Species in Stepwise Formed Complexes, 336
Chapter 10 Gravimetric Analysis and Precipitation Equilibria 281
8.1 Strong Acid versus Strong Base—The Easy Titrations, 282 8.2 The Charge Balance Method—An Excel Exercise for the Titration of a Strong Acid and a Strong Base, 285 8.3 Detection of the End Point: Indicators, 288 8.4 Standard Acid and Base Solutions, 290 8.5 Weak Acid versus Strong Base—A Bit Less Straightforward, 290 8.6 Weak Base versus Strong Acid, 295 8.7 Titration of Sodium Carbonate—A Diprotic Base, 296 8.8 Using a Spreadsheet to Perform the Sodium Carbonate—HCl Titration, 298 8.9 Titration of Polyprotic Acids, 300 8.10 Mixtures of Acids or Bases, 302 8.11 Equivalence Points from Derivatives of a Titration Curve, 304 8.12 Titration of Amino Acids—They Are Acids and Bases, 309 8.13 Kjeldahl Analysis: Protein Determination, 310 8.14 Titrations Without Measuring Volumes, 312
Chapter 9 Complexometric Reactions and Titrations
322
9.1 Complexes and Formation Constants—How Stable Are Complexes?, 322 9.2 Chelates: EDTA—The Ultimate Titrating Agent for Metals, 325 9.3 Metal–EDTA Titration Curves, 331 9.4 Detection of the End Point: Indicators—They Are Also Chelating Agents, 334
342
10.1 How to Perform a Successful Gravimetric Analysis, 343 10.2 Gravimetric Calculations—How Much Analyte Is There?, 349 10.3 Examples of Gravimetric Analysis, 353 10.4 Organic Precipitates, 353 10.5 Precipitation Equilibria: The Solubility Product, 355 10.6 Diverse Ion Effect on Solubility: Ksp and Activity Coefficients, 361
Chapter 11 Precipitation Reactions and Titrations
366
11.1 Effect of Acidity on Solubility of Precipitates: Conditional Solubility Product, 366 11.2 Mass Balance Approach for Multiple Equilibria, 368 11.3 Effect of Complexation on Solubility: Conditional Solubility Product, 372 11.4 Precipitation Titrations, 374
Chapter 12 Electrochemical Cells and Electrode Potentials 12.1 What Are Redox Reactions?, 384 12.2 Electrochemical Cells—What Electroanalytical Chemists Use, 384 12.3 Nernst Equation—Effects of Concentrations on Potentials, 390 12.4 Formal Potential—Use It for Defined Nonstandard Solution Conditions, 394 12.5 Limitations of Electrode Potentials, 395
383
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Chapter 13 Potentiometric Electrodes and Potentiometry
399
13.1 Metal Electrodes for Measuring the Metal Cation, 400 13.2 Metal–Metal Salt Electrodes for Measuring the Salt Anion, 401 13.3 Redox Electrodes—Inert Metals, 402 13.4 Voltaic Cells without Liquid Junction—For Maximum Accuracy, 404 13.5 Voltaic Cells with Liquid Junction—The Practical Kind, 405 13.6 Reference Electrodes: The Saturated Calomel Electrode, 407 13.7 Measurement of Potential, 409 13.8 Determination of Concentrations from Potential Measurements, 411 13.9 Residual Liquid-Junction Potential—It Should Be Minimized, 411 13.10 Accuracy of Direct Potentiometric Measurements—Voltage Error versus Activity Error, 412 13.11 Glass pH Electrode—Workhorse of Chemists, 413 13.12 Standard Buffers—Reference for pH Measurements, 418 13.13 Accuracy of pH Measurements, 420 13.14 Using the pH Meter—How Does It Work?, 421 13.15 pH Measurement of Blood—Temperature Is Important, 422 13.16 pH Measurements in Nonaqueous Solvents, 423 13.17 Ion-Selective Electrodes, 424 13.18 Chemical Analysis on Mars using Ion-Selective Electrodes, 432
Chapter 14 Redox and Potentiometric Titrations 14.1 First: Balance the Reduction–Oxidation Reaction, 437 14.2 Calculation of the Equilibrium Constant of a Reaction—Needed to Calculate Equivalence Point Potentials, 438 14.3 Calculating Redox Titration Curves, 441 14.4 Visual Detection of the End Point, 445 14.5 Titrations Involving Iodine: Iodimetry and Iodometry, 447
437
14.6 Titrations with Other Oxidizing Agents, 452 14.7 Titrations with Other Reducing Agents, 454 14.8 Preparing the Solution—Getting the Analyte in the Right Oxidation State before Titration, 454 14.9 Potentiometric Titrations (Indirect Potentiometry), 456
Chapter 15 Voltammetry and Electrochemical Sensors
466
15.1 Voltammetry, 467 15.2 Amperometric Electrodes—Measurement of Oxygen, 472 15.3 Electrochemical Sensors: Chemically Modified Electrodes, 472 15.4 Ultramicroelectrodes, 474 15.5 Microfabricated Electrochemical Sensors, 474 15.6 Micro and Ultramicroelectrode Arrays, 475
Chapter 16 Spectrochemical Methods
477
16.1 Interaction of Electromagnetic Radiation with Matter, 478 16.2 Electronic Spectra and Molecular Structure, 484 16.3 Infrared Absorption and Molecular Structure, 489 16.4 Near-Infrared Spectrometry for Nondestructive Testing, 491 16.5 Spectral Databases—Identifying Unknowns, 493 16.6 Solvents for Spectrometry, 493 16.7 Quantitative Calculations, 494 16.8 Spectrometric Instrumentation, 504 16.9 Types of Instruments, 519 16.10 Array Spectrometers—Getting the Entire Spectrum at Once, 522 16.11 Fourier Transform Infrared Spectrometers, 523 16.12 Near-IR Instruments, 525 16.13 Spectrometric Error in Measurements, 526 16.14 Deviation from Beer’s Law, 527 16.15 Fluorometry, 530 16.16 Chemiluminescence, 538 16.17 Fiber-Optic Sensors, 540
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Chapter 17 Atomic Spectrometric Methods
vii
548
17.1 Principles: Distribution between Ground and Excited States—Most Atoms Are in the Ground State, 550 17.2 Flame Emission Spectrometry, 553 17.3 Atomic Absorption Spectrometry, 556 17.4 Sample Preparation—Sometimes Minimal, 567 17.5 Internal Standard and Standard Addition Calibration, 567 17.6 Atomic Emission Spectrometry: The Induction Coupled Plasma (ICP), 569 17.7 Atomic Fluorescence Spectrometry, 574
Chapter 18 Sample Preparation: Solvent and Solid-Phase Extraction
20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9 20.10 20.11
579
Distribution Coefficient, 579 Distribution Ratio, 580 Percent Extracted, 581 Solvent Extraction of Metals, 583 Accelerated and Microwave-Assisted Extraction, 585 18.6 Solid-Phase Extraction, 586 18.7 Microextraction, 590 18.8 Solid-Phase Nanoextraction (SPNE), 593
19.1 Countercurrent Extraction: The Predecessor to Modern Liquid Chromatography, 598 19.2 Principles of Chromatographic Separations, 603 19.3 Classification of Chromatographic Techniques, 604 19.4 Theory of Column Efficiency in Chromatography, 607 19.5 Chromatography Simulation Software, 616
21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 21.10 21.11
596
649
High-Performance Liquid Chromatography, 651 Stationary Phases in HPLC, 654 Equipment for HPLC, 665 Ion Chromatography, 692 HPLC Method Development, 700 UHPLC and Fast LC, 701 Open Tubular Liquid Chromatography (OTLC), 702 Thin-Layer Chromatography, 702 Electrophoresis, 708 Capillary Electrophoresis, 711 Electrophoresis Related Techniques, 724
Chapter 22 Mass Spectrometry 22.1 22.2 22.3 22.4
619
Performing GC Separations, 620 Gas Chromatography Columns, 623 Gas Chromatography Detectors, 630 Temperature Selection, 638 Quantitative Measurements, 639 Headspace Analysis, 641 Thermal Desorption, 641 Purging and Trapping, 642 Small and Fast, 643 Separation of Chiral Compounds, 644 Two-Dimensional GC, 645
Chapter 21 Liquid Chromatography and Electrophoresis
18.1 18.2 18.3 18.4 18.5
Chapter 19 Chromatography: Principles and Theory
Chapter 20 Gas Chromatography
735
Principles of Mass Spectrometry, 735 Inlets and Ionization Sources, 740 Gas Chromatography–Mass Spectrometry, 741 Liquid Chromatography–Mass Spectrometry, 746 22.5 Laser Desorption/Ionization, 750 22.6 Secondary Ion Mass Spectrometry, 752 22.7 Inductively Coupled Plasma–Mass Spectrometry, 753
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22.8 Mass Analyzers and Detectors, 753 22.9 Hybrid Instruments and Tandem Mass Spectrometry, 764
Chapter 23 Kinetic Methods of Analysis
Available on textbook website: www.wiley.com/college/christian
Chapter G Century of the Gene—Genomics and Proteomics: DNA Sequencing and Protein Profiling G1 769
23.1 Kinetics—The Basics, 769 23.2 Catalysis, 771 23.3 Enzyme Catalysis, 772
Chapter 24 Automation in Measurements 24.1 24.2 24.3 24.4 24.5 24.6
784
Principles of Automation, 784 Automated Instruments: Process Control, 785 Automatic Instruments, 787 Flow Injection Analysis, 789 Sequential Injection Analysis, 791 Laboratory Information Management Systems, 792
G.7 G.8 G.9 G.10 G.11 G.12
Of What Are We Made?, G1 What Is DNA?, G3 Human Genome Project, G3 How Are Genes Sequenced?, G5 Replicating DNA: The Polymerase Chain Reaction, G6 Plasmids and Bacterial Artificial Chromosomes (BACs), G7 DNA Sequencing, G8 Whole Genome Shotgun Sequencing, G11 Single-Nucleotide Polymorphisms, G11 DNA Chips, G12 Draft Genome, G13 Genomes and Proteomics: The Rest of the Story, G13
APPENDIX A LITERATURE OF ANALYTICAL CHEMISTRY
Chapter 25 Clinical Chemistry
C1
25.1 Composition of Blood, C1 25.2 Collection and Preservation of Samples, C3 25.3 Clinical Analysis—Common Determinations, C4 25.4 Immunoassay, C6
EN1
Getting a Meaningful Sample, EN1 Air Sample Collection and Analysis, EN2 Water Sample Collection and Analysis, EN9 Soil and Sediment Sampling, EN11 Sample Preparation for Trace Organics, EN12 Contaminated Land Sites—What Needs to Be Analyzed?, EN12 26.7 EPA Methods and Performance-Based Analyses, EN13
794
APPENDIX B REVIEW OF MATHEMATICAL OPERATIONS: EXPONENTS, LOGARITHMS, AND THE QUADRATIC FORMULA 797 APPENDIX C TABLES OF CONSTANTS
Available on textbook website: www.wiley.com/college/christian
Chapter 26 Environmental Sampling and Analysis
G.1 G.2 G.3 G.4 G.5 G.6
Available on textbook website: www.wiley.com/college/christian
26.1 26.2 26.3 26.4 26.5 26.6
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Table C.1 Dissociation Constants for Acids, 801 Table C.2a Dissociation Constants for Basic Species, 802 Table C.2b Acid Dissociation Constants for Basic Species, 803 Table C.3 Solubility Product Constants, 803 Table C.4 Formation Constants for Some EDTA Metal Chelates, 805 Table C.5 Some Standard and Formal Reduction Electrode Potentials, 806
Available on textbook website: www.wiley.com/college/christian
APPENDIX D SAFETY IN THE LABORATORY
S1
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Available on textbook website: www.wiley.com/college/christian
APPENDIX E PERIODIC TABLES ON THE WEB
P1
Experiment 13 Determination of Chloride in a Soluble Chloride: Fajans’ Method, E23
Potentiometric Measurements
APPENDIX F ANSWERS TO PROBLEMS
808
Available on textbook website: www.wiley.com/college/christian
Experiments Use of Apparatus Experiment 1 Use of the Analytical Balance, E1 Experiment 2 Use of the Pipet and Buret and Statistical Analysis, E2 Experiment 3 Analysis of Volumetric Measurements Using Spectrophotometric Microplate Readers and Spreadsheet Calculations, E4
Gravimetry Experiment 4 Gravimetric Determination of Chloride, E6 Experiment 5 Gravimetric Determination of SO3 in a Soluble Sulfate, E9 Experiment 6 Gravimetric Determination of Nickel in a Nichrome Alloy, E11
Acid–Base Titrations Experiment 7 Determination of Replaceable Hydrogen in Acid by Titration with Sodium Hydroxide, E12 Experiment 8 Determination of Total Alkalinity of Soda Ash, E14 Experiment 9 Determination of Aspirin Using Back Titration, E16 Experiment 10 Determination of Hydrogen Carbonate in Blood Using Back-Titration, E18
Complexometric Titration Experiment 11 Determination of Water Hardness with EDTA, E19
Precipitation Titrations Experiment 12 Determination of Silver in an Alloy: Volhard’s Method, E21
E1
Experiment 14 Determination of the pH of Hair Shampoos, E24 Experiment 15 Potentiometric Determination of Fluoride in Drinking Water Using a Fluoride Ion-Selective Electrode, E25
Reduction–Oxidation Titrations Experiment 16 Analysis of an Iron Alloy or Ore by Titration with Potassium Dichromate, E27 Experiment 17 Analysis of Commercial Hypochlorite or Peroxide Solution by Iodometric Titration, E30 Experiment 18 Iodometric Determination of Copper, E32 Experiment 19 Determination of Antimony by Titration with Iodine, E34 Experiment 20 Microscale Quantitative Analysis of Hard-Water Samples Using an Indirect Potassium Permanganate Redox Titration, E36
Potentiometric Titrations Experiment 21 pH Titration of Unknown Soda Ash, E38 Experiment 22 Potentiometric Titration of a Mixture of Chloride and Iodide, E40
Spectrochemical Measurements Experiment 23 Spectrophotometric Determination of Iron, E41 Experiment 24 Spectrophotometric Determination of Iron in Vitamin Tablets Using a 96 Well Plate Reader, E43 Experiment 25 Determination of Nitrate Nitrogen in Water, E46 Experiment 26 Spectrophotometric Determination of Lead on Leaves Using Solvent Extraction, E47 Experiment 27 Spectrophotometric Determination of Inorganic Phosphorus in Serum, E48 Experiment 28 Spectrophotometric Determination of Manganese and Chromium in Mixture, E50
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Experiment 29 Spectrophotometric Determination of Manganese in Steel Using a 96 Well Plate Reader, E52 Experiment 30 Ultraviolet Spectrophotometric Determination of Aspirin, Phenacetin, and Caffeine in APC Tablets Using Solvent Extraction, E54 Experiment 31 Infrared Determination of a Mixture of Xylene Isomers, E56 Experiment 32 Fluorometric Determination of Riboflavin (Vitamin B2 ), E57
Atomic Spectrometry Measurements Experiment 33 Determination of Calcium by Atomic Absorption Spectrophotometry, E57 Experiment 34 Flame Emission Spectrometric Determination of Sodium, E60
Solid-Phase Extraction and Chromatography Experiment 35 Solid-Phase Extraction with Preconcentration, Elution, and Spectrophotometric Analysis, E61 Experiment 36 Thin-Layer Chromatography Separation of Amino Acids, E67 Experiment 37 Gas Chromatographic Analysis of a Tertiary Mixture, E69 Experiment 38 Qualitative and Quantitative Analysis of Fruit Juices for Vitamin C Using High-Performance Liquid Chromatography, E70
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Experiment 39 Analysis of Analgesics Using High-Performance Liquid Chromatography, E71
Mass Spectrometry Experiment 40 Capillary Gas Chromatography-Mass Spectrometry, E72
Kinetic Analysis Experiment 41 Enzymatic Determination of Glucose in Blood, E74
Flow Injection Analysis Experiment 42 Characterization of Physical Parameters of a Flow Injection Analysis System, E76 Experiment 43 Single-Line FIA: Spectrophotometric Determination of Chloride, E79 Experiment 44 Three-Line FIA: Spectrophotometric Determination of Phosphate, E80
Team Experiments Experiment 45 Method Validation and Quality Control Study, E82 Experiment 46 Proficiency Testing: Determination of z Values of Class Experiments, E84
Index
815
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Preface “Teachers open the door, but it is up to you to enter” —Anonymous
T
his edition has two new coauthors, Purnendu (Sandy) Dasgupta and Kevin Schug, both from the University of Texas at Arlington. So the authorship now spans three generations of analytical chemists who have each brought their considerable expertise in both teaching and research interests to this book. While all chapters have ultimately been revised and updated by all authors, the three authors have spearheaded different tasks. Among the most notable changes are the following: The addition of a dedicated chapter on mass spectrometry (Chapter 22) by Kevin. Sandy provided complete rewrites of the chapters on spectrochemical methods (Chapter 16) and atomic spectrometric methods (Chapter 17), and gas and liquid chromatography (Chapters 20 and 21), and added many new Excel problems and exercises. Gary compiled and organized all old and new supplementary materials for the textbook companion website and added QR codes for selected website materials, and he prepared the PowerPoint presentations of figures and tables. WHO SHOULD USE THIS TEXT? This text is designed for college students majoring in chemistry and in fields related to chemistry. It is written for an undergraduate quantitative analysis course. It necessarily contains more material than normally can be covered in a one-semester or one-quarter course, so that your instructor can select those topics deemed most important. Some of the remaining sections may serve as supplemental material. Depending on how a quantitative analysis and instrumental analysis sequence is designed, it may serve for both courses. In any event, we hope you will take time to read some sections that look interesting to you that are not formally covered. They can certainly serve as a reference for the future. WHAT IS ANALYTICAL CHEMISTRY? Analytical chemistry is concerned with the chemical characterization of matter, both qualitative and quantitative. It is important in nearly every aspect of our lives because chemicals make up everything we use. This text deals with the principles and techniques of quantitative analysis, that is, how to determine how much of a specific substance is contained in a sample. You will learn how to design an analytical method, based on what information is needed or requested (it is important to know what that is, and why!), how to obtain a laboratory sample that is representative of the whole, how to prepare it for analysis, what measurement tools are available, and the statistical significance of the analysis. xi
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PREFACE
xii
Analytical chemistry becomes meaningful when you realize that a blood analysis may provide information that saves a patient’s life, or that quality control analysis assures that a manufacturer does not lose money from a defective product. WHAT’S NEW TO THIS EDITION? This seventh edition is extensively rewritten, offering new and updated material. The goal was to provide the student with a foundation of the analytical process, tools, and computational methods and resources, and to illustrate with problems that bring realism to the practice and importance of analytical chemistry. We take advantage of digital technologies to provide supplementary material, including videos, website materials, spreadsheet calculations, and so forth (more on these below). We introduce the chapters with examples of representative uses of a technique, what its unique capabilities may be, and indicate what techniques may be preferred or limited in scope. The beginning of each chapter lists key learning objectives for the chapter, with page numbers for specific objectives. This will help students focus on the core concepts as they read the chapter. Here are some of the new things: ●
Professors Favorite Examples and Problems. We asked professors and practicing analytical chemists from around the world to suggest new analytical examples and problems, especially as they relate to real world practice, that we could include in this new edition. It is with appreciation and pleasure that we thank the many that have generously provided interesting and valuable examples and problems. We call these Professor’s Favorite Examples, and Professor’s Favorite Problems, and they are annotated within the text by a margin . We have included these in the text where appropriate and as element space allows, and have placed some on the text website. We hope you find these interesting and, as appropriate, are challenged by them. Our special thanks go to the following colleagues who have contributed problems, analytical examples, updates, and experiments:
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Christine Blaine, Carthage College
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Gary Hieftje, Indiana University
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Andre Campiglia, University of Central Florida
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Thomas Isenhour, Old Dominion University
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Peter Kissinger, Purdue University
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David Chen, University of British Columbia
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Samuel P. Kounaves, Tufts University
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Christa L. Colyer, Wake Forest University
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Ulrich Krull, University of Toronto
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Michael DeGranpre, University of Montana
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Thomas Leach, University of Washington
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Mary Kate Donais, Saint Anselm College
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Tarek Farhat, University of Memphis
Dong Soo Lee, Yonsei University, Seoul, Korea
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Carlos Garcia, The University of Texas at San Antonio
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Milton L. Lee, Brigham Young University
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Wen-Yee Lee, University of Texas at El Paso
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Steven Goates, BrighhamYoung University
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Shaorong Liu, University of Oklahoma
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Amanda Grannas, Villanova University
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Fred McLafferty, Cornell University
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Peter Griffiths, University of Idaho
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Christopher Harrison, San Diego State University
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Michael D. Morris, University of Michigan
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James Harynuk, University of Alberta
Noel Motta, University of Puerto Rico, R´ıo Piedras
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Fred Hawkridge, Virginia Commonwealth University
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Christopher Palmer, University of Montana
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Dimitris Pappas, Texas Tech University
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Aleeta Powe, University of Louisville
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Alberto Rojas-Hern´andez, Universidad Aut´onoma Metropolitana-Iztapalapa, Mexico
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Yi He, John Jay College of Criminal Justice, The City University of New York Charles Henry, Colorado State University
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Alexander Scheeline, University of Illinois
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Galina Talanova, Howard University
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W. Rudolph Seitz, University of New Hampshire
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Yijun Tang, University of Wisconsin, Oshkosh
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Paul S. Simone, Jr., University of Memphis
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Jon Thompson, Texas Tech University
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Nicholas Snow, Seton Hall University
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Kris Varazo, Francis Marion University
Wes Steiner, Eastern Washington University
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Akos Vertes, George Washington University
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Apryll M. Stalcup, City University of Dublin, Ireland
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Bin Wang, Marshall University
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George Wilson, University of Kansas
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Robert Synovec, University of Washington
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Richard Zare, Stanford University
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Mass spectrometry, especially when used as a hyphenated technique with chromatography, is increasingly a routine and powerful analytical tool, and a new chapter (Chapter 22) is dedicated to this topic. Likewise, liquid chromatography, including ion chromatography for anion determinations, is one of the most widely used techniques today, even surpassing gas chromatography. There are a wide variety of options of systems, instruments, columns, and detectors available, making selection of a suitable system or instrument a challenge for different applications. The present liquid chromatography chapter (Chapter 21) uniquely provides comprehensive coverage within the scope of an undergraduate text that not only gives the fundamentals of various techniques, how they evolved, and their operation, but also what the capabilities of different systems are and guidance for selecting a suitable system for a specific application. Revised chapters. All chapters have been revised, several extensively, especially those dealing with instrumentation to include recent technological innovations, as done for the liquid chromatography chapter. These include the spectrochemical chapter (16), the atomic spectrometric chapter (17), and the gas chromatography chapter (20). State-of-the-art technologies are covered. Some of this material and that of other chapters may be appropriate to use in an Instrumental Analysis course, as well as providing the basics for the quantitative analysis course; your instructor may assign selected sections for your course. Historical information is added throughout to put into perspective how the tools we have were developed and evolved. Some is this is included in margin pictures and notes, showing pioneers in development of our profession. Videos of Excel Programs. Major additions to the text and the text’s website supplemental material include powerful Excel programs to perform complicated calculations, and to create plots of titration curves, alpha vs. pH, logC vs. pH, etc. We have included video tutorials created by students of Professor Dasgupta to illustrate the use of many of these. The following videos, by chapter and in order of page appearance, with page numbers listed, are available on the text website. We have also created QR Codes for these in each chapter (see below) for those who want to access them on their smartphones. You will find these useful as you experiment with Excel and its power.
Chapter 3
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7. Error bars, 102
1. Solver, 87
8. Introduction to Excel, 113
2. Data Analysis Regression, 87, 120
9. Absolute Cell Reference, 115
3. F-test, 88
10. Average, 116
4. t-test for Paired Samples, 94
11. STDEV, 116
5. Paired t-test from Excel, 94
12. Intercept Slope and r-square, 119
6. Plotting in Excel, 102, 118
13. LINEST, 120
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Chapter 6 1. Goal Seek Equilibrium, 201 2. Goal Seek Problem 6.2, 219
Chapter 8 1. Excel H3 PO4 titration curve, 302 Chapter 9
Chapter 7 1. Goal Seek pH NH4 F, 238 2. Goal Seek mixture, 244
1. H4 Y alpha plot Excel 1, 328 2. H4 Y alpha plot Excel 2, 328 3. Example 9.6, 339
Thanks are due to the following students at the University of Texas as Arlington for their contributions: Barry Akhigbe, Jyoti Birjah, Rubi Gurung, Aisha Hegab, Akinde Kadjo, Karli Kirk, Heena Patel, Devika Shakya, and Mahesh Thakurathi. OTHER MODIFICATIONS TO EXISTING CONTENT It has been almost ten years since the last edition was published and since that time, much has changed! This seventh edition of Analytical Chemistry is extensively revised and updated, with new materials, new problems and examples, and new references. ●
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Spreadsheets. Detailed instructions are given on how to use and take advantage of spreadsheets in analytical calculations, plotting, and data processing. But the introductory material has been moved to the end of Chapter 3 as a separate unit, so that it can be assigned independently if desired, or treated as auxiliary material. The use of Excel Goal Seek and Excel Solver is introduced for solving complex problems and constructing titration curves (see below). Mastery of these powerful tools will allow students to tackle complex problems. Several useful programs introduced in the chapters are placed on the text website and instructions are given for applying these for plotting titration curves, derivative titrations, etc. by simply inputting equilibrium constant data, concentrations, and volumes. References. There are numerous recommended references given in each chapter, and we hope you will find them interesting reading. The late Tomas Hirschfeld said you should read the very old literature and the very new to know the field. We have deleted a number of outdated references, updating them with new ones. Many references are for classical, pioneering reports, forming the basis of current methodologies, and these remain. Material moved to the text website. As detailed elsewhere, we have moved certain parts to the text website as supplemental material and to make room for updating material on the techniques to be used. This includes: ● The single pan balance (Chapter 2) and normality calculations (Chapter 5), which may still be used, but in a limited capacity. ● The experiments. ● Auxiliary spreadsheet calculations from different chapters are posted on the website. ● Chapters dealing with specific applications of analytical chemistry are now on the text website for those interested in pursuing these topics. These are Clinical Chemistry (Chapter 25), and Environmental Sampling and Analysis (Chapter 26). ● Analytical chemistry played a key role in the completion of the historic Human Genome Project, and the Genomics and Proteomics chapter documents how. This material is not mainstream in the quantitative analysis course, so it has been moved to the website as Chapter G. It is available there for the interested student or for professor assignment.
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SPREADSHEETS Spreadsheets (using Excel) are introduced and used throughout the text for performing computations, statistical analysis, and graphing. Many titration curves are derived using spreadsheets, as are the calculations of α-values and plots of α-pH curves, and of logarithm concentration diagrams. The spreadsheet presentations are given in a “user-friendly” fashion to make it easier for you to follow how they are set up. We provide a list of the different types of spreadsheets that are used throughout the text, by topic, after the Table of Contents. GOAL SEEK We have introduced the use of Goal Seek, a powerful Excel tool, for solving complex problems. Goal Seek performs “trial and error” or successive approximation calculations to arrive at an answer. It is useful when one parameter needs to be varied in a calculation, as is the case for most equilibrium calculations. An introduction to Goal Seek is given in Section 6.11 in Chapter 6. Example applications are given on the text website, and we list these after the Table of Contents. SOLVER Excel Solver is an even more versatile tool. Goal Seek can only solve one parameter in a single equation, and does not allow for incorporating constraints in the parameter we want to solve. Solver, on the other hand, can solve for more than one parameter (or more than one equation) at a time. Example applications are given on the text website, with descriptions in the text. See the list after the Table of Contents. An introduction to its use is given in Example 7.21. REGRESSION FUNCTION IN EXCEL DATA ANALYSIS Possibly the most powerful tool to calculate all regression related parameters for a calibration plot is the “Regression” function in Data Analysis. It not only provides the results for r, r2 , intercept, and slope (which it lists as X variable 1), it also provides their standard errors and upper and lower limits at the 95% confidence level. It also provides an option for fitting the straight line through the origin (when you know for certain that the response at zero concentration is zero by checking a box “constant is zero”). A video illustrating its use is in the website of the book, Chapter 3, titled Data Analysis Regression. A description of how to use it is given in Chapter 16 at the end of Section 16.7, and example applications are given in Chapter 20, Section 20.5, and Chapter 23 for Examples 23.1 and Example 23.2. READY TO USE PROGRAMS As listed above, there are numerous supplemental materials on the text website, including Excel spreadsheets for different calculations. Many of these are for specific examples and are tutorial in nature. But several are suited to apply to different applications, simply by inputting data and not having to set up the calculation program. Examples include calculating titration curves and their derivatives, or for solving either quadratic or simultaneous equations. We list here a number that you should find useful. You can find them under the particular chapter on the website. Chapter 2 ●
Glassware calibration, Table 2.4
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Chapter 6 ● Calculate activity coefficients, equations 6.19 and 6.20 (Auxiliary data) ● Quadratic equation solution (Example 6.1) (See also Goal Seek for solving quadratic equations) Chapter 7 ● Stig Johannson pH calculator. For calculating pH of complex mixtures. Easy to use. ● CurtiPotpH calculator (Ivano Gutz) for calculating pH of complex mixtures, as well as constructing pH related curves. Learning curve higher, but very powerful. ● logC-pH Master Spreadsheet. See Section 7.16 on how to use it. Chapter 8 ● Derivative titrations—Easy method (Section 8.11) ● Universal Acid Titrator—Alex Scheeline—Easy method (Section 8.11). For polyprotic acid titration curves. ● Master Spreadsheet for titrations of weak bases—Easy method Chapter 10 ● Solving simultaneous equations (Example 10.5) Chapter 14 ● Derivative titration plots (for near the endpoint) Chapter 16 ● Calculation of unknown from calibration curve plot ● Standard deviation of sample concentration ● Two component Beer’s Law solution Chapter 17 ● Standard additions plot and unknown calculation Chapter 20 ● Internal standard calibration plot and unknown calculation (Section 20.5) EXPERIMENTS There are 46 experiments, grouped by topic, illustrating most of the measurement techniques presented in the text, and they can be downloaded from the text website. Each contains a description of the principles and chemical reactions involved, so the student gains an overview of what is being determined and how. Solutions and reagents to prepare in advance of the experiment are listed, so experiments can be performed efficiently. All experiments, particularly the volumetric ones, have been designed to minimize waste by preparing the minimum volumes of reagents, like titrants, required to complete the experiment. Two team experiments are included (Experiments 45 and 46) to illustrate the principles presented in Chapter 4 on statistical validation. One is on method validation and quality control, in which different members of teams perform different parts of the validation for a chosen experiment. The other is on proficiency testing in which students calculate the z-values for all the student results of one or more class experiments and each student compares their z-value to see how well they have performed.
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New experiments were contributed by users and colleagues. Included are three experiments from Professor Christopher Palmer, University of Montana using a spectrophotometric microplate reader (Experiments 3, 24, and 29). Experiment Video Resource. Professor Christopher Harrison from San Diego State University has a YouTube “Channel” of videos of different types of experiments, some illustrating laboratory and titration techniques: http://www.youtube.com/ user/crharrison. We would recommend that students be encouraged to look at the ones dealing with buret rinsing, pipetting, and aliquoting a sample, before they begin experiments. Also, they will find useful the examples of acid-base titrations illustrating methyl red or phenolphthalein indicator change at end points. There are a few specific experiments that may be related to ones from the textbook, for example, EDTA titration of calcium or Fajan’s titration of chloride. The video of glucose analysis gives a good illustration of the starch end point, which is used in iodometric titrations. SUPPLEMENTARY MATERIALS FOR THE INSTRUCTOR AND THE STUDENT WEBSITE URLs and QR CODES. There are some 200 website URLs, i.e., website addresses, given throughout the text for access to useful supplemental material. To efficiently access the websites, lists of all the URLs are posted on the text website for each chapter. These lists can be used to access the websites without typing the URLs. The lists of URLs for each chapter are also added as QR codes at the beginning of each chapter, facilitating access on smartphones. QR codes for selected ones are also given on the text pages where they appear (see below). We list in the QR code here all the chapter URL lists. QR codes are created for selected website materials in several chapters, as referred to in the chapter text. This will allow access to supplemental material using a smartphone, iPad, etc. So by accessing QR codes in a given chapter, one can browse for the videos and the selected URL links, alongside other valuable materials.
TEXT COMPANION WEBSITE John Wiley & Sons, Inc. maintains a companion website for your Analytical Chemistry textbook that contains additional valuable supplemental material. The website may be accessed at: www.wiley.com/college/christian Materials on the website include supplemental materials for different chapters that expand on abbreviated presentations in the text. Following is a list of the types of materials on the website: ● ● ●
Videos URLs Supplemental Material: WORD, PDFs, Excel, PowerPoint, JPEG
POWERPOINT SLIDES All figures and tables in the text are posted on the text website as PowerPoint slides for each chapter, with notes on each for the instructor, and can be downloaded for preparation of PowerPoint presentations. SOLUTIONS MANUAL A comprehensive saleable solutions manual is available for use by instructors and students in which all problems are completely worked out and all questions are
Complete URL list
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answered, a total of 824. More information on the solutions manual can be found at www.wiley.com, including where/how to purchase it. Answers for spreadsheet problems, which include the spreadsheets, are given on the text website. Answers to all problems are given in Appendix F. A WORD OF THANKS The production of your text involved the assistance and expertise of numerous people. Special thanks go to the users of the text who have contributed comments and suggestions for changes and improvements; these are always welcome. A number of colleagues served as reviewers of the text and manuscript and have aided immeasurably in providing specific suggestions for revision. They, naturally, express opposing views sometimes on a subject or placement of a chapter or section, but collectively have assured a near optimum outcome that we hope you find easy and enjoyable to read and study. First, Professors Louise Sowers, Stockton College; Gloria McGee, Xavier University; and Craig Taylor, Oakland University; and Lecturer Michelle Brooks, University of Maryland and Senior Lecturer Jill Robinson, Indiana University offered advice for revision and improvements of the 6th edition. Second, Professors Neil Barnett, Deakin University, Australia; Carlos Garcia, The University of Texas at San Antonio; Amanda Grannas, Villanova University; Gary Long, Virginia Tech; Alexander Scheeline, University of Illinois; and Mathew Wise, Condordia University, proofed the draft chapter manuscripts of this edition and offered further suggestions for enhancing the text. Dr. Ronald Majors, a leading chromatography expert from Agilent Technologies, offered advice on the liquid chromatography chapter. The professionals at John Wiley & Sons have been responsible for producing a high quality book. Petra Recter, Vice President, Publisher, Chemistry and Physics, Global Education, shepherded the whole process from beginning to end. Her Editorial Assistants Lauren Stauber, Ashley Gayle, and Katherine Bull were key in taking care of many details, with efficiency and accuracy. Joyce Poh was the production editor, arranging copyediting to printing, attending to many details, and assuring a quality final product. Laserwords Pvt Ltd was responsible for artwork in your text. We appreciate the efforts of Marketing Manager, Kristy Ruff, in making sure the text is available to all potential users. It has been a real pleasure for all of us working with this team of professionals and others in a long but rewarding process. We each owe special thanks to our families for their patience during our long hours of attention to this undertaking. Gary’s wife, Sue, his companion for over 50 years, has been through seven editions, and remains his strong supporter, even now. Purnendu owes his wife, Kajori, and his students, much for essentially taking off from all but the absolute essentials for the last three years. He also thanks Akinde Kadjo in particular for doing many of the drawings. Kevin’s wife, Dani, put up with yet another “interesting project” and lent her support in the form of keeping the kids at bay and making sure her husband was well fed while working on the text. GARY D. CHRISTIAN Seattle, Washington PURNENDU K. (SANDY) DASGUPTA KEVIN A. SCHUG Arlington, Texas September, 2013
“To teach is to learn twice.” —Joseph Joubert
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List of Spreadsheets Used Throughout the Text The use of spreadsheets for plotting curves and performing calculations is introduced in different chapters. Listed in the Preface are several that are ready to use for different applications. Following is a list of the various other applications of Microsoft Excel, by category, for easy reference for different uses. All spreadsheets are given in the text website. The Problem spreadsheets are only in the website; others are in the text but also in the website. You should always practice preparing assigned spreadsheets before referring to the website. You can save the spreadsheets in your website to your desktop for use. Use of Spreadsheets (Section 3.20) Filling the Cell Contents, 112 Saving the Spreadsheet, 113 Printing the Spreadsheet, 113 Relative vs. Absolute Cell References, 114 Use of Excel Statistical Functions (Paste functions), 115 Useful Syntaxes: LOG10; PRODUCT; POWER; SQRT; AVERAGE; MEDIAN; STDEV; VAR, 116 Statistics Calculations Standard Deviation: Chapter 3, Problems 14, 15, 16, 22, 24 Confidence Limit: Chapter 3, Problems 22, 24, 25, 29 Pooled Standard Deviation: Chapter 3, Problem 34 F-Test: Chapter 3, Problems 31, 33, 35 t-Test: Chapter 3, Problems 37, 38 t-Test, multiple samples: Chapter 3, Problem 53 Propagation of Error: Chapter 3, Problems 18 (add/subtract), 19 (multiply/divide) Using Spreadsheets for Plotting Calibration Curves Trendline; Least squares equation; R2 (Section 3.21, Figure 3.10)
Slope, Intercept and Coefficient of Determination (without a plot) (Section 3.22; Chapter 3, Problems 47, 51, 52) LINEST for Additional Statistics (Section 3.23, Figure 3.11) Ten functions: slope, std. devn., R2 , F, sum sq. regr., intercept, std. devn., std. error of estimate, d.f., sum sq. resid. Plotting α vs. pH Curves (Figure 7.2, H3 PO4 ), 251 . Plotting log C vs. pH Curves Chapter 7, Problem 66 (HOAc) Plotting log C vs. pH Curves Using Alpha Values (Section 7.16) Chapter 7, Problem 69 (Malic acid, H2 A) Chapter 7, Problem 73 (H3 PO4 , H3 A) Plotting Titration Curves HCl vs. NaOH (Figure 8.1), 283, 285 HCl vs. NaOH, Charge Balance (Section 8.2), 285 HOAc vs. NaOH (Section 8.5), 293 Hg2+ vs. EDTA: Chapter 9, Problem 24 SCN− and Cl− vs. AgNO3 : Chapter 11, Problem 12 Fe2+ vs. Ce4+ (Figure 14.1): Example 14.3 Derivative Titrations (Section 8.11), 305; Chapter 14, 458 . Plotting log K’ vs. pH (Figure 9.2): Chapter 9, Problem 23 . Plotting β-values vs. [ligand] (Ni(NH3 )6 2+ betavalues vs. [NH3 ]): Chapter 9, Problem 25 Spreadsheet Calculations/Plots Glassware Calibration (Table 2.4), 38 xix
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Weight in Vacuum Error vs. Sample Density (Chapter 2) Gravimetric Calculations Spreadsheet Examples-Grav. calcn. %Fe, 378 Chapter 10, Problem 40 (Example 10.2, %P2 O5 ) Solubility BaSO4 vs. [Ba2+ ] Plot (Figure 10.3): Chapter 10, Problem 41 Solubility vs. Ionic Strength Plot (Figure 10.4): Chapter 10, Problem 42 Van Deemter Plot: Chapter 19, Problem 13 EXCEL SOLVER FOR PROBLEM SOLVING This program can be used to solve several parameters or equations at a time. An introduction is given in Example 7.21. Chapter 3 video Solver (solving quadratic equation, Example 6.1) Example 7.21 Solver pH calculations of multiple solutions (H3 PO4 , NaH2 PO4 , Na2 HPO4 , Na3 PO4 ); 258 Example 7.24 Solver calculation (buffer composition), 264 Solubility from Ksp : Chapter 10, Problem 43 (Example 10.9) GOAL SEEK FOR PROBLEM SOLVING The spreadsheets listed below are on the text website for the particular chapter. The page numbers refer to corresponding discussions on setting up the programs. See Section 6.11 for introduction to and application of Goal Seek. It can be used to solve one parameter in an equation, as in most equilibrium problems. Excel Goal Seek for Trial and Error Problem Solving (Section 6.11): Equilibrium problem—introduction to Goal Seek, 197; Practice Goal Seek—setup, answer Goal Seek to Solve an Equation (Example 6.1—quadratic equation), 199 Solving a quadratic equation by Goal Seek—setup Goal Seek answer quadratic equation Chapter 6 video Goal Seek Equilibrium, 201 Goal Seek shortcomings (how to get around them)—setup (Example 6.4); 202 Goal Seek answer Example 6.4 Solving Example 6.13 Using Goal Seek (charge balance); 210 Chapter 6 video Goal Seek Problem 6.2 Goal Seek answer Problem 26 (quadratic equation), Chapter 6
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LIST OF SPREADSHEETS USED THROUGHOUT THE TEXT
Example 7.7 Goal Seek solution (pH HOAc) Example 7.8 Goal Seek solution (pH NH3 ) Example 7.9 Goal Seek solution (pH NaOAc) Example 7.10 Goal Seek solution (pH NH4 Cl) Chapter 7 video Goal Seek pH NH4 F, 238 Chapter 7 video Goal Seek mixture (NaOH + H2 CO3 ), 244 Example 7.19 Charge balance and Goal Seek to calc H3 PO4 pH (See the example for details of setting up the spreadsheet) Example 7.19b Goal Seek solution (pH H3 PO4 + NaOAc + K2 HPO4 ) (See Example 7.19 discussion for spreadsheet setup) 77PFP Goal Seek calculations—there are three tabs (Chapter 7, Problem 77). See 77PFP solution on the website for a detailed description of the problem solution and appropriate equations. Example 9.6—Goal Seek (complexation equilibria); (Section 9.6), 339 (See the example for the equation setup) Example 11.1 Goal Seek (solubility of CaC2 O4 in 0.001M HCl) Example 11.2 Goal Seek (charge balance, solubility of MA in 0.1M HCl) Example 11.5 Goal Seek (solubility of MX in presence of complexing ligand L)
REGRESSION FUNCTION IN EXCEL DATA ANALYSIS This Excel tool calculates all regression related parameters for a calibration plot. It provides the results for r, r2 , intercept, and slope, and also provides their standard errors and upper and lower limits at the 95% confidence level. Chapter 3 video Data Analysis Regression; 87, 120 Chapter 16, end of Section 16.7, Excel Exercise. Describes the use of the Excel Regression function in Data Analysis to readily calculate a calibration curve and its uncertainty, and then apply this to calculate an unknown concentration and its uncertainty from its absorbance; 502 Section 20.5, GC internal standard determination, 640 Chapter 20, Problem 11. GC internal standard determination Example 23.1, Lineweaver-Burk Km determination Example 23.2, Calculating unknown concentration from reaction rate Problem 23.17, Lineweaver-Burk Km determination
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About the Authors Gary Christian grew up Oregon, and has had a lifelong interest in teaching and research, inspired by great teachers throughout his education. He received his B.S. degree from the University of Oregon and Ph. D. degree from the University of Maryland. He began his career at Walter Reed Army Institute of Research, where he developed an interest in clinical and bioanalytical chemistry. He joined the University of Kentucky in 1967, and in 1972 moved to the University of Washington, where he is Emeritus Professor, and Divisional Dean of Sciences Emeritus. Gary wrote the first edition of this text in 1971. He is pleased that Professors Dasgupta and Schug have joined him in this new edition. They bring expertise and experience that markedly enhance and update the book in many ways. Gary is the recipient of numerous national and international awards in recognition of his teaching and research activities, including the American Chemical Society (ACS) Division of Analytical Chemistry Award for Excellence in Teaching and the ACS Fisher Award in Analytical Chemistry, and received an Honorary Doctorate Degree from Chiang Mai University. The University of Maryland inducted him into their distinguished alumni Circle of Discovery. He has authored five other books, including Instrumental Analysis, and over 300 research papers, and has been Editor-in-Chief of the international analytical chemistry journal, Talanta, since 1989. Purnendu K. (Sandy) Dasgupta is a native of India and was educated in a college founded by Irish missionaries and graduated with honors in Chemistry in 1968. After a MSc in Inorganic Chemistry in 1970 from the University of Burdwan and a brief stint as a researcher at the Indian Association for the Cultivation of Science (where Raman made his celebrated discovery), he came as a graduate student to Louisiana State University at Baton Rouge in 1973. Sandy received his PhD in Analytical Chemistry with a minor in Electrical Engineering from LSU in 1977 and managed to get a diploma as a TV mechanic while a graduate student. He joined the California Primate Research Center at the University of California at Davis as an Aerosol research Chemist in 1979 to be part of a research team studying inhalation toxicology of air pollutants. In his mother tongue, Bengali, he was once a well-published poet and a fledgling novelist but seemingly finally found his love of analytical chemistry as salvation. He joined Texas Tech in 1981 and was designated a Horn Professor in 1992, named after the first president of the University, the youngest person to be so honored at the time. He remained at Texas Tech for 25 years, joining the University of Texas at Arlington in 2007 as the Department Chair. He has stepped down as Chair, and currently holds the Jenkins Garrett Professorship. Sandy has written more than 400 papers/book chapters, and holds 23 US patents, many of which have been commercialized. His work has been recognized by the Dow Chemical Traylor Creativity Award, the Ion Chromatography Symposium xxi
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Outstanding Achievement Award (twice), the Benedetti-Pichler Memorial Award in Microchemistry, American Chemical Society Award in Chromatography, Dal Nogare Award in the Separation Sciences, Honor Proclamation of the State of Texas Senate and so on. He is the one of the Editors of Analytica Chimica Acta, a major international journal in analytical chemistry. He is best known for his work in atmospheric measurements, ion chromatography, the environmental occurrence of perchlorate and its effect on iodine nutrition, and complete instrumentation systems. He is a big champion of the role of spreadsheet programs in teaching analytical chemistry. Kevin Schug was born and raised in Blacksburg, Virginia. The son of a physical chemistry Professor at Virginia Tech, he grew up running around the halls of a chemistry building and looking over his father’s shoulder at chemistry texts. He pursued and received his B.S. degree in Chemistry from the College of William & Mary in 1998, and his Ph.D. degree in Chemistry under the direction of Professor Harold McNair at Virginia Tech in 2002. Following two years as a post-doctoral fellow with Professor Wolfgang Lindner at the University of Vienna (Austria), he joined the faculty in the Department of Chemistry & Biochemistry at The University of Texas at Arlington in 2005, where he is currently the Shimadzu Distinguished Professor of Analytical Chemistry. The research in Kevin’s group spans fundamental and applied aspects of sample preparation, separation science, and mass spectrometry. He also manages a second group, which focuses their efforts on chemical education research. He has been the recipient of several awards, including the Eli Lilly ACACC Young Investigator in Analytical Chemistry award, the LCGC Emerging Leader in Separation Science award, and the American Chemical Society Division of Analytical Chemistry Award for Young Investigators in Separation Science. At present, he has authored or coauthored 65 scientific peer-reviewed manuscripts. Kevin is a member of the Editorial Advisory Boards for Analytica Chimica Acta and LCGC Magazine, and is a regular contributor to LCGC on-line articles. He is also Associate Editor of the Journal of Separation Science.
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Chapter One ANALYTICAL OBJECTIVES, OR: WHAT ANALYTICAL CHEMISTS DO “Unless our knowledge is measured and expressed in numbers, it does not amount to much.” —Lord Kelvin
Chapter 1 URLs
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Analytical science deals with the chemical characterization of matter—what, how much?, p. 2
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You must select the appropriate method for measurement, p. 12
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The analyst must know what information is really needed, and obtain a representative sample, pp. 6, 9
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Validation is important, p. 15
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Few measurements are specific, so operations are performed to achieve high selectivity, p. 11
There are many useful websites dealing with analytical chemistry, p. 16
Analytical chemistry is concerned with the chemical characterization of matter and the answer to two important questions: what is it (qualitative analysis) and how much is it (quantitative analysis). Chemicals make up everything we use or consume, and knowledge of the chemical composition of many substances is important in our daily lives. Analytical chemistry plays an important role in nearly all aspects of chemistry, for example, agricultural, clinical, environmental, forensic, manufacturing, metallurgical, and pharmaceutical chemistry. The nitrogen content of a fertilizer determines its value. Foods must be analyzed for contaminants (e.g., pesticide residues) and for essential nutrients (e.g., vitamin content). The air we breathe must be analyzed for toxic gases (e.g., carbon monoxide). Blood glucose must be monitored in diabetics (and, in fact, most diseases are diagnosed by chemical analysis). The presence of trace elements from gun powder on a perpetrator’s hand will prove a gun was fired by that hand. The quality of manufactured products often depends on proper chemical proportions, and measurement of the constituents is a necessary part of quality assurance. The carbon content of steel will influence its quality. The purity of drugs will influence their efficacy. In this text, we will describe the tools and techniques for performing these different types of analyses. There is much useful supplemental material on the text website, including Excel programs that you can use, and videos to illustrate their use. You should first read the Preface to learn what is available to you, and then take advantage of some of the tools.
Lord Kelvin (William Thomson, 1824–1907)
Everything is made of chemicals. Analytical chemists determine what and how much.
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CHAPTER 1 ANALYTICAL OBJECTIVES, OR: WHAT ANALYTICAL CHEMISTS DO
1.1 What Is Analytical Science? The above description of analytical chemistry provides an overview of the discipline of analytical chemistry. There have been various attempts to more specifically define the discipline. The late Charles N. Reilley said: “Analytical chemistry is what analytical chemists do” (Reference 2). The discipline has expanded beyond the bounds of just chemistry, and many have advocated using the name analytical science to describe the field. This term is used in a National Science Foundation report from workshops on “Curricular Developments in the Analytical Sciences.” Even this term falls short of recognition of the role of instrumentation development and application. One suggestion is that we use the term analytical science and technology (Reference 3). The Federation of European Chemical Societies held a contest in 1992 to define analytical chemistry, and the following suggestion by K. Cammann was selected [Fresenius’ J. Anal. Chem., 343 (1992) 812–813]. Analytical Chemistry provides the methods and tools needed for insight into our material world . . . for answering four basic questions about a material sample: ● ● ● ●
What? Where? How much? What arrangement, structure or form?
These cover qualitative, spatial, quantitative, and speciation aspects of analytical science. The Division of Analytical Chemistry of the American Chemical Society developed a definition of analytical chemistry, reproduced in part here: Analytical Chemistry seeks ever improved means of measuring the chemical composition of natural and artificial materials. The techniques of this science are used to identify the substances which may be present in a material and to determine the exact amounts of the identified substance. Analytical chemists serve the needs of many fields: ●
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In medicine, analytical chemistry is the basis for clinical laboratory tests which help physicians diagnose disease and chart progress in recovery. In industry, analytical chemistry provides the means of testing raw materials and for assuring the quality of finished products whose chemical composition is critical. Many household products, fuels, paints, pharmaceuticals, etc. are analyzed by the procedures developed by analytical chemists before being sold to the consumer. Environmental quality is often evaluated by testing for suspected contaminants using the techniques of analytical chemistry. The nutritional value of food is determined by chemical analysis for major components such as protein and carbohydrates and trace components such as vitamins and minerals. Indeed, even the calories in food are often calculated from its chemical analysis.
Analytical chemists also make important contributions to fields as diverse as forensics, archaeology, and space science.
An interesting article published by a leading analytical chemist, G. E. F. Lundell, from the National Bureau of Standards in 1935 entitled “The Analysis of Things As They Are”, describes why we do analyses and the analytical process (Industrial and Engineering Chemistry, Analytical Edition, 5(4) (1933) 221–225). The article is posted on the text website.
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1.2 QUALITATIVE AND QUANTITATIVE ANALYSIS: WHAT DOES EACH TELL US?
A brief overview of the importance of analytical chemistry in society, with examples that affect our lives, and the tools and capabilities, is given in the article, “What Analytical Chemists Do: A Personal Perspective,” by Gary Christian, Chiang Mai Journal of Science, 32(2) (2005) 81–92: http://it.science.cmu .ac.th/ejournal/journalDetail.php?journal_id=202 Reading this before beginning this course will help place in context what you are learning. A reprint of the article is posted on the text website.
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What Analytical Chemists Do
1.2 Qualitative and Quantitative Analysis: What Does Each Tell Us? The discipline of analytical chemistry consists of qualitative analysis and quantitative analysis. The former deals with the identification of elements, ions, or compounds present in a sample (we may be interested in whether only a given substance is present), while the latter deals with the determination of how much of one or more constituents is present. The sample may be solid, liquid, gas, or a mixture. The presence of gunpowder residue on a hand generally requires only qualitative knowledge, not of how much is there, but the price of coal will be determined by the percent of undesired sulfur impurity present.
Qualitative analysis tells us what chemicals are present. Quantitative analysis tells us how much.
How Did Analytical Chemistry Originate? That is a very good question. Actually, some tools and basic chemical measurements date back to the earliest recorded history. Fire assays for gold are referred to in Zechariah 13:9, and the King of Babylon complained to the Egyptian Pharoah, Ammenophis the Fourth (1375–1350 BC), that gold he had received from the pharaoh was “less than its weight” after putting it in a furnace. The perceived value of gold, in fact, was probably a major incentive for acquiring analytical knowledge. Archimedes (287–212 BC) did nondestructive testing of the golden wreath of King Hieron II. He placed lumps of gold and silver equal in weight to the wreath in a jar full of water and measured the amount of water displaced by all three. The wreath displaced an amount between the gold and silver, proving it was not pure gold! The balance is of such early origin that it was ascribed to the gods in the earliest documents found. The Babylonians created standard weights in 2600 BC and considered them so important that their use was supervised by the priests. The alchemists accumulated the chemical knowledge that formed the basis for quantitative analysis as we know it today. Robert Boyle coined the term analyst in his 1661 book, The Sceptical Chymist. Antoine Lavoisier has been considered the “father of analytical chemistry” because of the careful quantitative experiments he performed on conservation of mass (using the analytical balance). (Lavoisier was actually a tax collector and dabbled in science on the side. He was guillotined on May 8, 1793, during the French Revolution because of his activities as a tax collector.) Gravimetry was developed in the seventeenth century, and titrimetry in the eighteenth and nineteenth centuries. The origin of titrimetry goes back to Geoffroy in 1729; he evaluated the quality of vinegar by noting the quantity of solid K2 CO3 that could be added before effervescence ceased (Reference 4). Gay-Lussac, in 1829, assayed silver by titration with 0.05% relative accuracy and precision!
Robert Boyle coined the term “analyst” in his book The Sceptical Chymist in 1661
Antoine Lavoisier used a precision balance for quantitative experiments on the conservation of mass. He is considered the “father of quantitative analysis.”
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Karl Remigius Fresenius (1818–1897) published a textbook on quantitative analysis in 1846, which went through six editions and became a standard in the field. He also founded the first journal in analytical chemistry, Zeitschrift Fur Analytische Chemie in 1862. A 2000-year-old balance. Han Dynasty 10 AD. Taiwan National Museum, Taipei. From collection of G. D. Christian.
Wilhelm Ostwald (1853–1932) published the influential text, Die Wissenschaflichen Grundlagen Der Analytischem Chemie (The scientific fundamentals of analytical chemistry) in 1894. He introduced theoretical explanations of analytical phenomena and equilibrium constants.
Textbooks of analytical chemistry began appearing in the 1800s. Karl Fresenius published Anleitung zur Quantitaven Chemischen Analyse in Germany in 1845. Wilhelm Ostwald published an influential text on the scientific fundamentals of analytical chemistry in 1894 entitled Die wissenschaflichen Grundagen der analytischen Chemie, and this book introduced theoretical explanations of analytical phenomena using equilibrium constants (thank him for Chapter 6 and applications in other chapters). The twentieth century saw the evolution of instrumental techniques. Steven Popoff’s second edition of Quantitative Analysis in 1927 included electroanalysis, conductimetric titrations, and colorimetric methods. Today, of course, analytical technology has progressed to include sophisticated and powerful computer-controlled instrumentation and the ability to perform highly complex analyses and measurements at extremely low concentrations. This text will teach you the fundamentals and give you the tools to tackle most analytical problems. Happy journey. For more on the evolution of the field, see Reference 8. Qualitative tests may be performed by selective chemical reactions or with the use of instrumentation. The formation of a white precipitate when adding a solution of silver nitrate in dilute nitric acid to a dissolved sample indicates the presence of a halide. Certain chemical reactions will produce colors to indicate the presence of classes of organic compounds, for example, ketones. Infrared spectra will give “fingerprints” of organic compounds or their functional groups. A clear distinction should be made between the terms selective and specific: ●
Few analyses are specific. Selectivity may be achieved through proper preparation and measurement.
●
A selective reaction or test is one that can occur with other substances but exhibits a degree of preference for the substance of interest. A specific reaction or test is one that occurs only with the substance of interest.
Unfortunately, very few reactions are truly specific but many exhibit selectivity. Selectivity may be also achieved by a number of strategies. Some examples are:
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● ● ● ●
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Sample preparation (e.g., extractions, precipitation) Instrumentation (selective detectors) Target analyte derivatization (e.g., derivatize specific functional groups) Chromatography, which separates the sample constituents
For quantitative analysis, the typical sample composition will often be known (we know that blood contains glucose), or else the analyst will need to perform a qualitative test prior to performing the more difficult quantitative analysis. Modern chemical measurement systems often exhibit sufficient selectivity that a quantitative measurement can also serve as a qualitative measurement. However, simple qualitative tests are usually more rapid and less expensive than quantitative procedures. Qualitative analysis has historically been composed of two fields: inorganic and organic. The former is usually covered in introductory chemistry courses, whereas the latter is best left until after the student has had a course in organic chemistry. In comparing qualitative versus quantitative analysis, consider, for example, the sequence of analytical procedures followed in testing for banned substances at the Olympic Games. The list of prohibited substances includes about 500 different active constituents: stimulants, steroids, beta-blockers, diuretics, narcotics, analgesics, local anesthetics, and sedatives. Some are detectable only as their metabolites. Many athletes must be tested rapidly, and it is not practical to perform a detailed quantitative analysis on each. There are three phases in the analysis: the fast-screening phase, the identification phase, and possible quantification. In the fast-screening phase, urine samples are rapidly tested for the presence of classes of compounds that will differentiate them from “normal” samples. Techniques used include immunoassays, gas chromatography–mass spectrometry, and liquid chromatography–mass spectrometry. About 5% of the samples may indicate the presence of unknown compounds that may or may not be prohibited but need to be identified. Samples showing a suspicious profile during the screening undergo a new preparation cycle (possible hydrolysis, extraction, derivatization), depending on the nature of the compounds that have been detected. The compounds are then identified using the highly selective combination of
(Courtesy of Merck KGaA. Reproduced by permission.)
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gas chromatography/mass spectrometry (GC/MS). In this technique, complex mixtures are separated by gas chromatography, and they are then detected by mass spectrometry, which provides molecular structural data on the compounds. The MS data, combined with the time of elution from the gas chromatograph, provide a high probability of the presence of a given detected compound. GC/MS is expensive and time consuming, and so it is used only when necessary. Following the identification phase, some compounds must be precisely quantified since they may normally be present at low levels, for example, from food, pharmaceutical preparations, or endogenous steroids, and elevated levels must be confirmed. This is done using quantitative techniques such as spectrophotometry or gas chromatography. This text deals principally with quantitative analysis. In the consideration of applications of different techniques, examples are drawn from the life sciences, clinical chemistry, environmental chemistry, occupational health and safety applications, and industrial analysis. We describe briefly in this chapter the analytical process. More details are provided in subsequent chapters.
See the text website for useful chapters from The Encyclopedia of Analytical Chemistry (Reference 9 at the end of the chapter) on literature searching and selection of analytical methods.
“To many . . . , the object of chemical analysis is to obtain the composition of a sample . . . . It may seem a small point . . . that the analysis of the sample is not the true aim of analytical chemistry . . . . the real purpose of the analysis is to solve a problem . . . ” H. A. Laitinen, Editorial: The Aim of Analysis, Anal. Chem., 38 (1966) 1441.
The way an analysis is performed depends on the information needed.
1.3 Getting Started: The Analytical Process The general analytical process is shown in Figure 1.1. The analytical chemist should be involved in every step. The analyst is really a problem solver, a critical part of the team deciding what, why, and how. The unit operations of analytical chemistry that are common to most types of analyses are considered in more detail below. DEFINING THE PROBLEM——WHAT DO WE REALLY NEED TO KNOW? (NOT NECESSARILY EVERYTHING) Before the analyst can design an analysis procedure, he or she must know what information is needed, by whom, for what purpose, and what type of sample is to be analyzed. As the analyst, you must have good communication with the client. This stage of an analysis is perhaps the most critical. The client may be the Environmental Protection Agency (EPA), an industrial chemist, an engineer, or your grandmother—each of which will have different criteria or needs, and each having their own understanding of what a chemical analysis involves or means. It is important to communicate in language that is understandable by both sides. If someone puts a bottle on your desk and asks, “What is in here?” or “Is this safe?”, you may have to explain that there are 10 million known compounds and substances. A client who says, “I want to know what elements are in here” needs to understand that at perhaps $20 per analysis for 85 elements it will cost $1700 to test for them all, when perhaps only a few elements are of interest. Laypersons might come to analytical chemists with cosmetics they wish to “reverse engineer” so they can market them and make a fortune. When they realize it may cost a small fortune to determine the ingredients, requiring a number of sophisticated analyses, they always rethink their goals. On the other hand, a mother may come to you with a white pill that her teenage son insists is vitamin C and she fears is an illicit drug. While it is not trivial to determine what it is, it is rather straightforward to determine if it undergoes the same reactions that ascorbic acid (vitamin C) does. You may be able to greatly alleviate the concerns of an anxious mother. The concept of “safe” or “zero/nothing” is one that many find hard to define or understand. Telling someone their water is safe is not for the analyst to say. All you
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Define the Problem Factors
Prepare the Sample for Analysis
• What is the problem—what needs to be found? Qualitative and/or quantitative?
Factors
• What will the information be used for? Who will use it?
• Dissolve?
• When will it be needed?
• Ash or digest?
• How accurate and precise does it have to be?
• Chemical separation or masking of interferences needed?
• What is the budget?
• Need to concentrate the analyte?
• The analyst (the problem solver) should consult with the client to plan a useful and efficient analysis, including how to obtain a useful sample.
• Need to change (derivatize) the analyte for detection? • Need to adjust solution conditions (pH, add reagents)?
Select a Method Factors • Sample type • Size of sample • Sample preparation needed • Concentration and range (sensitivity needed) • Selectivity needed (interferences) • Accuracy/precision needed
• Solid, liquid, or gas?
Perform Any Necessary Chemical Separations • Distillation • Precipitation • Solvent extraction • Solid phase extraction • Chromatography (may include the measurement step) • Electrophoresis (may include the measurement step)
• Tools/instruments available • Expertise/experience • Cost
Perform the Measurement
• Speed • Does it need to be automated? • Are methods available in the chemical literature? • Are standard methods available? • Are there regulations that need to be followed?
Obtain a Representative Sample
Factors • Calibration • Validation/controls/blanks • Replicates
Calculate the Results and Report
Factors
• Statistical analysis (reliability)
• Sample type/homogeneity/size
• Report results with limitations/accuracy information
• Sampling statistics/errors
• Interpret carefully for intended audience. Critically evaluate results. Are iterations needed?
Fig. 1.1.
Steps in an analysis.
can do is present the analytical data (and give an indication of its range of accuracy). The client must decide whether it is safe to drink, perhaps relying on other experts. Also, never report an answer as “zero,” but as less than the detection limit, which is based on the measurement device/instrument. We are limited by our methodology and equipment, and that is all that can be reported. Some modern instruments, though, can measure extremely small amounts or concentrations, for example, parts per trillion. This presents a dilemma for policy makers (often political in nature). A law may be passed that there should be zero concentration of a chemical effluent in water. In practice, the acceptable level is defined by how low a concentration can be detected; and the very low detectability may be far below the natural occurrence of the chemical
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The way you perform an analysis will depend on your experience, the equipment available, the cost, and the time involved.
The analyte is the substance analyzed for. Its concentration is determined.
Chemical Abstracts is a good source of literature.
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or below the levels to which it can be reasonably reduced. We analysts and chemists need to be effective communicators of what our measurements represent. Once the problem is defined this will dictate how the sample is to be obtained, how much is needed, how sensitive the method must be, how accurate and precise1 it must be, and what separations may be required to eliminate interferences. The determination of trace constituents will generally not have to be as precise as for major constituents, but greater care will be required to eliminate trace contamination during the analysis. Once the required measurement is known, the analytical method to be used will depend on a number of factors, including the analyst’s skills and training in different techniques and instruments; the facilities, equipment, and instrumentation available; the sensitivity and precision required; the cost and the budget available; and the time for analysis and how soon results are needed. There are often one or more standard procedures available in reference books for the determination of an analyte (constituent to be determined) in a given sample type. This does not mean that the method will necessarily be applicable to other sample types. For example, a standard EPA method for groundwater samples may yield erroneous results when applied to the analysis of sewage water. The chemical literature (journals) contains many specific descriptions of analyses. Chemical Abstracts (http://info.cas.org), published by the American Chemical Society, is a good place to begin a literature search. It contains abstracts of all articles appearing in the major chemical journals of the world. Yearly and cumulative indices are available, and many libraries have computer search facilities. If your library subscribes to Scifinder from Chemical Abstracts Service, this is the best place to start your search (www.cas.org/products/scifindr/index.html). The Web of Science, a part of the Web of Knowledge (www.isiwebofknowledge.com) is an excellent place to search the literature and provides also the information as to where a particular article has been cited and by whom. Another excellent source, available to anyone, is Google Scholar, which allows you to search articles, authors, etc. (http://scholar.google.com). The major analytical chemistry journals may be consulted separately. Some of these are: Analytica Chimica Acta, Analytical Chemistry, Analytical and Bioanalytical Chemistry, Analytical Letters, Analyst, Applied Spectroscopy, Clinica Chimica Acta, Clinical Chemistry, Journal of the Association of Official Analytical Chemists, Journal of Chromatography, Journal of Separation Science, Spectrochimica Acta, and Talanta. While the specific analysis of interest may not be described, the analyst can often use literature information on a given analyte to devise an appropriate analytical scheme. Finally, the analyst may have to rely upon experience and knowledge to develop an analytical method for a given sample. The literature references in Appendix A describe various procedures for the analysis of different substances. Examples of the manner in which the analysis of particular types of samples are made are given in application Chapters 25 and 26 on the text’s website. These chapters describe commonly performed clinical, biochemical, and environmental analyses. The various techniques described in this text are utilized for the specific analyses. Hence, it will be useful for you to read through these applications chapters both now and after completing the majority of this course to gain an appreciation of what goes into analyzing real samples and why the analyses are made. Once the problem has been defined, the following steps can be started.
1 Accuracy is the degree of agreement between a measured value and a true value. Precision is the degree of
agreement between replicate measurements of the same quantity and does not necessarily imply accuracy. These terms are discussed in more detail in Chapter 3.
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OBTAINING A REPRESENTATIVE SAMPLE——WE CAN’T ANALYZE THE WHOLE THING A chemical analysis is usually performed on only a small portion of the material to be characterized. If the amount of material is very small and it is not needed for future use, then the entire sample may be used for analysis. The gunshot residue on a hand may be an example. More often, though, the characterized material is of value and must be altered as little as possible in sample collection. For example, sampling of a Rembrandt painting for authenticity would need to be done with utmost care for sample quantity, so as not to deface the artwork. The material to be sampled may be solid, liquid, or gas. It may be homogeneous or heterogeneous in composition. In the former case, a simple “grab sample” taken at random will suffice for the analysis. In the latter, we may be interested in the variation throughout the sample, in which case several individual samples will be required. If the gross composition is needed, then special sampling techniques will be required to obtain a representative sample. For example, in analyzing for the average protein content of a shipment of grain, a small sample may be taken from each bag, or tenth bag for a large shipment, and combined to obtain a gross sample. Sampling is best done when the material is being moved, if it is large, in order to gain access. The larger the particle size, the larger should be the gross sample. The gross sample must be reduced in size to obtain a laboratory sample of several grams, from which a few grams to milligrams will be taken to be analyzed (analysis sample). The size reduction may require taking portions (e.g., two quarters) and mixing, in several steps, as well as crushing and sieving to obtain a uniform powder for analysis. Methods of sampling solids, liquids, and gases are discussed in Chapter 2. If one is interested in spatial structure, then homogenization must not be carried out, but spatially resolved sampling must be done. In the case of biological fluids, the conditions under which the sample is collected can be important, for example, whether a patient has just eaten. The composition of blood varies considerably before and after meals, and for many analyses a sample is collected after the patient has fasted for a number of hours. Persons who have their blood checked for cholesterol levels are asked to fast for up to twelve hours prior to sampling. Preservatives such as sodium fluoride for glucose preservation and anticoagulants for blood samples may be added when samples are collected; these may affect a particular analysis. Blood samples may be analyzed as whole blood, or they may be separated to yield plasma or serum according to the requirements of the particular analysis. Most commonly, the concentration of the substance external to the red cells (the extracellular concentration) will be a significant indication of physiological condition, and so serum or plasma is taken for analysis. If whole blood is collected and allowed to stand for several minutes, the soluble protein fibrinogen will be converted by a complex series of chemical reactions (involving calcium ion) into the insoluble protein fibrin, which forms the basis of a gel, or clot. The red and white cells of the blood become caught in the meshes of the fibrin network and contribute to the clot, although they are not necessary for the clotting process. After the clot forms, it shrinks and squeezes out a straw-colored fluid, serum, which does not clot but remains fluid indefinitely. The clotting process can be prevented by adding a small amount of an anticoagulant, such as heparin or a citrate salt (i.e., a calcium complexor). Blood collection vials are often color-coded to provide a clear indication of the additives they contain. An aliquot of the unclotted whole blood can be taken for analysis, or the red cells can be centrifuged to the bottom, and the light pinkish-colored
The gross sample consists of several portions of the material to be tested. The laboratory sample is a small portion of this, taken after homogenization. The analysis sample is that actually analyzed. See Chapter 2 for methods of sampling.
Serum is the fluid separated from clotted blood. Plasma is the fluid separated from unclotted blood. It is the same as serum, but contains fibrinogen, the clotting protein.
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Care must be taken not to alter or contaminate the sample.
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plasma remaining can be analyzed. Plasma and serum are essentially identical in chemical composition, the chief difference being that fibrinogen has been removed from the latter. Details of sampling other materials are available in reference books on specific areas of analysis. See the references at the end of the chapter for some citations. Certain precautions should be taken in handling and storing samples to prevent or minimize contamination, loss, decomposition, or matrix change. In general, one must prevent contamination or alteration of the sample by (1) the container, (2) the atmosphere, (3) heat/temperature, or (4) light. Also, a chain of custody should be established and will certainly be required for any analysis that may be involved in legal proceedings. In the O.J. Simpson case, there were television news clips of people handling samples, purportedly without proper custody, placing them in the hot trunk of a car, for example. While this may not have affected the actual analyses and correctness of samples analyzed, it provided arguments for the defense to discredit analyses. The sample may have to be protected from the atmosphere or from light. It may be an alkaline substance, for example, which will react with carbon dioxide in the air. Blood samples to be analyzed for CO2 must be protected from the atmosphere. The stability of the sample must be considered. To minimize degradation of glucose, for example, a preservative such as sodium fluoride is added to blood samples. The preservative must not, of course, interfere in the analysis. Proteins and enzymes tend to denature on standing and should be analyzed without delay. Trace constituents may be lost during storage by adsorption onto the container walls. Urine samples are unstable, and calcium phosphate precipitates out, entrapping metal ions or other substances of interest. Precipitation can be prevented by keeping the urine acidic (pH 4.5), usually by adding 1 or 2 mL glacial acetic acid per 100-mL sample and stored under refrigeration. Urine, as well as whole blood, serum, plasma, and tissue samples, can also be frozen for prolonged storage. Deproteinized blood samples are more stable than untreated ones. Corrosive gas samples will often react with the container. Sulfur dioxide, for example, is troublesome. In automobile exhaust, SO2 is also lost by dissolving in condensed water vapor from the exhaust. In such cases, it is best to analyze the gas by an in situ analyzer that operates at a temperature in which condensation does not occur. PREPARING THE SAMPLE FOR ANALYSIS——IT PROBABLY NEEDS TO BE ALTERED
The first thing you must do is measure the size of sample to be analyzed.
The first step in analyzing a sample is to measure the amount being analyzed (e.g., volume or weight of sample). This will be needed to calculate the percent composition from the amount of analyte found. The analytical sample size must be measured to the degree of precision and accuracy required for the analysis. An analytical balance sensitive to 0.1 mg is usually used for weight measurements and balances that can weigh down to 0.01 mg are becoming increasingly common. Solid samples are often analyzed on a dry basis and must be dried in an oven at 110 to 120◦ C for 1 to 2 h and cooled in a dessicator before weighing, if the sample is stable at the drying temperatures. Some samples may require higher temperatures and longer heating time (e.g., overnight) because of their great affinity for moisture. The amount of sample taken will depend on the concentration of the analyte and how much is needed for isolation and measurement. Determination of a major constituent may require only 100 mg of sample, while a trace constituent may require several grams. Usually replicate samples are taken for analysis, in order to obtain statistical data on the precision of the analysis and thus provide more reliable results.
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Analyses may be nondestructive in nature, for example, in the measurement of lead in paint by X-ray fluorescence in which the sample is bombarded with an X-ray beam and the characteristic reemitted X-radiation is measured. More often, the sample must be in solution form for measurement, and solids must be dissolved. Inorganic materials may be dissolved in various acids, redox, or complexing media. Acid-resistant material may require fusion with an acidic or basic flux in the molten state to render it soluble in dilute acid or water. Fusion with sodium carbonate, for example, forms acid-soluble carbonates. Organic materials that are to be analyzed for inorganic constituents, for example, trace metals, may be destroyed by dry ashing. The sample is slowly combusted in a furnace at 400 to 700◦ C, leaving behind an inorganic residue that is soluble in dilute acid. Alternately, the organic matter may be destroyed by wet digestion by heating with oxidizing acids. A mixture of nitric and sulfuric acids is common. Perchloric acid digestion is used for complete oxidative digestion; this is a last resort as special extraction or fume hoods are required due to potential explosion hazards. Biological fluids may sometimes be analyzed directly. Often, however, proteins interfere and must be removed. Dry ashing and wet digestion accomplish such removal. Or proteins may be precipitated with various reagents and filtered or centrifuged away, to give a protein-free filtrate (PFF). If the analyte is organic in nature, these oxidizing methods cannot be used. Rather, the analyte may be extracted away from the sample or dialyzed, or the sample dissolved in an appropriate solvent. It may be possible to measure the analyte nondestructively. An example is the direct determination of protein in feeds by near-infrared spectrometry. Once a sample is in solution, the solution conditions must be adjusted for the next stage of the analysis (separation or measurement step). For example, the pH may have to be adjusted, or a reagent added to react with and “mask” interference from other constituents. The analyte may have to be reacted with a reagent to convert it to a form suitable for measurement or separation. For example, a colored product may be formed that will be measured by spectrometry. Or the analyte will be converted to a form that can be volatilized for measurement by gas chromatography. The gravimetric analysis of iron as Fe2 O3 requires that all the iron be present as iron(III), its usual form. A volumetric determination by reaction with dichromate ion, on the other hand, requires that all the iron be converted to iron(II) before reaction, and the reduction step will have to be included in the sample preparation. The solvents and reagents used for dissolution and preparation of the solution should be of high purity (reagent grade). Even so, they may contain trace impurities of the analyte. Hence, it is important to prepare and analyze replicate blanks, particularly for trace analyses. A blank theoretically consists of all chemicals in the unknown and used in an analysis in the same amounts (including water), run through the entire analytical procedure. The blank result is subtracted from the analytical sample result to arrive at a net analyte concentration in the sample solution. If the blank is appreciable, it may invalidate the analysis. Oftentimes, it is impossible to make a perfect blank for an analysis. PERFORMING NECESSARY CHEMICAL SEPARATIONS In order to eliminate interferences, to provide suitable selectivity in the measurement, or to preconcentrate the analyte for more sensitive or accurate measurement, the analyst must often perform one or more separation steps. It is preferable to separate the analyte away from the sample matrix, in order to minimize losses of the analyte. Separation steps may include precipitation, extraction into an immiscible solvent, chromatography, dialysis, and distillation.
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Solid samples usually must be put into solution.
Ashing is the burning of organic matter. Digestion is the wet oxidation of organic matter.
The pH of the sample solution will usually have to be adjusted.
Always run a blank!
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PERFORMING THE MEASUREMENT——YOU DECIDE THE METHOD
Instruments are more selective and sensitive than volumetric and gravimetric methods. But they may be less precise.
The method employed for the actual quantitative measurement of the analyte will depend on a number of factors, not the least important being the amount of analyte present and the accuracy and precision required. Many available techniques possess varying degrees of selectivity, sensitivity, accuracy and precision, cost, and rapidity. Analytical chemistry research often deals with the optimization of one or more of these parameters, as they relate to a particular analysis or analysis technique. Gravimetric analysis usually involves the selective separation of the analyte by precipitation, followed by the very nonselective measurement of mass (of the precipitate). In volumetric, or titrimetric, analysis, the analyte reacts with a measured volume of reagent of known concentration, in a process called titration. A change in some physical or chemical property signals the completion of the reaction. Gravimetric and volumetric analyses can provide results accurate and precise to a few parts per thousand (tenth of 1 percent) or better. However, they require relatively large (millimole or milligram) quantities of analyte and are only suited for the measurement of major constituents, although microtitrations may be performed. Volumetric analysis is more rapid than gravimetric analysis and is therefore preferred when applicable. Instrumental techniques are used for many analyses and constitute the discipline of instrumental analysis. They are based on the measurement of a physical property of the sample, for example, an electrical property or the absorption of electromagnetic radiation. Examples are spectrophotometry (ultraviolet, visible, or infrared), fluorimetry, atomic spectroscopy (absorption, emission), mass spectrometry, nuclear magnetic resonance spectrometry (NMR), X-ray spectroscopy (absorption, fluorescence), electroanalytical chemistry (potentiometric, voltammetric, electrolytic), chromatography (gas, liquid), and radiochemistry. Instrumental techniques are generally more sensitive and selective than the classical techniques but are less precise, on the order of 1 to 5% or so. These techniques are usually much more expensive, especially in terms of initial capital investment. But depending on the numbers of analyses, they may be less expensive when one factors in personnel costs. They are usually more rapid, may be automated, and may be capable of measuring more than one analyte at a time. Chromatography techniques are particularly powerful for analyzing complex mixtures. They integrate the separation and measurement steps. Constituents are separated as they are pushed through (eluted from) a column of appropriate material that interacts with the analytes to varying degrees, and these are sensed with an appropriate detector as they emerge from the column, to give a transient peak signal, proportional to the amount of each. Table 1.1 compares various analytical methods to be described in this text with respect to sensitivity, precision, selectivity, speed, and cost. The numbers given may be exceeded in specific applications, and the methods may be applied to other uses, but these are representative of typical applications. The lower concentrations determined by titrimetry require the use of an instrumental technique for measuring the completion of the titration. The selection of a technique, when more than one is applicable, will depend, of course, on the availability of equipment, and personal experience, and preference of the analyst. As examples, you might use spectrophotometry to determine the concentration of nitrate in river water at the sub parts-per-million level, by first reducing to nitrite and then using a diazotization reaction to produce a color. Fluoride in toothpaste may be determined potentiometrically using a fluoride ion-selective electrode. A complex mixture of hydrocarbons in gasoline can be separated using gas chromatography and determined by flame ionization detection. Glucose in blood can be determined kinetically by the rate of the enzymatic reaction between glucose and oxygen, catalyzed by the enzyme glucose oxidase, with measurement of the rate of oxygen depletion or the rate of hydrogen peroxide production. The purity of a silver
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Table 1.1
Comparison of Different Analytical Methods Method Gravimetry Titrimetry Potentiometry Electrogravimetry, coulometry Voltammetry Spectrophotometry Fluorometry Atomic spectroscopy
Approx. Range (mol/L)
Approx. Precision (%)
Selectivity
Speed
Cost
Principal Uses
10 –1 –10 –2 10 –1 –10 –4 10 –1 –10 –6 10 –1 –10 –4
0.1 0.1–1 2 0.01–2
Poor–moderate Poor–moderate Good Moderate
Slow Moderate Fast Slow–moderate
Low Low Low Moderate
Inorg. Inorg., org. Inorg. Inorg., org.
10 –3 –10 –10 10 –3 –10 –6 10 –6 –10 –9 10 –3 –10 –9
2–5 2 2–5 2–10
Good Good–moderate Moderate Good
Moderate Fast–moderate Moderate Fast
Moderate Low–moderate Moderate Moderate–high
2–5
Good
Fast–moderate
Moderate–high
2–10
Good–moderate
Fast–moderate
Moderate
Inorg., org. Inorg., org. Org. Inorg., multielement Org., multicomponent Inorg., org., enzymes
Chromatography—Mass10 –4 –10 –9 Spectrometry Kinetics methods 10 –2 –10 –10
bar can be determined gravimetrically by dissolving a small sample in nitric acid and precipitating AgCl with chloride and weighing the purified precipitate. The various methods of determining an analyte can be classified as either absolute or relative. Absolute methods rely upon accurately known fundamental constants for calculating the amount of analyte, for example, atomic weights. In gravimetric analysis, for example, an insoluble derivative of the analyte of known chemical composition is prepared and weighed, as in the formation of AgCl for chloride determination. The precipitate contains a known fraction of the analyte, in this case, fraction of Cl = at wt Cl/f wt AgCl = 35.453/143.32 = 0.24737.2 Hence, it is a simple matter to obtain the amount of Cl contained in the weighed precipitate. Gravimetry, titrimetry and coulometry are examples of absolute methods. Most other methods, however, are relative in that they require comparison against some solution of known concentration (also called calibration or standardization, see below).
Most methods require calibration with a standard.
INSTRUMENT STANDARDIZATION Most instrumental methods of analysis are relative. Instruments register a signal due to some physical property of the solution. Spectrophotometers, for example, measure the fraction of electromagnetic radiation from a light source that is absorbed by the sample. This fraction must be related to the analyte concentration by comparison against the fraction absorbed by a known concentration of the analyte. In other words, the instrumentation must be standardized. Instrument response may be linearly or nonlinearly related to the analyte concentration. Calibration is accomplished by preparing a series of standard solutions of the analyte at known concentrations and measuring the instrument response to each of these (usually after treating them in the same manner as the samples) to prepare an analytical calibration curve of response versus concentration. Figure 1.2 shows examples of calibration curves obtained in a mass spectrometry experiment. The concentration of an unknown can then be determined from 2 at wt = atomic weight; f wt = formula weight.
A calibration curve is the instrument response as a function of concentration.
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12
1.4 1.2 1 0.8 0.6 0.4 0.2 0
Calibration curves for the measurement of proteins using matrix-assisted laser desorption ionization (MALDI)—mass spectrometry and an ionic liquid matrix (Courtesy of Prof. Michael Gross, Washington University in St. Louis. Reprinted with permission).
Signal intensity ratio
10
Fig. 1.2.
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8
R2 = 0.999
2
4
6
8
10
6
R2 = 0.999 4
R2 = 0.998
2 0 0
20
40
60 80 Analyte amount (pmol)
100
120
the response, using the calibration curve. With modern computer-controlled instruments, this is done electronically or digitally, and direct readout of concentration is obtained.
METHOD OF STANDARD ADDITIONS Standard additions calibration is used to overcome sample matrix effects.
The sample matrix may affect the instrument response to the analyte. In such cases, calibration may be accomplished by the method of standard additions. A portion of the sample is spiked with a known amount of standard, and the increase in signal is due to the standard. In this manner, the standard is subjected to the same environment as the analyte. These calibration techniques are discussed in more detail when describing the use of specific instruments. See Section 17.5 and the website supplement for that section for a detailed description of the standard additions method and calculations using it. Section 20.5 illustrates its use in gas chromatography, and Example 14.8 illustrates how it is used in potentiometry. Experiments 33 (atomic spectrometry) and 35 (solid-phase extraction) on the text website employ the method of standard additions.
INTERNAL STANDARD CALIBRATION An instrumental response is often subject to variations from one measurement to the next due to changing instrument conditions, resulting in imprecision. For example, in gas chromatography, the volume of injected sample or standard from a Hamilton microliter syringe (see Chapter 2) may vary. In atomic absorption spectrometry, fluctuations in gas flows and aspiration rates for sample introduction may occur. In order to compensate for these types of fluctuations, internal standard calibration may be used. Here, a fixed concentration of a different analyte, that is usually chemically similar to the sample analyte, is added to all solutions to be measured. Signals for both substances are recorded, and the ratio of the sample to internal standard signals is plotted versus sample analyte concentration. So, if say the volume of injected sample is 10% lower than assumed, each signal is reduced 10%, and the ratio at a given sample analyte concentration remains constant. See Sections 17.5 (atomic spectrometry) and 20.5 (gas chromatography) for illustrations of internal standard use.
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15
CALCULATING THE RESULTS AND REPORTING THE DATA Once the concentration of analyte in the prepared sample solution has been determined, the results are used to calculate the amount of analyte in the original sample. Either an absolute or a relative amount may be reported. Usually, a relative composition is given, for example, percent or parts per million, along with the mean value for expressing accuracy. Replicate analyses can be performed (three or more), and a precision of the analysis may be reported, for example, standard deviation. A knowledge of the precision is important because it gives the degree of uncertainty in the result (see Chapter 3). The analyst should critically evaluate whether the results are reasonable and relate to the analytical problem as originally stated. Remember that the customer often does not have a scientific background so will take a number as gospel. Only you, as analyst, can put that number in perspective, and it is important that you have good communication and interaction with the “customer” about what the analysis represents.
The analyst must provide expert advice on the significance of a result.
1.4 Validation of a Method—You Have to Prove It Works! Great care must be taken that accurate results are obtained in an analysis. Two types of error may occur: random and systematic. Every measurement has some imprecision associated with it, which results in random distribution of results, for example, a Gaussian distribution. The experiment can be designed to narrow the range of this, but it cannot be eliminated. A systematic error is one that biases a result consistently in one direction. Such errors may occur when the sample matrix suppresses the instrument signal, a weight of an analytical balance may be in error, skewed either high or low, or a sample may not be sufficiently dried. Proper calibration of an instrument is only the first step in assuring accuracy. In developing a method, samples should be spiked with known amounts of the analyte (above and beyond what is already in the sample). The amounts determined (recovered) by the analysis procedure (after subtraction of the amount apparently present in the sample as determined by the same procedure) should be close to what was added. This is not a foolproof approach, however, and only assures that the intended analyte is measured. It cannot assure that some interferent present in the sample is not measured. A new method is better validated by comparison of sample results with those obtained with another accepted method. There are various sources of certified standards or reference materials that may be analyzed to assure accuracy by the method in use. For example, environmental quality control standards for pesticides in water or priority pollutants in soil are commercially available. The National Institute of Standards and Technology (NIST) prepares standard reference materials (SRMs) of different matrix compositions (e.g., steel, ground leaves) that have been certified for the content of specific analytes, by careful measurement by at least two independent techniques. Values are assigned with statistical ranges. Different agencies and commercial concerns can provide samples for round-robin or blind tests in which control samples are submitted to participating laboratories for analysis at random; the laboratories are not informed of the control values prior to analysis. Standards should be run intermittently with samples. A control sample should also be run at least daily and the results plotted as a function of time to prepare a quality control chart, which is compared with the known standard deviation of the method. The measured quantity is assumed to be constant with time, with a Gaussian distribution, and there is a 1 in 20 chance that values will fall outside two standard deviations from the known value, and a 1 in 100 chance it will be 2.5 standard deviations away. Numbers exceeding these suggest uncompensated errors, such as instrument malfunction, reagent deterioration, or improper calibration.
The best way to validate a method is to analyze a standard reference material of known composition.
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Good laboratory practice (validation) is required to assure accuracy of analyses.
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Government regulations require careful established protocol and validation of methods and analyses when used for official or legal purposes. Guidelines of good laboratory practice (GLP) have been established to assure validation of analyses. They, of course, ideally apply to all analyses. These are discussed in detail in Chapter 4.
1.5 Analyze Versus Determine—They Are Different You analyze a sample to determine the amount of analyte.
The terms analyze and determine have two different meanings. We say a sample is analyzed for part or all of its constituents. The substances measured are called the analytes. The process of measuring the analyte is called a determination. Hence, in analyzing blood for its chloride content, we determine the chloride concentration. The constituents in the sample may be classified as major (>1% of the sample), minor (0.1 to 1%), or trace ( 1000◦ C. Alloys with gold, silver, and other metals Fused samples contaminated with the metal Not attacked by alkalis and acids except conc. HCl, dil. H2 SO4 , and boiling conc. HNO3 Not attacked by alkali solutions or HF. Attacked by many organic solvents (acetone, ethanol OK) Translucent. Has replaced polyethylene for many purposes
Not attacked by HF. Attacked by many organic solvents Inert to most chemicals
Usually alloyed with iridium or rhodium to increase hardness. Platinum crucibles for fusions and treatment with HF
Ni and Fe crucibles used for peroxide fusions
Flexible plastic
Somewhat brittle Useful for storage of solutions and reagents for trace metal analysis. Is permeable to oxygen
device must be both accurate and sensitive. There are various sophisticated ways of achieving this, but the most useful and versatile device used is the analytical balance. Most analytical balances used today are electronic balances. The mechanical single-pan balance is seldom used in the modern analytical laboratory anymore. The calibration of electronic or digital balances is based on comparison of one weight against another. Factors such as zero-point drift and air buoyancy must be considered for all balance types. We really deal with masses rather than weights. The weight of an object is the force exerted on it by the gravitational attraction. This force will differ at different locations on Earth. Mass, on the other hand, is the quantity of matter of which the object is composed and is invariant.
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Modern electronic balances offer convenience in weighing and are subject to fewer errors or mechanical failures than are mechanical balances, which have become largely obsolete. The operation of dialing weights, turning and reading micrometers, and beam and pan arrest of mechanical balances are eliminated, greatly speeding the measurement. A digital-display electronic balance is shown in Figure 2.1, and the operating principle of an electronic balance is illustrated in Figure 2.2. There are no weights or knife edges as with mechanical balances. The pan sits on the arm of a movable hanger (2), and this movable system is compensated by a constant electromagnetic force. The position of the hanger is monitored by an electrical position scanner (1), which brings the weighing system back to the zero position. The compensation current is proportional to the mass placed on the pan. This is sent in digital form to a microprocessor that converts it into the corresponding weight value, which appears as a digital display. The weight of the container can be automatically subtracted. These balances use the principle of electromagnetic force compensation first described by Angstrom in 1895. But they still use the principle of comparing one weight with another. The balance is “zeroed,” or calibrated, with a known weight. When the sample is placed on the pan, its weight is electronically compared with the known. This is a form of self-calibration. Modern balances may have such features as compensating for wandering from true zero and averaging variations due to building vibrations. A single control bar is used to switch the balance on and off, to set the display to zero, and to automatically tare a container on the pan. Since results are available as an electrical signal, they can be readily processed by a personal computer and stored. Weighing statistics can be automatically calculated. Electronic analytical balances can be purchased with different weighing ranges and readabilities. A standard analytical balance typically has a maximum capacity of 160–300 g and a readability of 0.1 mg. Semimicro balances (readability 0.01 mg, capacity up to 200 g), microbalances (readability 1 μg, capacity up to 30 g) and ultramicro balances (readability 0.1 μg, capacity up to 2 g) are currently commercially available. Electrochemical quartz balances are available with 100-μg range that can detect 1 ng (10−9 g) changes! The balance utilizes a thin quartz crystal disk oscillating at,
Greater precision equals greater cost. A balance readable to 0.1 mg costs $10,000.
1
2 3
Fig. 2.2.
Fig. 2.1.
Electronic analytical balance. (Courtesy of Denver Instrument Co. Denver Instrument Company owns all images.)
4
Operating principle of electronic balance: 1, position scanner; 2, hanger; 3, coil; 4, temperature sensor. (From K. M. Lang, American Laboratory, March, 1983, p. 72. Reproduced by permission of American Laboratory, Inc.)
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Older mechanical balances used the lever principle: M1 L1 = M2 L2 , where L1 and L2 are the lengths of the two arms of the lever and M1 and M2 are the corresponding masses. If L1 and L2 are constructed to be as nearly equal as possible then at balance, M1 = M 2 . Burns’ Hog Weighing Method: “1) Get a perfectly straight plank and balance it across a sawhorse. 2) Put the hog on one end of the plank. 3) Pile rocks on the other end until the plank is again perfectly balanced. 4) Carefully guess the weight of the rocks.”—Robert Burns
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for example, 10 MHz. The frequency of oscillation changes with any change in mass, and the frequency change measured by the instrument is converted to mass units. A film of gold is evaporated on the quartz, and the gold substrate can be coated with the material of interest. Mass changes as small as a few percent of a monolayer coverage of atoms or molecules on the gold surface can be measured. Mass changes with time can be recorded. Such balances are incorporated in air particle mass monitors, see for example, http://www.kanomax-usa.com/dust/3521/3521.html. SINGLE-PAN MECHANICAL BALANCE Electronic balances have largely replaced mechanical balances. But they are still used some, so we have placed on the Chapter 2 text website a description of the single-pan mechanical balance. Such balances are based on the first-class lever, like a teeter totter, that compares two masses at each end of the lever, one the unknown and the other standard weights, and the relationship M1 L1 = M2 L2 holds. The single-pan balance actually has unequal lever lengths, and the operation is based on removing weights from the lever end on which the unknown is placed, equal in value to the unknown mass. See the website for details. SEMIMICRO- AND MICROBALANCES
We make most quantitative weighings to 0.1 mg.
The discussion thus far has been limited to conventional macro or analytical balances. These perform weighings to the nearest 0.1 mg, and loads of up to 160-300 g can be handled. These are satisfactory for most routine analytical weighings. All of the above classes of balances can be made more sensitive by changing the parameters affecting the sensitivity, such as decreasing the mass of the beam (for mechanical balances) and pans, increasing the length of the beam, and changing the center of gravity of the beam. Lighter material can be used for the beam since it need not be as sturdy as the beam of a conventional balance. The semimicrobalance is sensitive to about 0.01 mg, and the microbalance is sensitive to about 0.001 mg (1 μg). The load limits of these balances are correspondingly less than the conventional balance, and greater care must be taken in their use. ZERO-POINT DRIFT The zero setting of a balance is not a constant that can be determined or set and forgotten. It will drift for a number of reasons, including temperature changes, humidity, and static electricity. The zero setting should therefore be checked at least once every half-hour during the period of using the balance. WEIGHT IN A VACUUM——THIS IS THE ULTIMATE ACCURACY
An object of 1-mL volume will be buoyed up by 1.2 mg!
The weighings that are made on a balance will, of course, give the weight in air. When an object displaces its volume in air, it will be buoyed up by the weight of air displaced (Archimedes’ principle—see the box in Chapter 1 on how analytical chemistry originated). The density of air is 0.0012 g (1.2 mg) per milliliter. If the density of the weights and the density of the object being weighed are the same, then they will be buoyed up by the same amount, and the recorded weight will be equal to the weight in a vacuum, where there is no buoyancy. If the densities are markedly different, the differences in the buoyancies will lead to a small error in the weighing: One will be buoyed up more than the other, and an unbalance will result. Such a situation arises in the weighing of very dense objects [e.g., platinum vessels (density = 21.4)
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or mercury (density = 13.6)] or light, bulky objects [e.g., water (density ≈ 1)]; and in very careful work, a correction should be made for this error. For comparison, the density of weights used in balances is about 8. See Reference 14 for air buoyancy corrections with a single-pan balance. (Reference 10 describes the calibration of the weights in a single-pan balance.) Note that in most cases, a correction is not necessary because the error resulting from the buoyancy will cancel out in percent composition calculations. The same error will occur in the numerator (as the concentration of a standard solution or weight of a gravimetric precipitate) and in the denominator (as the weight of the sample). Of course, all weighings must be made with the materials in the same type of container (same density) to keep the error constant. An example where correction in vacuum is used is in the calibration of glassware. The mass of water or mercury delivered or contained by the glassware is measured. From a knowledge of the density of the liquid at the specified temperature, its volume can be calculated from the mass. Even in these cases, the buoyancy correction is only about one part per thousand. For most objects weighed, buoyancy errors can be neglected. Weights of objects in air can be corrected to the weight in vacuum by 0.0012 0.0012 Wvac = Wair + Wair − (2.1) Do Dw where
27
The buoyancy of the weighing vessel is ignored, since it is subtracted.
Buoyancy corrections are usually significant in glassware calibration.
Wvac = weight in vacuum, g Wair = observed weight in air, g Do = density of object Dw = density of standard weights 0.0012 = density of air The density of brass weights is 8.4 and that of stainless steel weights is 7.8. A calculation with water as the object will convince you that even here the correction will amount to only about one part per thousand.
Example 2.1 A convenient way to calibrate pipets is to weigh water delivered from them. From the exact density of water at the given temperature, the volume delivered can then be calculated. Suppose a 20-mL pipet is to be calibrated. A stoppered flask when empty weighs 29.278 g. Water is delivered into it from the pipet, and it now weighs 49.272 g. If brass weights are used, what is the weight of water delivered, corrected to weight in vacuum? Solution
The increase in weight is the weight of water in air: 49.272 − 29.278 = 19.994 g The density of water is 1.0 g/mL (to 2 significant figures from 10 to 30◦ C—see Table 2.4). Therefore, 0.0012 0.0012 − = 20.015 g Wvac = 19.994 + 19.994 1.0 8.4
The same buoyancy corrections apply for mechanical or electronic balances (which are calibrated with weights of known density).
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Example 2.2 Recalculate the weight of the water delivered by the pipet in Example 2.1, using stainless steel weights at density 7.8 g/cm3 . Solution
Do not round off until the end of the calculation. Then the same value results:
Wvac
0.0012 0.0012 = 19.994 + 19.994 − 1.0 7.8
= 20.015 g
This illustrates that the buoyancy corrections in Table 2.4 are valid for either type of weight (see Calibration of Glassware below).
SOURCES OF ERROR IN WEIGHING Several possible sources of error have been mentioned, including zero-point drift and buoyancy. Changes in ambient temperature or temperature of the object being weighed are probably the biggest sources of error, causing a drift in the zero or rest point due to convection-driven air currents. Hot or cold objects must be brought to ambient temperature before being weighed. Hygroscopic samples may pick up moisture, particularly in a high-humidity atmosphere. Exposure of the sample to air, prior to and during weighing, must be minimized.
GENERAL RULES FOR WEIGHING
Learn these rules!
The specific operation of your particular balance will be explained by your instructor. The main objectives are to protect all parts from dust and corrosion, avoid contamination or change in load (of sample or container), and avoid draft (air convection) errors. Some general rules you should familiarize yourself with before weighing with any type of analytical balance are:
1. Never handle objects to be weighed with the fingers. A piece of clean paper or tongs should be used. 2. Weigh at room temperature, and thereby avoid air convection currents. 3. Never place chemicals directly on the pan, but weigh them in a vessel (weighing bottle, weighing dish) or on powder paper. Always brush spilled chemicals off immediately with a soft brush. 4. Always close the balance case door before making the weighing. Air currents will cause the balance to be unsteady.
Although modern digital balances do not have user-manipulable weights, corrosion can still cause problems. Volatile corrosive substances (e.g., iodine or conc. HCl) should never be weighed in open containers in a balance.
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WEIGHING OF SOLIDS Solid chemical (nonmetal) materials are usually weighed and dried in a weighing bottle. Some of these are shown in Figure 2.3. They have standard tapered groundglass joints, and hygroscopic samples (which take on water from the air) can be weighed with the bottle kept tightly capped. Replicate weighings can be conveniently carried out by difference. The sample in the weighing bottle is weighed, and then a portion is removed (e.g., by tapping) and quantitatively transferred to a vessel appropriate for dissolving the sample. The weighing bottle and sample are reweighed, and from the difference in weight, the weight of sample is calculated. The next sample is removed and the weight is repeated to get its weight by difference, and so on. This is illustrated in the Laboratory Notebook example for the soda ash experiment. It is apparent that by this technique an average of only one weighing for each sample, plus one additional weighing for the first sample, is required. However, each weight represents the difference between two weighings, so that the total experimental error is given by the combined error of both weighings. Weighing by difference with the bottle capped must be used if the sample is hygroscopic or cannot otherwise be exposed to the atmosphere before weighing. If there are no effects from atmospheric exposure, the bottles need not be capped. For direct weighing, a weighing dish, weighing paper, or a weighing boat (all typically disposable) is used. The dish, paper, or boat is weighed empty and then with the added sample. This requires two weighings for each sample. The weighed sample is transferred by tapping. Direct weighing is satisfactory only if the sample is nonhygroscopic. When making very careful weighings (e.g., to a few tenths of a milligram or less), you must take care not to contaminate the weighing vessel with extraneous material that may affect its weight. Special care should be taken not to get perspiration from the hands on the vessel because this can be quite significant. It is best to handle the vessel with a piece of paper. Alternatively, finger cots may be used. These are similar to just the fingertip region of protective gloves. Solid samples must frequently be dried to a constant weight (e.g., ±0.5 mg for a 0.5 g sample). Highly insulating material, for example laboratory ware made from fluorocarbons, easily acquire static charge that affect weigh readings. Brushes with a built in source of ionizing radiation (http://www.amstat.com/solutions/staticmaster.html) that help dissipate such charges are recommended for gently swiping such objects before weighing.
Weighing by difference is required for hygroscopic samples.
Fig. 2.3.
Weighing bottles.
WEIGHING OF LIQUIDS Weighing of liquids is usually done by direct weighing. The liquid is transferred to a weighed vessel (e.g., a weighing bottle), which is capped to prevent evaporation during weighing, and is then weighed. If a liquid sample is weighed by difference by pipetting out an aliquot from the weighing bottle, the inside of the pipet must be rinsed several times after transferring. Care should be taken not to lose any sample from the tip of the pipet during transfer. TYPES OF WEIGHING——WHAT ACCURACY DO YOU NEED? There are two types of weighing done in analytical chemistry, rough or accurate. Rough weighings to two or three significant figures are normally used when the amount of substance to be weighed need only be known to within a few percent. Examples are reagents to be dissolved and standardized later against a known standard, or the apportioning of reagents that are to be dried and then later weighed accurately, or simply added as is, as for adjusting solution conditions. That is, only rough weighings
Only some weighings have to be done on an analytical balance, those involved in the quantitative calculations.
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are needed when the weight is not involved in the computation of the analytical result. Rough weighings need not be done on analytical balances but may be completed on a top-loading balance. Accurate weighings are reserved for obtaining the weight of a sample to be analyzed, the weight of the dried product in gravimetric procedures, or the weight of a dried reagent being used as a standard in a determination, all of which must generally be known to four significant figures or better to be used in calculating the analytical result. These are performed only on an analytical balance, usually to the nearest 0.1 mg. An exact predetermined amount of reagent is rarely weighed (e.g., 0.5000 g), but rather an approximate amount (about 0.5 g) is weighed accurately (e.g., to give 0.5129 g). Some chemicals are never weighed on an analytical balance. Sodium hydroxide pellets, for example, are so hygroscopic that they continually absorb moisture. The weight of a given amount of sodium hydroxide is not reproducible (and its purity is not known). To obtain a solution of known sodium hydroxide concentration, the sodium hydroxide is weighed on a rough balance and dissolved, and the solution is standardized against a standard acid solution.
2.4 Volumetric Glassware—Also Indispensible Although accurate volume measurements of solutions can be avoided in gravimetric methods of analysis, they are required for almost any other type of analysis involving solutions. VOLUMETRIC FLASKS Volumetric flasks contain an accurate volume.
Fig. 2.4.
Volumetric flask.
Volumetric flasks are used for the dilution of solutions to a certain volume. They come in a variety of sizes, from 1 mL to 2 L or more. A typical flask is shown in Figure 2.4. These flasks are designed to contain an accurate volume at the specified temperature (20 or 25◦ C) when the bottom of the meniscus (the concave curvature of the upper surface of water in a column caused by capillary action—see Figure 2.10) just touches the etched “fill” line across the neck of the glass. The coefficient of expansion of glass is small, and for ambient temperature fluctuations the volume can be considered constant. These flasks are marked with “TC” to indicate “to contain.” Other, less accurate containers, such as graduated cylinders, are also marked “TC.” Many of these are directly marked on the face by the manufacturer as to the uncertainty of the container measurement; for example, a 250 mL volumetric flask is “±0.24 mL,” or roughly a 0.1% error. Initially, a small amount of diluent (usually distilled water) is added to the empty flask. Reagent chemicals should never be added directly to a dry glass surface, as glass is highly absorbant. When using a volumetric flask, a solution should be prepared stepwise. The desired reagent chemical (either solid or liquid) to be diluted is added to the flask, and then diluent is added to fill the flask about two-thirds (taking care to rinse down any reagent on the ground glass lip). It helps to swirl the solution before diluent is added to the neck of the flask to obtain most of the mixing (or dissolving in the case of a solid). Finally, diluent is added so that the bottom of the meniscus is even with the middle of the calibration mark (at eye level). If there are any droplets of water on the neck of the flask above the meniscus, take a piece of tissue and blot these out. Also, dry the ground-glass stopper joint. The solution is finally thoroughly mixed as follows. Keeping the stopper on securely by using the thumb or palm of the hand, invert the flask and swirl or shake it vigorously for 5 to 10 s. Turn right side up and allow the solution to drain from the neck of the flask. Repeat at least 10 times.
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Note. When preparing the solution of an expensive chemical, should the volume of liquid go over the calibration mark, it is still possible to save the solution as follows. Paste against the neck of the flask a thin strip of paper and mark on it with a sharp pencil the position of the meniscus, avoiding parallax error. After removing the thoroughly mixed solution from the flask, fill the flask with water to the calibration mark. Then by means of a buret or small volume graduated pipet, add water to the flask until the meniscus is raised to the mark on the strip of paper. Note and record the volume so added and use it to mathematically correct the concentration calculation. For an inexpensive chemical, start over. If the volume goes over the mark, you cannot accurately calculate concentration without determining how far over the mark you went. Be very careful and patient when filling volumetric flasks, especially when the components in the flask are irreplaceable or expensive.
V
V
PIPETS
Fig. 2.5.
Transfer or volumetric
pipets.
Volumetric pipets deliver an accurate volume.
Tekk
N0. 37020
1/10IN1/100ml
M
TO 20° C
The pipet is used to transfer a particular volume of solution. As such, it is often used to deliver a certain fraction (aliquot) of a solution. To ascertain the fraction, the original volume of solution from which the aliquot is taken must be known, but it need not all be present, so long as it has not evaporated or been diluted. There are two common types of pipets, the volumetric, or transfer, pipet and the measuring or graduated pipet (see Figures 2.5 and 2.6). Variations of the latter are also called clinical, or serological, pipets. Pipets are designed to deliver a specified volume at a given temperature, and they are marked “TD.” Again, the volume can be considered to be constant with small changes in temperature. Pipets are calibrated to account for the drainage film remaining on the glass walls. This drainage film will vary somewhat with the time taken to deliver, and usually the solution is allowed to drain under the force of gravity and the pipet is removed shortly after the solution is delivered. A uniform drainage time should be adopted. The volumetric pipet is used for accurate measurements since it is designed to deliver only one volume and is calibrated at that volume. Accuracy to four significant figures is generally achieved, although with proper calibration, five figures may be obtained if necessary. See the table on the back cover for tolerances of class A transfer pipets. Measuring pipets are straight-bore pipets that are marked at different volume intervals. These are not as accurate because nonuniformity of the internal diameter of the device will have a relatively larger effect on total volume than is the case for pipets with a bulb shape. Also, the drainage film will vary with the volume delivered. At best, accuracy to three significant figures can be expected from these pipets, unless you make the effort to calibrate the pipet for a given volume delivered. Most volumetric pipets are calibrated to deliver with a certain small volume remaining in the tip. This should not be shaken or blown out. In delivering, the pipet is held vertically and the tip is touched on the side of the vessel to allow smooth delivery without splashing and so that the proper volume will be left in the tip. The forces of attraction of the liquid on the wall of the vessel will draw out a part of this. Some pipets are blowout types (including measuring pipets calibrated to the entire tip volume). The final volume of solution must be blown out from the tip to deliver the calibrated amount. These pipets are easy to identify, as they will always have one or two ground bands or rings around the top. (These are not to be confused with a colored ring that is used only as a color coding for the volume of the pipet.) The solution is not blown out until it has been completely drained by gravity. Blowing to increase the rate of delivery will change the volume of the drainage film. Volumetric pipets are available in sizes of 100 to 0.5 mL or less. Measuring and serological pipets range from a total capacity of 25 to 0.1 mL. Measuring pipets can be
ml 0
1
0 2
.01 7
.10 8
9
10
Fig. 2.6.
Measuring pipets.
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used for accurate measurements, especially for small volumes, if they are calibrated at the particular volume wanted. The larger measuring pipets usually deliver too quickly to allow drainage as fast as the delivery, and they have too large a bore for accurate reading. In using a pipet, one should always wipe the outside of the tip dry after filling. If a solvent other than water is used, or if the solution is viscous, pipets must be recalibrated for the new solvent or solution to account for difference in drainage rate. Pipets are filled by suction, using a rubber pipet bulb, a pipet pump, or other such pipetting device. Before using a pipet, practice filling and dispensing with water. No solution should be pipetted by mouth. SYRINGE PIPETS
Syringe pipets are useful for delivering microliter volumes.
The volume may not be accurately known, but it is reproducible.
Joseph Gay-Lussac (1778–1850) designed the first buret and named the buret and pipet.
These can be used for both macro and micro volume measurements. The calibration marks on the syringes may not be very accurate, but the reproducibility can be excellent if automatic delivery is used, such as a spring-loaded device that draws the plunger up to the same preset level each time. The volume delivered in this manner is free from drainage errors because the solution is forced out by the plunger. The volume delivered can be accurately calibrated. Microliter syringe pipets are used for introduction of samples into gas chromatographs. A typical syringe is illustrated in Figure 2.7. They are fitted with a needle tip, and the tolerances are as good as those found for other micropipets. In addition, any desired volume throughout the range of the syringe can be delivered. Syringes are available with a total volume as small as 0.5 μL, in which a wire plunger travels within the needle and so the entire syringe volume is within the needle. The above syringe pipets are useful for accurate delivery of viscous solutions or volatile solvents; with these materials the drainage film would be a problem in conventional pipets. Syringe pipets are well suited to rapid delivery and also for thorough mixing of the delivered solution with another solution as a result of the rapid delivery. A second type of syringe pipet is that shown in Figure 2.8. This type is convenient for rapid, one-hand dispensing of fixed (or variable) volumes in routine procedures and is widely used in the analytical chemistry laboratory. It contains a disposable nonwetting plastic tip (e.g., polypropylene) to reduce both film error and contamination. A thumb button operates a spring-loaded plunger, which stops at an intake or a discharge stop; the latter stop is beyond the former to ensure complete delivery. The sample never contacts the plunger and is contained entirely in the plastic tip. These pipets are available in volumes of 0.1 to 5000 μL and are reproducible to 1 to 2% or better, depending on the volume. Variable volume pipettes of this type typically deliver volumes across a defined range, such as 0.1 to 10 μL, 20 to 200 μL, 100 to 1000 μL, and 500 to 5000 μL. It is important to be consistent in your use of different pipettes for related procedures (e.g., always use a single draw from the 20 to 200 μL pipette to dispense 100 μL) to maintain consistency. Sometimes, the actual volume delivered by these and other micropipets does not need to be known because they are used in relative measurements. For example, the same pipet may be used to deliver a sample and an equal volume of a standard solution for calibrating the instrument used for the measurement. Precision in delivery is usually more important than the absolute volume delivered. The European standard for pipet calibration in Europe is the German DIN 126650 (or a similar international standard ISO 8655). Calibrations are based on gravimetric testing (weighing of water). The DIN standard does not give separate limits for accuracy and precision, but rather uses a combined error limit equal to percent accuracy plus 2 times the standard deviation, that is, it gives a range in which we are 95% confident the delivery will fall (see
Fig. 2.7. syringe.
Hamilton microliter
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Fig. 2.8.
Single-channel and multichannel digital displacement pipets and microwell plates. (Courtesy of Thermo Fisher Scientific.)
Chapter 3 for a discussion of standard deviation and confidence limits). Table 2.2 lists the DIN error limits for single-channel displacement pipets. Table 2.3 lists accuracies and precisions for a typical variable volume single-channel pipet. Besides the manually operated syringes, there are electronically controlled and variable-volume motor-driven syringes available for automated repetitive deliveries. When syringes are used in electrically driven syringe pumps for very slow infusion of solution, there can be stick–slip behavior, resulting in pulsed flow. Inexpensive
Table 2.2
DIN 12650 Error Limits for Single-Channel Fixed-Volume Air Displacement Pipetsa Nominal Volume (μL)
Maximum Error (μL)
Relative Error (%)
1 2 5 10 20 50 100 200 500 1000 2000 5000 10000
±0.15 ±0.20 ±0.30 ±0.30 ±0.40 ±0.80 ±1.50 ±2.00 ±5.00 ±10.00 ±20.00 ±50.00 ±100.00
±15.0 ±10.0 ±6.0 ±3.0 ±2.0 ±1.6 ±1.5 ±1.0 ±1.0 ±1.0 ±1.0 ±1.0 ±1.0
a These limits apply to manufacturers with a controlled environment. If the tests are performed by a user in a normal laboratory environment, the limits in the table may be doubled. Courtesy of Thermo Labsystems Oy, Finland.
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Table 2.3
Accuracy and Precision Limits for Single-Channel Variable-Volume Finnpipettes Model F1 Range (μL)
Increment (μL)
Volume (μL)
0.2–2 μL
0.002 μL
0.5–5 μL
0.01 μL
1–10 μL
0.02 μL
1–10 μL
0.02 μL
2–20 μL
0.02 μL
2–20 μL
0.02 μL
5–50 μL
0.1 μL
5–50 μL
0.1 μL
10–100 μL
0.2 μL
20–200 μL
0.2 μL
30–300 μL
1 μL
100–1000 μL
1 μL
0.5–5 ml
0.01 ml
1–10 ml
0.02 ml
2 1 0.2 5 2.5 0.5 10 5 1 10 5 1 20 10 2 20 10 2 50 25 5 50 25 5 100 50 10 200 100 20 300 150 30 1000 500 100 5000 2500 500 10000 5000 1000
Precisiona
Accuracy (μL)
(%)
s.d. (μL)
CV (%)
±0.050 ±0.040 ±0.024 ±0.080 ±0.0625 ±0.030 ±0.100 ±0.080 ±0.025 ±0.100 ±0.080 ±0.035 ±0.20 ±0.15 ±0.06 ±0.20 ±0.15 ±0.06 ±0.30 ±0.25 ±0.15 ±0.30 ±0.25 ±0.15 ±0.80 ±0.60 ±0.30 ±1.2 ±1.0 ±0.36 ±1.8 ±1.5 ±0.45 ±6.0 ±4.0 ±1.0 ±25.0 ±17.5 ±10.0 ±50.0 ±40.0 ±20.0
±2.50 ±4.00 ±12.00 ±1.50 ±2.50 ±6.00 ±1.00 ±1.50 ±2.50 ±1.00 ±1.50 ±3.50 ±1.00 ±1.50 ±3.00 ±1.00 ±1.50 ±3.00 ±0.60 ±1.00 ±3.00 ±0.60 ±1.00 ±3.00 ±0.80 ±1.20 ±3.00 ±0.60 ±1.00 ±1.80 ±0.60 ±1.00 ±1.50 ±0.60 ±0.80 ±1.00 ±0.50 ±0.70 ±2.00 ±0.50 ±0.80 ±2.00
0.040 0.040 0.020 0.050 0.0375 0.025 0.050 0.040 0.020 0.080 0.040 0.030 0.08 0.06 0.05 0.08 0.06 0.05 0.15 0.13 0.125 0.15 0.13 0.125 0.20 0.20 0.10 0.4 0.4 0.14 0.6 0.6 0.18 2.0 1.5 0.6 10.0 7.5 4.0 20.0 15.0 8.0
2.00 3.50 10.00 1.00 1.50 5.00 0.50 0.80 2.00 0.80 0.80 3.00 0.40 0.60 2.50 0.40 0.60 2.50 0.30 0.50 2.50 0.30 0.50 2.50 0.20 0.40 1.00 0.20 0.40 0.70 0.20 0.40 0.60 0.20 0.30 0.60 0.20 0.30 0.80 0.20 0.30 0.80
a s.d. = standard deviation, CV = coefficient of variation. From https://fscimage.fishersci.com/images/D11178∼.pdf
automated dispensers can be laboratory made from syringes [see for example, “Inexpensive Automated Electropneumatic Syringe Dispenser”, P. K. Dasgupta and J. R. Hall, Anal. Chim. Acta 221 (1989) 189.] Also, you may purchase pipets with multiple syringes for simultaneous delivery, with for example, 12 or 16 channels. These are useful for delivering solutions into microwell plates used in biotechnology
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or clinical chemistry laboratories that process thousands of samples (Figure 2.8). You may find more information on displacement pipets from representative manufacturers, for example, www.thermoscientific.com/finnpipette or www.eppendorf.com. BURETS A buret is used for the accurate delivery of a variable amount of solution. Its principal use is in titrations, where a standard solution is added to the sample solution until the end point (the detection of the completion of the reaction) is reached. The conventional buret for macrotitrations is marked in 0.1-mL increments from 0 to 50 mL; one is illustrated in Figure 2.9. The volume delivered can be read to the nearest 0.01 mL by interpolation actually (good to about ±0.02 or ±0.03 mL). Burets are also obtainable in 10-, 25-, and 100-mL capacities, and microburets are available in capacities of down to 2 mL, where the volume is marked in 0.01-mL increments and can be estimated to the nearest 0.001 mL. Ultramicroburets of 0.1-mL capacity graduated in 0.001-mL (1-μL) intervals are used for microliter titrations. Drainage film is a factor with conventional burets, as with pipets, and this can be a variable if the delivery rate is not constant. The usual practice is to deliver at a fairly slow rate, about 15 to 20 mL per minute, and then to wait several seconds after delivery to allow the drainage to “catch up.” In actual practice, the rate of delivery is only a few drops per minute near the end point, and there will be no time lag between the flow rate and the drainage rate. As the end point is approached, fractions of a drop are delivered by just opening, or “cracking,” the stopcock and then touching the tip of the buret to the wall of the titration vessel. The fraction of the drop is then washed down into the solution with distilled water.
ml 0 1 2 3 4 5 6 7 8 9
48 49 50
DELIVERS AT
20°C
CARE AND USE OF VOLUMETRIC GLASSWARE We have mentioned a few precautions in the use of volumetric flasks, pipets, and burets. Your laboratory instructor will supply you with detailed instructions in the use of each of these tools. A discussion of some general precautions and good laboratory technique follows. Cleanliness of glassware is of the utmost importance. If films of dirt or grease are present, liquids will not drain uniformly and will leave water breaks or droplets on the walls. Under such conditions the calibration will be in error. Initial cleaning should be by repeated rinses with laboratory detergent and then water. Then try cleaning with dilute nitric acid and rinse with more water. Use of a buret or test tube brush aids the cleaning of burets and necks of volumetric flasks—but be careful of scratching the interior walls. Pipets should be rotated to coat the entire surface with detergent. There are commercial cleaning solutions available that are very effective. Pipets and burets should be rinsed at least twice with the solution with which they are to be filled. If they are wet, they should be rinsed first with water, if they have not been already, and then a minimum of three times with the solution to be used; about one-fifth the volume of the pipet or buret is adequate for each rinsing. A volumetric flask, if it is wet from a previously contained solution, is rinsed with three portions of water only since later it will be filled to the mark with water. It need not be dry. Note that analytical glassware should not be subjected to the common practice employed in organic chemistry laboratories of drying either in an oven (this can affect the volume of calibrated glassware) or by drying with a towel or by rinsing with a volatile organic solvent such as acetone (which can cause contamination). The glassware usually does not have to be dried. The preferred procedure is to rinse it with the solution that will fill it. Care in reading the volume will avoid parallax error, that is, error due to incorrect alignment of the observer’s eye, the meniscus, and the scale. Correct position is with your eye at the same level as the menicus. If the eye level is above the
Fig. 2.9.
Typical buret.
Rinse pipets and burets with the solution to be measured.
Avoid parallax error in reading buret or pipet volumes.
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meniscus, the volume read will be smaller than that taken; the opposite will be true if the eye level is too low. After glassware is used, it can usually be cleaned sufficiently by immediate rinsing with water. If the glassware has been allowed to dry, it should be cleaned with detergent. Volumetric flasks should be stored with the stopper inserted, and preferably filled with distilled water. Burets should be filled with distilled water and stoppered with a rubber stopper when not in use. There are commercial glassware washing machines to automate the cleaning of glassware. These use detergents and deionized water for cleaning and rinsing. See, for example, L. Choplo, “The Benefits of Machine Washing Laboratory Glassware Versus Hand Washing,” Amer. Lab. October (2008) 6 (http://new.americanlaboratory.com/ 914-Application-Notes/34683-The-Benefits-of-Machine-Washing-Laboratory-Glass ware-Versus-Hand-Washing/), and M. J. Felton, “Labware Washers,” Today’s Chemist at Work, November (2004) 43 (http://pubs.acs.org/subscribe/archive/tcaw/13/ i11/pdf/1104prodprofile.pdf). GENERAL TIPS FOR ACCURATE AND PRECISE TITRATING If the buret contains a Teflon stopcock, it does not require lubrication.
0 1
Fig. 2.10.
Meniscus illuminator
for buret.
Fig. 2.11. titration.
Proper technique for
Your buret probably has a Teflon stopcock, and this will not require lubrication. Make sure it is secured tightly enough to prevent leakage, but not so tight as to make rotation hard. If your buret has a ground-glass stopcock, you may have to grease the stopcock. A thin layer of stopcock grease (not silicone lubricant) is applied uniformly to the stopcock, using very little near the hole and taking care not to get any grease in the hole. The stopcock is inserted and rotated. There should be a uniform and transparent layer of grease, and the stopcock should not leak. If there is too much lubricant, it will be forced into the barrel or may work into the buret tip and clog it. Grease can be removed from the buret tip and the hole of the stopcock by using a fine wire. Next, we fill the buret with the solution it will deliver. The buret is filled above the zero mark and the stopcock is opened to fill the tip. Check the tip for air bubbles. If any are present, they may work out of the tip during the titration, causing an error in reading. Work air bubbles out by rapid opening and closing of the stopcock to squirt the titrant through the tip or tapping the tip while solution is flowing. No bubbles should be in the barrel of the buret. If there are, the buret is probably dirty. The initial reading of the buret is taken by allowing it to drain slowly to the zero mark. Wait a few seconds to make certain the drainage film has caught up to the meniscus. Read the buret to the nearest 0.01 mL (for a 50-mL buret). The initial reading may be 0.00 mL or greater. The reading is best taken by placing your finger just in back of the meniscus or by using a meniscus illuminator (Figure 2.10). The meniscus illuminator has a white and a black field, and the black field is positioned just below the meniscus. Avoid parallax error by making the reading at eye level. The titration is performed with the sample solution in an Erlenmeyer flask. The flask is placed on a white background, and the buret tip is positioned within the neck of the flask. The flask is swirled with the right hand while the stopcock is manipulated with the left hand (Figure 2.11), or whatever is comfortable. This grip on the buret maintains a slight inward pressure on the stopcock to ensure that leakage will not occur. The solution can be more efficiently stirred by means of a magnetic stirrer and stirring bar. As the titration proceeds, the indicator changes color in the vicinity where the titrant is added, owing to local excesses; but it rapidly reverts to the original color as the titrant is dispersed through the solution to react with the sample. As the end point is approached, the return to the original color occurs more slowly, since the dilute solution must be mixed more thoroughly to consume all the titrant. At this point, the titration should be stopped and the sides of the flask washed down with distilled water from the wash bottle. A drop from the buret is about 0.02 to 0.05 mL, and the volume
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is read to the nearest 0.02 mL. It is therefore necessary to split drops near the end point. This can be done by slowly turning the stopcock until a fraction of a drop emerges from the buret tip and then closing it. The fraction of drop is touched off onto the wall of the flask and is washed into the flask with the wash bottle, or it is transferred with a glass stirring rod. There will be a sudden and “permanent” (lasting at least 30 s) change in the color at the end point when a fraction of a drop is added. The titration is usually performed in triplicate. After performing the first titration, you can calculate the approximate volume for the replicate titrations from the weights of the samples and the molarity of the titrant. This will save time in the titrations. The volume should not be calculated to nearer than 0.1 mL in order to avoid bias in the reading. After a titration is complete, unused titrant should never be returned to the original bottle but should be discarded. If the titrant isn’t between pH 4 and 8 and on the short list of substances cleared to go down the drain, it should be disposed of in a recycle container. If a physical property of the solution, such as potential, is measured to detect the end point, the titration is performed in a beaker with magnetic stirring so electrodes can be placed in the solution.
37
Subsequent titrations can be speeded up by using the first to approximate the end-point volumes.
TOLERANCES AND PRECISION OF GLASSWARE The National Institute of Standards and Technology (NIST) has prescribed certain tolerances, or absolute errors, for different volumetric glassware, and some of these are listed on the back cover of the text. For volumes greater than about 25 mL, the tolerance is within 1 part per thousand relative, but it is larger for smaller volumes. The letter “A” stamped on the side of a volumetric flask, buret, or pipet indicates that it complies with class A tolerances. This says nothing about the precision of delivery. Volumetric glassware that meets NIST specifications or that is certified by NIST can be purchased, but at a significantly higher price than uncertified glassware. Less expensive glassware may have tolerances double those specified by NIST. It is a simple matter, however, to calibrate this glassware to an accuracy as good as or exceeding the NIST specifications (see Experiment 2). The precision of reading a 50-mL buret is about ±0.02 mL. Since a buret is always read twice, the total absolute uncertainty may be as much as ±0.04 mL. The relative uncertainty will vary inversely with the total volume delivered. It becomes apparent that a titration with a 50-mL buret should involve about 40 mL titrant to achieve a precision of 1 ppt. Smaller burets can be used for increased precision at smaller volumes. Pipets will also have a certain precision of reading, but only one reading is required for volumetric pipets. CALIBRATION OF GLASSWARE— —FOR THE ULTIMATE ACCURACY Example 2.1 illustrated how Equation 2.1 may be used in the calibration of glassware, to correct for buoyancy of the water used for calibration, that is, to correct to weight in vacuum. Dividing the weight of the water in vacuum by its density at the given temperature will convert it to volume. Table 2.4 lists the calculated volumes for a gram of water in air at atmospheric pressure for different temperatures, corrected for buoyancy with stainless steel weights of density 7.8 g/cm3 . These are used to give the volume of the glassware being calibrated, from the weight of water contained or delivered by the glassware. (The values are not significantly different for brass weights of 8.4 g/cm3 density. See Example 2.2.) The glass volumes calculated for the standard temperature of 20◦ C include slight adjustments for borosilicate glass (Pyrex or Kimax) container expansion or contraction with temperature changes (volumetric glassware has a cubical coefficient of expansion of about 0.000025 per degree centigrade, resulting in changes of about
Class A glassware is accurate enough for most analyses. It can be calibrated to NIST specifications.
The variances or the uncertainties in each reading are additive. See propagation of error, Chapter 3.
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Table 2.4
Glassware Calibration 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44
A B C D E F G Table 2.4 Glassware Calibration Weight in vacuum assuming stainless steel weights, density 7.8 g/mL. Glass expansion for borosilicate glass, 0.000025 mL/mL/ oC. The actual spreadsheet is available on the website (Table 2.4). Save it to your desktop, and use it to calculate calibrated volumes of glassware. Substitute into the appropriate Cell B, at temperature t, the weight of water in air, obtained from the glassware at the temperature of the measurement (Cell A). o The calibration volume at the temperature, t (Cell D), and at 20 C (Cell F), is calculated. Round the calculated values to the appropriate number of significant figures, usually four or five. t, oC
Wt. H 2O
Wt. in
Vol. at t,
Glass expansion, Vol. at 20°C Density,
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
in air, g 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
vacuum, g 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010 1.0010
mL 1.0013 1.0014 1.0015 1.0017 1.0018 1.0019 1.0021 1.0023 1.0025 1.0026 1.0028 1.0031 1.0033 1.0035 1.0038 1.0040 1.0043 1.0045 1.0048 1.0051 1.0054
at 20oC, mL -0.000250 -0.000225 -0.000200 -0.000175 -0.000150 -0.000125 -0.000100 -0.000075 -0.000050 -0.000025 0.000000 0.000025 0.000050 0.000075 0.000100 0.000126 0.000151 0.000176 0.000201 0.000226 0.000251
mL 1.0016 1.0017 1.0017 1.0018 1.0020 1.0021 1.0022 1.0023 1.0025 1.0027 1.0028 1.0030 1.0032 1.0034 1.0037 1.0039 1.0041 1.0044 1.0046 1.0049 1.0052
H
g/mL 0.9997026 0.9996081 0.9995004 0.9993801 0.9992474 0.9991026 0.9989460 0.9987779 0.9985896 0.9984082 0.9982071 0.9979955 0.9977735 0.9975415 0.9972995 0.9970479 0.9967867 0.9965162 0.9962365 0.9959478 0.9956502
Formulas are entered into the boldface cells above as indicated below. They are copied down for all temperatures. See Chapter 3 for setting up a spreadsheet. Cell C14: W vac = W air + W air (0.0012/D o – 0.0012/D w) = W air (0.0012/1.0 + 0.0012/7.8) = B14+B14*(0.0012/1.0-0.0012/7.8) Copy down Cell D14: V t (mL)= W vac ,t (g)/Dt (g/mL) = C14/G14 Copy down Cell E14: Glass expans. = (t – 20) (deg) x 0.000025 (mL/mL/deg) x Vt (mL) = (A14-20)*0.000025*D14 Copy down Cell F14: V 20o = V t – Glass exp = D14-E14 Copy down
0.0025% per degree; for 1 mL, this is 0.000025 mL per degree). Water expands about 0.02% per degree around 20◦ C. Volume (concentration) corrections may be made using the water density data in Table 2.4, taking the ratios of the relative densities. In the textbook website, the Table 2.4 spreadsheet is available, with formulas as indicated in the table. You can substitute specific weights of water in air, obtained from a flask, pipet, or buret, in cell B at the temperature of measurement to obtain the calculated calibration volume at temperature, t, and for 20◦ C. We describe the use of spreadsheets in Chapter 3. The book website also has a table and figure of the percent error for weight in vacuum as a function of sample density. For those of you who live at high elevations, the density of air is slightly less than 0.0012 g/mL (at sea level), for example, about 0.0010 g/mL at 5,000 feet elevation. You may, in your downloaded spreadsheet of Table 2.4, substitute the appropriate value in the the formula in cell C14, and copy the new formula down the column.
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Example 2.3 (a) Use Table 2.4 to calculate the volume of the 20-mL pipet in Example 2.2 (steel weights), from its weight in air. Assume the temperature is 23◦ C. (b) Give the corresponding volume at 20◦ C as a result of glass contraction. (c) Compare with the volume calculated using the weight in air with that calculated using the weight in vacuum and the density in water (Example 2.2). Solution
(a) From Table 2.4, the volume per gram in air is 1.0035 mL at 23◦ C: 19.994 g × 1.0035 mL/g = 20.064 mL (b) The glass contraction at 20◦ C relative to 23◦ C is 0.0015 mL (0.000025 mL/mL/◦ C × 20 mL × 3◦ C), so the pipet volume at 20◦ C is 20.062 mL. (c) The density of water at 23◦ C is 0.99754 g/mL, so from the weight in vacuum: 20.015 g/0.99754 g/mL = 20.064 mL The same value is obtained.
Example 2.4 You prepared a solution of hydrochloric acid and standardized it by titration of primary standard sodium carbonate. The temperature during the standardization was 23◦ C, and the concentration was determined to be 0.11272 M. The heating system in the laboratory malfunctioned when you used the acid to titrate an unknown, and the temperature of the solution was 18◦ C. What was the concentration of the titrant? Solution
M18◦ = M23◦ × (D18◦ /D23◦ ) = 0.11272 × (0.99859/0.99754) = 0.11284 M (See Chapter 3 for significant figures and the meaning of the subscript numbers.)
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TECHNIQUES FOR CALIBRATING GLASSWARE You generally calibrate glassware to five significant figures, the maximum precision you are likely to attain in filling or delivering solutions. Hence, your net weight of water needs to be five figures. If the glassware exceeds 10 mL, this means weighing to 1 mg is all that is needed. This can be readily and conveniently accomplished using a top-loading balance, rather than a more sensitive analytical balance. [Note: If the volume number is large without regard to the decimal, e.g., 99, then four figures will suffice—see Chapter 3 discussion on significant figures. For example, a 10-mL pipette may be calibrated and shown to actually deliver 9.997 mL. This is as accurate as if the pipette was determined to deliver 10.003 mL (the last significant figure in both cases is one part in 10,000)]. 1. Volumetric Flask Calibration. To calibrate a volumetric flask, first weigh the clean, dry flask and stopper. Then fill it to the mark with distilled water. There should be no droplets on the neck. If there are, blot them with tissue paper. The flask and
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water should be equilibrated to room temperature. Weigh the filled flask, and then record the temperature of the water to 0.1◦ C. The increase in weight represents the weight in air of the water contained by the flask. 2. Pipet Calibration. To calibrate a pipet, weigh a dry Erlenmeyer flask with a rubber stopper or a weighing bottle with a glass stopper or cap, depending on the volume of water to be weighed. Fill the pipet with distilled water (whose temperature you have recorded) and deliver the water into the flask or bottle, using proper pipetting technique, and quickly stopper the container to avoid evaporation loss. Reweigh to obtain the weight in air of water delivered by the pipet. 3. Buret Calibration. Calibrating a buret is similar to the procedure for a pipet, except that several volumes will be delivered. The internal bore of the buret is not perfectly cylindrical, and it will be a bit “wavy,” so the actual volume delivered will vary both plus and minus from the nominal volumes marked on the buret, as increased volumes are delivered. You will ascertain the volume at 20% full-volume increments (e.g., each 10 mL, for a 50-mL buret) by filling the buret each time and then delivering the nominal volume into a dry flask. (The buret is filled each time to minimize evaporation errors. You may also make successive deliveries into the same flask, i.e., fill the buret only once. Make rapid deliveries.) Since the delivered volume does not have to be exact, but close to the nominal volume, you can make fairly fast deliveries, but wait about 10 to 20 s for film drainage. Prepare a plot of volume correction versus nominal volume and draw straight lines between each point. Interpolation is made at intermediate volumes from the lines. Typical volume corrections for a 50-mL buret may range up to ca. 0.05 mL, plus or minus.
Professor’s Favorite Experiment Contributed by Professor Alex Scheeline, University of Illinois This illustrates a possibly more precise alternative to buret calibration, as described here, or determines the optimal approach. Place a beaker of water in the weighing compartment of the balance so that the air becomes saturated with water vapor. Fill the buret. Place it through the top of the balance housing so the effluent will enter a beaker on the weighing table. Tare the balance and record the reading on the buret. Now drain a few tenths of a milliliter from the buret. Record the volume, and once the balance has settled, record the mass. Continue to do this so that one has ca. 100 points from the drainage of the buret. Repeat three times. Compute the volume delivered for each reading. Plot. Now answer these questions: ●
●
●
● ●
Which is more precise: the fiducial marking of the buret, as manufactured, or your attempt to calibrate the buret? Is there a smooth curve through the data, or is there significant variation from a smooth curve? Is there any indication as to what the appropriate measurement interval should be to obtain calibration to ±0.01 mL while minimizing the number of calibration points? How does the calibration of your buret compare with that of two other students? What is the best precision one can get if one ignores the results of buret calibaration?
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Example 2.5 You calibrate a 50-mL buret at 10-mL increments, filling the buret each time and delivering the nominal volume, with the following results: Buret Reading (mL) Weight H2 O Delivered (g) 10.02 10.03 20.08 20.03 29.99 29.85 40.06 39.90 49.98 49.86 Construct a plot of volume correction versus volume delivered. The temperature of the water is 20◦ C and stainless steel weights are used. Solution
From Table 2.4 (or use Table 2.4 from the website for automatic calculation of volumes): Wvac = 10.03 + 10.03(0.00105) = 10.03 + 0.01 = 10.04 g Vol. = 10.04 g/0.9982 g/mL = 10.06 mL Likewise, for the others, we construct the table: Nominal Volume (mL)
Actual Volume (mL)
Correction (mL)
10.02 20.08 29.99 40.06 49.98
10.06 20.09 29.93 40.01 50.00
+0.04 +0.01 –0.06 –0.05 +0.02
Prepare a graph of nominal volume (y axis) versus correction volume. Use 10, 20, 30, 40, and 50 mL as the nominal volumes. SELECTION OF GLASSWARE— —HOW ACCURATE DOES IT HAVE TO BE? As in weighing operations, there will be situations where you need to accurately know volumes of reagents or samples measured or transferred (accurate measurements), and others in which only approximate measurements are required (rough measurements). If you wish to prepare a standard solution of 0.1 M hydrochloric acid, it can’t be done by measuring an accurate volume of concentrated acid and diluting to a known volume because the concentration of the commercial acid is not known adequately. Hence, an approximate solution is prepared that is then standardized. We see in the table on the inside back cover that the commercial acid is about 12.4 M. To prepare 1 L of a 0.1 M solution, about 8.1 mL needs to be taken and diluted. It would be a waste of time to measure this (or the water used for dilution) accurately. A 10-mL graduated cylinder or 10-mL measuring pipet will suffice, and the acid can be diluted in an ungraduated 1-L bottle. If, on the other hand, you wish to dilute a stock standard solution accurately, then a transfer pipet must be used and the dilution must be done in a volumetric flask. Any volumetric measurement that is a part of the actual analytical measurement must be done with the accuracy and precision required of the analytical measurement. This generally means four-significant-figure accuracy, and transfer pipets and volumetric flasks are required. This includes taking an accurate portion of a sample, preparation of
Only certain volumes need to be measured accurately, those involved in the quantitative calculations.
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a standard solution from an accurately weighed reagent, and accurate dilutions. Burets are used for accurate measurement of variable volumes, as in a titration. Preparation of reagents that are to be used in an analysis just to provide proper solution conditions (e.g., buffers for pH control) need not be prepared highly accurately, and less accurate glassware can be used, for example, graduated cylinders.
2.5 Preparation of Standard Base Solutions
Remove Na2 CO3 by preparing a saturated solution of NaOH.
See Experiment 7 for preparing and standardizing sodium hydroxide.
Sodium hydroxide is usually used as the titrant when a base is required. It contains significant amounts of water and sodium carbonate, and so it cannot be used as a primary standard. For accurate work, the sodium carbonate must be removed from the NaOH because it reacts to form a buffer that decreases the sharpness of the end point. In addition, an error will result if the NaOH is standardized using a phenolphthalein end point (in which case the CO3 2− is titrated only to HCO3 − ), and then a methyl orange end point is used in the titration of a sample (in which case the CO3 2− is titrated to CO2 ). In other words, the effective molarity of the base has been increased, owing to further reaction of the HCO3 − . Sodium carbonate is essentially insoluble in nearly saturated sodium hydroxide. It is conveniently removed by dissolving the weighed NaOH in a volume (milliliters) of water equal to its weight in grams. The insoluble Na2 CO3 can be allowed to settle for several days, and then the clear supernatant liquid can be carefully decanted,1 or it can be filtered in a Gooch crucible with a quartz fiber filter mat (do not wash the filtered Na2 CO3 ). This procedure does not work with KOH because K2 CO3 remains soluble. Water dissolves CO2 from the air. In many routine determinations not requiring the highest degree of accuracy, carbonate or CO2 impurities in the water will result in an error that is small enough to be considered negligible. For the highest accuracy, however, CO2 should be removed from all water used to prepare solutions for acid–base titrations, particularly the alkaline solutions. This is conveniently done by boiling the water and then cooling it under a cold-water tap. Sodium hydroxide is usually standardized by titrating a weighed quantity of primary standard potassium acid phthalate (KHP), which is a moderately weak acid (Ka = 4 × 10−6 ), approximately like acetic acid; a phenolphthalein end point is used. The sodium hydroxide solution should be stored in a plastic-lined glass bottle to prevent absorption of CO2 from the air. If the bottle must be open (e.g., a siphon bottle), the opening is protected with an Ascarite II (fibrous silicate impregnated with NaOH) or soda-lime [Ca(OH)2 and NaOH] tube.
2.6 Preparation of Standard Acid Solutions Hydrochloric acid is the usual titrant for the titration of bases. Most chlorides are soluble, and few possible side reactions with this acid. It is convenient to handle. It is not a primary standard (although constant-boiling HCl, which is a primary standard, can be prepared), and an approximate concentration is prepared simply by diluting the concentrated acid. For most accurate work, the water used to prepare the solution should be boiled, although use of boiled water is not so critical as with NaOH; CO2 will have a low solubility in strongly acidic solutions and will tend to escape during shaking of the solution. 1 Concentrated alkali attacks glass whereas carbon dioxide can permeate through most organic polymers. One
solution out of this dilemma is to insert a polyethylene bag inside a glass bottle and use a rubber stopper.
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Primary standard sodium carbonate is usually used to standardize HCl solutions. The disadvantage is that the end point is not sharp unless an indicator such as methyl red or methyl purple is used and the solution is boiled at the end point. A modified methyl orange end point may be used without boiling, but this is not so sharp. Another disadvantage is the low formula weight of Na2 CO3 . Tris-(hydroxymethyl)aminomethane (THAM), (HOCH2 )3 CNH2 , is another primary standard that is more convenient to use. It is nonhygroscopic, but it is still a fairly weak base (Kb = 1.3 × 10−6 ) with a low molecular weight. The end point is not complicated by released CO2 , and it is recommended as the primary standard unless the HCl is being used to titrate carbonate samples. If a standardized NaOH solution is available, the HCl can be standardized by titrating an aliquot with the NaOH. The end point is sharp and the titration is more rapid. The NaOH solution is a secondary standard. Any error in standardizing this will be reflected in the accuracy of the HCl solution. The HCl is titrated with the base, rather than the other way around, to minimize absorption of CO2 in the titration flask. Phenolphthalein or bromothymol blue can be used as indicator.
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See Experiment 8 for preparing and standardizing hydrochloric acid.
A secondary standard is less accurate than a primary standard.
2.7 Other Apparatus—Handling and Treating Samples Besides apparatus for measuring mass and volume, there are a number of other items of equipment commonly used in analytical procedures. BLOOD SAMPLERS Syringes/needles are used to collect blood samples typically into evacuated glass vials, (Vacutainers).2 Stainless steel needles are generally used with glass or plastic syringes. These usually present no problem of contamination, although special precautions may be required for analysis of trace elements (e.g., metals) in the sample. Vacutainers are evacuated test tubes with a rubber cap. The needle is pushed through the cap after the other end has been inserted into the vein, and the blood is drawn into the evacuated tube. The tube may contain an anticoagulating agent to prevent clotting of the blood if plasma or whole blood samples are to be analyzed. A finger puncture, instead of a venipuncture, is used when small quantities of blood are to be collected for microprocedures. As much as ∼0.5 mL blood can be squeezed from the finger into a small collection tube by puncturing the finger with a sterilized sharp-pointed knifelike object. Finger puncture is commonly used in conjunction with sampling for glucose monitoring devices. DESICCATORS A desiccator is used to keep samples dry while they are cooling and before they are weighed and, in some cases, to dry a wet sample. Dried or ignited samples and vessels are cooled in the desiccator. A typical glass desiccator is shown in Figure 2.12. A desiccator is an airtight container that maintains an atmosphere of low humidity. A desiccant such as calcium chloride is placed in the bottom to absorb the moisture. This desiccant will have to be changed periodically as it becomes “spent.” It will usually become wet in appearance or caked from the moisture when it is time to be changed. A porcelain plate is usually placed in the desiccator to support weighing bottles, crucibles, and other vessels. An airtight seal is made by application of stopcock grease to the ground-glass rim on the top of the desiccator. A vacuum desiccator has a side arm on the top for evacuation so that the contents can be kept in a vacuum rather than just an atmosphere of dry air. 2 You should not attempt to collect a blood sample unless you have been specifically trained to do so. A trained
technician will generally be assigned to this job.
Oven-dried samples or reagents are cooled in a desiccator before weighing.
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Fig. 2.12.
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Fig. 2.13.
Muffle furnace. (Courtesy of Arthur H. Thomas Company.)
Desiccator and desiccator plate.
The top of a desiccator should not be removed any more than necessary since the removal of moisture from the air introduced is rather slow, and continued exposure will limit the lifetime of the desiccant. A red-hot crucible or other vessel should be allowed to cool in the air about 60 s before it is placed in the desiccator. Otherwise, the air in the desiccator will be heated appreciably before the desiccator is closed, and as the air cools, a partial vacuum will be created. This will result in a rapid inrush of air when the desiccator is opened and in possible spilling or loss of sample as a consequence. A hot weighing bottle should not be stoppered when placed in a desiccator because on cooling, a partial vacuum is created and the stopper may seize. The stopper should be placed in the desiccator with the weighing bottle. Table 2.5 lists some commonly used desiccants and their properties. Aluminum oxide, magnesium perchlorate, calcium oxide, calcium chloride, and silica gel can be regenerated by heating at 150, 240, 500, 275, and 150◦ C, respectively. FURNACES AND OVENS A muffle furnace (Figure 2.13) is used to ignite samples to high temperatures, either to convert precipitates to a weighable form or to burn organic materials prior to inorganic analysis. There should be some means of regulating the temperature since losses of some metals may occur at temperatures in excess of 500◦ C. Temperatures up to about 1200◦ C can be reached with muffle furnaces.
Table 2.5
Some Common Drying Agents Agent
Capacity
Deliquescenta
CaCl2 (anhydrous) CaSO4 CaO MgClO4 (anhydrous) Silica gel Al2 O3 P2 O5
High Moderate Moderate High
Yes No No Yes
Low Low Low
No No Yes
a Becomes
Trade Name Drierite (W. A. Hammond Drierite Co.) Anhydrone (J. T. Baker Chemical Co.); Dehydrite (Arthur H. Thomas Co.)
liquid by absorbing moisture. Take care of liquids generated. For example, P2 O5 generates H3 PO4 .
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A drying oven is used to dry samples prior to weighing. A typical drying oven is shown in Figure 2.14. These ovens are well ventilated for uniform heating. The usual drying temperature is about 110◦ C, but many laboratory ovens can be heated up to temperatures of 200 to 300◦ C. HOODS A fume hood is used when chemicals or solutions are to be evaporated. When perchloric acid or acid solutions of perchlorates are to be evaporated, the fumes should be collected, or the evaporation should be carried out in fume hoods specially designed for perchloric acid work (i.e., constructed from components resistant to attack by perchloric acid). When performing trace analysis, as in trace metal analysis, care must be taken to prevent contamination. The conventional fume hood is one of the “dirtiest” areas of the laboratory since laboratory air is drawn into the hood and over the sample. Laminarflow hoods or workstations are available for providing very clean work areas. Rather than drawing unfiltered laboratory air into the work area, the air is prefiltered and then flows over the work area and out into the room to create a positive pressure and prevent unfiltered air from flowing in. A typical laminar-flow workstation is shown in Figure 2.15. The high-efficiency particulate air (HEPA) filter removes all particles larger than 0.3 μm from the air. Vertical laminar-flow stations are preferred when fumes are generated that should not be blown over the operator. Facilities are available to exhaust noxious fumes. Biohazard hoods can also be found in many modern analytical and clinical laboratories. They are different from exhaust hoods. Their main purpose is to provide a safe place to work with potentially infectious material and particulates. They are often outfitted with high-intensity UV lamps that can be turned on prior to or after working (never during) in order to disinfect the workspace. All analysts should receive proper training prior to handling biohazard materials.
Fig. 2.14.
Drying oven. (Courtesy of Arthur H. Thomas Company.)
Laminar-flow hoods provide clean work areas.
High-efficiency particulate air final filter
Work area
WASH BOTTLES A wash bottle of some sort should be handy in any analytical laboratory, to be used for quantitative transfer of precipitates and solutions and to wash precipitates. These are commercially available in a variety of shapes and sizes, as seen in Figure 2.16. Alternatively they may be constructed from a Florence flask and glass tubing, as in Figure 2.16b.
Fig. 2.15.
CENTRIFUGES AND FILTERS A centrifuge has many useful applications, particularly in the clinical laboratory, where blood may have to be separated into fractions such as serum or plasma, and proteins may have to be separated by precipitation followed by centrifuging. Many
Fig. 2.16. (a)
Prefilter Blower
(b)
Wash bottles: (a) polyethylene, squeeze type; (b) glass, blow type.
Laminar-flow workstation. (Courtesy of Dexion, Inc., 344 Beltine Boulevard, Minneapolis, MN.)
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Fig. 2.17.
Filtering crucibles: (a) Gooch crucible; (b) sintered-glass crucible; (c) porcelain filter crucible.
(a)
( b)
(c)
To vacuum
Trap
Fig. 2.18.
Crucible holders.
laboratories also have an ultracentrifuge. Such an instrument has a larger capacity and can achieve higher speeds (gravitational force) in order to more readily separate sample components. Centrifugal filters are available with different molecular weight (MW) cutoff filter elements. They consist of disposable centrifuge tubes, separated by a horizontally placed filter element. A sample (typically biological) is put in the top compartment and the device is then subjected to centrifugation. Either the low MW material (often a clear filtrate) or the higher MW material retained by the filter can be of interest. Filters for filtering precipitates (e.g., in gravimetric analysis) are of various types. The Gooch crucible, sintered-glass crucible, and porcelain filter crucible are illustrated in Figure 2.17. The Gooch crucible is porcelain and has holes in the bottom; a glass fiber filter disk is supported on top of it. The glass fiber filter disk will handle fine precipitates. The sintered-glass crucible contains a sintered-glass bottom, which is available in fine (F), medium (M), or coarse (C) porosity. The porcelain filter crucible contains a porous unglazed bottom. Glass filters are not recommended for concentrated alkali solutions because of the possibility of attack by these solutions. See Table 2.1 for maximum working temperatures for different types of crucible materials. Gelatinous precipitates such as hydrous iron oxide should not be filtered in filter crucibles because they clog the pores. Even with filter paper, the filtration of the precipitates can be slow. Filter crucibles are used with a crucible holder mounted on a filtering flask (Figure 2.18). A safety bottle is connected between the flask and the aspirator. Ashless filter paper is generally used for quantitative work in which the paper is ignited away and leaves a precipitate suitable for weighing (see Chapter 10). There are various grades of filter papers for different types of precipitates. These are listed in Table 2.6 for Whatman papers. TECHNIQUES OF FILTRATION
Fig. 2.19. paper.
Properly folded filter
By proper fitting of the filter paper, the rate of filtration can be increased. A properly folded filter paper is illustrated in Figure 2.19. The filter paper is folded in the shape of a cone, with the overlapped edges of the two quarters not quite meeting (0.5 cm apart).
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Table 2.6
Whatman Filter Papers Precipitate Very fine (e.g., BaSO4 ) Small or medium (e.g., AgCl) Gelatinous or large crystals (e.g., Fe2 O3 · xH2 O)
Whatman No. 50 (2.7 μm) 52 (7 μm) 54 (22 μm)
See http://www.whatman.com/QuantitativeFilterPapersHardenedLowAshGrades.aspx
About 1 cm is torn away from the corner of the inside edge. This will allow a good seal against the funnel to prevent air bubbles from being drawn in. After the folded paper is placed in the funnel, it is wetted with distilled water. The stem is filled with water and the top of the wet paper is pressed against the funnel to make a seal. With a proper fit, no air bubbles will be sucked into the funnel, and the suction supplied by the weight of the water in the stem will increase the rate of filtration. The filtration should be started immediately. The precipitate should occupy not more than one-third to one-half of the filter paper in the funnel because many precipitates tend to “creep.” Do not allow the water level to go over the top of the paper. The precipitate should be allowed to settle in the beaker before filtration is begun. The bulk of the clear liquid can then be decanted and filtered at a rapid rate before the precipitate fills the pores of the filter paper. Care must be taken in the decanting and the transferring of the precipitate to avoid losses. This is properly done by use of a stirring rod and a wash bottle, as illustrated in Figure 2.20. Note: the wash liquid is not distilled water—see Chapter 10. The solution is decanted by pouring it down the glass rod, which guides it into the filter without splashing. The precipitate is most readily washed while still in the beaker. After the mother liquor has been decanted off, wash the sides of the beaker down with several milliliters of the wash liquid, and then allow the precipitate to settle as before. Decant the wash liquid into the filter and repeat the washing operation two or three times. Then transfer the precipitate to the filter by holding the glass rod and beaker in one hand, as illustrated, and wash it out of the beaker with wash liquid from the wash bottle. If the precipitate must be collected quantitatively, as in gravimetric analysis, the last portions of precipitate are removed by scrubbing the walls with a moistened rubber policeman, which contains a flexible rubber scraper attached to a glass rod (Figure 2.21). [For a description of the origin of its name, see J. W. Jensen, J. Chem. Ed., 85 (6) (2008) 776.] Wash the remainder of loosened precipitate from the beaker and from the policeman. If the precipitate is being collected in a filter paper, then instead of a rubber policeman, a small piece of the ashless filter paper can be rubbed on the beaker walls to remove the last bits of precipitate and added to the filter. This should be held with a pair of forceps. After the precipitate is transferred to the filter, it is washed with five or six small portions of wash liquid. This is more effective than adding one large volume. Divert the liquid around the top edge of the filter to wash the precipitate down into the cone. Each portion should be allowed to drain before adding the next one. Check for completeness of washing by testing for the precipitating agent in the last few drops of the washings. Note: there will always be some precipitate in the wash due to finite solubility, i.e., finite Ksp , but it will be undetectable after sufficient washing.
Let the precipitate settle before filtering.
Wash the precipitate while it is in the beaker.
Test for completeness of washing.
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Fig. 2.21.
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Rubber policeman.
Fig. 2.20.
Proper technique for transfer of a precipitate.
2.8 Igniting Precipitates—Gravimetric Analysis
Fig. 2.22.
Crucible and cover supported on a wire triangle for charring off paper.
If a precipitate is to be ignited in a porcelain filter crucible, the moisture should be driven off first at a low heat. The ignition may be done in a muffle furnace or by heating with a burner. If a burner is to be used, the filter crucible should be placed in a porcelain or platinum crucible to prevent reducing gases of the flame diffusing through the pores of the filter. When precipitates are collected on filter paper, the cone-shaped filter containing the precipitate is removed from the funnel, the upper edge is flattened, and the corners are folded in. Then, the top is folded over and the paper and contents are placed in a crucible with the bulk of the precipitate on the bottom. The paper must now be dried and charred off. The crucible is placed at an angle on a triangle support with the crucible cover slightly ajar, as illustrated in Figure 2.22. The moisture is removed by low heat from the burner, with care taken to avoid splattering. The heating is gradually increased as moisture is evolved and the paper begins to char. Care should be taken to avoid directing the reducing portion of the flame into the crucible. A sudden increase in the volume of smoke evolved indicates that the paper is about to burst into flame, and the burner should be removed. If it does burst into flame, it should be smothered
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quickly by replacing the crucible cover. Carbon particles will undoubtedly appear on the cover, and these will ultimately have to be ignited. Finally, when no more smoke is detected, the charred paper is burned off by gradually increasing the flame temperature. The carbon residue should glow but should not flame. Continue heating until all the carbon and tars on the crucible and its cover are burned off. The crucible and precipitate are now ready for igniting. The ignition can be continued with the burner used at highest temperature or with the muffle furnace. Before a precipitate is collected in a filter crucible or transferred to a crucible, the crucible should be dried to constant weight (e.g., 1 h of heating, followed by cooling and weighing and repeating the cycle) if the precipitate is to be dried, or it should be ignited to constant weight if the precipitate is to be ignited. Constant weight is considered to have been achieved with an analytical balance when successive weighings agree within about 0.3 or 0.4 mg. The crucible plus the precipitate are heated to constant weight in a similar manner. After the first heating, the time of heating can be reduced by half. The crucible should be allowed to cool in a desiccator for at least 1 2 h before weighing. Red-hot crucibles should be allowed to cool below redness before placing them in the desiccator (use crucible tongs—usually nickel plated or stainless steel to minimize contamination from rust). Before weighing a covered crucible, check for any radiating heat by placing your hand near it (don’t touch).
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Do the initial ignition slowly.
Dry and weigh the crucible before adding the precipitate!
2.9 Obtaining the Sample—Is It Solid, Liquid, or Gas? Collecting a representative sample is an aspect of analytical chemistry that the beginning analytical student is often not concerned with because the samples handed to him or her are assumed to be homogeneous and representative. Yet this process can be the most critical aspect of an analysis. The significance and accuracy of measurements can be limited by the sampling process. Unless sampling is done properly, it becomes the weak link in the chain of the analysis. A life could sometimes depend on the proper handling of a blood sample during and after sampling. If the analyst is given a sample and does not actively participate in the sampling process, then the results obtained can only be attributed to the sample “as it was received.” And the chain of custody as mentioned earlier must be documented. Many professional societies have specified definite instructions for sampling given materials [e.g., the American Society for Testing and Materials (ASTM: www.astm.org), the Association of Official Analytical Chemists International (AOAC International: www.aoac.org), and the American Public Health Association (APHA: www.apha.org)]. By appropriate application of experience and statistics, these materials can be sampled as accurately as the analysis can be performed. Often, however, the matter is left up to the analyst. The ease or complexity of sampling will, of course, depend on the nature of the sample. The problem involves obtaining a sample that is representative of the whole. This sample is called the gross sample. Its size may vary from a few grams or less to several pounds, depending on the type of bulk material. Once a representative gross sample is obtained, it may have to be reduced to a sufficiently small size to be handled. This is called the sample. Once the sample is obtained, an aliquot, or portion, of it will be analyzed. This aliquot is called the analysis sample. Several replicate analyses on the same sample may be performed by taking separate aliquots. In the clinical laboratory, the gross sample is usually satisfactory for use as the sample because it is not large and it is homogeneous (e.g., blood and urine samples). The analysis sample will usually be from a few milliliters to a fraction of a drop (a few microliters) in quantity.
See Chapter 3 for important statistical considerations in sampling.
Replication in sampling and analysis are key considerations.
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Some of the problems associated with obtaining gross samples of solids, liquids, and gases are considered below.
Sampling (From the journals collection of the Chemical Heritage Foundations’ Othmer Library.)
1. Solids. Inhomogeneity of the material, variation in particle size, and variation within the particle make sampling of solids more difficult than other materials. The easiest but usually most unreliable way to sample a material is the grab sample, which is one sample taken at random and assumed to be representative. The grab sample will be satisfactory only if the material from which it is taken is homogeneous. For most reliable results, it is best to take 1/50 to 1/100 of the total bulk for the gross sample, unless the sample is fairly homogeneous. The larger the particle size, the larger the gross sample should be. The easiest and most reliable time to sample large bodies of solid materials is while they are being moved. In this way any portion of the bulk material can usually be exposed for sampling. Thus, a systematic sampling can be performed to obtain aliquots representing all portions of the bulk. Some samples follow. In the loading or unloading of bags of cement, a representative sample can be obtained by taking every fiftieth or so bag or by taking a sample from each bag. In the moving of grain by wheelbarrow, representative wheelbarrow loads or a shovelful from each wheelbarrow can be taken. All of these aliquots are combined to form the gross sample.
See Chapter 25 on the text’s website for more on sampling biological fluids.
2. Liquids. Liquid samples tend to be homogeneous and representative samples are much easier to get. Liquids mix by diffusion only very slowly and must be shaken to obtain a homogeneous mixture. If the material is indeed homogeneous, a simple grab (single random) sample will suffice. For all practical purposes, this method is satisfactory for taking blood samples. The composition of some samples vary on when it is taken. This is the case for urine samples, Therefore 24-h urine sample collections are generally more representative than a single “spot sample”. The timing of sampling of biological fluids is, however, very important. The composition of blood varies considerably before and after meals, and for many analyses a sample is collected after the patient has fasted for a number of hours. Preservatives such as sodium fluoride for glucose preservation and anticoagulants may be added to blood samples when they are collected. Blood samples may be analyzed as whole blood, or they may be separated to yield plasma or serum according to the requirements of the particular analysis. Most commonly, the concentration of the substance external to the red cells (the extracellular concentration) will be a significant indication of physiological condition, and so serum or plasma is taken for analysis. If liquid samples are not homogeneous, and if they are small enough, they can be shaken and sampled immediately. For example, there may be particles in the liquid that have tended to settle. Large bodies of liquids are best sampled after a transfer or, if in a pipe, after passing through a pump when they have undergone thorough mixing. Large stationary liquids can be sampled with a “thief” sampler, which is a device for obtaining aliquots at different levels. It is best to take the sample at different depths at a diagonal, rather than straight down. The separate aliquots of liquids can be analyzed individually and the results combined, or the aliquots can be combined into one gross sample and replicate analyses performed. This latter procedure is probably preferred because the analyst will then have some idea of the precision of the analysis.
See Chapter 26 on the text’s website for more on sampling environmental samples.
3. Gases The usual method of sampling gases involves sampling into an evacuated container, often a specially treated stainless steel canister or an inert polyvinyl fluoride (Tedlar) bag is commonly used. The sample may be collected rapidly (a grab sample)
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or over a long period of time, using a small orifice to slowly fill the bag. A grab sample is satisfactory in many cases. To collect a breath sample, for example, the subject could blow into an evacuated bag or blow up a mylar balloon. Auto exhaust could be collected in a large evacuated plastic bag. The sample may be supersaturated with moisture relative to ambient temperature at which the sample container is. Moisture will condense in the sampling container after sample collection and the analyte of interest (e.g., ammonia in breath or nitrous acid in car exhaust) will be removed by the condensed moisture. The sample container must be heated and the sample transferred through a heated transfer line if the analyte is to be recovered. The volume of gross gas sample collected may or may not need to be known. Often, the concentration of a certain analyte in the gas sample is measured, rather than the amount. The temperature and pressure of the sample will, of course, be important in determining the volume and hence the concentration. Gas sampling techniques mentioned here does not concern gases dissolved in liquids, such CO2 or O2 in blood. These are treated as liquid samples and are then handled accordingly to measure the gas in the liquid or to release it from the liquid prior to measurement.
2.10 Operations of Drying and Preparing a Solution of the Analyte After a sample has been collected, a solution of the analyte must usually be prepared before the analysis can be continued. Drying of the sample may be required, and it must be weighed or the volume measured. If the sample is already a solution (e.g., serum, urine, or water), then extraction, precipitation, or concentration of the analyte may be in order, and this may also be true with other samples. In this section we describe common means for preparing solutions of inorganic and organic materials. Included are the dissolution of metals and inorganic compounds in various acids or in basic fluxes (fusion), the destruction of organic and biological materials for determination of inorganic constituents (using wet digestion or dry ashing), and the removal of proteins from biological materials so they do not interfere in the analysis of organic or inorganic constituents. DRYING THE SAMPLE Solid samples will usually contain variable amounts of adsorbed water. With inorganic materials, the sample will generally be dried before weighing. This is accomplished by placing it in a drying oven at 105 to 110◦ C for 1 or 2 h. Other nonessential water, such as that entrapped within the crystals, may require higher temperatures for removal. Decomposition or side reactions of the sample must be considered during drying. Thermally unstable samples can be dried in a desiccator; using a vacuum desiccator will hasten the drying process. A lyophilizer (freeze dryer) can also be used to remove a fairly large amount of water from a sample that contains thermally labile material; the sample must be frozen prior to placing it in a vessel within, or attached to, the apparatus. If the need to sample is weighed without drying, the results will be reported on an “as is” basis. Plant and tissue samples can usually be dried by heating. See Chapter 1 for a discussion of the various weight bases (wet, dry, ash) used in connection with reporting analytical results for these samples.
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SAMPLE DISSOLUTION Before the analyte can be measured, some sort of sample manipulation is generally necessary to get the analyte into solution or, for biological samples, to rid it of interfering substances, such as proteins. Complex samples can be subjected to centrifugal filtration prior to analysis (e.g., perchlorate and iodide in milk have been chromatographically determined after centrifugal filtration). There are two types of sample preparation: those that totally destroy the sample matrix and those that are nondestructive or only partially destructive. The former type can generally be used only when the analyte is inorganic or can be converted to an inorganic derivative for measurement (e.g., Kjeldahl analysis, in which organic nitrogen is converted to ammonium ion—see below). Iodine in food is similarly determined after total oxidative digestion to HIO3 . Destructive digestion typically must be used if trace element analysis must be conducted in a largely organic matrix. DISSOLVING INORGANIC SOLIDS
Fusion is used when acids do not dissolve the sample.
Fusion is used when acids do not dissolve the sample.
Strong mineral acids are good solvents for many inorganics. Hydrochloric acid is a good general solvent for dissolving metals that are above hydrogen in the electromotive series. Nitric acid is a strong oxidizing acid and will dissolve most of the common metals, nonferrous alloys, and the “acid-insoluble” sulfides. Perchloric acid, when heated to drive off water, becomes a very strong and efficient oxidizing acid in the dehydrated state. It dissolves most common metals and destroys traces of organic matter. It must be used with extreme caution because it will react explosively with many easily oxidizable substances, especially organic matter. Some instruments today are extraordinarily sensitive. An inductively coupled plasma mass spectrometer (ICP-MS) is essential for measuring trace impurities in semiconductor grade silicon, for example. The silicon sample is dissolved in a mixture of nitric and hydrofluoric acids before analysis. To carry out such ultra trace analysis the acids must also be ultra-pure. Such ultra-pure “semiconductor grade” acids are also very expensive. Some inorganic materials will not dissolve in acids, and fusion with an acidic or basic flux in the molten state must be used to solubilize them. The sample is mixed with the flux in a sample-to-flux ratio of about 1 to 10 or 20, and the combination is heated in an appropriate crucible until the flux becomes molten. When the melt becomes clear, usually in about 30 min, the reaction is complete. The cooled solid is then dissolved in dilute acid or in water. During the fusion process, insoluble materials in the sample react with the flux to form a soluble product. Sodium carbonate is one of the most useful basic fluxes, and acid-soluble carbonates are produced. DESTRUCTION OF ORGANIC MATERIALS FOR INORGANIC ANALYSIS— —BURNING OR ACID OXIDATION Animal and plant tissue, biological fluids, and organic compounds are usually decomposed by wet digestion with a boiling oxidizing acid or mixture of acids, or by dry ashing at a high temperature (400 to 700◦ C) in a muffle furnace. In wet digestion, the acids oxidize organic matter to carbon dioxide, water, and other volatile products, which are driven off, leaving behind salts or acids of the inorganic constituents. In dry ashing, atmospheric oxygen serves as the oxidant; that is, the organic matter is burned off, leaving an inorganic residue. Auxiliary oxidants (e.g., NaNO3 ) can be used as a flux during dry ashing.
In dry ashing, the organic matter is burned off.
1. Dry Ashing. Although various types of dry ashing and wet digestion combinations are used with about equal frequency by analysts for organic and biological materials,
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simple dry ashing with no chemical aids is probably the most commonly employed technique. Lead, zinc, cobalt, antimony, chromium, molybdenum, strontium, and iron traces can be recovered with little loss by retention or volatilization. Usually a porcelain crucible can be used. Lead is volatilized at temperatures in excess of about 500◦ C, especially if chloride is present, as in blood or urine. Platinum crucibles are preferred for lead for minimal retention losses. If an oxidizing material is added to the sample, the ashing efficiency is enhanced. Magnesium nitrate is one of the most useful aids, and with this it is possible to recover arsenic, copper, and silver, in addition to the above-listed elements. Liquids and wet tissues are dried on a steam bath or by gentle heat before they are placed in a muffle furnace. The heat from the furnace should be applied gradually up to full temperature to prevent rapid combustion and foaming. After dry ashing is complete, the residue is usually leached from the vessel with 1 or 2 mL hot concentrated or 6 M hydrochloric acid and transferred to a flask or beaker for further treatment. Another dry technique is that of low-temperature ashing. A radio-frequency discharge is used to produce activated oxygen radicals, which are very reactive and will attack organic matter at low temperatures. Temperatures of less than 100◦ C can be maintained, and volatility losses are minimized. Introduction of elements from the container and the atmosphere is reduced, and so are retention losses. Radiotracer studies have demonstrated that 17 representative elements are quantitatively recovered after complete oxidation of organic substrate. Elemental analysis in the case of organic compounds (e.g., for carbon or hydrogen) is usually performed by oxygen combustion in a tube, followed by an absorption train. Oxygen is passed over the sample in a platinum boat, which is heated and quantitatively converts carbon to CO2 and hydrogen to H2 O. These combustion gases pass into the absorption train, where they are absorbed in preweighed tubes containing a suitable absorbent. For example, Ascarite II is used to absorb the CO2 , and Dehydrite (magnesium perchlorate) is used to absorb the H2 O. The gain in weight of the absorption tubes is a measure of the CO2 and H2 O liberated from the sample. Details of this technique are important, and, should you have occasion to use it, you are referred to more comprehensive texts on elemental analysis. Modern elemental analyzers are more automated and may be based on chromatographic separation of the combustion gases followed by detection with a thermal conductivity detector—Chapter 20 (see, e.g., http://www-odp.tamu.edu/publications/tnotes/tn30/tn30_10.htm). 2. Wet Digestion. Next to dry ashing, wet digestion with a mixture of nitric and sulfuric acids is the second most frequently used oxidation procedure. Usually a small amount (e.g., 5 mL) of sulfuric acid is used with larger volumes of nitric acid (20 to 30 mL). Wet digestions are usually performed in a Kjeldahl flask (Figure 2.24). The nitric acid destroys the bulk of the organic matter, but it does not get hot enough to destroy the last traces. It is boiled off during the digestion process until only sulfuric acid remains and dense, white SO3 fumes are evolved and begin to reflux in the flask. At this point, the solution gets very hot, and the sulfuric acid acts on the remaining organic material. Charring may occur at this point if there is considerable or very resistant organic matter left. If the organic matter persists, more nitric acid may be added. Digestion is continued until the solution clears. All digestion procedures must be performed in a fume hood. A much more efficient digestion mixture uses a mixture of nitric, perchloric, and sulfuric acids in a volume ratio of about 3:1:1. Ten milliliters of this mixture will usually suffice for 10 g fresh tissue or blood. The perchloric acid is an extremely efficient oxidizing agent when it is dehydrated and hot and will destroy the last traces of organic matter with relative ease. Samples are heated until nitric acid is boiled off and perchloric acid fumes appear; these, are less dense than SO3 but fill the flask
In wet ashing, the organic matter is oxidized with an oxidizing acid.
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Perchloric acid must be used with caution!
In Kjeldahl digestions, nitrogen is converted to ammonium ion, which is then distilled as ammonia and titrated.
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more readily. The hot perchloric acid is boiled, usually until fumes of SO3 appear, signaling the evaporation of all the perchloric acid. Sufficient nitric acid must be added at the beginning to dissolve and destroy the bulk of organic matter, and there must be sulfuric acid present to prevent the sample from going to dryness, or else there is danger of explosion from the perchloric acid. A hood specially designed for perchloric acid work must be used for all digestions incorporating perchloric acid. Typically, a nitric-sulfuric acid digestion is first carried out to remove the more easily oxidizable material before digestion with the perchloric acid cocktail is carried out. Perchloric acid digestion is even more efficient if a small amount of molybdenum(VI) catalyst is added. As soon as water and nitric acid are evaporated, oxidation proceeds vigorously with foaming, and the digestion is complete in a few seconds. The digestion time is reduced considerably. A mixture of nitric and perchloric acids is also commonly used. The nitric acid boils off first, and care must be taken to prevent evaporation of the perchloric acid to near dryness, or a violent explosion may result; this procedure is not recommended unless you have considerable experience in digestion procedures. Perchloric acid should never be added directly to organic or biological material. Always add an excess of nitric acid first. Explosions with perchloric acid are generally associated with formation of peroxides, and the acid turns dark in color (e.g., yellowish brown) prior to explosion. Certain organic compounds such as ethanol, cellulose, and polyhydric alcohols can cause hot concentrated perchloric acid to explode violently; this is presumably due to formation of ethyl perchlorate. A mixture of nitric, perchloric, and sulfuric acids allows zinc, selenium, arsenic, copper, cobalt, silver, cadmium, antimony, chromium, molybdenum, strontium, and iron to be quantitatively recovered. Lead is often lost if sulfuric acid is used. The mixture of nitric and perchloric acids can be used for lead and all the above elements. Perchloric acid must be present to prevent losses of selenium. It maintains strong oxidizing conditions and prevents charring that would result in formation of volatile compounds of lower oxidation states of selenium. Samples containing mercury cannot be dry ashed. Wet digestion that involves the simultaneous application of heat must be done using a reflux apparatus because of the volatile nature of mercury and its compounds. Cold or room temperature procedures are often preferred to obtain partial destruction of organic matter. For example, in urine samples, which contain a relatively small amount of organic matter compared with blood, mercury can be reduced to the element with copper(I) and hydroxylamine hydrochloride and the organic matter destroyed by potassium permanganate at room temperature. The mercury can then be dissolved and the analysis continued. Urinary mercury can be mineralized using Fenton’s reagent (Fe(II) and H2 O2 ) and determined as the elemental vapor after addition of sodium borohydride, a powerful reducing agent. Many nitrogen-containing compounds can be determined by Kjeldahl digestion to convert the nitrogen to ammonium sulfate. The digestion mixture consists of sulfuric acid plus potassium sulfate to increase the boiling point of the acid and thus increase its efficiency. A catalyst is also added (such as copper or selenium). After destruction of the organic matter, sodium hydroxide is added to make the solution alkaline, and the ammonia is distilled into an excess of standard hydrochloric acid. The excess acid is back-titrated with standard alkali to determine the amount of ammonia collected. With a knowledge of the percent nitrogen composition in the compound of interest, the amount of the compound can be calculated from the amount of ammonia determined. This is the most accurate method for determining protein content. Protein contains a definite percentage of nitrogen, which is converted to ammonium sulfate during the digestion. See Chapter 8 for further details. Note: of course, if other nitrogen-containing species are present, the nitrogen determination will not accurately reflect the protein content. This was amply demonstrated in China when melamine, an inexpensive
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organic amine base containing six nitrogen atoms, was added to milk formula to boost the apparent “protein” content. This caused the death of many infants in China. The relative merits of various oxidation methods have been studied extensively. However, there may be no universal and generally applicable dry ashing method. Dry ashing is recommended for its simplicity and relative freedom from positive errors (contamination) since few or no reagents are added. The potential errors of dry oxidation are volatilization of elements and losses by retention on the walls of the vessel. Adsorbed metals on the vessel may in turn contaminate future samples. Wet digestion is considered superior in terms of rapidity (although it does require more operator attention), low level of temperature maintained, and freedom from loss by retention. The chief error attributed to wet digestion is the introduction of impurities from the reagents necessary for the reaction. This problem has been minimized as commercial reagentgrade acids have become available in greater purity and specially prepared high-purity acids can now be obtained commercially, albeit not inexpensively. The time required for ashing or digestion will vary with the sample and the technique employed. Two to four hours are common for dry ashing and half to one hour is common for wet digestion.
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Dry and wet ashing each has advantages and limitations.
MICROWAVE PREPARATION OF SAMPLES Microwave ovens are now widely used for rapid and efficient drying and acid decomposition of samples. Laboratory microwave ovens are specially designed to overcome limitations of household ovens, and these are discussed below. Advantages of microwave digestions include reduction of dissolution times from hours to minutes and low blank levels due to reduced amounts of reagents required. 1. How Do Microwaves Heat? The microwave region is between infrared radiation and radio waves in the electromagnetic spectrum, in the frequency range of 300 to 300,000 MHz (3 × 108 to 3 × 1011 Hz, or beginning at about 1000 μm wavelength—see Figure 16.2). Microwaves consist of an electric field and a magnetic field perpendicular to the electric field. The electric field is responsible for energy transfer between the microwave source and the irradiated sample. Microwave energy affects molecules in two ways: dipole rotation and ionic conduction. The first is the more generally important. When the microwave energy passes through the sample, molecules with non-zero dipole moments will try to align with the electric field, and the more polar ones will have the stronger interaction with the field. This molecular motion (rotation) results in heating. The energy transfer, a function of the dipole moment and the dielectric constant, is most efficient when the molecules are able to relax quickly, that is, when the relaxation time matches the microwave frequency. Large molecules such as polymers relax slowly, but once the temperature increases and they relax more rapidly, they can absorb the energy more efficiently. Small molecules such as water, though, relax more quickly than the resonating microwave energy, and they move farther away from the resonance frequency and absorb less energy as they heat up. The ionic conduction effect arises because ionic species in the presence of an electric field will migrate in one direction or the other. Energy is transferred from the electric field, causing ionic interactions that speed up the heating of a solution. Ionic absorbers become stronger absorbers of microwave energy as they are heated since ionic conductance increases with temperature. Deionized water heats slowly, but if salt is added, it heats rapidly. Acids, of course, are good conductors and heat rapidly. So microwave energy heats by causing movement of molecules due to dipole rotation and movement of ions due to ionic conductance. The microwave energy interacts with different materials in different ways. Reflective materials such as metals are good heat conductors: They do not heat and instead will reflect the microwave energy. (It is not good practice to put metals in microwave ovens due to electrical discharge occurring between one very highly charged metal segment to another.) Transparent materials are insulators because they transmit the microwave energy and
Microwaves heat by causing molecules to rotate and ions to migrate.
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Isolator Wave guide Microwaves
Mode Stirrer
Reflected microwaves Magnetron Vessel
Fig. 2.23.
Schematic of a microwave system. [From G. Le Blanc, LC/GC Suppl., 17(6S) (1999) S30.] (Courtesy of LC/GC Magazine).
Microwave cavity
also do not heat. The absorptive materials, the molecules and ions discussed above, are the ones that receive microwaves and are heated. Microwave energy is too low to break chemical bonds (a feature that has generated interest in using microwave energy to speed up chemical reactions in syntheses). The properties of reflective and insulator materials are utilized in designing microwave digestion systems. Household microwave ovens don’t work for small sample heating.
2. Design of Laboratory Microwave Ovens. Home microwave ovens were initially used for laboratory purposes, but it soon became apparent that modifications were needed. Laboratory samples are usually much smaller than food samples that are cooked and absorb only a small fraction of the energy produced by the magnetron of the oven. The energy not absorbed by the sample is bounced back to the magnetron, causing it to overheat and burn out. Also, arcing could occur. So laboratory ovens are designed to protect the magnetron from stray energy. The main components of these ovens (Figure 2.23) include the magnetron, an isolator, a waveguide, the cavity, and a mode stirrer. Microwave energy generated by the magnetron is propagated down the waveguide into the cavity. The stirrer distributes the energy in different directions. The isolator, made of a ferromagnetic material and placed between the magnetron and the waveguide, deflects the microwave energy returning from the cavity into a fan-cooled ceramic load, keeping it away from the magnetron. The frequency used for cooking turns out to be good for chemistry as well, and the standard is 2450 MHz. Powers of 1200 W are typically used. 3. Acid Digestions. Digestions are normally done in closed plastic containers, either Teflon PFA (perfluoroalkoxy ethylene) or polycarbonate (insulators). This is to avoid acid fumes in the oven. It provides additional advantages. Pressure is increased and the boiling point of the acid is raised (the acid is superheated). So digestions occur more rapidly. Also, volatile metals are not lost. Modern ovens provide for control of pressure and temperature. Fiber-optic temperature probes are used that are transparent to microwave energy. Temperature control has enabled the use of the oven for microwave-assisted molecular extractions, by maintaining the temperature low enough to avoid molecular decomposition. PARTIAL DESTRUCTION OR NONDESTRUCTION OF SAMPLE MATRIX Obviously, when the substance to be determined is organic in nature, nondestructive means of preparing the sample must be used. For the determination of metallic elements, it is also sometimes unnecessary to destroy the molecular structure of the
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sample, particularly with biological fluids. For example, several metals in serum or urine can be determined by atomic absorption spectroscopy by direct aspiration of the sample or a diluted sample into a flame. Constituents of solid materials such as soils can sometimes be extracted by an appropriate reagent. Thorough grinding, mixing, and refluxing are necessary to extract the analyte. Many trace metals can be extracted from soils with 1 M ammonium chloride or acetic acid solution. Some, such as selenium, can be distilled as the volatile chloride or bromide. PROTEIN-FREE FILTRATES Proteins in biological fluids interfere with many analyses and must be removed nondestructively. Several reagents will precipitate (coagulate) proteins. Trichloroacetic acid (TCA), tungstic acid (sodium tungstate plus sulfuric acid), and barium hydroxide plus zinc sulfate (a neutral mixture) are some of the common ones. A measured volume of sample (e.g., serum) is usually treated with a measured volume of reagent. Following precipitation of the protein (approximately 10 min), the sample is filtered through dry filter paper without washing, or else it is centrifuged. An aliquot of the protein-free filtrate (PFF) is then taken for analysis. Molecular weight selective centrifugal filtration can sometimes be used. Details for preparing specific types of protein-free filtrates are given in the text’s website in Chapter 25 (under Collection and Preservation of Samples) as well as in experiments requiring them.
See Chapter 25 on the text’s website for the preparation of protein-free filtrates.
LABORATORY TECHNIQUES FOR DRYING AND DISSOLVING When a solid sample is to be dried in a weighing bottle, the cap is removed from the bottle and, to avoid spilling, both are placed in a beaker and covered with a ribbed watch glass. Some form of identification should be placed on the beaker. The weighed sample may be dissolved in a beaker or Erlenmeyer flask. If there is any fizzing action, cover the vessel with a watch glass. After dissolution is complete, wash the walls of the vessel down with distilled water. Also wash the watch glass so the washings fall into the vessel. You may have to evaporate the solution to decrease the volume. This is best done by covering the beaker with a ribbed watch glass to allow space for evaporation. Low heat should be applied to prevent bumping; a steam bath or variable-temperature hot plate is satisfactory. Use of a Kjeldahl flask for dissolution will avoid some of the difficulties of splattering or bumping. Kjeldahl flasks are also useful for performing digestions. They derive their name from their original use in digesting samples for Kjeldahl nitrogen analysis. They are well suited to all types of wet digestions of organic samples and acid dissolution of metals. Kjeldahl flasks come in assorted sizes from 10 to 800 mL. Some of these are shown in Figure 2.24. The sample and appropriate acids are placed in the round bottom of the flask and the flask is tilted while it is heated. In this way the acid can be boiled or refluxed with little danger of loss by “bumping.” The flask may be heated with a flame or in special electrically heated Kjeldahl digestion racks, which heat several samples simultaneously.
Take care in drying or dissolving samples.
2.11 Laboratory Safety
You must familiarize yourself with laboratory safety rules and procedures before conducting experiments! Read Appendix D and the material provided by your instructor. Get a free copy of Reference 31.
Before beginning any of the experiments, you must familiarize yourself with laboratory safety procedures. Appendix D on the text website discusses general safety rules. You should read this material before beginning experiments. Your instructor will provide you with specific guidelines and rules for operation in the laboratory and the disposal of
10 ml
Fig. 2.24.
30 ml
500 ml
Kjeldahl flasks.
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chemicals. For a more complete discussion of safety in the laboratory, you are referred to Safety in Academic Chemistry Laboratories, published by The American Chemical Society (Reference 31). This guide discusses personal protection and laboratory protocol, recommended laboratory techniques, chemical hazards, instructions on reading and understanding material safety data sheets (MSDSs), and safety equipment and emergency procedures. Rules are given for waste disposal, waste classification terminology, Occupational Safety and Health Administration (OSHA) laboratory standards for exposures to hazardous chemicals, and EPA requirements. The handling and treatment of inorganic and organic peroxides are discussed in detail, and an extensive list of incompatible chemicals is given, and maximum allowable container capacities for flammable and combustible liquids are listed. This resourceful booklet is recommended reading for students and instructors. It is available for free (one copy) from The American Chemical Society, Washington, DC (1-800-227-5558).
(Courtesy of Merck KGaA. Reproduced by permission.)
The Waste Management Manual for Laboratory Personnel, also published by The American Chemical Society, provides an overview of government regulations (Reference 32).
Professor’s Favorite Example Contributed by Professor Akos Vertes, George Washington University The Shroud of Turin Dating—A Sampling Problem
Various means have been enlisted to ascertain the validity and age of the Shroud of Turin (http://en.wikipedia.org/wiki/Shroud_of_Turin). One such study was radiocarbon dating done in 1988. Samples were obtained to be given to three separate laboratories for independent analyses. Details are given in: http://www.shroud.com/nature.htm. The paper describes in detail how these important samples were obtained and different sample preparation procedures used by the different laboratories. Three control samples of other known ancient textiles were treated and analyzed in the same way. Check the paper to see the results! Is the controversy concluded?
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Questions 1. Describe the basic pieces of apparatus used for volumetric measurements. List whether each is designed to contain or to deliver the specified volume. 2. Describe the principle and operation of the analytical balance. 3. Why is a microbalance more sensitive than an analytical balance? 4. What does TD on glassware mean? TC? 5. Explain weighing by difference. 6. List the general rules for the use of the balance. 7. Describe the preparation of a standard HCl solution and a standard NaOH solution. 8. Describe the principles of dry ashing and wet digestion of organic and biological materials. List the advantages of each. 9. What are the two principal means of dissolving inorganic materials? 10. What is a PFF? How would you prepare it? 11. What set of conditions must be carefully avoided to use perchloric acid safely for digesting organic materials? 12. What is a gross sample? Sample? Analysis sample? Grab sample? 13. What happens when microwave energy heats samples?
Problems GLASSWARE CALIBRATION/TEMPERATURE CORRECTIONS 14. You calibrate a 25-mL volumetric flask by filling to the mark with distilled water, equilibrated at 22◦ C. The dry stoppered flask weighs 27.278 g and the filled flask and stopper is 52.127 g. The balance uses stainless steel weights. What is the volume of the flask? What is it at the standard 20◦ C. Also insert the weight in air at 22◦ C into Table 2.4 (available on the textbook website), and compare the values obtained. 15. You calibrate a 25-mL pipet at 25◦ C using steel weights. The weight of the delivered volume of water is 24.971 g. What is the volume of the pipet at 25 and 20◦ C? 16. You calibrate a 50-mL buret in the winter time at 20◦ C, with the following corrections: Buret Reading (mL) 10 20 30 40 50
Correction (mL) +0.02 +0.03 0.00 –0.04 –0.02
You use the buret on a hot summer day at 30◦ C. What are the corrections then? 17. You prepare a standard solution at 21◦ C, and use it at 29◦ C. If the standardized concentration is 0.05129 M, what is it when you use the solution?
PROFESSOR’S FAVORITE PROBLEMS Contributed by Professor Bin Wang, Marshall University 18. For an electronic analytical balance with 0.1 mg readability, what is the maximum mass that can be weighed (capacity)? (a) 500–1000 g; (b) 100–300 g; (c) 10–20 g; (d) several g or less 19. The densities of air, calibration weights, and salt are 0.0012 g/mL, 7.8 g/cm3 , and 2.16 g/mL, respectively. If the apparent mass of salt (i.e., NaCl) weighed in air is 15.914 g, what is its true mass?
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Recommended References REAGENTS AND STANDARDS 1. Reagent Chemicals, Specifications and Procedures, 10th ed. Washington, DC: American Chemical Society, 2005. 2. Dictionary of Analytical Reagents. London: Chapman & Hall/CRC, 1993. Data on over 14,000 reagents. 3. J. R. Moody and E. S. Beary, “Purified Reagents for Trace Metal Analysis,” Talanta, 29 (1982) 1003. Describes sub-boiling preparation of high-purity acids. 4. R. C. Richter, D. Link, and H. M. Kingston, “On-Demand Production of High-Purity Acids in the Analytical Laboratory,” Spectroscopy, 15(1) (2000) 38. www.spectroscopyonline.com. Describes preparation using a commercial sub-boiling system. 5. www.thornsmithlabs.com. Thorn Smith Laboratories provides prepackaged materials for student analytical chemistry experiments (unknowns and standards); www.sigma-aldrich.com/analytical. A supplier of standard solutions.
ANALYTICAL BALANCES 6. D. F. Rohrbach and M. Pickering, “A Comparison of Mechanical and Electronic Balances,” J. Chem. Ed., 59 (1982) 418. 7. R. M. Schoonover, “A Look at the Electronic Analytical Balance,” Anal. Chem., 54 (1982) 973A. 8. J. Meija, “Solution to Precision Weighing Challenge,” Anal. Bioanal. Chem., 394 (2009) 11. Discusses, in detail, quantitative corrections for minor variations in the apparent mass of a stainless steel weight relative to a platinum weight as air density changes (the apparent mass of the stainless steel weight increases relative to that of the platinum weight as the air density is reduced).
CALIBRATION OF WEIGHTS 9. W. D. Abele, “Laboratory Note: Time-Saving Applications of Electronic Balances,” Am. Lab., 13 (1981) 154. Calibration of weights using mass standards calibrated by the National Institute of Standards and Technology is discussed. 10. D. F. Swinehart, “Calibration of Weights in a One-Pan Balance,” Anal. Lett., 10 (1977) 1123.
CALIBRATION OF VOLUMETRIC WARE 11. G. D. Christian, “Coulometric Calibration of Micropipets,” Microchem. J., 9 (1965) 16. 12. M. R. Masson, “Calibration of Plastic Laboratory Ware,” Talanta, 28 (1981) 781. Tables for use in calibration of polypropylene vessels are presented. 13. W. Ryan, “Titrimetric and Gravimetric Calibration of Pipettors: A Survey,” Am. J. Med. Technol., 48 (1982) 763. Calibration of pipettors of 1 to 500 μL is described. 14. M. R. Winward, E. M. Woolley, and E. A. Butler, “Air Buoyancy Corrections for Single-Pan Balances,” Anal. Chem., 49 (1977) 2126. 15. R. M. Schoonover and F. E. Jones, “Air Buoyancy Corrections in High-Accuracy Weighing on Analytical Balances,” Anal. Chem., 53 (1981) 900.
CLEAN LABORATORIES 16. J. R. Moody, “The NBS Clean Laboratories for Trace Element Analysis,” Anal. Chem., 54 (1982) 1358A.
SAMPLING 17. J. A. Bishop, “An Experiment in Sampling,” J. Chem. Ed., 35 (1958) 31. 18. J. R. Moody, “The Sampling, Handling and Storage of Materials for Trace Analysis,” Phil. Trans. Roy. Soc. London, Ser. A, 305 (1982) 669.
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19. G. E. Baiulescu, P. Dumitrescu, and P. Gh. Zugravescu, Sampling. Chichester: Ellis Horwood, 1991.
SAMPLE PREPARATION AND DISSOLUTION 20. G. D. Christian, “Medicine, Trace Elements, and Atomic Absorption Spectroscopy,” Anal. Chem., 41 (1) (1969) 24A. Describes the preparation of solutions of biological fluids and tissues. 21. G. D. Christian, E. C. Knoblock, and W. C. Purdy, “Polarographic Determination of Selenium in Biological Materials,” J. Assoc. Offic. Agric. Chemists, 48 (1965) 877; R. K. Simon, G. D. Christian, and W. C. Purdy, “Coulometric Determination of Arsenic in Urine,” Am. J. Clin. Pathol., 49 (1968) 207. Describes use of Mo(VI) catalyst in digestions. 22. T. T. Gorsuch, “Radiochemical Investigations on the Recovery for Analysis of Trace Elements in Organic and Biological Materials,” Analyst, 84 (1959) 135. 23. S. Nobel and D. Nobel, “Determination of Mercury in Urine,” Clin. Chem., 4 (1958) 150. Describes room temperature digestion. 24. G. Knapp, “Mechanical Techniques for Sample Decomposition and Element Preconcentration,” Mikrochim. Acta, II (1991) 445. Lists acid mixtures and uses for digestions and other decomposition methods.
MICROWAVE DIGESTION 25. H. M. Kingston and L. B. Jassie, eds., Introduction to Microwave Sample Preparation: Theory and Practice. Washington, D.C.: American Chemical Society, 1988. 26. H. M. Kingston and S. J. Haswell, eds., Microwave-Enhanced Chemistry: Fundamentals, Sample Preparation, and Applications. Washington, DC: American Chemical Society, 1997. 27. H. M. Kingston and L. B. Jassie, “Microwave Energy for Acid Decomposition at Elevated Temperatures and Pressures Using Biological and Botanical Samples,” Anal. Chem., 58 (1986) 2534. 28. R. A. Nadkarni, “Applications of Microwave Oven Sample Dissolution in Analysis,” Anal. Chem., 56 (1984) 2233. 29. B. D. Zehr, “Development of Inorganic Microwave Dissolutions,” Am. Lab., December (1992) 24. Also lists properties of useful acid mixtures. 30. www.cem.com. CEM is a manufacturer of laboratory microwave ovens.
LABORATORY SAFETY 31. Safety in Academic Chemistry Laboratories, 7th ed., Vol. 1 (Student), Vol. 2 (Teacher), American Chemical Society, Committee on Chemical Safety. Washington, DC: American Chemical Society, 2003. Available online at http://portal.acs.org/portal/PublicWebSite/about/governance/ committees/chemicalsafety/publications/WPCP_012294 (vol. 1) and http://portal.acs.org/portal/ PublicWebSite/about/governance/committees/chemicalsafety/publications/WPCP_012293 (vol. 2). 32. The Waste Management Manual for Laboratory Personnel, Task Force on RCRA, American Chemical Society, Department of Government Relations and Science Policy. Washington, DC: American Chemical Society, 1990. 33. A. K. Furr, ed., CRC Handbook of Laboratory Safety, 4th ed. Boca Raton, FL: CRC Press, 1993. 34. R. H. Hill and D. Finster, Laboratory Safety for Chemistry Students. Hoboken, NJ: Wiley, 2010. 35. M.-A. Armour, Hazardous Laboratory Chemicals Disposal Guide. Boca Raton, FL: CRC Press, 1990.
MATERIAL SAFETY DATA SHEETS 36. http://siri.org/msds. Online searchable database, with links to other MSDS and hazardous chemical sites. 37. www.env-sol.com. Solutions Software Corporation. MSDS database available on DVD or CD-ROM.
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Chapter Three STATISTICS AND DATA HANDLING IN ANALYTICAL CHEMISTRY “Facts are stubborn, but statistics are much more pliable.” —Mark Twain “43.8% of all statistics are worthless.” —Anonymous Chapter 3 URLs
“Oh, people can come up with statistics to prove anything. 14% of people know that.” —Homer Simpson
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Accuracy and precision, p. 62
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Rejection of a result, p. 95
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Types of errors in measurements, p. 63, 64
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Significant figures in measurements and calculations, p. 65, 81
Least-squares p. 100, 104
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Detection limits, p. 105
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Absolute and relative uncertainty, p. 66, 71
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Statistics of sampling, p. 107
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Standard deviation, p. 72
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Power analysis, p. 110
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Propagation of errors, p. 75
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How to use spreadsheets, p. 112
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Control charts, p. 83
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Using spreadsheets for plotting calibration curves, p. 117
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Statistics: confidence limits, t-tests, F-tests, p. 84, 86, 88
plots
and
coefficient
of
determination,
Although data handling normally follows the collection of analytical data, it is treated early in this textbook because a knowledge of statistical analysis will be required as you perform experiments in the laboratory. Also, statistical analysis is necessary to understand the significance of the data that are collected and thus sets limits on each step of the analysis. Experimental design (including required sample size, measurement accuracy, and number of analyses needed) relies on a proper understanding of what the data represent. The availability of spreadsheets to process data has made statistical and other calculations very efficient. You will first be presented with the details of various calculations throughout the text, which are necessary for full understanding of the principles. A variety of pertinent spreadsheet calculations will also be introduced at the end of the chapter to illustrate how to take advantage of this software for routine calculations.
Accuracy is how close you get to the bull’s-eye. Precision is how close the repetitive shots are to one another. It is nearly impossible to have accuracy without good precision.
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3.1 Accuracy and Precision: There Is a Difference Accuracy is the degree of agreement between the measured value and the true value. An absolute true value is seldom known. A more realistic definition of accuracy, then, would assume it to be the agreement between a measured value and the accepted true value.
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3.2 DETERMINATE ERRORS— —THEY ARE SYSTEMATIC
We can, by good analytical technique, such as making comparisons against a known standard sample of similar composition, arrive at a reasonable assumption about the accuracy of a method, within the limitations of the knowledge of the “known” sample (and of the measurements). The accuracy to which we know the value of the standard sample is ultimately dependent on some measurement that will have a given limit of certainty in it. Precision is defined as the degree of agreement between replicate measurements of the same quantity. That is, it is the repeatability of a result. The precision may be expressed as the standard deviation, the coefficient of variation, the range of the data, or as a confidence interval (e.g., 95%) about the mean value. Good precision does not assure good accuracy. This would be the case, for example, if there were a systematic error in the analysis. The volume of a pipet used to dilute each of the samples may be in error. This error does not affect the precision, but it does affect the accuracy. On the other hand, the precision can be relatively poor and the accuracy may be good; admittedly, this is very rare. Since all real analyses are unknown, the higher the degree of precision, the greater the chance of obtaining the true value. It is fruitless to hope that a value is accurate despite the precision being poor; and the analytical chemist strives for repeatable results to assure the highest possible accuracy. These concepts can be illustrated with targets, as in Figure 3.1. Suppose you are at target practice and you shoot the series of bullets that all land in the bull’s-eye (left target). You are both precise and accurate. In the middle target, you are precise (steady hand and eye), but inaccurate. Perhaps the sight on your gun is out of alignment. In the right target you are imprecise and therefore probably inaccurate. So we see that good precision is needed for good accuracy, but it does not guarantee it. As we shall see later, reliability increases with the number of measurements made. The number of measurements required will depend on the level of uncertainty that is acceptable and on the known reproducibility of the method.
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Good precision does not guarantee accuracy.
“To be sure of hitting the target, shoot first, and call whatever you hit the target.”—Ashleigh Brilliant
3.2 Determinate Errors—They Are Systematic Two main classes of errors can affect the accuracy or precision of a measured quantity. Determinate errors are those that, as the name implies, are determinable and that presumably can be either avoided or corrected. They may be constant, as in the case of an uncalibrated pipet that is used in all volume deliveries. Or, they may be variable but of such a nature that they can be accounted for and corrected, such as a buret whose volume readings are in error by different amounts at different volumes. The error can be proportional to sample size or may change in a more complex manner. More often than not, the variation is unidirectional, as in the case of solubility loss of a precipitate due to its solubility (negative error). It can, however, be random in sign, i.e., a positive or negative error. Such an example is the change in solution volume and concentration occurring with changes in temperature. This can be corrected for by measuring the solution temperature. Such measurable determinate errors are classed as systematic errors.
Good precision, good accuracy
Good precision, poor accuracy
Poor precision, poor accuracy (could be accurate if shots symmetircal)
Determinate or systematic errors are nonrandom and occur when something is intrinsically wrong in the measurement.
Fig. 3.1. precision.
Accuracy versus
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Some common determinate errors are:
It is always a good idea to run a blank.
1. Instrumental errors. These include faulty equipment such as uncalibrated glassware. 2. Operative errors. These include personal errors and can be reduced by experience and care of the analyst in the physical manipulations involved. Operative errors can be minimized by having a checklist of operations. Operations in which these errors may occur include transfer of solutions, effervescence and “bumping” during sample dissolution, incomplete drying of samples, and so on. These are difficult to correct for. Other personal errors include mathematical errors in calculations and prejudice in estimating measurements. 3. Errors of the method. These are the most serious errors of an analysis. Most of the above errors can be minimized or corrected for, but errors that are inherent in the method cannot be changed unless the conditions of the determination are altered. Some sources of methodical errors include coprecipitation of impurities, slight solubility of a precipitate, side reactions, incomplete reactions, and impurities in reagents. Sometimes correction can be relatively simple, for example, by running a reagent blank. A blank determination is an analysis on the added reagents only. It is standard practice to run such blanks and to subtract the results from those for the sample. But a good blank analysis alone cannot guarantee correct measurements. If the method, for example, responds to an analyte present in the sample other than the intended analyte, the method must be altered. Thus, when errors become intolerable, another approach to the analysis must be made. Sometimes, however, we are forced to accept a given method in the absence of a better one.
Determinate errors may be additive or multiplicative, depending on the nature of the error or how it enters into the calculation. In order to detect systematic errors in an analysis, it is common practice to add a known amount of standard to a sample (a “spike”) and measure its recovery (see Validation of a Method in Chapter 1) and note that good spike recovery cannot also correct for response from an unintended analyte (i.e., an interference). The analysis of reference samples helps guard against method errors or instrumental errors.
3.3 Indeterminate Errors—They Are Random Indeterminate errors are random and cannot be avoided.
The second class of errors includes the indeterminate errors, often called accidental or random errors, which represent the experimental uncertainty that occurs in any measurement. These errors are revealed by small differences in successive measurements made by the same analyst under virtually identical conditions, and they cannot be predicted or estimated. These accidental errors will follow a random distribution; therefore, mathematical laws of probability can be applied to arrive at some conclusion regarding the most probable result of a series of measurements. It is beyond the scope of this text to go into mathematical probability, but we can say that indeterminate errors should follow a normal distribution, or Gaussian curve. Such a curve is shown in Figure 3.2. The symbol σ represents the standard deviation of an infinite population of measurements, and this measure of precision
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Mean (X )
68.3%
99.7% −3σ
−2σ
95.4%
−σ +σ 0 Standard deviations from the mean
+2σ
+3σ
defines the spread of the normal population distribution as shown in Figure 3.2. It is apparent that there should be few very large errors and that there should be an equal number of positive and negative errors. Indeterminate errors really originate in the limited ability of the analyst to control or make corrections for external conditions, or the inability to recognize the appearance of factors that will result in errors. Some random errors stem from the intrinsic nature of things, for example, consider that a sample of the radionuclide 129 I is taken. The isotope is long lived, and in a short time there will not be a perceptible change in its number. But, if a sufficient amount is taken, then based on the half-life, you can expect a decay to occur every 60 s. In reality, this may not occur every 60 s, but can fluctuate, with an average of 60 s. Sometimes, by changing conditions, some unknown error will disappear. Of course, it will be impossible to eliminate all possible random errors in an experiment, and the analyst must be content to minimize them to a tolerable or insignificant level.
Fig. 3.2.
Normal error curve.
“Undetectable errors are infinite in variety, in contrast to detectable errors, which by definition are limited.”—Tom Gibb
3.4 Significant Figures: How Many Numbers Do You Need? The weak link in the chain of any analysis is the measurement that can be made with the least accuracy or precision. It is useless to extend an effort to make the other measurements of the analysis more accurately than this limiting measurement. The number of significant figures can be defined as the number of digits necessary to express the results of a measurement consistent with the measured precision. Since there is uncertainty (imprecision) in any measurement of at least ±1 in the last significant figure, the number of significant figures includes all of the digits that are known, plus the first uncertain one. In reported answers, it generally does not make sense to include additional digits beyond the first uncertain one. Each digit denotes the actual quantity it specifies. For example, in the number 237, we have 2 hundreds, 3 tens, and 7 units. If this number is reported as a final answer, it implies that uncertainty lies in the units digit (e.g., ±1). The digit 0 can be a significant part of a measurement, or it can be used merely to place the decimal point. The number of significant figures in a measurement is independent of the placement of the decimal point. Take the number 92,067. This number has five significant figures, regardless of where the decimal point is placed. For example, 92, 067 μm, 9.2067 cm, 0.92067 dm, and 0.092067 m all have the same number of significant figures. They merely represent different ways (units) of expressing one measurement. The zero between the decimal point and the 9 in the last number is used only to place the decimal point. There is no doubt whether any zero that follows a decimal point is significant or is used to place the decimal point. In
The last digit of a measurement has some uncertainty. You can’t include any more digits.
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Writing a value in scientific notation avoids ambiguities about the number of significant figures it contains.
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the number 727.0, the zero is not used to locate the decimal point but is a significant part of the figure. Ambiguity can arise if a zero precedes a decimal point. If it falls between two other nonzero integers, then it will be significant. Such was the case with 92,067. In the number 936,600, it is impossible to determine whether one or both or neither of the zeros is used merely to place the decimal point or whether they are a part of the measurement. It is best in cases like this to write only the significant figures you are sure about and then to locate the decimal point by scientific notation. Thus, 9.3660 × 105 has five significant figures, but 936,600 contains six digits, one to place the decimal. Sometimes, the number may be written with a period at the end to denote all digits are significant, e.g., 936,000. to avoid amiguity.
Example 3.1 List the proper number of significant figures in the following numbers and indicate which zeros are significant. 0.216; 90.7; 800.0; 0.0670 Solution
0.216 90.7 800.0 0.0670
three significant figures three significant figures; zero is significant four significant figures; all zeros are significant three significant figures; only the last zero is significant
If a number is written as 500, it could represent 500 ± 100. If it is written as 5.00 × 102 , then it is 500 ± 1. The significance of the last digit of a measurement can be illustrated as follows. Assume that each member of a class measures the width of a classroom desk, using the same meter stick. Assume further that the meter stick is graduated in 1-mm increments. The measurements can be estimated to the nearest 0.1 division (0.1 mm) by interpolation, but the last digit is uncertain since it is only an estimation. A series of class readings, for example, might be 565.4 mm 565.8 mm 565.0 mm 566.1 mm 565.6 mm (average) ABSOLUTE AND RELATIVE UNCERTAINTY The uncertainty of a number is often called the absolute uncertainty. The relative uncertainty is the absolute uncertainty divided by the value of the number.
Whenever a measurement or the result of a calculation is numerically represented, it is implicit that the last digit of the number is uncertain to ±1. In the above example, when we say that the average width of the desk is 565.6 mm, it is understood that the value can be between 565.5 to 565.7 mm. Here the uncertainty of the number, also called the absolute uncertainty is in the tenth digit, 0.1 mm. If we say the distance between Seattle, Washington, and Arlington, Texas, is 2.99 × 103 km, the uncertainty is in the
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tens digit, the absolute uncertainty is 10 km—this number denotes a range between 2980 to 3000 km. The relative uncertainty is the absolute uncertainty divided by the value of the number, in the first case, this is 0.1/565.6 or 1 part in 5656; in the second case, it is 10/2.99 × 103 or 1 part in 299. PROPAGATION OF UNCERTAINTIES We carry out arithmetic operations like addition, subtraction, division, multiplication, and so forth with numbers. Much like a chain can be no stronger than its weakest link, the result of such operations can be no more certain than the number with the greatest uncertainty involved in the calculations. However, in addition and subtraction, the greatest absolute uncertainty is preserved; the final result is no more certain than the number with the greatest absolute uncertainty. If I have 56 quarters (read: this means 55 to 57) and you have 2300 pennies (means 2299 to 2301) and we add our resources, the net result cannot have the uncertainty less than a quarter. On the other hand, when we were driving from Arlington, Texas, to Seattle, Washington, Kevin was reading the odometer and said the distance was 2990 km (read again: 2980 to 3000, relative uncertainty 1 part in 299) and Gary was keeping the time and said it took 27 hours (read: 26 to 28, relative uncertainty 1 part in 27). If we want to calculate the speed, we will have to divide the distance traveled by the time, and the final result can be no more accurate than the number with the greatest relative uncertainty, 1 part in 27. ADDITION AND SUBTRACTION——THINK ABSOLUTE In addition and subtraction, the placement of the decimal point is important in determining how many figures will be significant. Suppose you wish to calculate the formula weight of Ag2 MoO4 from the individual atomic weights (Ag = 107.870 amu, Mo = 95.94 amu, O = 15.9994 amu); amu = atomic mass unit. Note that the atomic weight of molybdenum is known only to the nearest 0.01 amu, while that for Ag and O are known to 0.001 and 0.0001 amu, respectively. We cannot justifiably say that we know the formula weight of a compound containing molybdenum to any closer than 0.01 atomic unit. Therefore, the most accurately known value for the atomic weight of Ag2 MoO4 is 375.68. All numbers being added or subtracted can be rounded to the least significant unit before adding or subtracting. But for consistency in the answer, one additional figure should be carried out and then the answer rounded to one less figure. Ag Ag Mo O O O O
107.87 107.87 95.94 15.99 15.99 15.99 15.99 375.67
| | | | | | | |
The answer of an addition or subtraction is known to the same number of units as the number containing the least significant unit.
0 0 94 94 94 94 76
MULTIPLICATION AND DIVISION——THINK RELATIVE In all measurements, the last reported digit is uncertain. This is the last significant figure in the measurement; any digits beyond it are meaningless. In multiplication and division, the uncertainty of this digit is carried through the mathematical operations, thereby limiting the number of certain digits in the answer. The final answer of a multiplication and division operation has the same relative uncertainty as the number with greatest relative uncertainty involved in the operation. The answer should contain
The answer of a multiplication or division can be no more accurate than the least accurately known operator.
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no more significant digits than a number, involved in the operation, which has the least number of significant digits. If more than one number share the distinction of having the same least number of significant digits, the number with the lowest magnitude has the greatest relative uncertainty and it governs the final result. For example, 0.0344 and 5.39 both have the same number of significant figures; but the first has the greater relative uncertainty (1 in 344 compared to 1 in 539).
Example 3.2 Give the answer of the following computation to the correct number of significant figures: 97.7 × 100.0 + 36.04 32.42 Solution 687 337.4 301.36 + 36.04 = = 0.04911 687 687 In the first operation, the number with the greatest relative uncertainty is 97.7 and the result is 301.36 . We carried an additional fifth figure until the addition step. Then we rounded the added numbers to four figures before the division since the divisor has only three significant figures. In the division step, the number with the greatest relative uncertainty is 687. If we keep the answer as 0.04911, it has more significant digits than 687. If we make the answer 0.0491, we have a greater uncertainty in the final result than in 687. Therefore, we keep all the digits and put the last one as a subscript. Note that if in the first step we had rounded to 301.4 , the numerator would have become 337.5 and the final answer would be 0.4913 (still within the experimental uncertainty).
Example 3.3 In the following pairs of numbers, pick the number with the greatest relative uncertainty that would control the result of multiplication or division. (a) 42.67 or 0.0967; (b) 100.0 or 0.4570; (c) 0.0067 or 0.10. Solution
(a) 0.0967 (has three significant figures) (b) 100.0 (both have four significant figures, but the uncertainty here is 1 part per thousand versus about 1 part in 4600) (c) 0.10 [both have two significant figures, but the uncertainty here is 10% (1 part in 10) versus about 1 part in 70]
Example 3.4 Give the answer of the following operation to the maximum number of significant figures and indicate the number with the greatest relative uncertainty. 35.63 × 0.5481 × 0.05300 × 100% = 88.5470578% 1.1689
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Solution
The number with the greatest relative uncertainty is 35.63. The answer is therefore 88.55%, and it is meaningless to carry the operation out to more than five figures (the fifth figure is used to round off the fourth). The 100% in this calculation is an absolute number since it is used only to move the decimal point, and it has an infinite number of significant figures. (Numerical coefficients in a formula such as the multipliers 2 and 4 or the atomic weights of Ag and O while calculating the FW of Ag2 Mo4 also are absolute numbers, with an infinite number of significant digits.) Note that 35.63 has a relative uncertainty at best of 1 part in 3600, and so the answer has a relative uncertainty at best of 1 part in 3600; thus, the answer has a relative uncertainty at least of this magnitude (i.e., about 2.5 parts in 8900). The objective in a calculation is to express the answer to at least the precision of the least certain number, but to recognize the magnitude of its uncertainty. The final number is determined by the measurement of significant figures. (Similarly, in making a series of measurements, one should strive to make each measurement to about the same degree of relative uncertainty.) There can be situations, where carrying the same number of significant digits in the final answer as the number(s) with least significant digits result in a substantially smaller relative uncertainty in the final result than the number with the greatest relative uncertainty involved in the operation. In such cases, the same number of significant digits is preserved in the results, but the last digit is written as a subscript to connote uncertainty.
Example 3.5 Rationalize the following operation 1.001 × 103 × 99.89 = 8.874 × 10−1 2 1.006 × 10 × 1120 Solution
All the numbers involved in the operation have the same number of significant digits, 4. The number 1.001 × 103 has the greatest relative uncertainty, 1 part in 1001. If we keep the same number of significant digits in the final result, the relative uncertainty is almost an order of magnitude smaller, 1 part in 8874. The last digit is therefore indicated as a subscript. In multiplication and division, the answer from each step of a series of operations can statistically be rounded to the number of significant figures to be retained in the final answer. But for consistency in the final answer, it is preferable to carry one additional figure until the end and then round off.
PUTTING IT ALL TOGETHER Summarizing the importance of significant figures, there are two questions to ask. First, how accurately do you have to know a particular result? If you only want to learn whether there is 12 or 13% of a substance in the sample, then you need only make all required measurements to achieve a final error of ±1% (including in the case when the measurement includes multiple steps and operations). If the sample weighs about 2 g, there is no need to weigh it more precisely than to 0.1 g. The second question is, how accurately can you make each required measurement? Obviously, if you can read
A subscript number is used to indicate an added degree of uncertainty in the final result. It is used when the relative uncertainty of the result is less than that of the least certain number involved in the operation.
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It is good practice to keep an extra figure during stepwise calculations and then drop it in the final number. “Check the answer you have worked out once more—before you tell it to anybody.” —Edmund C. Berkely
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the absorbance of light by a colored solution to only three figures (e.g., A = 0.447), it would be useless to weigh the sample to more than three figures (e.g., 6.67 g). When a number in a measurement is small (without regard to the decimal point) compared with those of the other measurements, there is some justification in making the measurement to one additional figure. This can be visualized as follows. Suppose you wish to weigh two objects of essentially the same mass, and you wish to weigh them with the same precision, for example, to the nearest 0.1 mg, or 1 part per thousand. The first object weighs 99.8 mg, but the second weighs 100.1 mg. You have weighed both objects with equal accuracy, but you have retained an additional significant figure in one of them to keep the relative uncertainty of the same order. This is related to keeping the last significant digit as a subscript when the final result has a smaller relative uncertainty than the number with the greatest relative uncertainty involved in the calculations. When the number with the greatest relative uncertainty in a series of measurements is known, then the overall accuracy can be improved, if desired, either by making the number larger (e.g., by increasing the sample size) or by making the measurement to an additional figure if possible (e.g., by weighing more precisely to one additional significant figure). This would be desirable when the number has greater relative uncertainty compared to those of the other measurements in order to bring its uncertainty closer to that of the others. In carrying out analytical operations, then, you should try to measure quantities to the same absolute uncertainty when adding or subtracting and to the same relative uncertainty when multiplying or dividing. If a computation involves both multiplication/division and addition/subtraction, then the individual steps must be treated separately. As good practice, retain one extra figure in the intermediate calculations until the final result (unless it drops out in a subsequent step). When a calculator is used, all digits can be kept in the calculator until the end. Do not assume that a number spat out by a calculator is correct; it is common to make mistakes in entering numbers, especially when you are in a hurry. Always try to estimate the size of the answer you expect. If you expect 2% and you calculate 0.02%, you probably forgot to multiply by 100. Or if you expect 20% and the answer is 4.3, you probably made a calculation error or perhaps a measurement error. Always ask yourself: Does it make sense? Is this number reasonable?
LOGARITHMS——THINK MANTISSA In changing from logarithms to antilogarithms, and vice versa, the number being operated on and the logarithm mantissa have the same number of significant figures. (See Appendix B for a review of the use of logarithms.) All zeros in the mantissa are significant. Suppose, for example, we wish to calculate the pH of a 2.0 × 10−3 M solution of HCl from pH = −log[H+ ]. Then, pH = − log 2.0 × 10−3 = −(−3 + 0.30) = 2.70 For pH = 2.70, we call “2” the characteristic and “70” the mantissa. In logarithms, it is the number of significant figures in the mantissa that determines the number of significant figures in the final value. Zeros in mantissa count as significant figures.
The −3 is the characteristic (from 10−3 ), a pure number determined by the position of the decimal. The 0.30 is the mantissa from the logarithm of 2.0 and therefore has only two digits. So, even though we know the concentration to two figures, the pH (the logarithm) has three figures. If we wish to take the antilogarithm of a mantissa, the corresponding number will likewise have the same number of digits as the mantissa. The antilogarithm of 0.072 (contains three figures in mantissa .072) is 1.18, and the logarithm of 12.1 is 1.083 (1 is the characteristic, and the mantissa has three digits, .083).
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3.5 Rounding Off If the digit following the last significant figure is greater than 5, the number is rounded up to the next higher digit. If it is less than 5, the number is rounded to the present value of the last significant figure: 9.47 = 9.5 9.43 = 9.4 If the last digit is a 5, the number is rounded off to the nearest even digit: 8.65 = 8.6 8.75 = 8.8 8.55 = 8.6 This is based on the statistical prediction that there is an equal chance that the last significant figure before the 5 will be even or odd. That is, in a suitably large sampling, there will be an equal number of even and odd digits preceding a 5. All nonsignificant digits should be rounded off all at once. The even-number rule applies only when the digit dropped is exactly 5 (not . . . 51). (For example, if we want four significant figures, then 45.365 rounds to 45.36, whereas 45.3651 rounds to 45.37.)
3.6 Ways of Expressing Accuracy There are various ways and units in which the accuracy of a measurement can be expressed. In each case, it is assumed that a “true” value is available for comparison. ABSOLUTE ERRORS The difference between the true value and the measured value, with regard to the sign, is the absolute error, and it is reported in the same units as the measurement. If an analyst reports 2.62 g of copper as 2.51 g, the absolute error is −0.11 g. If the measured value is the average of several measurements, the error is called the mean error. The mean error can also be calculated by taking the average difference, with regard to sign, of the individual test results from the true value. RELATIVE ERROR The absolute or mean error expressed as a percentage of the true value is the relative error. The above analysis has a relative error of (−0.11/2.62) × 100% = −4.2%. The relative accuracy is the measured value or mean expressed as a percentage of the true value. The above analysis has a relative accuracy of (2.51/2.62) × 100% = 95.8%. We should emphasize that, more often than not, neither number is known to be “true,” and the relative error or accuracy is based on the mean of two sets of measurements; that the true value is “true” is an assumption, unless that value is a certified value in a reference standard. Relative error can be expressed in units other than percentages. In very accurate work, we are usually dealing with relative errors of much less than 1%, and it is convenient to use a smaller unit. A 1% error is equivalent to 1 part in 100. It is also equivalent to 10 parts in 1000. This latter unit is commonly used for expressing small uncertainties. That is, the uncertainty is expressed in parts per thousand. The number
Always round to the even number, if the last digit is a 5.
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0.11 expressed as parts per thousand of the number 2.62 would be 0.11 parts per 2.62, or 42 parts per thousand. Parts per thousand is often used in expressing precision of measurement. For even smaller relative amounts, parts per million (1 ppm = 1 part in 1,000,000) and parts per billion (1 ppb = 1 part in 1,000,000,000) are commonly used.
Example 3.6 The results of an analysis are 36.97 g, compared with the accepted value of 37.06 g. What is the relative error in parts per thousand? Which is a bigger relative error: 395 ppb or 0.412 ppm? Solution: Multiply ppm by 1000 to get the value in ppb. Thus, the latter value is 412 ppb, which is obviously greater than 395 ppb and constitutes the bigger error.
Solution
Absolute error = 36.97 g − 37.06 g = −0.09 g Relative error =
−0.09 × 1000‰ = −2.4 parts per thousand 37.06
‰ indicates parts per thousand, just as % indicates parts per hundred.
3.7 Standard Deviation—The Most Important Statistic “If reproducibility be a problem, conduct the test only once.”—Anonymous
average, x = (xi /N)
See Section 3.15 and Equation 3.17 for another way of estimating s for four or less numbers.
Each set of analytical results should be accompanied by an indication of the precision of the analysis. Various ways of indicating precision are acceptable. The standard deviation σ of an infinite set of experimental data is theoretically given by (xi − μ)2 σ = (3.1) N where xi represents the individual measurements and μ represents the mean of an infinite number of measurements (which should represent the “true” value). This equation holds strictly only as N → ∞, where N is the number of measurements. In practice, we must calculate the individual deviations from the mean of a limited number of measurements, x, in which it is anticipated that x → μ as N → ∞, although we have no assurance this will be so; x is given by (xi /N). For a set of N measurements, there are N (independently variable) deviations from some reference number. But if the reference number chosen is the estimated mean, x, the sum of the individual deviations (retaining signs) must necessarily add up to zero, and so values of N − 1 deviations are adequate to define the Nth value. That is, there are only N − 1 independent deviations from the mean; when N − 1 values have been selected, the last is predetermined. We have, in effect, used one degree of freedom of the data in calculating the mean, leaving N − 1 degrees of freedom for calculating the precision. As a result, the estimated standard deviation s of a finite set of experimental data (generally N < 30) more nearly approximates σ if N − 1, the number of degrees of freedom, is substituted for N (N − 1 adjusts for the difference between x and μ). s=
(xi − x)2 N−1
(3.2)
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The value of s is only an estimate of σ , and it will more nearly approach σ as the number of measurements increases. Since we deal with small numbers of measurements in an analysis, the precision is appropriately represented by s.
Example 3.7 Calculate the mean and the standard deviation of the following set of analytical results: 15.67, 15.69, and 16.03 g. Solution
xi
15.67 15.69 16.03
xi − x
(xi − x)2
0.13 0.11 0.23
0.0169 0.0121 0.0529
47.39
0.47
0.0819
47.39 xi = = 15.80 N 3 0.0819 = 0.20 g s= 3−1
x=
The standard deviation may be calculated also using the following equivalent equation: 2 2 xi − xi /N (3.3) s= N−1 This is useful for computations with a calculator. Many calculators, in fact, have a standard deviation program that automatically calculates the standard deviation from entered individual data. All spreadsheets can calculate the mean and standard deviations of a row or column of entered data. Microsoft Excel uses the respective functions AVERAGE and STDEV. These are discussed later, in more detail, in Section 3.19.
Example 3.8 Calculate the standard deviation for the data in Example 3.7 using Equation 3.3. Solution
xi 15.67 15.69 16.03 47.39 s=
xi2 245.55 246.18 256.96 748.69
748.69 − (47.39)2 /3 = 0.21 g 3−1
This result would be properly displayed as 15.8 ± 0.2 g (mean ± standard deviation) in a final report.
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The difference of 0.01 g from Example 3.7 is not statistically significant since the variation is at least ±0.2 g. In applying this formula, it is important to keep an extra digit or even two in xi2 for the calculation. The precision improves as the square root of the number of measurements.
The standard deviation calculation considered so far is an estimate of the probable error of a single measurement. The arithmetical mean of a series of N measurements taken from an infinite population will show less scatter from the “true value” than will an individual observation. The scatter will decrease as N is increased; as N gets very large the sample average will approach the population average μ, and the scatter approaches zero. The arithmetical mean derived from N measurements can be shown √ to be N times more reliable than a single measurement. Hence, the random error in the mean of a series of four observations is one-half that of a single observation. In other words, the precision of the mean of a series of N measurements is inversely proportional to the square root of N of the deviation of the individual values. Thus, s Standard deviation of the mean = smean = √ N
rsd = s/x; % rsd = (s/x) × 100%
(3.4)
The standard deviation of the mean is sometimes referred to as the standard error. The standard deviation is sometimes expressed as the relative standard deviation (rsd), which is just the standard deviation expressed as a fraction of the mean; usually it is given as the percentage of the mean (% rsd), which is often called the coefficient of variation.
Example 3.9 The following replicate weighings were obtained: 29.8, 30.2, 28.6, and 29.7 mg. Calculate the standard deviation of the individual values and the standard deviation of the mean. Express these as absolute (units of the measurement) and relative (% of the measurement) values. Solution
xi 29.8 30.2 28.6 29.7 118.3
x= s=
xi − x
(xi − x)2
0.2 0.6 1.0 0.1 1.9
0.04 0.36 1.00 0.01 1.41
118.3 = 29.6 4
0.69 1.41 = 0.69 mg(absolute); × 100% = 2.3%(coefficient of variation) 4−1 29.6 0.69 0.34 × 100% = 1.1%(relative) smean = √ = 0.34mg(absolute); 29.6 4
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The precision of a measurement can be improved by increasing the number of observations. In other words, the spread ±s of the normal curve in Figure 3.2 becomes smaller as the number of observations is increased and would approach zero as the number of observations approached infinity. However, as seen above (Equation 3.4), the deviation of the mean does not decrease in direct proportion to the number of observations, but instead it decreases as the square root of the number of observations. A point will be reached where a slight increase in precision will require an unjustifiably large increase in the number of observations. For example, to decrease the standard deviation by a factor of 10 requires 100 times as many observations. The practical limit of useful replication is reached when the standard deviation of the random errors is comparable to the magnitude of the determinate or systematic errors (unless, of course, these can be identified and corrected for). This is because the systematic errors in a determination cannot be removed by replication. The significance of s in relation to the normal distribution curve is shown in Figure 3.2. The mathematical treatment from which the curve was derived reveals that 68% of the individual deviations fall within one standard deviation (for an infinite population) from the mean, 95% are less than twice the standard deviation, and 99% are less than 2.5 times the standard deviation. So, a good approximation is that 68% of the individual values will fall within the range x ± s, 95% will fall within x ± 2s, 99% will fall within x ± 2.5s, and so on. Actually, these percentage ranges were derived assuming an infinite number of measurements. There are then two reasons why the analyst cannot be 95% certain that the true value falls within x ± 2s. First, one makes a limited number of measurements, and the fewer the measurements, the less certain one will be. Second, the normal distribution curve assumes no determinate errors, but only random errors. Determinate errors, in effect, shift the normal error curve from the true value. An estimate of the actual certainty a number falls within s can be obtained from a calculation of the confidence limit (see below). It is apparent that there are a variety of ways in which the precision of a number can be reported. Whenever a number is reported as x ± y, you should always qualify under what conditions this holds, that is, how you arrived at ±y. It may, for example, represent s, 2s, s (mean), or the coefficient of variation. A term that is sometimes useful in statistics is the variance. This is the square of the standard deviation, s2 . We shall use this in determining the propagation of error and in the F-test below (Section 3.12).
3.8 Propagation of Errors—Not Just Additive When discussing significant figures earlier, we stated that the relative uncertainty in the answer to a multiplication or division operation could be no better than the relative uncertainty in the operator that had the poorest relative uncertainty. Also, the absolute uncertainty in the answer of an addition or subtraction could be no better than the absolute uncertainty in the number with the largest absolute uncertainty. Without specific knowledge of the uncertainties, we assumed an uncertainty of at least ±1 in the last digit of each number. From a knowledge of the uncertainties in each number, it is possible to estimate the actual uncertainty in the answer. The errors in the individual numbers will propagate throughout a series of calculations, in either a relative or an absolute fashion, depending on whether the operation is a multiplication/division or whether it is an addition/subtraction.
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In Example 3.9, how many observations might you need to reduce the coefficient of variation (cv) to 1.0%? Solution: You presently have 4 observations and want to improve the cv by a factor of 2.3. The total number of observations needed = 4 × 2.32 21 “Randomness is required to make statistical calculations come out right.”—Anonymous
The true value will fall within x ± 2s 95% of the time for an infinite number of measurements. See the confidence limit and Example 3.5.
The variance equals s2 .
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ADDITION AND SUBTRACTION——THINK ABSOLUTE VARIANCES Consider the addition and subtraction of the following numbers: (65.06 ± 0.07) + (16.13 ± 0.01) − (22.68 ± 0.02) = 58.51 (±?) The absolute variances of additions and subtractions are additive.
The uncertainties listed represent the random or indeterminate errors associated with each number, expressed as standard deviations of the numbers. The maximum error of the summation, expressed as a standard deviation, would be ±0.10; that is, it could be either +0.10 or −0.10 if all uncertainties happened to have the same sign. The minimum uncertainty would be 0.00 if all combined by chance to cancel. Both of these extremes are not highly likely, and statistically the uncertainty will fall somewhere in between. For addition and subtraction, absolute uncertainties are additive. The most probable error is represented by the square root of the sum of the absolute variances. That is, the absolute variance of the answer is the sum of the individual variances. For a = b + c − d, s2a = s2b + s2c + s2d sa = s2b + s2c + s2d In the above example, sa = = =
(3.5) (3.6)
(±0.07)2 + (±0.01)2 + (±0.02)2 (49 × 10−4 ) + (1 × 10−4 ) + (4 × 10−4 ) 54 × 10−4 = ±7.3 × 10−2
So the answer is 58.51 ± 0.07. The number ±0.07 represents the absolute uncertainty. If we wish to express it as a relative uncertainty, this would be ±0.07 × 100% = ±0.12 % 58.51
Example 3.10 You have received three shipments of Monazite sand of equal weight that contain traces of europium. Analysis of the three ores provided europium concentrations of 397.8 ± 0.4, 253.6 ± 0.3, and 368.0 ± 0.3 ppm, respectively. What is the average europium content of the ores and what are the absolute and relative uncertainties? Solution
(397.8 ± 0.4) + (253.6 ± 0.3) + (368.0 ± 0.3) 3 The uncertainty in the summation is
sa = (±0.4)2 + (±0.3)2 + (±0.3)2 √ = 0.16 + 0.09 + 0.09 √ = 0.34 = ±0.58 ppm Eu x=
Hence, the absolute uncertainty is x=
1019.4 0.6 ppm ± = 339.8 ± 0.2 ppm Eu 3 3
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Note that since there is no uncertainty in the divisor 3, the relative uncertainty in the europium content is 0.2 ppm Eu = 6 × 10−4 or 0.06% 339.8 ppm Eu MULTIPLICATION AND DIVISION——THINK RELATIVE VARIANCES Consider the following operation: (13.67 ± 0.02) (120.4 ± 0.2) = 356.0 (±?) 4.623 ± 0.006 Here, the relative uncertainties are additive, and the most probable error is represented by the square root of the sum of the relative variances. That is, the relative variance of the answer is the sum of the individual relative variances. For a = bc/d, (s2a )rel = (s2b )rel + (s2c )rel + (s2d )rel (sa )rel = (s2b )rel + (s2c )rel + (s2d )rel
(3.7) (3.8)
In the above example, ±0.02 = ±0.0015 13.67 ±0.2 = ±0.0017 = 120.4 ±0.006 = ±0.0013 = 4.623
(sb )rel = (sc )rel (sd )rel (sa )rel = = =
(±0.0015)2 + (±0.0017)2 + (±0.0013)2 (2.2 × 10−6 ) + (2.9 × 10−6 ) + (1.7 × 10−6 ) (6.8 × 10−6 ) = ±2.6 × 10−3
The absolute uncertainty is given by sa = a × (sa )rel = 356.0 × (±2.6 × 10−3 ) = ±0.93 So the answer is 356.0 ± 0.9.
Example 3.11 Calculate the uncertainty in the number of millimoles of chloride contained in 250.0 mL of a sample when three equal aliquots of 25.00 mL are titrated with silver nitrate with the following results: 36.78, 36.82, and 36.75 mL. The molarity of the AgNO3 solution is 0.1167 ± 0.0002 M. Solution
The mean volume is
36.78 + 36.82 + 36.75 = 36.78 mL 3
The relative variances of multiplication and division are additive.
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The standard deviation is
s=
xi
xi − x
(xi − x)2
36.78 36.82 36.75
0.00 0.04 0.03
0.0000 0.0016 0.0009 0.0025
0.0025 = 0.035 3−1
Mean volume = 36.78 ± 0.04 mL
mmol Cl− titrated = (0.1167 ± 0.0002 mmol/mL)(36.78 ± 0.04 mL) = 4.292 (±?) ±0.0002 = ±0.0017 0.1167 ±0.035 = = ±0.00095 36.78
(sb )rel = (sc )rel
(±0.0017)2 + (±0.00095)2 = (2.9 × 10−6 ) + (0.90 × 10−6 )
= 3.8 × 10−6
(sa )rel =
= ±1.9 × 10−3 The absolute uncertainty in the millimoles of Cl− is 4.292 × (±0.0019) = ±0.0082 mmol mmol Cl− in 25 mL = 4.292 ± 0.0082 mmol mmol Cl− in 250 mL = 10(4.292 ± 0.0082) = 42.92 ± 0.08 mmol Note that we retained one extra figure in computations until the final answer. Here, the absolute uncertainty determined is proportional to the size of the sample; it would not remain constant for twice the sample size, for example. If there is a combination of multiplication/division and addition/subtraction in a calculation, the uncertainties of these must be combined. One type of calculation is done at a time and the uncertainties calculated at each step.
Example 3.12 You have received three shipments of iron ore of the following weights: 2852, 1578, and 1877 lb. There is an uncertainty in the weights of ±5 lb. Analysis of the ores gives 36.28 ± 0.04%, 22.68 ± 0.03%, and 49.23 ± 0.06%, respectively. You are to pay $300 per ton of iron. What should you pay for these three shipments and what is the uncertainty in the payment? Solution
We need to calculate the weight of iron in each shipment, with the uncertainties, and then add these together to obtain the total weight of iron and the uncertainty in this.
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The relative uncertainties in the weights are ±5 ±5 ±5 = ±0.0017 = ±0.0032 = ±0.0027 2852 1578 1877 The relative uncertainties in the analyses are ±0.04 ±0.03 ±0.06 = ±0.0011 = ±0.0013 = ±0.0012 36.28 22.68 49.23 The weights of iron in the shipments are (2852 ± 5 lb)(36.28 ± 0.04%) = 1034.7(±?) lb Fe 100 We calculate the relative standard deviation of the multiplication product as before:
(sa )rel = (±0.0017)2 + (±0.0011)2 = ±0.0020 sa = 1034.7 × (±0.0020) = ±2.1 lb lb Fe = 1034.7 ± 2.1 in the first shipment (We will carry an additional figure throughout.) (1578 ± 5 lb)(22.68 ± 0.03%) = 357.89 (±?) lb Fe 100
(sa )rel = (±0.0032)2 + (±0.0013)2 = ±0.0034 sa = 357.89 × (±0.0034) = ±1.2 lb lb Fe = 357.9 ± 1.2 lb in the second shipment (1877 ± 5 lb)(49.23 ± 0.06%) = 924.05 (±?) lb Fe 100
(sa )rel = (±0.0027)2 + (±0.0012)2 = ±0.0030 sa = 924.05 × (±0.0030) = ±2.8 lb lb Fe = 924.0 ± 2.8 lb in the third shipment Total Fe = (1034.7 ± 2.1 lb) + (357.9 ± 1.2 lb) + (924.0 ± 2.8 lb) = 2316.6 (±?) lb Here we use absolute uncertainties:
sa = (±2.1)2 + (±1.2)2 + (±2.8)2 = ±3.7 lb Total Fe = 2317 ± 4 lb A price of $300/ton is the same as $ 0.15/lb since 1 ton = 2000 lbs. Price = (2316.6 ± 3.7 lb)($0.15/lb) = $347.49 ± 0.56 Hence, you should pay $347.50 ± 0.60.
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Example 3.13 You determine the acetic acid (HOAc) content of vinegar by titrating with a standard (known concentration) solution of sodium hydroxide to a phenolphthalein end point. An approximately 5-mL sample of vinegar is weighed on an analytical balance in
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a weighing bottle (the increase in weight represents the weight of the sample) and is found to be 5.0268 g. The uncertainty in making a single weighing is ±0.2 mg. The sodium hydroxide must be accurately standardized (its concentration determined) by titrating known weights of high-purity potassium acid phthalate, and three such titrations give molarities of 0.1167, 0.1163, and 0.1164 M. A volume of 36.78 mL of sodium hydroxide is used to titrate the sample. The uncertainty in reading the buret is ±0.02 mL. What is the percent acetic acid in the vinegar? Include the standard deviation of the final result.
Solution A
B
1
N1
0.1167
2
N2
0.1163
3
N3
4
STDEV:
5
Cell B4:
6
STDEV(B1:B3)
(See Section 3.20)
0.1164 0.000208
Two weighings are required to obtain the weight of the sample: that of the empty weighing bottle and that of the bottle plus sample. Each has an uncertainty of ±0.2 mg, and so the uncertainty of the net sample weight (the difference of the two weights) is
swt = (±0.2)2 + (±0.2)2 = ±0.3 mg The mean of the molarity of the sodium hydroxide is 0.1165 M, and its standard deviation is ±0.0002 M. Similarly, two buret readings (initial and final) are required to obtain the volume of base delivered, and the total uncertainty is
svol = (±0.02)2 + (±0.02)2 = ±0.03 mL The moles of acetic acid are equal to the moles of sodium hydroxide used to titrate it, so the amount of acetic acid in mmol is mmol HOAc = (0.1165 ± 0.0002)mmol mL−1 (36.78 ± 0.03)mL = 4.2849 (±?) mmol As before, we calculate the relative standard deviation of the product as the root mean sum of squares of the individual relative standard deviations:
(sproduct )rel = [(±0.0002/0.1165)2 + (±0.03/36.78)2 ] = ±0.0019 Multiplying by 4.285, we obtain the standard deviation s: s = ±4.285 × 0.0019 = 0.0081 4.285 ± 0.0081 mmol acetic acid is converted to mg HOAc by multiplying by the formula weight of 60.05 mg/mmol (we assume that there is no significant uncertainty in the formula weight), to obtain 257.31 ± 0.49 mg HOAc. To calculate the % acetic acid content, we must divide by the sample weight, 5026.8 ± 0.3 mg: %HOAc = 257.31 ± 0.49 mg/5026.8 ± 0.3 mg × 100% = 5.119 (±?) %acetic acid Once again we calculate via the relative standard deviation:
(sproduct )rel = [(±0.49/257.3)2 + (±0.3/5026.8)2 ] = ±0.0019 Multiplying by 5.119, we obtain the standard deviation to be 0.01, the final answer thus being 5.12 ± 0.01 wt% acetic acid. The factor that increased the uncertainty the most was the variance in the molarity of the sodium hydroxide solution. This illustrates the importance of careful calibration, which is discussed in Chapter 2.
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3.9 Significant Figures and Propagation of Error We noted earlier that the total uncertainty in a computation determines how accurately we can know the answer. In other words, the uncertainty sets the number of significant figures. Take the following example: (73.1 ± 0.2) (2.245 ± 0.008) = 164.1 ± 0.7 We are justified in keeping four figures, even though the first number has three. Here, we don’t have to carry the additional figure as a subscript since we have indicated the actual uncertainty in it. Note that the greatest relative uncertainty in the multipliers is 0.008/2.245 = 0.0036, while that in the answer is 0.7/164.1 = 0.0043; so, due to the propagation of error, we know the answer somewhat less accurately than the least certain number. When actual uncertainties are known, the number with the greatest uncertainty may not necessarily be the one with the smallest number of digits. For example, the relative uncertainty in 78.1 ± 0.2 is 0.003, while that in 11.21 ± 0.08 is 0.007. Suppose we have the following calculation: (73.1 ± 0.9) (2.245 ± 0.008) = 164.1 ± 2.1 = 164 ± 2 Now the uncertainty in the answer is the units place, and so figures beyond that are meaningless. In this instance, the uncertainty in the original number with the highest relative uncertainty (73.1) and the answer are similar (±0.012) since the uncertainty in the other multiplier is significantly smaller.
Example 3.14 Provide the uncertainties in the following calculations to the proper number of significant figures: (38.68 ± 0.07) − (6.16 ± 0.09) = 32.52 (12.18 ± 0.08) (23.04 ± 0.07) = 86.43 3.247 ± 0.006
(a) (b) Solution
(a) The calculated absolute uncertainty in the answer is ±0.11. Therefore, the answer is 32.5 ± 0.1. (b) The calculated relative uncertainty in the answer is 0.0075, so the absolute uncertainty is 0.0075 × 86.43 = 0.65. Therefore, the answer is 86.4 ± 0.6, even though we know all the other numbers to four figures; there is substantial uncertainty in the fourth digit, which leads to the uncertainty in the answer. The relative uncertainty in that answer is 0.0075, and the largest relative uncertainty in the other numbers is 0.0066, very similar.
PROFESSOR’S FAVORITE EXAMPLE Contributed by Professor Nicholas H. Snow, Seton Hall University Why Propagate All Those Errors? Propagation of errors is one of the most useful and straightforward tools for determining and understanding the precision of an analytical method. In industry, precision is
The number of significant figures in an answer is determined by the uncertainty due to propagation of error.
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critical, as analysts may be called upon to perform the same procedure hundreds or even thousands of times to determine whether a product has been manufactured correctly. In consulting with industrial scientists on how to improve precision in their chemical analysis methods, I use propagation of errors as one of my first tools for evaluating the method and for leading to suggested improvements. For example, in a method for the analysis of β-carotene, the following steps are part of the procedure. A precision of less than ±2% is required of analysts for assays performed using this procedure. (1) Transfer about 50 mg of β-carotene into a 100-mL volumetric flask and make up to volume. (2) Pipet 5 mL of this preparation into another 100-mL volumetric flask and dilute to volume. (3) Pipet 5 mL of this preparation into a 25-mL volumetric flask and dilute to volume. Glassware and balance uncertainties—see the inside back cover for Class A glassware (we will assume use of either Class A or Class B glassware, and compare the uncertainties): Glassware
Uncertainty
Relative Uncertainty (%)
100-mL Vol. Flask Class A 100-mL Vol. Flask Class B 25-mL Vol. Flask Class A 25-mL Vol. Flask Class B 5-mL Vol. Pipet class A 5-mL Vol. Pipet Class B Analytical Balance
±0.08 mL ±0.16 mL ±0.03 mL ±0.06 mL ±0.01 mL ±0.02 mL ±0.0002 g
0.08 0.16 0.12 0.24 0.2 0.4 0.4
For procedures involving glassware transfers, the relative uncertainties are additive in a similar manner as if they were multiplied in a formula. For this portion of the procedure, there are two 100-mL volumetric flasks, one 25-mL volumetric flask, two 5-mL volumetric pipets and two weighing steps (remember, weighing is always by difference). The relative uncertainty using Class A glassware is:
0.082 + 0.082 + 0.122 + 0.22 + 0.22 + 0.42 + 0.42 = 0.65% The relative uncertainty using Class B glassware is:
0.162 + 0.162 + 0.242 + 0.42 + 0.42 + 0.42 + 0.42 = 0.86% The full method includes preparing all standards and samples in this manner, plus instrumental analysis and calculation steps, probably giving total uncertainty of approximately ±1.3% for Class A glassware and ±1.7% for Class B glassware. With a requirement of better than ±2%, this leaves little room for mistakes, poor or variable technique. Note that a significant improvement can be made by weighing the initial sample on a microbalance, which would add one decimal place (reduce the uncertainty to 0.04%) at the cost of a more difficult to operate and expensive balance. This would give an uncertainty in the sample preparation step of 0.33% when using Class A glassware, and 0.66% for the Class B glassware (do the calculations!). Propagation of errors tells us the best a procedure and associated calculations CAN do. Our own standard deviations generated by running the procedures with our
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own samples tell us what the procedure IS DOING. If there is a big difference, usually our performance is less precise than predicted using propagation of errors, then there is likely something wrong that requires further investigation. In business, propagation of errors is one of the simplest and most useful tools for evaluating procedures that must be run numerous times with cost and precision in mind.
3.10 Control Charts A quality control chart is a time plot of a measured quantity that is assumed to be constant (with a Gaussian distribution) for the purpose of ascertaining that the measurement remains within a statistically acceptable range. It may be a day-to-day plot of the measured value of a standard that is run intermittently with samples. The control chart consists of a central line representing the known or assumed value of the control and either one or two pairs of limit lines, the inner and outer control limits. Usually the standard deviation of the procedure is known (a good estimate of σ ), and this is used to establish the control limits. An example of a control chart is illustrated in Figure 3.3, representing a plot of day-to-day results of the analysis of calcium in a pooled serum sample or a control sample that is run randomly and blindly with samples each day. A useful inner control limit is two standard deviations since there is only 1 chance in 20 that an individual measurement will exceed this purely by chance. This might represent a warning limit. The outer limit might be 2.5 or 3σ , in which case there is only 1 chance in 100 or 1 chance in 500 a measurement will fall outside this range in the absence of systematic error. Usually, one control is run with each batch of samples (e.g., 20 samples), so several control points may be obtained each day. The mean of these may √ be plotted each day. The random scatter of this would be expected to be smaller by N, compared to individual points. Particular attention should be paid to trends in one direction; that is, the points lie largely on one side of the central line. This would suggest that either the control is in error or there is a systematic error in the measurement. A tendency for points to lie outside the control limits would indicate the presence of one or more determinate errors in the determination, and the analyst should check for deterioration of reagents, instrument malfunction, or environmental and other effects. Trends should signal contamination of reagents, improper calibration or erroneous standards, or change in the control lot.
A control chart is constructed by periodically running a “known” control sample.
Calcium quality control chart for October 2013
meq/dL
Control
+2.5σ 5.2 +2σ 5.1
−2σ −2.5σ
5.0 4.9 4.8
Oct. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Date
Fig. 3.3. chart.
Typical quality control
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3.11 The Confidence Limit—How Sure Are You?
The true value falls within the confidence limit, estimated using t at the desired confidence level.
Calculation of the standard deviation for a set of data provides an indication of the precision inherent in a particular procedure or analysis. But unless there is a large amount of data, it does not by itself give any information about how close the experimentally determined mean x might be to the true mean value μ. Statistical theory, though, allows us to estimate the range within which the true value might fall, within a given probability, defined by the experimental mean and the standard deviation. This range is called the confidence interval, and the limits of this range are called the confidence limit. The likelihood that the true value falls within the range is called the probability, or confidence level, usually expressed as a percent. The confidence limit, in terms of the standard deviation, s (Equation 3.2), is given by ts Confidence limit = x ± √ N
(3.9)
where t is a statistical factor that depends on the number of degrees of freedom and the confidence level desired. The number of degrees of freedom is one less than the number of measurements. Values of t at different confidence levels and degrees of freedom ν are given in Table 3.1. Note that the √ confidence limit is simply the product of t and the standard deviation of the mean (s/ N), also called the standard error of the mean. (The confidence limit for a single observation √ (N = 1), x, is given by x ± ts. This is larger than that of the mean by a factor N.)
Example 3.15 A soda ash sample is analyzed in the analytical chemistry laboratory by titration with standard hydrochloric acid. The analysis is performed in triplicate with the
Table 3.1
Values of t for ν Degrees of Freedom for Various Confidence Levelsa Confidence Level ν
90%
95%
99%
1 2 3 4 5 6 7 8 9 10 15 20 25 ∞
6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.753 1.725 1.708 1.645
12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.306 2.262 2.228 2.131 2.086 2.060 1.960
63.657 9.925 5.841 4.604 4.032 3.707 3.500 3.355 3.250 3.169 2.947 2.845 2.787 2.576
av
= N − 1 = degrees of freedom.
99.5% 127.32 14.089 7.453 5.598 4.773 4.317 4.029 3.832 3.690 3.581 3.252 3.153 3.078 2.807
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following results: 93.50, 93.58, and 93.43 wt% Na2 CO3 . Within what range are you 95% confident that the true value lies? Solution
The mean is 93.50 wt%. The standard deviation s is calculated to be 0.075 wt% Na2 CO3 (absolute—calculate it with a spreadsheet). At the 95% confidence level and two degrees of freedom, t = 4.303 and ts Confidence limit = x ± √ N 4.303 × 0.075 = 93.50 ± √ 3 = 93.50 ± 0.19 wt%
Too high a confidence level will give a wide range that may encompass nonrandom numbers. Too low a confidence level will give a narrow range and exclude valid random numbers. Confidence levels of 90 to 95% are generally accepted as reasonable.
So you are 95% confident that, in the absence of a determinate error, the true value falls within 93.31 to 93.69 wt%. Note that for an infinite number of measurements, we would have predicted with 95% confidence that the true value falls within two standard deviations (Figure 3.2); we see that for v = ∞, t is actually 1.96 (Table 3.1), and so the confidence limit would indeed be about twice the standard deviation of the mean (which approaches σ for large N).
Compare with Figure 3.2 where 95% of the values fall within 2σ .
PROFESSOR’S FAVORITE EXAMPLE Contributed by Professor Dimitri Pappas, Texas Tech University How to Calculate the Minimum Number of Measurements to Get a Given Error from the Standard Error of the Mean Note that the standard error of the mean (SE) has an inverse, square root relationship to the number of samples. The result is that there is a point in any analysis where a modest improvement in measurement statistics requires a prohibitively large sample. One of the most useful applications of the standard error of the mean is estimation of sample size number, especially if a desired confidence interval or tolerance is known or specified. One would necessarily have to make the approximation that standard deviation does not much change with the number of samples, this is not always defensible. Example
A series of nominally identical samples are taken for analysis. The mean weight of the samples is 9.78 g, with a standard deviation being 0.09 g for very few samples having been weighed. How many samples must be weighed so that the sample mean has an error 0.02 g from the population mean? Note that from Equation 3.9 and Table 3.1, the 95% confidence interval (CI) for an infinite population is CI = x ± 1.96 (SE) We want
CI = 9.78 ± 0.02 g at the 95% confidence level
If 1.96 (SE) is 0.02, then
0.02 g = 0.0102 g 1.96 s 2 0.09 2 s Since SE = √ , then N = = 78 SE 0.0102 N SE =
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In this case, a sample size of 78 measurements would yield the appropriate standard error of the mean. This simple exercise is useful for experiment planning, especially for experiments where the sample is difficult to obtain or the analysis is time consuming. Compare this with Example 3.26 using an iterative procedure. If you use that alternate (but essentially equivalent) approach, this method gives the first iteration value. We can’t do a second iteration using Table 3.1 since 78 is between n = 25 and n = ∞ in the table. You would need to find a more extensive table. For example, at www.jeremymiles.co.uk (go to the Other Stuff link on this website), t at the 95% level for 70 to 95 samples is 1.99. This would give a second (and final) iteration of 80 samples. So for a large number of samples, this approach (and the first iteration) gives a good approximation of the minimum number of samples required. Remember from Section 3.7 and Figure 3.2 that we are 68% confident that the true value falls within ±1σ , 95% confident it will fall within ±2σ , and 99% confident it will fall within ±2.5σ . Note that it is possible to estimate a standard deviation from a stated confidence interval, and vice versa a confidence interval from a standard deviation. If a mean value is 27.37 ± 0.06 g at the 95% confidence interval, then since this is two standard deviations for a suitably large number of measurements, the standard deviation is 0.03 g. If we know the standard deviation is 0.03 g, then this is the confidence interval at the 68% confidence level, or it is 0.06 g at the 95% confidence level. For small numbers of measurements, t will be larger, which proportionately changes these numbers. √ As the number of measurements increases, both t and s/ N decrease, with the result that the confidence interval is narrowed. So the more measurements you make, the more confident you will be that the true value lies within a given range or, conversely, that the range will be narrowed at a given confidence level. However, t decreases exponentially with an increase in N, just as the standard deviation of the mean does (see Table 3.1), so a point of diminishing returns is eventually reached in which the increase in confidence is not justified by the increase in the multiple of samples analyses required.
3.12 Tests of Significance—Is There a Difference? In developing a new analytical method, it is often desirable to compare the results of that method with those of an accepted (perhaps standard) method. How, though, can one tell if there is a significant difference between the new method and the accepted one? Again, we resort to statistics for the answer. Deciding whether one set of results is significantly different from another depends not only on the difference in the means but also on the amount of data available and the spread. There are statistical tables available that show how large a difference needs to be in order to be considered not to have occurred by chance. The F-test evaluates differences between the spread of results, while the t-test looks at differences between means. The F-test is used to determine if two variances are statistically different.
THE F-TEST This is a test designed to indicate whether there is a significant difference between two methods based on their standard deviations. F is defined in terms of the variances of the two methods, where the variance is the square of the standard deviation: F=
s21 s22
(3.10)
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Table 3.2
Values of F at the 95% Confidence Level
v2 = 2 3 4 5 6 7 8 9 10 15 20 30
v1 = 2
3
4
5
6
7
8
9
10
15
20
30
19.0 9.55 6.94 5.79 5.14 4.74 4.46 4.26 4.10 3.68 3.49 3.32
19.2 9.28 6.59 5.41 4.76 4.35 4.07 3.86 3.71 3.29 3.10 2.92
19.2 9.12 6.39 5.19 4.53 4.12 3.84 3.63 3.48 3.06 2.87 2.69
19.3 9.01 6.26 5.05 4.39 3.97 3.69 3.48 3.33 2.90 2.71 2.53
19.3 8.94 6.16 4.95 4.28 3.87 3.58 3.37 3.22 2.79 2.60 2.42
19.4 8.89 6.09 4.88 4.21 3.79 3.50 3.29 3.14 2.71 2.51 2.33
19.4 8.85 6.04 4.82 4.15 3.73 3.44 3.23 3.07 2.64 2.45 2.27
19.4 8.81 6.00 4.77 4.10 3.68 3.39 3.18 3.02 2.59 2.39 2.21
19.4 8.79 5.96 4.74 4.06 3.64 3.35 3.14 2.98 2.54 2.35 2.16
19.4 8.70 5.86 4.62 3.94 3.51 3.22 3.01 2.85 2.40 2.20 2.01
19.4 8.66 5.80 4.56 3.87 3.44 3.15 2.94 2.77 2.33 2.12 1.93
19.5 8.62 5.75 4.50 3.81 3.38 3.08 2.86 2.70 2.25 2.04 1.84
where s21 > s22 . There are two different degrees of freedom, v1 and v2 , where degrees of freedom is defined as N − 1 for each case. If the calculated F value from Equation 3.10 exceeds a tabulated F value at the selected confidence level, then there is a significant difference between the variances of the two methods. A list of F values at the 95% confidence level is given in Table 3.2. The F-test is available in Excel. The preferred method is to install the “Data Analysis” Add-In. The videos Solver, and Data Analysis Regression on the website of the book discusses installation of Add-Ins. It is also discussed at the end of Section 3.24. After installation, the Data Analysis icon will appear in the top right-hand corner of the tool bar. Click on this icon. A drop-down menu will appear. Select “F-test Two samples for Variance”. Another box will appear. Select your first set of data as “Variable 1 Range” and the second set as “Variable 2 Range”. You can select the confidence level you want by choosing the “alpha” value. An alpha value of 0.05 means a 95% confidence level, 0.01 connotes 99% confidence level; the default is an alpha of 0.05. The F-test results will appear in a box that will occupy 3 columns and 10 rows, giving Mean, Variance, df, F, P one tail, and F Critical one tail for Variables 1 and 2 (see also 4. t-test to Compare Different Samples on page 94). Click on the output range button and specify a cell in the box that will constitute the top left corner of the output. Your F-value should always be >1; if it is 1. The tabulated F value for v1 = 6 and v2 = 5 is 4.95. Since the calculated value of 1.73 is less than this, we conclude that there is no significant difference in the precision of the two methods, that is, the standard deviations are from random error alone and do not depend on the sample. In short, statistically, your method does as well as the established procedure. For a more critical look at the F-test, see the web supplements for Chapter 3, contributed by Professor Michael D. Morris, University of Michigan, on the textbook website. See the text website video illustrating the use of Excel for calculating F values. F=
If Fcalc > Ftable , then the variances being compared are significantly different. If Fcalc < Ftable , they are statistically the same.
Video: F-test
The t-test is used to determine if two sets of measurements are statistically different. If tcalc > ttable , then the two data sets are significantly different at the chosen confidence level.
THE STUDENT t-TEST— —ARE THERE DIFFERENCES IN THE METHODS? Frequently, the analyst wishes to decide whether there is a statistical difference between the results obtained using two different procedures, that is, whether they both indeed measure the same thing. The t-test is very useful for such comparisons. In this method, comparison is made between two sets of replicate measurements made by two different methods; one of them will be the test method, and the other will be an accepted or benchmark method. In a similar manner, we may be comparing the blood concentration of a particular analyte in a population of diabetic patients vs. those measured in a control group. A statistical t value is calculated and compared with a tabulated value for the given number of tests at the desired confidence level (Table 3.1). If the calculated t value exceeds the tabulated t value, then there is a significant difference between the results by the two methods at that confidence level. If it does not exceed the tabulated value, then we can predict that there is no significant difference between the methods at the confidence level we have chosen. This in no way implies that the two results are identical. There are several ways and several different types of situations in which a t-test can be used. Consider the following cases: 1. You have taken a certified single sample for which the analytical result is exactly known or is known with a degree of certainty much higher than you expect from your test method. You analyze the same sample by the test method a number of times with
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an objective to determine if there is no difference between the certified value and the mean value obtained by your method at a specified degree of certainty. 2. The situation is the same as above except that the uncertainty of the certified value or the standard deviation of measurements by a reference method is not negligible. You compare repeated measurements of the same single sample made by the reference method with repeated measurements made by the test method. This is often referred to as t-test by comparison of the means. The number of measurements in the two measurement sets need not be the same. 3. Often the intent is to compare a newly developed method with another, and a certified reference standard is not available. Further, even if a reference standard is available, it can only check a method at only one concentration and it will be desirable to check the applicability of the method spanning the entire range of concentrations in which the method is to be used. You take a number of different samples spanning the concentration range of interest. All samples are divided in two parts; one set is analyzed by the benchmark method and the other by the test method. The pairs of analytical results thus generated are subjected to the paired t-test to determine if the two methods produce results that are statistically different at a specified confidence level. This test is also useful in other situations. For example, it can be used to answer the question: Does a new drug cause a statistically different change in blood pressure compared to another? A number of people are studied and each person is examined twice for a change in blood pressure: once after administration of drug 1 and another time after administration of drug 2. Necessarily, equal numbers of sampled data exist for each measurement set in a paired t-test. 4. You want to compare two sample populations that are unrelated to each other. Is coal from West Virginia statistically different in its sulfur content compared to coal from Pennsylvania? Are post-mortem brain tissue data for aluminum content statistically different in Alzheimer affected subjects compared to control subjects? Note that unlike the preceding example, in this case, two different sample populations are tested and the numbers in each population do not have to be equal. This type of t-test can be subdivided in two groups: (a) when the variance or standard deviations of the two sample sets being compared are statistically the same (as determined, e.g., by the F-test), the two sample sets are said to be homoscedastic, (b) when the variance of the two sample sets are statistically different, the sample populations are heteroscedastic. Pooled standard deviation. The concept of pooled standard deviation is often used in performing t-tests and this is discussed first. The pooled standard deviation is used to obtain an improved estimate of the precision of a method, and it is used for calculating the precision of the two sets of data. That is, rather than relying on a single set of data to describe the precision of a method, it is sometimes preferable to perform several sets of analyses, for example, on different days, or on different samples with slightly different compositions. If the indeterminate (random) error is assumed to be the same for each set, then the data of the different sets can be pooled. This provides a more reliable estimate of the precision of a method than is obtained from a single set. The pooled standard deviation sp is given by
sp =
(xi1 − x1 )2 +
(xi2 − x2 )2 + · · · + N−k
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(xik − xk )2
(3.11)
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where x1 , x2 , . . . , xk are the means of each of k sets of analyses, and xi1 , xi2 , . . . , xik are the individual values in each set. N is the total number of measurements and is equal to (N1 + N2 + · · · + Nk ). If five sets of 20 analyses each are performed, k = 5 and N = 100. (The number of samples in each set need not be equal.) N − k is the degrees of freedom obtained from (N1 − 1) + (N2 − 1) + · · · + (Nk − 1); one degree of freedom is lost for each subset. This equation represents a combination of the equations for the standard deviations of each set of data. 1. t-Test When a Reference Value Is Known. Note that Equation 3.9 is a representation of the true value μ. We can write that ts μ=x± √ N
(3.12)
√ Using Equation 3.12, you can calculate the confidence interval (i.e., x + ts/ N to √ x − ts/ N) and see if the interval contains the expected value. It follows that √ N tcalc = (x − μ) (3.13) s If the determined confidence interval contains the expected value (e.g., μ), then the determination can be considered statistically not different from the known value.
Occasionally, a good estimate of the “true” value μ for the sample or reference standard is known with a great degree of certainty, with a greater degree of relative accuracy than is possible in typical laboratory analyses. Many metals and some pure compounds are available as highly pure form, with specified purities of 99.999, even 99.9999%. If analyzing for the metal or the compound in such samples, Equation 3.12 can be used to determine whether the value obtained from a test method is statistically indistinguishable from the certified value. The same approach can also be used even when some uncertainty is specified in the certified value, provided that this uncertainty is much smaller than the standard deviation of the results of the method under test. Our goal is to see if the result from our method differs from the certified value. Another application of this method is to determine if a set of measured values exceeds a prescribed regulation limit, for example, that of fluoride in water.
Example 3.17 You are developing a procedure for determining traces of copper in biological materials using a wet digestion followed by measurement by atomic absorption spectrophotometry. In order to test the validity of the method, you obtain a certified reference material and analyze this material. Five replicas are sampled and analyzed, and the mean of the results is found to be 10.8 ppm with a standard deviation of ±0.7 ppm. The listed reference value is 11.7 ppm. Does your method give a statistically correct value relative to the certified value at the 95% confidence level? √ Solution N tcalc = (x − μ) s √ 5 = (10.8 − 11.7) 0.7 = 2.9 The absolute value of tcalc is taken, since x − μ may be negative.
There are five measurements, so there are four degrees of freedom (N − 1). From Table 3.1, we see that the tabulated value of t at the 95% confidence level is 2.776.
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Because tcalc > ttable , one is 95% sure that your procedure is producing a value that is statistically different from the true value (look at Table 3.1—we aren’t quite 99% sure ttable at 99% is 4.6); but most work is done at the 95% confidence level). Your procedure will be considered unacceptable until you can find the source of the discrepancy and fix it. It is possible that you may not be able to do either one. For example, your method may have a negative interference from iron in the sample which is not accounted for. Note from Equation 3.13 that as the precision is improved, that is, as s becomes smaller, the calculated t becomes larger. Thus, there is a greater chance that the tabulated t value will be less than this. That is, as the precision improves, it is easier to distinguish nonrandom differences. Looking again at Equation 3.13, this means as s decreases, so must the difference between the two methods (x − μ) in order for the difference to be ascribed only to random error. What this means is that when one has a very large set of samples, which usually results in a reduced value of s, one is more likely to find a statistically significant difference. 2. Comparison of the Means of Two Methods. When the t-test is applied to two sets of data, μ in Equation 3.13 is replaced by√the mean of the second set. The reciprocal of the standard deviation of the mean ( N/s) is replaced by that of the differences between the two, which is readily shown to be
N1 N2 N1 + N2
sp
where sp is the pooled standard deviation of the individual measurements of two sets as defined in Equation 3.11 and N1 and N2 are the number of samples measured in each set:
tcalc
x − x2 = 1 sp
N1 N2 N1 + N2
(3.14)
We can use this when comparing two methods, that is, when analyzing a sample by two different methods, as illustrated below. To apply the comparison of means t-test between two methods, it is necessary that both methods have statistically the same standard deviation. This must first be verified by using the F-test.
Example 3.18 A new gravimetric method is developed for iron(III) in which the iron is precipitated in crystalline form with an organoboron “cage” compound. The accuracy of the method is checked by determining the iron in an ore sample and comparing with the results using the standard precipitation with ammonia and weighing the Fe2 O3 formed after ignition of the Fe(OH)3 precipitated. The results, reported as % Fe for each analysis, were as follows:
The F-test can be applied to the variances of the two methods rather than assuming they are statistically equal before applying the t-test.
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Test Method
Reference Method
20.10% 20.50 18.65 19.25 19.40 19.99 x1 = 19.65%
18.89% 19.20 19.00 19.70 19.40 x2 = 19.24%
Is there a significant difference between the two methods? Solution
xi1 20.10 20.50 18.65 19.25 19.40 19.99
xi1 − x1
(xi1 − x1 )2
0.45 0.202 0.85 0.722 1.00 1.000 0.40 0.160 0.25 0.062 0.34 0.116 (xi1 − x1 )2 = 2.262 F=
xi2 18.89 19.20 19.00 19.70 19.40
xi2 − x2
(xi2 − x2 )2
0.35 0.122 0.04 0.002 0.24 0.058 0.46 0.212 0.16 0.026 (xi2 − x2 )2 = 0.420
s21 2.262/5 = 4.31 = 2 0.420/4 s2
This is less than the tabulated value (6.26), so the two methods have comparable standard deviations and the t-test can be applied (xi1 − x1 )2 + (xi2 − x2 )2 sp = N1 + N2 − 2 2.262 + 0.420 = 0.546 = 6+5−2 19.65 − 19.24 (6)(5) ±t = = 1.23 0.546 6+5 The tabulated t for nine degrees of freedom (N1 + N2 − 2) at the 95% confidence level is 2.262, so there is no statistical difference in the results by the two methods. Rather than comparing two methods using one sample, two samples could be examined for comparability using a single analysis method in a manner identical to the above examples. The Excel Data Analysis Add-in can perform the comparison of means ttest, but requires paired data. We illustrate its use with the following problem. NIST stainless steel standard reference material (SRM) 73c certificate of analysis (https://www-s.nist.gov/srmors/certificates/73c.pdf) reports the following values for % Mo when analyzed by four analysts: 0.089, 0.092, 0.095, 0.087. You are trying to validate a new analytical method you have developed and you analyze four portions of the same SRM and get values of 0.090, 0.094, 0.098, and 0.096 % Mo, respectively. The NIST reported average is 0.091, while yours is 0.094. Is this difference significant? Enter both sets of data in an Excel Worksheet, putting the NIST results in columns A1:A4 and your results in columns B1:B4. Go to Data/Data Analysis and
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scroll down the drop-down menu and select “t-test: Paired two sample for means”. In the new data entry box we enter A1:A4 for “Variable 1 Range” and B1:B4 for “Variable 2 Range”. We do not have any reason to believe that there should be a difference between the two means, so in the box “Hypothesized Mean Difference”, allow the default value of zero to remain. If we had labels, e.g., “NIST Data”, “My Data” put in as column headings, we would tick the labels box, so Excel will ignore the top row. Since we have no headings, we leave the labels box unchecked. Alpha has the same connotation as in the F-test, the default value α = 0.05 connotes a confidence limit of 95%. You are now ready to do the test. Excel will require three columns and 14 rows to write the results. Click on the “Output Range” button and designate D1 as the cell where the output will begin. The first column has entries that are too wide to fit the default column width. Go to the column heading D and double-click on the right border. The column will expand to automatically fit the entries. The output lists mean, variance, degrees of freedom (df), the calculated t value (t-Stat), and the critical t value, both for a one-tailed test and for a two-tailed test. Our calculated t value is negative, but we really need to consider the magnitude. If we entered the data in column B as variable 1 and those in column A as variable 2, the t value would be calculated to be positive, with the same magnitude (try!). The one- and two-tailed test designations simply connote whether there are both halves of a Normal Distribution or just one need to be considered. In cases of comparison, where the one data set can only be higher (or lower) than the other data set, the one-tailed test is applicable. In analytical chemistry, one data set can generally either be higher or lower than the other set, there are no restrictions. So you must use the t-critical two-tailed value to compare with the calculated value. In the present case, tcalc of 2.08 is significantly smaller than the tcrit value of 3.18. At the 95% confidence level, your results therefore are not statistically different from the NIST results. 3. Paired t-Test. The paired t-test below compares the results of a series of different samples by two different methods. For the validation of a new method or intercomparison between methods, the paired t-test, described here, is preferred. A new method is frequently tested against an accepted method by analyzing several different samples of varying analyte content by both methods. In this case, the t value is calculated in a slightly different form. Often each sample has only been measured once. The difference between each of the paired measurements on each sample is computed. An average difference D is calculated and the individual deviations of each from D are used to compute a standard deviation, sd . The t value is calculated from |D| √ N sd (Di − D)2 sd = N−1 t=
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(3.15)
(3.16)
where Di is the individual difference between the two methods for each sample, with regard to sign; and D is the mean of all the individual differences.
Example 3.19 You are developing a new analytical method for the determination of blood urea nitrogen (BUN). You want to ascertain whether your method differs significantly from a standard method for determining a range of sample concentrations expected to be found
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in the routine laboratory. It has been ascertained that the two methods have comparable precision. Following are two sets of results for a number of individual samples:
Video: t-Test for Paired Samples
Sample
Your Method (mg/dL)
Standard Method (mg/dL)
A B C D E F
10.2 12.7 8.6 17.5 11.2 11.5
10.5 11.9 8.7 16.9 10.9 11.1
Di −0.3 0.8 −0.1 0.6 0.3 0.4 1.7 D = 0.28
Di − D −0.6 0.5 −0.4 0.3 0.0 0.1
(Di − D)2 0.36 0.25 0.16 0.09 0.00 0.01 0.87
Solution
0.87 = 0.42 6−1 0.28 √ t= × 6 = 1.63 0.42 The tabulated t value at the 95% confidence level for five degrees of freedom is 2.571. Therefore, tcalc < ttable , and there is no significant difference between the two methods at this confidence level. sd =
Video: Paired t-Test from Excel
Two examples of a problem using a paired t-test applied to a series of samples is shown in two videos in the website, t-Test for Paired Samples (demonstrating a manual calculation method) and Paired t-Test from Excel. Usually, a test at the 95% confidence level is considered significant, while one at the 99% level is highly significant. That is, the smaller the calculated t value, the more confident you are that there is no significant difference between the two methods. If you employ too low a confidence level (e.g., 80%), you are likely to conclude erroneously that there is a significant difference between two methods (sometimes called type I error). On the other hand, too high a confidence level will require too large a difference to detect (called type II error). See Section 3.19 on Powering a Study, for what is meant by a type I and a type II error. If a calculated t value is near the tabulated value at the 95% confidence level, more tests should be run to ascertain definitely whether the two methods are significantly different. 4. t-Test to Compare Different Samples. We will illustrate this using Excel. Consider the following problem. Well-water fluoride concentrations (in mg/L) have been determined in two adjoining counties, and are as follows. Does County B have a statistically different amount of fluoride in their water compared to County A? County A
County B
0.76 0.81 0.77 0.79 0.80 0.78 0.76
1.20 1.41 1.69 0.91 0.50 1.80 1.53
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We first determine by the F-test if the two sets of data have a difference in variance that is statistically significant. Do the F-test as described in Section 3.12. Enter column B data as variable 1 and column A data as variable 2 (if you do the opposite, you will find F < 1). The F-test output is shown below left. You clearly have very different variances in the two sets of data (if not, you would have used next: Data/Data Analysis/ t-test: Two Sample: Assuming Equal Variance). So we go to Data/Data Analysis/ t-test: Two Sample: Assuming Unequal Variances and again enter column B data as variable 1 and column A data as variable 2. The t-test output is below right. Since the tcalc value of 2.93 exceeds the critical two-tailed t value of 2.44, the fluoride concentrations in the wells of county B are statistically different at the 95% confidence level. F Test Two-Sample for Variances
Mean Variance Observations df F P(F < f ) one-tail F Critical one-tail
Variable 1
Variable 2
1.2914286 0.2114476 7 6
0.781429 0.000381 7 6 555.05 5.8E-08 4.283866
t-Test: Two-Sample Assuming Unequal Variances
Mean Variance Observations Hypothesized Mean Difference df t Stat P(T < t) one-tail t Critical one-tail P(T < t) two-tail t Critical two-tail
Variable 1
Variable 2
1.291429 0.211448 7 0
0.781429 0.000381 7
6 2.93175 0.013114 1.94318 0.026227 2.446912
3.13 Rejection of a Result: The Q Test When a series of replicate analyses is performed, it is not uncommon that one of the results will appear to differ markedly from the others. A decision will have to be made whether to reject the result or to retain it. Unfortunately, there are no uniform criteria that can be used to decide if a suspect result can be ascribed to accidental error rather than chance variation. It is tempting to delete extreme values from a data set because they alter the calculated statistics in an unfavorable way, that is, increase the standard deviation and variance (measures of spread), and they may substantially alter the reported mean. The only reliable basis for rejection occurs when it can be decided that some specific error may have been made in obtaining the doubtful result. No result should be retained in cases where a known error has occurred in its collection. Experience and common sense may provide just as practical a basis for judging the validity of a particular observation as a statistical test would be. Frequently, the experienced analyst will gain a good idea of the precision to be expected in a particular method and will recognize when a particular result is suspect. Additionally, an analyst who knows the standard deviation expected of a method may reject a data point that falls outside 2s or 2.5s of the mean because there is about 1 chance in 20 or 1 chance in 100 this will occur. A wide variety of statistical tests have been suggested and used to determine whether an observation should be rejected. In all of these, a range is established within which statistically significant observations should fall. The difficulty with all of them
Finagle’s third law: In any collection of data, the figure most obviously correct, beyond all checking, is the mistake.
The Q test is used to determine if an “outlier” is due to a determinate error. If it is not, then it falls within the expected random error and should be retained.
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Table 3.3
Rejection Quotient, Q, at Different Confidence Limitsa No. of Observations
If Qcalc > Qtable , then the data point may be an outlier and could be discarded. In practice, it is a good idea to make more measurements to be sure, especially if the decision is a close one.
“ And now the sequence of events in no particular order.”—Dan Rather, television news anchor
3 4 5 6 7 8 9 10 15 20 25 30 a
Confidence Level Q90
Q95
Q99
0.941 0.765 0.642 0.560 0.507 0.468 0.437 0.412 0.338 0.300 0.277 0.260
0.970 0.829 0.710 0.625 0.568 0.526 0.493 0.466 0.384 0.342 0.317 0.298
0.994 0.926 0.821 0.740 0.680 0.634 0.598 0.568 0.475 0.425 0.393 0.372
Adapted from D. B. Rorabacher, Anal. Chem., 63 (1991) 139.
is determining what the range should be. If it is too small, then perfectly good data will be rejected; and if it is too large, then erroneous measurements will be retained too high a proportion of the time. The Q test is, among the several suggested tests, one of the most statistically correct for a fairly small number of observations and is recommended when a test is necessary. The ratio Q is calculated by arranging the data in increasing or decreasing order of numbers. If you have a large data set, you will find the Data/Sort function in Excel very helpful to arrange the data either in ascending or descending order. The difference between the suspect number and its nearest neighbor (a) is divided by the range (w), the range being the difference between the highest number and the lowest number. Referring to the figure in the margin, Q = a/w. This ratio is compared with tabulated values of Q. If it is equal to or greater than the tabulated value, the suspected observation can be rejected. The tabulated values of Q at the 90, 95, and 99% confidence levels are given in Table 3.3. If Q exceeds the tabulated value for a given number of observations and a given confidence level, the questionable measurement may be rejected with, for example, 95% confidence that some definite error is in this measurement.
Example 3.20 The following set of chloride determinations on separate aliquots of a pooled serum were reported: 103, 106, 107, and 114 meq/L. One value appears suspect. Determine if it can be ascribed to accidental error, at the 95% confidence level. Solution
The suspect result is 114 meq/L. It differs from its nearest neighbor, 107 meq/L, by 7 meq/L. The range is 114 to 103, or 11 meq/L. Q is therefore 7/11 = 0.64. The tabulated value for four observations is 0.829. Since the calculated Q is less than the tabulated Q, the suspected number may be ascribed to random error and should not be rejected.
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For a small number of measurements (e.g., three to five), the discrepancy of the measurement must be quite large before it can be rejected by this criterion, and it is likely that erroneous results may be retained. This would cause a significant change in the arithmetic mean because the mean is greatly influenced by a discordant value. For this reason, it has been suggested that the median rather than the mean be reported when a discordant number cannot be rejected from a small number of measurements. The median is the middle result of an odd number of results, or the average of the central pair for an even numbered set, when they are arranged in order of their values. The median has the advantage of not being unduly influenced by an outlying value. In the above example, the median could be taken as the average of the two middle values [= (106 + 107)/2 = 106]. This compares with a mean of 108, which is influenced more by the suspected number. The following procedure is suggested for interpretation of the data of three to five measurements if the precision is considerably poorer than expected and if one of the observations is considerably different from the others of the set.
1. Estimate the precision that can reasonably be expected for the method in deciding whether a particular number actually is questionable. Note that for three measurements with two of the points very close, the Q test is likely to fail. (See the paragraph below.) 2. Check the data leading to the suspected number to see if a definite error can be identified. 3. If new data cannot be collected, run a Q test. 4. If the Q test indicates retention of the outlying number, consider reporting the median rather than the mean for a small set of data. 5. As a last resort, run another analysis. Agreement of the new result with the apparently valid data previously collected will lend support to the opinion that the suspected result should be rejected. You should avoid, however, continually running experiments until the “right” answer is obtained.
The Q test should not be applied to three data points if two are identical. In that case, the test always indicates rejection of the third value, regardless of the magnitude of the deviation, because a is equal to w and Qcalc is always equal to 1. The same obviously applies for three identical data points in four measurements, and so forth. A calculator/applet for performing the Q-test is available on the web (e.g., at http://asdlib.org/onlineArticles/ecourseware/Harvey/Outliers/OutlierProb1.html). There are other useful tests for outliers. One that is recommended by ASTM International (formerly known as the American Society for Testing and Materials) is the Grubbs Test for Outliers. This test is often preferred for detecting outliers in a univariate data set. It assumes the data follow a normal error curve (this is difficult to determine for a limited data set and the Grubbs test is only of value when n ≥ 7. However, gcritical values (see below) are available for g = 3–6 and can be used as an estimate). One determines which datum is furthest away from the mean (i.e., | Xi − X | is maximum) and divides this difference by the standard deviation, s. This quotient is called the Grubbs test statistic, g. If the calculated g value exceeds the tabulated critical value of g (gcritical ) at the desired confidence level, the datum can be rejected. A description of the use of the Grubbs test, including tabulated Grubbs Critical Values for the 95% and 99% confidence levels, is given in the Chapter 3 website.
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Consider reporting the median when an outlier cannot quite be rejected.
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Example calculations are given along with references for further information on applicability. A useful online calculator is given at: http://www.graphpad.com/quickcalcs/ Grubbs1.cfm.
3.14 Statistics for Small Data Sets Large population statistics do not strictly apply for small populations.
We have discussed, in previous sections, ways of estimating, for a normally distributed population, the central value (mean, x), the spread of results (standard deviation, s), and the confidence limits (t-test). These statistical values hold strictly for a large population. In analytical chemistry, we typically deal with fewer than 10 results, and for a given analysis, perhaps 2 or 3. For such small sets of data, other estimates may be more appropriate. The Q test in the previous section is designed for small data sets, and we have already mentioned some rules for dealing with suspect results. THE MEDIAN MAY BE BETTER THAN THE MEAN
The median may be a better representative of the true value than the mean, for small numbers of measurements.
The mean is the arithmetic average, whereas the median is the middle value in a set of values. The mean can be more dramatically affected by outliers than the median in a set of data with only a few values and significant imprecision.
If you have a set of N values and put them in either ascending or descending order, for an odd value of N, the center number in the series is the median value. For an even value of N, it is the average of the two central numbers. The median is denoted by M and may be used as an estimate of the central value. It has the advantage that it is not markedly influenced by extraneous (outlier) values, while the mean, x is. The efficiency of M, defined as the ratio of the variances of sampling distributions of these two estimates of the “true” mean value and denoted by EM , is given in Table 3.4. It varies from 1 for only two observations (where the median is necessarily identical with the mean) to 0.64 for large numbers of observations. The numerical value of the efficiency implies that the median from, for example, 100 observations where the efficiency is essentially 0.64, conveys as much information about the central value of the population as does the mean calculated from 64 observations. The median of 10 observations is as efficient conveying the information as is the mean from 10 × 0.71 = 7 observations. It may be desirable to use the median in order to avoid deciding whether a gross error is present, that is, using the Q test. It has been shown
Table 3.4
Efficiencies and Conversion Factors for 2 to 10 Observationsa
No. of Observations
Of Median, EM
Of Range, ER
Range Deviation Factor, KR
2 3 4 5 6 7 8 9 10 ∞
1.00 0.74 0.84 0.69 0.78 0.67 0.74 0.65 0.71 0.64
1.00 0.99 0.98 0.96 0.93 0.91 0.89 0.87 0.85 0.00
0.89 0.59 0.49 0.43 0.40 0.37 0.35 0.34 0.33 0.00
a Adapted
Efficiency
from R. B. Dean and W. J. Dixon, Anal. Chem., 23 (1951) 636.
Range Confidence Factor (t) tr 0.95
tr 0.99
6.4 1.3 0.72 0.51 0.40 0.33 0.29 0.26 0.23 0.00
31.83 3.01 1.32 0.84 0.63 0.51 0.43 0.37 0.33 0.00
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that for three observations from a normal population, the median is better than the mean of the best two out of three (the two closest) values. RANGE INSTEAD OF THE STANDARD DEVIATION The range R for a small set of measurements is highly efficient for describing the spread of results. The range is computed by subtracting the smallest value recorded from the largest value recorded. The efficiency of the range, ER , shown in Table 3.4, is virtually identical to that of the standard deviation for four or fewer measurements. This high relative efficiency arises from the fact that the standard deviation is a poor estimate of the spread for a small number of observations, although it is still the best known estimate for a given set of data. To convert the range to a measure of spread that is independent of the number of observations, we must multiply it by the deviation factor, K, given in Table 3.4. This factor adjusts the range R so that on average it reflects the standard deviation of the population, which we represent by sr : sr = RK R
The range is as good a measure of the spread of results as is the standard deviation for four or fewer measurements.
(3.17)
In Example 3.9 the standard deviation of the four weights was calculated to be 0.69 mg. The range of the data was 1.6 mg. Multiplying by KR for four observations, sr = 1.6 mg × 0.49 = 0.78 mg; s and sr are obviously quite comparable. As N increases, the efficiency of the range decreases relative to the standard deviation. The median M may be used in computing the standard deviation, in order to minimize the influence of extraneous values. Taking Example 3.9 again, the standard deviation calculated using the median, 29.8, in place of the mean in Equation 3.2, is 0.73 mg, instead of 0.69 mg. CONFIDENCE LIMITS USING THE RANGE Confidence limits could be calculated using sr obtained from the range, in place of s in Equation 3.9, and a corresponding but different t table, denoted as tr . It is more convenient, though, to calculate the limits directly from the range as Confidence limit = x ± Rtr
(3.18)
The factor for converting R to sr has been included in the quantity, tr , which is tabulated in Table 3.4 for 99 and 95% confidence levels. The calculated confidence limit at the 95% confidence level in Example 3.15 using Equation 3.18 is 93.50 ± 0.19 (1.3) = 93.50 ± 0.25% Na2 CO3 .
3.15 Linear Least Squares—How to Plot the Right Straight Line The analyst is frequently confronted with plotting data that fall on a straight line, as in an analytical calibration curve. Graphing, that is, curve fitting, is critically important in obtaining accurate analytical data. It is the calibration graph that is used to calculate the unknown concentration. Straight-line predictability and consistency will determine the accuracy of the unknown calculation. All measurements will have a degree of uncertainty, and so will the plotted straight line. Graphing is often done intuitively, that is, by simply “eyeballing” the best straight line by placing a ruler through the points, which invariably have some scatter. A better approach is to apply statistics to define the most probable straight-line fit of the data. The availability of statistical functions in spreadsheets today make it straightforward to prepare straight-line, or
“If a straight line fit is required, obtain only two data points.”—Anonymous
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y (Dependent variable)
y = mx + b
residual = yi − (mxi + b)
yi ∆y
slope =
∆x
∆y y−b =m= x ∆x
y-intercept = b = y − mx 0
Fig. 3.4.
xi
x (Independent variable)
Straight-line plot.
even nonlinear, fits. We will first learn the computations that are involved in curve fitting and statistical evaluation. We should note that a straight line is a model of the relationship between observations and amount of an analyte. One can always blindly apply least squares fitting (see below) to any random set of numbers. That does not necessarily mean a linear model is appropriate. Perhaps one should be fitting logarithms, or should be fitting sigmoids. We fit models to data, not data to models. To some extent, we prefer systems that are linear because they’re easier to deal with. However, with facile availability of computation, we need not always avoid nonlinearity. If a straight-line relationship is assumed, then the data fit the equation y = mx + b
(3.19)
where y is the dependent variable, x is the independent variable, m is the slope, of the curve, and b is the intercept on the ordinate (y axis); y is usually the measured variable, plotted as a function of changing x (see Figure 3.4). In a spectrophotometric calibration curve, y would represent the measured absorbances and x would be the concentrations of the standards. Our problem, then, is to establish values for m and b. LEAST-SQUARES PLOTS It can be shown statistically that the best straight line through a series of experimental points is that line for which the sum of the squares of the deviations (the residuals) of the points from the line is minimum. This is known as the method of least squares. If x is the fixed variable (e.g., concentration) and y is the measured variable (absorbance in a spectrophotometric measurement, the peak area in a chromatographic measurement, etc.), then the deviation of y vertically from the line at a given value of x (xi ) is of interest. If yl is the value on the line, it is equal to mxi + b. The square of the sum of the differences, S, is then S = (yi − yl )2 = [yi − (mxi + b)]2 The least-squares slope and intercept define the most probable straight line.
(3.20)
This equation assumes no error in x, the independent variable. The best straight line occurs when S is minimum (Smin ). Smin is obtained by use of differential calculus by setting the derivatives of S with respect to m and b equal to zero and solving for m and b. The result is (xi − x)(yi − y) m= (3.21) (xi − x)2 b = y − mx
(3.22)
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where x is the mean of all the values of xi and y is the mean of all the values of yi . The use of differences in calculations is cumbersome, and Equation 3.21 can be transformed into an easier to use form: xi yi − xi yi /n m= (3.23) 2 2 xi − [ xi /n] where n is the number of data points. Aside from spreadsheets, many calculators can also perform linear regression on a set of x-y data and give the values of m and b.
Example 3.21 Riboflavin (vitamin B2 ) is determined in a cereal sample by measuring its fluorescence intensity in 5% acetic acid solution. A calibration curve was prepared by measuring the fluorescence intensities of a series of standards of increasing concentrations. The following data were obtained. Use the method of least squares to obtain the best straight line for the calibration curve and to calculate the concentration of riboflavin in the sample solution. The sample fluorescence intensity was 15.4.
Riboflavin, μg/mL(xi ) 0.000 0.100 0.200 0.400 0.800 xi = 1.500
Fluorescence Intensity, Arbitrary Units (yi ) 0.0 5.8 12.2 22.3 43.3 yi = 83.6
xi2 0.0000 0.0100 0.0400 0.1600 2 0.6400 xi = 0.8500
2 xi = 2.250 x=
xi n
= 0.3000
xi yi 0.00 0.58 2.44 8.92 34.6 4 xi yi = 46.58
n=5 y=
yi n
= 16.72
Solution
Using Equations 3.23 and 3.22. 46.58 − [(1.500 × 83.6)/5] = 53.75 fluor. units/ppm m= 0.8500 − 2.250/5 b = 16.72 − (53.75 × 0.3000 ) = 0.60 fluor. units We have retained the maximum number of significant figures in computation. Since the experimental values of y are obtained to only the first decimal place, we can round m and b to the first decimal. The equation of the straight line is (FU = fluorescence units; ppm = μg/mL) y(FU) = 53.8(FU/ppm)x(ppm) + 0.6(Fu) The sample concentration is 15.4 = 53.8x + 0.6 x = 0.275 μg/mL
Take note of the units assigned to each variable in the y = mx + b linear equation. This will help you apply the equation for determination of unknowns.
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50.0 45.0
Fluorescence intensity
40.0 35.0
y = 53.75x + 0.595 R2 = 0.9989
30.0 25.0 20.0 15.0 10.0 5.0
Fig. 3.5.
Least-squares plot of data from Example 3.21.
Video: Plotting in Excel
0.0 0.0
0.1
0.2
0.3
0.4 0.5 Riboflavin, ppm
0.6
0.7
0.8
0.9
To prepare an actual plot of the line, take two arbitrary values of x sufficiently far apart and calculate the corresponding y values (or vice versa) and use these as points to draw the line. The intercept y = 0.6 (at x = 0) could be used as one point. At 0.500 μg/mL, y = 27.5. A plot of the experimental data and the least-squares line drawn through them is shown in Figure 3.5. This was plotted using Excel, with the equation of the line and the square of the correlation coefficient (a measure of agreement between the two variables—ignore this for now, we will discuss it later). The program automatically gives additional figures, but note the agreement with our calculated values for the slope and intercept. Note that all points are treated equally in a standard least-squares fit. For the same amount of relative deviation, such a procedure effectively gives greater weight to the higher x, y values compared to those at the low end. Weighted least-squares fitting can be used by the more adventurous. For a reference, consult, for example, Strutz, T.: Data Fitting and Uncertainty. Vieweg + Teubner Verlag, 2010. Also J. A. Irvin and T. I. Quickenden, “Linear Least Squares Treatment When There are Errors in Box x and y,” J. Chem. Ed., 60 (1983) 711–712. See the video clips on the website illustrating using Excel to perform leastsquares plots (Plotting in Excel), and for plotting error bars (Error bars). First, review the videos in Section 3.20 on the use of Excel. STANDARD DEVIATIONS OF THE SLOPE AND INTERCEPT——THEY DETERMINE THE UNKNOWN UNCERTAINTY
Video: Error Bars
The standard deviations of m and b give an equation from which the uncertainty in the unknown is calculated, using propagation of errors.
Each data point on the least-squares line exhibits a normal (Gaussian) distribution about the line on the y axis. The deviation of each yi from the line is yi − yl = y − (mx + b), as in Equation 3.20. The standard deviation of each of these y-axis deviations is given by an equation analogous to Equation 3.2, except that there are two less degrees of freedom since two are used in defining the slope and the intercept: 2 2 2 2 2 − y /N − m − xi /N y x i i i [yi − (mxi + b)]2 = sy = N −2 N −2 (3.24)
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This quantity is also called the standard deviation of regression, sr . The sy value can be used to obtain uncertainties for the slope, m, and intercept, b, of the least-squares line since they are related to the uncertainty in each value of y. For the slope: sm =
s2y (x − xi
)2
=
s2y 2 xi /N xi2 −
(3.25)
where x is the mean of all xi values. For the intercept: sb = sy
xi2 1 2 = sy 2 2 2 xi xi / xi N xi − N− sy
(3.26)
In calculating an unknown concentration, xi , from Equation 3.19, representing the least-squares line, the uncertainties in y, m, and b are all propagated in the usual manner, from which we can determine the uncertainty in the unknown concentration.
Example 3.22 Estimate the uncertainty in the slope, intercept, and y for the least-squares plot in Example 3.21, and the uncertainty in the determined riboflavin concentration for the sample with fluorescence intensity of 15.4 FU.
Solution
2 2 2 y , In order to solve for all the uncertainties, we need values for yi , xi , 2 2 2i 2 2 yi = (83.6) = 6989.0; xi = 0.8500 ; x , and m . From Example 3.21, i 2 xi = 2.250, and m2 = (53.75 )2 = 2.889 . The (yi )2 values are (0.0)2 , (5.8)2 , (12.2)2 , (22.3)2 , and (43.3)2 = 0.0, 33.6, 148.8, 497.3, and 1874.9, and y2i = 2554.6 (carrying extra figures). From Equation 3.24, sy =
(2554.6 − 6989.0/5) − (53.75 )2 (0.8500 − 2.250/5) = ±0.63 FU 5−2
From Equation 3.25, sm =
(0.63 )2 = ±1.0 FU/ppm 0.8500 − 2.250/5
From Equation 3.26, sb = 0.63
0.8500 = ±0.41 FU 5(0.8500 ) − 2.250
Therefore, m = 53.8 + 1.0 and b = 0.6 ± 0.4.
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The unknown riboflavin concentration is calculated from x=
(y ± sy ) − (b ± sb ) m ± sm
=
(15.4 ± 0.6) − (0.6 ± 0.4) = 0.275 ±? 53.8 ± 1.0
Applying the principles of propagation of error (absolute variances in numerator additive, relative variances in the division step additive), we calculate that x = 0.275 ± 0.014 ppm. See Chapter 16 for the spreadsheet calculation of the standard deviation of regression and the standard deviation of an unknown for this.
3.16 Correlation Coefficient and Coefficient of Determination The correlation coefficient is used as a measure of the correlation between two variables. When variables x and y are correlated rather than being functionally related (i.e., are not directly dependent upon one another), we do not speak of the “best” y value corresponding to a given x value, but only of the most “probable” value. The closer the observed values are to the most probable values, the more definite is the relationship between x and y. This postulate is the basis for various numerical measures of the degree of correlation. The Pearson correlation coefficient is one of the most convenient to calculate. This is given by r=
(xi − x)(yi − y) nsx sy
(3.27)
where r is the correlation coefficient, n is the number of observations, sx is the standard deviation of x, sy is the standard deviation of y, xi and yi are the individual values of the variables x and y, respectively, and x and y are their means. The use of differences in the calculation is frequently cumbersome, and the equation can be transformed to a more convenient form:
xi yi − nxy 2 xi2 − nx2 yi − ny2 yi n xi yi − xi = 2 2 n xi2 − n y2i − xi yi
r =
A correlation coefficient near 1 means there is a direct relationship between two variables, for example, absorbance and concentration.
(3.28)
Despite its formidable appearance, Equation 3.28 is probably the most convenient for calculating r. Calculators that can perform linear regression calculate r directly; the formula is embedded. The maximum value of r is 1. When this occurs, there is exact correlation between the two variables. When the value of r is zero (this occurs when xi yi is equal to zero), there is complete independence of the variables. The minimum value of r is −1. A negative correlation coefficient indicates that the assumed dependence is opposite to what exists and is therefore a positive coefficient for the reversed relation.
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Example 3.23 Calculate the correlation coefficient for the data in Example 3.19, taking your method as x and the standard method as y. Solution
2 2 We calculate that xi = 903.2, yi = 855.2, x = 12.0, y = 11.7, and xi yi = 878.5. Therefore, from Equation 3.28, 878.5 − (6)(12.0)(11.7) r=
= 0.991 [903.2 − (6)(12.0)2 ][855.2 − (6)(11.7)2 ]
A correlation coefficient can be calculated for a calibration curve to ascertain the degree of correlation between the measured instrumental variable and the sample concentration. As a general rule, 0.90 < r < 0.95 indicates a fair curve, 0.95 < r < 0.99 a good curve, and r > 0.99 indicates excellent linearity. An r > 0.999 can be obtained with care and is not uncommon. The correlation coefficient gives the dependent and independent variables equal weight, which is usually not true in scientific measurements. The r value tends to give more confidence in the goodness of fit than warranted. The fit must be quite poor before r becomes smaller than about 0.98 and is really very poor when less than 0.9. A more conservative measure of closeness of fit is the square of the correlation coefficient, r2 , and this is what most statistical programs calculate (including Excel—see Figure 3.5). An r value of 0.90 corresponds to an r2 value of only 0.81, while an r of 0.95 is equivalent to an r2 of 0.90. The goodness of fit is judged by the number of 9’s. So three 9’s (0.999) or better represents an excellent fit. We will use r2 as a measure of fit. This is also called the coefficient of determination. It should be mentioned that it is possible to have a high degree of correlation between two methods (r2 near unity) but to have a statistically significant difference between the results of each according to the t-test. This would occur, for example, if there were a constant determinate error in one method. This would make the differences significant (not due to chance), but there would be a direct correlation between the results [r2 would be near unity, but the slope (m) may not be near unity or the intercept (b) not near zero]. In principle, an empirical correction factor (a constant) could be applied to make the results by each method the same over the concentration range analyzed.
The coefficient of determination (r2 ) is a better measure of fit. The designations “r2 ” and “R2 ” are often used interchangeably, depending only on preference. Excel uses “R2 ”.
3.17 Detection Limits—There Is No Such Thing as Zero The previous discussions have dealt with statistical methods to estimate the reliability of analyses at specific confidence levels, these being ultimately determined by the precision of the method. All instrumental methods have a degree of noise associated with the measurement that limits the amount of analyte that can be detected. The noise is reflected in the precision of the blank or background signal, and noise may be apparent even when there is no significant blank signal. This may be due to fluctuation in the dark current of a photomultiplier tube, flame flicker in an atomic absorption instrument, and other factors. The limit of detection (LOD) is the lowest concentration level that can be determined to be statistically different from an analyte blank. There are numerous
The concentration that gives a signal equal to three times the standard deviation of the background above the mean background is generally taken as the detection limit.
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Fig. 3.6.
Peak-to-peak noise level as a basis for detection limit. The background fluctuations represent continuously recorded background signals, with the analyte measurement represented by the peak signal. A “detectable” analyte signal would be 12 divisions above a line drawn through the average of the baseline fluctuations.
12.0 div
6.0 div
Mean baseline
ways that detection limits have been defined. For example, the concentration that gives twice the peak-to-peak noise of a series of background signal measurements (or of a continuously recorded background signal) is sometimes taken as the detection limit (see Figure 3.6). A more generally accepted definition of the LOD is as follows. The LOD is the concentration that gives a signal that is above the background signal by three times the standard deviation of the background signal. It is instructive to compare the twice the peak-to-peak baseline noise criterion with the three times the standard deviation of the baseline criterion. Note that a continuous trace, as depicted in Figure 3.6, is tantamount to an infinite number of observations. According to the t-table (Table 3.1), 99% of the observations will fall within an interval of 2.576 standard deviations. The peak-to-peak noise is thus equivalent to ∼2.6 standard deviations, and twice this value is ∼5 standard deviations, a more stringent criterion than the 3 standard deviations criterion.
Example 3.24 A series of sequential baseline absorbance measurements are made in a spectrophotometric method, for determining the purity of aspirin in tablets using a blank solution. The absorbance readings are 0.002, 0.000, 0.008, 0.006, and 0.003. A standard 1 ppm aspirin solution gives an absorbance reading of 0.051. What is the detection limit? Solution
The standard deviation of the blank readings is ±0.0032 absorbance units, and the mean of the blank readings is 0.004 absorbance units. The detection limit is that concentration of analyte that gives a reading of 3 × 0.0032 = 0.0096 absorbance reading, above the blank signal. The net reading for the standard is 0.051 − 0.004 = 0.047. The detection limit would correspond to 1 ppm (0.0096/0.047) = 0.2 ppm and would give a total absorbance reading of 0.0096 + 0.004 = 0.014. The precision at the detection limit is by definition about 33%. For quantitative measurements, the limit of quantitation (LOQ) is generally considered to be that value corresponding to 10 standard deviations above the baseline, approximately 3.3 times the LOD, about 0.7 ppm in the above example. There have been various attempts to place the concept of detection limit on a more firm statistical ground. The International Conference on Harmonization (ICH; see Chapter 4) of Technical Requirements for Registration of Pharmaceuticals for Human Use has proposed guidelines for analytical method validation (Reference 18). The ICH Q2B guideline on validation methodology suggests calculation based on the
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standard deviation, s, of the response and the slope or sensitivity, S, of the calibration curve at levels approaching the limit. For the limit of detection (LOD), LOD = 3.3(s/S)
(3.29)
And for limit of quantitation (LOQ) LOQ = 10(s/S)
(3.30)
The standard deviation of the response can be determined based on the standard deviation of either the blank, the residual standard deviation of the least-squares regression line, or the standard deviation of the y intercept of the regression line. The Excel statistical function can be used to obtain the last two. The International Union of Pure and Applied Chemistry (IUPAC) uses a value of 3 in Equation 3.29 (for blank measurements), derived from a confidence level of 95% for a reasonable number of measurements. The confidence level, of course, varies with the number of measurements, and 7 to 10 measurements should be taken. The bottom line is that one should regard a detection limit as an approximate guide to performance and not make efforts to determine it too precisely. The excellent article by Long and Winefordner on a closer look at the IUPAC definition of the LOD [Anal. Chem. 55:712A (1983)] is highly recommended.
3.18 Statistics of Sampling—How Many Samples, How Large? Acquiring a valid analytical sample is perhaps the most critical part of any analysis. The physical sampling of different types of materials (solids, liquids, gases) is discussed in Chapter 2. We describe here some of the statistical considerations in sampling.
THE PRECISION OF A RESULT——SAMPLING IS THE KEY More often than not, the accuracy and precision of an analysis is limited by the sampling rather than the measurement step. The overall variance of an analysis is the sum of the sampling variance and the variance of the remaining analytical operations, that is, s2o = s2s + s2a (3.31) If the variance due to sampling is known (e.g., by having performed multiple samplings of the material of interest and analyzing it using a precise measurement technique), then there is little to be gained by reduction of sa to less than 1/3ss . For example, if the relative standard deviation for sampling is 3.0% and that of the analysis is 1.0%, then s2o = (1.0)2 + (3.0)2 = 10.0, or so = 3.2%. Here, 94% of the imprecision is due to sampling and only 6% is due to measurement (so is increased from 3.0 to 3.2%, so 0.2% is due to the measurement). If the sampling imprecision is relatively large, it is better to use a rapid, lower precision method and analyze more samples. We are really interested in the value and variance of the true value. The total variance is s2total = s2g + s2s + s2a , where s2g describes the “true” variability of the analyte in the system, the value of which is the goal of the analysis. For reliable interpretation of the chemical analysis, the combined sampling and analytical variance should not exceed 20% of the total variance. [See M. H. Ramsey, “Appropriate Precision: Matching Analytical Precision Specifications to the Particular Application,” Anal. Proc., 30 (1993) 110.]
Little is gained by improving the analytical variance to less than one-third the sampling variance. It is better to analyze more samples using a faster, less precise method.
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THE “TRUE VALUE” The range in which the true value falls for the analyte content in a bulk material can be estimated from a t-tests at a given confidence level (Equation 3.12). Here, x is the average of the analytical results for the particular material analyzed, and s is the standard deviation that is obtained previously from analysis of similar material samples or from the present analysis if there are sufficient samples. MINIMUM SAMPLE SIZE Statistical guidelines have been developed for the proper sampling of heterogeneous materials, based on the sampling variance. The minimum size of individual increments for a well-mixed population of different kinds of particles can be estimated from Ingamell’s sampling constant, Ks : wR2 = Ks The greater the sample size, the smaller the variance.
(3.32)
where w is the weight of sample analyzed and R is the desired percent relative standard deviation of the determination. KS represents the weight of sample for 1% sampling uncertainty at a 68% confidence level and is obtained by determining the standard deviation from the measurement of a series of samples of weight w. This equation, in effect, says that the sampling variance is inversely proportional to the sample weight.
Example 3.25 Ingamell’s sampling constant for the analysis of the nitrogen content of wheat samples is 0.50 g. What weight sample should be taken to obtain a sampling precision of 0.2% rsd in the analysis? Solution
w (0.2)2 = 0.50 g w = 12.5 g Note that the entire sample is not likely to be analyzed. The 12.5-g gross sample will be finely ground, and a few hundred milligrams of the homogeneous material analyzed. If the sample were not made homogeneous, then the bulk of it would have to be analyzed.
MINIMUM NUMBER OF SAMPLES The number of individual sample increments needed to achieve a given level of confidence in the analytical results is estimated by n=
t2 s2s g2 x2
(3.33)
where t is the Student t value for the confidence level desired, s2s is the sampling variance, g is the acceptable relative standard deviation of the average of the analytical results, x; ss is the absolute standard deviation, in the same units as x, and so n is
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unitless. Values of ss and x are obtained from preliminary measurements or prior knowledge. Since g is equal to sx /x, we can write that n=
t2 s2s s2x
(3.34)
ss and sx can then be expressed in either absolute or relative standard deviations, so long as they are both expressed the same. Since n is initially unknown, the t value for the given confidence level is initially estimated and an iterative procedure is used to calculate n.
Example 3.26 The iron content in a blended lot of bulk ore material is about 5% (wt/wt), and the relative standard deviation of sampling, ss , is 0.021 (2.1% rsd). How many samples should be taken in order to obtain a relative standard deviation g of 0.016 (1.6% rsd) in the results at the 95% confidence level [i.e., the standard deviation, sx , for the 5% iron content is 0.08% (wt/wt)]? Solution
We can use either Equation (3.33) or (3.34). We will use the latter. Set t = 1.96 (for n = ∞, Table 3.1) at the 95% confidence level. Calculate a preliminary value of n. Then use this n to select a closer t value, and recalculate n; continue iteration to a constant n. (1.96)2 (0.021)2 = 6.6 n= (0.016)2 For n = 7, t = 2.365. n= For n = 10, t = 2.23 n=
(2.365)2 (0.021)2 = 9.6 (0.016)2
(2.23)2 (0.021)2 = 8.6 ≡ 9 (0.016)2
See if you get the same result using Equation (3.33).
Challenge: Can you apply Professor Pappas’ approach (Section 3.11) to this problem? What requisite number of samples do you get? Equation 3.33 holds for a Gaussian distribution of analyte concentration within the bulk material, that is, it will be centered around x with 68% of the values falling within one standard deviation, or 95% within two standard deviations. In this case, the variance of the population, σ 2 , is small compared to the true value. If the concentration follows a Poisson distribution, that is, follows a random distribution in the bulk material such that the true or mean value x approximates the variance, s2s , of the population, then Equation 3.33 is somewhat simplified: n=
t2 t2 s2s · = g2 x x g2 x
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(3.35)
Note that since s2s is equal to x, the right-hand part of the expression becomes equal to 1, but the units do not cancel. In this case, when the concentration distribution is broad
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rather than narrow, many more samples are required to get a representative result from the analysis. If the analyte occurs in clumps or patches, the sampling strategy becomes more complicated. The patches can be considered as separate strata and sampled separately. If bulk materials are segregated or stratified, and the average composition is desired, then the number of samples from each stratum should be in proportion to the size of the stratum.
3.19 Powering a Study: Power Analysis Contributed by Professor Michael D. Morris, University of Michigan How many samples are needed for a statistically valid result? The answer of course depends on variability of the population and confidence level needed. Chemists rarely ask this question, they tend to assume that nominally identical materials being tested are identical or nearly identical—only instrument accuracy and precision matter. But these assumptions do not hold for people or animals or plants. Some claim that there is emerging iodine nutrition deficiency in the United States—how many people must we test for iodine intake to determine the veracity of this hypothesis? We also need to introduce some new terms: Statistical tests of significance always use a null hypothesis, the t-test for example has the null hypothesis that the two populations are not statistically different. Tests are usually described as confirming or rejecting the null hypothesis, that is, that there is no difference between the measured parameter and some reference parameter. A test is said to have a Type I error—when the null hypothesis is rejected when in fact it is true (probability = α). Type II error constitutes failing to reject the null hypothesis when it is false (probability = β). The power of a statistical test is the probability that it will reject a false null hypothesis (i.e., that it will not make a Type II error). The probability of a Type II error is referred to as the false negative rate (β). Therefore, power is equal to 1 − β. The larger the power, the less probable a Type II error. Power analysis can be used to calculate the minimum sample size needed to accept the statistical test result at a given level of confidence. A simple example for a null hypothesis: The average height of all 5-year-old boys in your town is the same as the average height of all 2-year-old boys in your town. An exhaustive test will involve that you measure the height of every 5-year-old boy and every 2-year-old boy, calculate the averages and calculate the difference. A more practical approach may be to measure a random sample from each age group and measure the difference. Either way, if the difference isn’t zero, the null hypothesis is refuted—maybe. Important issues are: How many 5-year-old and 2-year-old boys are needed. It is expensive and time-consuming to use too many, but the result may be meaningless with too few. What do we mean by non-zero difference—with what confidence level? Some conventional (but arbitrary) levels are used: We want the probability of a type I error to be low, so α = 0.05 is commonly chosen. We want the probability of a type II error to be low, i.e., the probability of correctly detecting a difference (1 − β), to be high. The power of a study is (1 − β). Power 0.8 is commonly chosen; it is often referred to as the study having 80% power and is the minimum that the National Institutes of Health will deem acceptable. Typically the same number n of measurements (subjects) is used in the two groups. The mean values are X1 and X2 and s2 is the mean of the variances (heights, in our example) of the two groups. A simple example will suffice.
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There are several cases, but we consider only two of the most important ones: A given difference, d = dmin is significant at p = 0.05 (95% confidence level) in both positive and negative directions, so then 2α = 0.05. The number of standard deviations encompassed in either direction is z2α = 1.96. In this case, n, the size of the sample needed in each of the two groups, is given by: (3.36) n > 2(sZ2α /d)2 In general, we don’t know what dmin is, but we can specify it here as, for example, 1 cm. We can use child growth data published by the Centers for Disease Control (http://www.cdc.gov/growthcharts/) to estimate the standard deviations for each of the two age groups. From the growth charts, we estimate s ≈ 2.5 cm for 2-year-old boys and s ≈ 3.5 cm for 5-year-old boys. We can take s to be the average standard deviation in the two populations, i.e., 1/2(2.5 + 3.5) = 3 cm. n > 2(3 ∗ 1.96/1)2
(3.37)
In that case, using Equation 3.37, we calculate that n > 69.1. So, a minimum sample size is 70. Because the sample is typically divided equally between the two classes, we need at least thirty-five 2-year-old boys and at least thirty-five 5-year-old boys. It is more common to specify the statistical power required to achieve a specified type II error rate. A given statistical power (usually 80%) has a specified true difference, δ > δ1 . Here δ is a true difference and δ1 is the threshold true difference sought. So, if the measured difference between the means is d, then it is significant if : δ1 > (z2α + z2β )SE(d)
(3.38)
s n SE(d) = √ 1 − (3.39) N n Because β is the type II error (false negative probability), 1 − β is the 95% probability of correctly reporting a difference. So for β = 0.2, z2β = 1.64. The final result is that 2 z2α + z2β s n>2 (3.40) δ1
and
Using the data above and taking δ1 = 2 cm, we calculate that n > 2[(1.96 + 1.64)3/2]2 = 58.3
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(3.41)
or n > 59. In summary, if the total population size is known and the measurement error can be estimated, and the significance (or confidence interval) and power are defined, then the number of subjects needed can be specified. P. Armitage, G. Berry and J. N. S. Matthews, Statistical Methods in Medical Research, 4th Ed. Malden, MA: Blackwell Science, Inc., 2002 is a good reference on powering studies.
See the book’s website for a PowerPointTM presentation contributed by Professor Morris on diverse statistical topics including box-and-whisker plots, signal-to-noise ratio, histograms, non-Gaussian histograms and histogram equalization/normalization, sensitivity and specificity in binary tests, and receiver operating characteristics (ROC). See the website also for Dr. Morris’s statistical view of the recent decision on questionable utility of mammograms in preventing breast cancer.
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3.20 Use of Spreadsheets in Analytical Chemistry A spreadsheet is a powerful software program that can be used for a variety of functions, such as data analysis and plotting. Spreadsheets are useful for organizing data, doing repetitive calculations, and displaying the calculations graphically or in chart form. They have built-in functions, for example, standard deviation and other statistical functions, for carrying out computations on data that are input by the user. Currently the most popular spreadsheet program is Microsoft Excel. We will use Excel 2010 in our illustrations. You probably have used a spreadsheet program before and are familiar with the basic functions. But we will summarize here the most useful aspects for analytical chemistry applications. Also, the Excel Help on the tool bar provides specific information. You are referred to the excellent tutorial on using the Excel spreadsheet prepared by faculty at California State University at Stanislaus: http://science.csustan.edu/tutorial/Excel/index.htm. The basic functions in the spreadsheet are described, including entering data and formulas, formatting cells, graphing, and regression analysis. You are likely to find this helpful, even though this presentation is based on an older version of Excel. Introductory lessons on how to use Excel 2010 are available at http://www.gcflearnfree.org/excel2010/1. On the website of this book, you will find a video titled Introduction to Excel. This will familiarize you with the contents of the following several paragraphs. The website http://www.ncsu.edu/labwrite/res/gt/gt-menu.html at North Carolina State University also provides instructions for graphing in Excel. A spreadsheet consists of cells arranged in columns (labeled A, B, C, . . . ) and rows (numbered 1, 2, 3, . . . ). An individual cell is identified by its column letter and row number, for example, B3. Figure 3.7 has the identifiers typed into some of the cells to illustrate. When the mouse pointer (the cross) is clicked on an individual cell, it becomes the active cell (dark lines around it), and the active cell is indicated at the top left of the formula bar, and the contents of the cell are listed to the right of the fx sign on the bar.
Use Excel Help if you are having trouble. It provides step-by-step instructions on how to carry out most of the functions.
Spreadsheet tutorial csustan: Basic functions
Spreadsheet tutorial ncsu: Graphing in Excel
FILLING THE CELL CONTENTS You may enter text, numbers, or formulas in specific cells. Formulas are the key to the utility of spreadsheets, allowing the same calculation to be applied to many numbers. We will illustrate with calculations of the weights of water delivered by two different 20-mL pipets, from the difference in the weights of a flask plus water and the empty flask. Refer to Figure 3.8 as you go through the steps.
A
A 1 A1 2 A2 3 A3 4 5
Fig. 3.7.
B B1 B2 B3
C C1 C2 C3
Spreadsheet cells.
D etc. etc. etc.
E
1 2 3 4 5 6 7 8 9 10
B
C
Net weights Pipet Weight of flask + water, g Weight of flask, g Weight of water, g cell B6=B4-B5 cell C6=C4-C5
Fig. 3.8.
Filling cell contents.
1 47.702 27.687 20.015
2 49.239 29.199 20.040
D
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Open an Excel spreadsheet by clicking on the Excel icon (or the Microsoft Excel program under Start/All Programs/Microsoft Office/Microsoft Excel 2010). You will enter text, numbers, and formulas. Double-click on the specific cell to activate it. Enter as follows (information typed into a cell is entered by depressing the Enter key): Cell A1: Net weights Cell A3: Pipet Cell A4: Weight of flask + water, g Cell A5: Weight of flask, g Cell A6: Weight of water, g You may make corrections by double-clicking on a cell; then edit the text. (You can also edit the text in the formula bar.) If you single click, new text replaces the old text. You will have to widen the A cells to accommodate the lengthy text. Do so by placing the mouse pointer on the line between A and B on the row at the top, and dragging it to the right until all the text shows. You can also adjust the column width by selecting the entire column (click on the top title column marked A, B, etc.) and on the Home tab click on Format/Autofit Column Width. Cell B3: 1 Cell C3: 2 Cell B4: 47.702 Cell C4: 49.239 Cell B5: 27.687 Cell C5: 29.199 Cell B6: =B4-B5 You can also enter the formula by typing =, then click on B4, then type—, and click on B5. You need to format the cells B4 to C6 to three decimal places. Highlight that block of cells by clicking on one corner and dragging to the opposite corner of the block. In the Menu bar, click on Number/number, adjust decimal places to 3, and click OK. You need to add the formula to cell C6. You can retype it. But there is an easier way, by copying (filling) the formula in cell B6. Place the mouse pointer on the lower right corner of cell B6 and drag it to cell C6. This fills the formula into C6 (or additional cells to the right if there are more pipet columns). You may also fill formulas into highlighted cells in other ways. Double-click on B6. This shows the formula in the cell and outlines the other cells contained in the formula. Do the same for C6. Note that when you activate the cell by either single or double-clicking on it, the formula is shown in the formula bar. See the website video Introduction to Excel for an illustration of setting up an Excel spreadsheet.
You can also change the number of digits shown after the decimal point, and do many other operations, by clicking on the appropriate buttons on the toolbar.
SAVING THE SPREADSHEET Save the spreadsheet you have just created by clicking on File:SaveAs. Give the document a File Name at the bottom, for example, Pipet Calibration. Then click Save. Video: Introduction to Excel
PRINTING THE SPREADSHEET Click Page Layout/Orientaion. Normally, a sheet is printed in the Portrait format, that is, vertically on the 8 1/2- × 11-inch paper. If there are many columns, you may wish to print in Landscape, that is, horizontally. If you want gridlines to print, click on Page
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Layout/Gridlines/Print. Now you are ready to print. Click on the MS Office icon and then Print. Just the working area of the spreadsheet will print, not the column and row identifiers. You can also set the desired print area by highlighting the cells you want to appear on a page. After you elect to Print, Excel will ask you if you want to print just the selected area or the entire worksheet. RELATIVE VS. ABSOLUTE CELL REFERENCES In the example above, we used relative cell references in copying the formula. The formula in cell B6 said subtract the cell above from the one above it. The copied formula in C6 said the same for the cells above it. Sometimes we need to include a specific cell in each calculation, containing, say, a constant. To do this, we need to identify it in the formula as an absolute reference. This is accomplished by placing a $ sign in front of the column and row cell identifiers, for example, $B$2. Placing the sign in front of both assures that whether we move across columns or rows, it will remain an absolute reference. We can illustrate this by creating a spreadsheet to calculate the means of different series of numbers. Fill in the spreadsheet as follows (refer to Figure 3.9): A1: Titration means A3: Titn. No. B3: Series A, mL C3: Series B, mL B4: 39.27 B5: 39.18 B6: 39.30 B7: 39.20 C4: 45.59 C5: 45.55 C6: 45.63 C7: 45.66 A4: 1 We type in each of the titration numbers (1 through 4), but there are automatic ways of incrementing a string of numbers. Click on Fill/Series. Click the Columns and Linear radio buttons, and leave Step Value at 1. For Stop Value, enter 4 and click OK. The numbers 2 through 4 are inserted in the spreadsheet. You could also first highlight the cells you want filled (beginning with cell A4). Then you do not have to insert a
Fig. 3.9.
Relative and absolute cell references.
1 2 3 4 5 6 7 8 9 10 11 12 13
A Titration means Titn. No. 1 2 3 4 Mean: Std.Dev.
B
C
D
Series A, mL Series B, mL 45.59 39.27 45.55 39.18 45.63 39.30 45.66 39.22 45.61 39.24 0.053150729 0.047871355
Cell B8= SUM(B4:B7)/$A$7 Copy right to Cell C8. Cell B9= STDEV(B4:B7) Copy right to Cell C9. We have boldfaced the cells with formulas entered.
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Stop Value. Another way of incrementing a series is to do it by formula. In cell A5, type = A4 + 1. Then you can fill down by highlighting from A5 down, and clicking on Fill/Down. (This is a relative reference.) Or, you can highlight cell A5, click on its lower right corner, and drag it to cell A7. This automatically copies the formula in the other cells or increments numbers in a series. Now we wish to insert a formula in cell B8 to calculate the mean. This will be the sum divided by the number of titrations (cell A7). B8 := sum(B4:B7)/$A$7 We place the $ signs in the denominator because it will be an absolute reference that we wish to copy to the right in cell C8. Placing a $ before both the column and row addresses assures that the cell will be treated as absolute whether it is copied horizontally or vertically. The sum(B4:B7) is a syntax in the program for summing a series of numbers, from cell B4 through cell B7. Instead of typing in the cell addresses, you can also type “=sum(”, then click on cell B4 and drag to cell B7, and type “)”. We have now calculated the mean for series A. [In B8 we could also enter =AVERAGE(B4:B7) to calculate the mean, see later.] We wish to do the same for series B. Highlight cell B8, click on its lower right corner, and drag it to cell C8. Voil`a, the next mean is calculated! Double-click on cell C8, and you will see that the formula has the same divisor (absolute reference), but the sum is a relative reference. If we had not typed in the $ signs to make the divisor absolute, the formula would have assumed it was relative, and the divisor in cell C8 would be cell B7. In the book’s website, you will find a video titled Absolute Cell Reference that illustrates this specific problem above.
Video: Absolute Cell Reference
USE OF EXCEL STATISTICAL FUNCTIONS Excel has a large number of mathematical and statistical functions that can be used for calculations in lieu of writing your own formulas. Let’s try the statistical functions to automatically calculate the mean. Highlight an empty cell and click Formulas/Insert Function. The Insert Function window appears. Select Statistical in the select a category box. The following window appears:
For advanced statistical functions, you can also install the free Analysis ToolPak add-in. See section 3.23.
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Select AVERAGE for the Function name. Click OK, and then in the newly appeared windows, in the Number 1 window, type B4:B7, and click OK. The same average is calculated as you obtained with your own formula. You can also type in the activated cell the syntax =average(B4:B7). Try it. Let’s calculate the standard deviation of the results. Highlight cell B9. Under the Statistical function, select STDEV for the Function name. Alternatively, you can type the syntax into cell B9, =stdev(B4:B7). Now copy the formula to cell C9. Perform the standard deviation calculation using Equation 3.2 and compare with the Excel values. The calculation for series A is ±0.05 mL. The value in the spreadsheet, of course, should be rounded to ±0.05 mL. In the website of the book, you will find two video clips, titled AVERAGE and STDEV, that illustrate the use of these functions.
USEFUL SYNTAXES Excel has numerous mathematical and statistical functions or syntaxes that can be used to simplify setting up calculations. Peruse the Function names for the Math & Trig and the Statistical function categories under fx in the formulas submenu. Some you will find useful for this text are: Math and trig functions LOG10 PRODUCT POWER
SQRT
Video: Average
Calculates the base-10 logarithm of a number Calculates the products of a series of numbers Calculates the result of a number raised to a power Calculates the square root of a number
Statistical functions AVERAGE Calculates the mean of a series of numbers MEDIAN Calculates the median of a series of numbers STDEV Calculates the standard deviation of a series of numbers VAR Calculates the variance of a series of numbers
The syntaxes may be typed, followed by the range of cells in parentheses, as we did above. This tutorial should provide you the basics for other spreadsheet applications. You can write any formula that is in this book into an active cell, and insert appropriate data for calculations. And, obviously, we can perform a variety of data analyses. We can prepare plots and charts of the data, for example, a calibration curve of instrument response versus concentration, along with statistical information. We will illustrate this later in the chapter. See the videos Average and STDEV on the text website for illustrations of using these functions.
TEXT WEBSITE EXAMPLES
Video STDEV
Professor Steven Goates, Brigham Young University, provides tutorial examples, given on your text website, of using Excel for calculating confidence limits, for calculating least-squares regression lines, and for t-test comparisons of the means of two sets of data. Also, Professor Wen-Yee Lee, University of Texas at El Paso, gives an example for plotting and calculating least-squares regression lines for finding concentrations of unknowns using working curves.
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3.21 Using Spreadsheets for Plotting Calibration Curves While independent graphing software may provide greater flexibility, most spreadsheet programs, including Excel, allow graphing and charting, aside from statistical and least-squares calculations already mentioned. We will use the data in Example 3.21 to prepare the plot shown in Figure 3.5, using Excel. Open a new spreadsheet and enter: Cell A1: Riboflavin, ppm (adjust the column width to incorporate the text) Cell B1: Fluorescence intensity Cell A3: 0.000 Cell A4: 0.100 Cell A5: 0.200 Cell A6: 0.400 Cell A7: 0.800 Cell B3: 0.0 Cell B4: 5.8 Cell B5: 12.2 Cell B6: 22.3 Cell B7: 43.3 Format the cell numbers to have three decimal places for column A and one for column B. If you click on the “Insert” tab on the menu bar of Excel 2010, the chart options will appear in the middle. Most often you will use scatter plots (the submenu
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shown on bottom left) appears upon clicking on “Scatter”. The top left selection (see the figure) in the scatter plots will result in the data points being plotted without a line connecting them, the top right in smooth (often called spline-fit) lines connecting them, the middle left in smooth lines but without data points, the middle right in points connected with straight lines between them, and the bottom left with straight lines between points that are invisible. Here is how to prepare a plot using Excel 2010: Highlight the numbers in columns A and B. To the right of the Home bar, click on Insert, Column, All Chart Types. Selct XY (scatter), and then Scatter with only markers. Click OK, and the plot appears. (You may also directly click on Scatter under Insert.) Add statistical data by clicking on the line in order to display Chart Tools, and then select Trendline. Select Linear Trendline, and check Display Equation on chart and Display R-squared value on chart (these are under More Trendline Options). Click Close, and the data are added to the chart. To add axis labels, go to Layout under Chart Tools and click on Axis Titles. Clicking on Primary Horizontal Axis Title and Vertical Axis Title adds these to the chart, and you can type in the actual titles. You can delete the Series notation at the right. You can place the chart on a separate sheet under Chart Tools/Design, and select Move Chart. One example of plotting data is given in the video titled Plotting in Excel in the website of the book. Video: Plotting in Excel
3.22 Slope, Intercept, and Coefficient of Determination We can use the Excel statistical functions to calculate the slope and intercept for a series of data, and the R2 value, without a plot. Open a new spreadsheet and enter the calibration data from Example 3.21, as in Figure 3.10, in cells A3:B7. In cell A9 type Intercept, in cell A10, Slope, and in cell A11, R2 . Highlight cell B9, click on formulas/insert function, select statistical, and scroll down to INTERCEPT under Function name, and click OK. For Known_x’s, enter the array A3:A7, and for Known_y’s, enter B3:B7. Click OK, and the intercept is displayed in cell B9. Now A
B
Riboflavin, ppm
C
D
E
F
G
Fluorescence intensity
3
0.000
0.0
4
0.100
5.8
5
0.200
12.2
6
0.400
22.3
7
0.800
43.3
8 9 10 11 12 13 14 15
Calibration Curve
Fluorescence intensity
1 2
50.0 45.0 40.0 y = 53.75x + 0.595 35.0 R2 = 0.9989 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.000 0.200 0.400 0.600 Riboflavin, ppm
16 17 18
Fig. 3.10.
Calibration graph inserted in spreadsheet (Sheet 1).
0.800
1.000
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repeat, highlighting cell B10, scrolling to Slope, and entering the same arrays. The slope appears in cell B10. Repeat again, highlighting cell B11, and scrolling to RSQ. R2 appears in cell B11. Compare with the values in Figure 3.10. The above example is illustrated in a video titled Intercept Slope and r2 in the website of the book. Video: Intercept Slope and r-square
3.23 LINEST for Additional Statistics The LINEST program of Excel allows us to quickly obtain several statistical functions for a set of data, in particular, the slope and its standard deviation, the intercept and its standard deviation, the coefficient of determination, and the standard error of the estimate, besides others we will not discuss now. LINEST will automatically calculate a total of 10 functions in 2 columns of the spreadsheet. Open a new spreadsheet, and enter the calibration data from Example 3.21 as you did above, in cells A3:B7. Refer to Figure 3.11. The statistical data will be placed in 10 cells, so let’s label them now. We will place them in cells B9:C13. Type labels as follows: Cell A9: slope Cell A10: std. devn. Cell A11: R2 Cell A12: F Cell A13: sum sq. regr. Cell D9: intercept Cell D10: std. devn. Cell D11: std. error of estim. Cell D12: d.f. Cell D13: sum sq. resid. Highlight cells B9:C13, and click on formulas fx . Select Statistical function, scroll down to LINEST and click OK. For Known_y’s, enter the array B3:B7, and for Known_x’s, enter A3:A7. Then in each of the boxes labeled Const and Stats, type “true”. Now we have to use the keyboard to execute the calculations. Depress Control, Shift, and Enter, and release. The statistical data are entered into the highlighted cells. This keystroke combination must be used whenever performing a function on an array of cells, like here. The slope is in cell B9 and its standard deviation in cell B10.
1 2 3 4 5 6 7 8 9 10 11 12 13
A B C Riboflavin, ppm Fluorescence intensity 0.000 0.100 0.200 0.400 0.800 slope std. devn. R2 F sum sq. regr.
D
E
0.0 5.8 12.2 22.3 43.3 53.75 0.595 intercept 1.017759 0.419633 std. devn. 0.998926 0.643687 std. error of estim. 2789.119 3 d.f. 1155.625 1.243 sum sq. resid.
Fig. 3.11. statistics.
Using LINEST for
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Video: LINEST
Video: Data Analysis Regression
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The intercept is in cell C9 and its standard deviation in cell C10. The coefficient of determination is in cell B11. Compare the standard deviations with those calculated in Example 3.22, and the slope, intercept, and R2 with Example 3.21 or Figure 3.5. Cell C11 contains the standard error of the estimate (or standard deviation of the regression) and is a measure of the error in estimating values of y. The smaller it is, the closer the numbers are to the line. The other cells contain data we will not consider here: Cell B12 is the F value, cell C12 the degrees of freedom (used for F), cell B13 the sum of squares of the regression, and cell C13 the sum of squares of the residuals. How many significant figures should we keep for the least-squares line? The standard deviations give us the answer. The slope has a standard deviation of 1.0, and so we write the slope as 53.8 ± 1.0 at best. The intercept standard deviation is ±0.42, so for the slope we write 0.6 ± 0.4. See also Example 3.22. The video LINEST in the website of the book illustrates the above example. Possibly the most powerful tool to calculate all regression-related parameters is to use the “Regression” function in Data Analysis. It not only provides the results for r, r2 , intercept, and slope (which it lists as X variable 1), it also provides their standard errors and upper and lower limits at a 95% confidence level. It also provides an option for fitting the straight line through the origin (when you know for certain that the response at zero concentration is zero by checking a box “constant is zero”. A video illustrating its use is in the website of the book, titled Data Analysis Regression. A description of its use is given in Chapter 16 at the end of Section 16.7, and example applications are given in Chapter 20, Section 20.5, and Chapter 23 for Example 23.1 and Example 23.2.
3.24 Statistics Software Packages Excel offers a number of statistical functions through the Analysis ToolPak Add-in. Click on the MS Office icon on the top left corner of the screen and click through Excel Options/Add-Ins and Analysis ToolPak. Click OK. Return to the spreadsheet. Now when you go to the Data menu, you will see Data Analysis available on the top right corner. The Solver Add-In, another very useful routine, can be installed in the same way. Click on the Data Analysis link, and you will see 18 programs listed, including the F test, various types of t-tests, and Regression (this provides the best fit, along with the r2 and uncertainties). As you experiment with the programs available under Data Analysis, you will find some very useful. The use of Solver is described in Chapter 6. See also the text website for a list of some commercial software packages for performing basic as well as more advanced statistical calculations.
PROFESSOR’S FAVORITE EXAMPLE Contributed by Professor James Harynuk, University of Alberta An excellent online Flash video tutorial is available in the Journal of Chemical Education (JCE) illustrating how to use Excel, using linear least-squares calculations as an example. It is available at: http://jchemed.chem.wisc.edu/JCEDLib/WebWare/collec tion/reviewed/JCE2009p0879WW/index.html. It is available only to subscribers of JCE, and requires a username and password. If your university/college is a subscriber, you may be able to access the site on a college computer without the username/password, or may be able to get them. Otherwise, ask your instructor if she/he has access.
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The specific equations used are based on those in earlier editions of another text, but the calculations can be adjusted for those in this textbook once you have learned how to create the spreadsheet. For example, the equation given for the slope is a rearrangement of Equation 3.23 (multiply both numerator and denominator by n, and you get Equation 3.23). Try setting up a spreadsheet and apply it to Example 3.21 and see if you get the same answers.
Questions 1. Distinguish between accuracy and precision. 2. What is determinate error? An indeterminate error? 3. The following is a list of common errors encountered in research laboratories. Categorize each as a determinate or an indeterminate error, and further categorize determinate errors as instrumental, operative, or methodological: (a) An unknown being weighed is hygroscopic. (b) One component of a mixture being determined quantitatively by gas chromatography reacts with the column packing. (c) A radioactive sample being counted repeatedly without any change in conditions yields a slightly different count at each trial. (d) The tip of the pipet used in the analysis is broken.
Problems For the statistical problems, do the calculations manually first and then use the Excel statistical functions and see if you get the same answers. See the textbook companion website, Problems 14–18, 20, 21, 25–30, and 37–40.
SIGNIFICANT FIGURES 4. How many significant figures does each of the following numbers have? (a) 200.06, (b) 6.030 × 10−4 , and (c) 7.80 × 1010 . 5. How many significant figures does each of the following numbers have? (a) 0.02670, (b) 328.0, (c) 7000.0, and (d) 0.00200. 6. Calculate the formula weight of LiNO3 to the correct number of significant figures. 7. Calculate the formula weight of PdCl2 to the correct number of significant figures. 8. Give the answer to the following problem to the maximum number of significant figures: 50.00 × 27.8 × 0.1167. 9. Give the answer of the following to the maximum number of significant figures: (2.776 × 0.0050) − (6.7 × 10−3 ) + (0.036 × 0.0271). 10. An analyst wishes to analyze spectrophotometrically the copper content in a bronze sample. If the sample weighs about 5 g and if the absorbance (A) is to be read to the nearest 0.001 absorbance unit, how accurately should the sample be weighed? Assume the volume of the measured solution will be adjusted to obtain minimum error in the absorbance, that is, so that 0.1 < A < 1.
EXPRESSIONS OF RESULTS 11. A standard serum sample containing 102 meq/L chloride was analyzed by coulometric titration with silver ion. Duplicate results of 101 and 98 meq/L were obtained. Calculate (a) the mean value, (b) the absolute error of the mean, and (c) the relative error in percent. 12. A batch of nuclear fuel pellets was weighed to determine if they fell within control guidelines. The weights were 127.2, 128.4, 127.1, 129.0, and 128.1 g. Calculate (a) the mean, (b) the median, and (c) the range. 13. Calculate the absolute error and the relative error in percent and in parts per thousand in the following:
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Measured Value (a) (b) (c) (d)
22.62 g 45.02-mL 2.68% 85.6 cm
Accepted Value 22.57 g 45.31-mL 2.71% 85.0 cm
STANDARD DEVIATION 14. The tin and zinc contents of a brass sample are analyzed with the following results: (a) Zn: 33.27, 33.37, and 33.34% and (b) Sn: 0.022, 0.025, and 0.026%. Calculate the standard deviation and the coefficient of variation for each analysis. 15. Replicate water samples are analyzed for water hardness with the following results; 102.2, 102.8, 103.1, and 102.3 ppm CaCO3 . Calculate (a) the standard deviation, (b) the relative standard deviation, (c) the standard deviation of the mean, and (d) the relative standard deviation of the mean. 16. Replicate samples of a silver alloy are analyzed and determined to contain 95.67, 95.61, 95.71, and 95.60% Ag. Calculate (a) the standard deviation, (b) the standard deviation of the mean, and (c) the relative standard deviation of the mean (in percent) of the individual results.
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, University of Texas at El Paso 17. The mileage at which 10,000 sets of car brakes had been 80% worn through was recorded: the average was 64,700 and the standard deviation was 6,400 miles. (a) What fraction of brakes is expected to be 80% worn in less than 50,000 miles? (b) If the brake manufacturer offers free replacement for any brake that is 80% worn in less than 50,000 miles, how many extra brakes should be kept available for every 1 million product sold? You will need to calculate a Z-value, from normal probability distribution, and from this, an area under the Gaussian curve. See www.intmath.com/Counting-probability/14_Normalprobability-distribution.php for a discussion of this. The following website contains a Standard Normal (Z) Table that can be used to obtain the area from the Z-value: www.statsoft.com/textbook/distribution-tables.
PROPAGATION OF ERROR 18. Calculate the uncertainty in the answers of the following: (a) (128 ± 2) + (1025 ± 8) − (636 ± 4), (b) (16.25 ± 0.06) − (9.43 ± 0.03), (c) (46.1 ± 0.4) + (935 ± 1). 19. Calculate the absolute uncertainty in the answers of the following: (a) (2.78 ± 0.04) (0.00506 ± 0.00006), (b) (36.2 ± 0.4)/(27.1 ± 0.6), (c) (50.23 ± 0.07)(27.86 ± 0.05)/(0.1167 ± 0.0003). 20. Calculate the absolute uncertainty in the answer of the following: [(25.0 ± 0.1) (0.0215 ± 0.0003) − (1.02 ± 0.01)(0.112 ± 0.001)](17.0 ± 0.2)/(5.87 ± 0.01).
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Jon Thompson, Texas Tech University 21. Climate Change and Propagation of Uncertainty. Many factors can cause changes in Earth’s climate. These range from well-known greenhouse gases such as carbon dioxide and methane to changes in ozone levels, the effective reflectivity of Earth’s surface, and the presence of nm - μm sized aerosol particles that can scatter and absorb sunlight in the atmosphere. Some effects lead to a warming influence on climate, while others may cool the Earth and atmosphere. Climate scientists attempt to keep score and determine the net effect of all competing processes by assigning radiative forcing values (W/m2 ) to each effect independently and then summing
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them. Positive radiative forcing values warm climate, while negative values cool the Earth and atmosphere. This approach is insightful since the net radiative forcing (Fnet ) can be linked to expected mean temperature change (Tsurface ) via the climate sensitivity parameter (λ), which is often assigned values of 0.3 − 1.1 K/(W/m2 ). Tsurface = λ × Fnet
(3.42)
The Intergovernmental Panel on Climate Change (IPCC) has studied the work of many scientists to provide current best estimates of radiative forcings and associated uncertainties for each effect. These are described in the figure and table shown below.
Radiative Forcing Terms
Radiative Forcing of Climate between 1750 – 2005 Cooling Warming Climate influence influence efficacy CO2
Long-lived greenhouse gases
Spatial Scale
1.0
(see caption)
Global
High
1.0−1.2
~ 10− 100 yrs
Global
High
N2O CH4 Halocarbons
Anthropogenic
Ozone
Tropospheric
Stratospheric (−0.05)
0.5−2.0
Stratospheric water vapour from CH4
~1.0
10 years
.....
10− 100 yrs
Local to Med continental - Low
Direct effect
0.7−1.1
Days
Continental Med to global - Low
Cloud albedo effect
1.0−2.0
HoursDays
Continental to global
Low
~0.6
Hours
Continental
Low
0.7−1.0
10− 100 yrs
Global
Low
Surface albedo
Total Aerosol
Black carbon on snow
Land use
(0.01)
Linear contrails Natural
Weeks to Continental Med to global 100 yrs
Solar irradiance −2
−1
1
2
Timescale
Global
Low
Scientific Understanding
1.61±??? Radiative Forcing (W m−2)
Figure Credit: “Climate Change 2007: The Physical Science Basis” Intergovernmental Panel on Climate Change (IPCC).
Table 1. Best estimates of climate forcings and uncertainties. Climate Effect Long-lived Greenhouse Gases Tropospheric and Stratospheric Ozone Surface Albedo Direct Aerosol Effect Indirect Aerosol Effect
Best Estimate ± Uncertainty (W / m2 ) 2.61 ± 0.26 0.30 ± 0.22 −0.10 ± 0.20 −0.50 ± 0.36 −0.70 ± 0.7
Use the rules for propagation of uncertainty to Ask Yourself: (a) The sum of the radiative forcing terms is 1.61 W/m2 . What is the uncertainty associated with this estimate? Which of the individual terms seem to dominate the magnitude of the overall uncertainty?
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(b) If a value of 0.7 ± 0.4 K/(W/m2 ) is used as the climate sensitivity parameter (λ), what is the expected surface temperature change? What is the associated uncertainty in this estimate? (c) Instrumental temperature records from the 1850s to present day suggest the global mean surface temperature has increased by roughly 0.8◦ C. Is this estimate consistent with your calculation from b?
CONFIDENCE LIMIT For an Excel-based problem related to confidence limits, courtesy of Professor Steven Goates, Brigham Young University, see the website supplement. 22. The following molarities were calculated from replicate standardization of a solution: 0.5026, 0.5029, 0.5023, 0.5031, 0.5025, 0.5032, 0.5027, and 0.5026 M. Assuming no determinate errors, within what range are you 95% certain that the true mean value of the molarity falls?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Amanda Grannas, Villanova University 23. You work in an analytical testing lab and have been asked by your supervisor to purchase new pH meters to replace your old supply. You have identified four potential suppliers and now need to evaluate the performance of each pH meter. You obtained a solution with a known pH of 5.5. You performed 10 replicate measurements of the pH of that known solution using each of the four different pH meters and obtained the data shown below. pH meter Brand A
pH meter Brand B
pH meter Brand C
pH meter Brand D
5.6 5.8 5.6 5.5 5.6 5.7 5.6 5.7 5.1 5.6
5.5 5.6 5.5 5.6 5.6 5.6 5.4 5.5 5.5 5.4
5.8 5.9 6.0 5.9 5.3 5.6 5.7 5.8 5.9 5.1
7.0 6.9 6.8 7.0 6.9 6.9 7.0 6.8 6.9 6.9
All other factors being equal (cost, ease of use, etc), which brand would you suggest the lab purchase? Explain your reasoning. 24. Determination of the sodium level in separate portions of a blood sample by ion-selective electrode measurement gave the following results: 139.2, 139.8, 140.1, and 139.4 meq/L. What is the range within which the true value falls, assuming no determinate error (a) at the 90% confidence level, (b) at the 95% confidence level, and (c) at the 99% confidence level? 25. Lead on leaves by a roadside was measured spectrophotometrically by reaction with dithizone. The standard deviation for a triplicate analysis was 2.3 ppm. What is the 90% confidence limit?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, University of Texas at El Paso 26. The blood cell counts on 7 normal subjects are 5.2, 4.8, 5.4, 5.3, 5.1, 4.9, 5.5 million cells per microliter. The lab work showed that my blood cell count is 4.5 million cells per microliter. Is my blood cell count too low? How confident are you? Show the statistics that supports your answer. 27. The standard deviation established for the determination of blood chloride by coulometric titration is 0.5 meq/L. What is the 95% confidence limit for a triplicate determination?
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PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, The University of Texas at El Paso 28. A standard reference material is certified to contain 94.6 ppm of an organic pesticide in soil. Your analysis gives values of 99.3, 92.5, 96.2, and 93.6 ppm. Do your results significantly differ from the expected result at 95% confidence level? Show your work to explain your answer. 29. Estimate the range of the true molarity of the solution at the 90% confidence level from the standardization in Problem 40.
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Bin Wang, Marshall University 30. A repeated analysis of Cl in a given compound resulted in the following results for %Cl: 2.98, 3.16, 3.02, 2.99, and 3.07. (a) Can any of these results be rejected for statistical reasons at the 90% confidence level? (b) If the true value was 3.03%, can you be 95% confident that your results agree with the known value?
TESTS OF SIGNIFICANCE 31. A study is being performed to see if there is a correlation between the concentration of chromium in the blood and a suspected disease. Blood samples from a series of volunteers with a history of the disease and other indicators of susceptibility are analyzed and compared with the results from the analysis of samples from healthy control subjects. From the following results, determine whether the differences between the two groups can be ascribed to chance or whether they are real. Control group (ppb Cr): 15, 23, 12, 18, 9, 28, 11, 10. Disease group: 25, 20, 35, 32, 15, 40, 16, 10, 22, 18.
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, The University of Texas at El Paso 32. This is a comparison of peak expiratory flow rate (PEFR) before and after a walk on a cold winter’s day for a random sample of 9 asthmatics. As shown in the following table, data in one column indicate the PEFRs before the walk and the other of PEFRs after the walk. Each row represents the same subject. Is the PEFR significantly different before and after a walk at 95% confidence level? Expiratory flow rate (L/min) Subject 1 2 3 4 5 6 7 8 9
Before walk 312 242 340 388 296 254 391 402 290
After walk 300 201 232 312 220 256 328 330 231
33. An enzymatic method for determining alcohol in wine is evaluated by comparison with a gas-chromatographic (GC) method. The same sample is analyzed several times by both methods with the following results (% ethanol). Enzymatic method: 13.1, 12.7, 12.6, 13.3, 13.3. GC method: 13.5, 13.3, 13.0, 12.9. Does the enzymatic method give the same value as the GC method at the 95% confidence level? 34. Your laboratory is evaluating the precision of a colorimetric method for creatinine in serum in which the sample is reacted with alkaline picrate to produce a color. Rather than perform one set of analyses, several sets with different samples are performed over several days, in order to get
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a better estimate of the precision of the method. From the following absorbance data, calculate the pooled standard deviation. Day 1 (Sample A)
Day 2 (Sample B)
Day 3 (Sample C)
0.826 0.810 0.880 0.865 xA = 0.845
0.682 0.655 0.661
0.751 0.702 0.699 0.724 xC = 0.719
xB = 0.666
35. The following replicate calcium determinations on a blood sample using atomic absorption spectrophotometry (AAS) and a new colorimetric method were reported. Is there a significant difference in the precision of the two methods? AAS (mg/dL)
Colorimetric (mg/dL)
10.9 10.1 10.6 11.2 9.7 10.0 Mean 10.4
9.2 10.5 9.7 11.5 11.6 9.3 10.1 11.2 Mean 10.4
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Mary K. Donais, Saint Anselm College 36. This question addresses the following: Use of quality control sample to evaluate method accuracy, use of a significance test to compare two methods, sample homogeneity and data precision. The question is based on actual data collected by students for a senior research project. The lead content of ancient Roman bronze coins can be used to approximate their age. Lead concentrations in individual coins can vary from single percents up to over 20%. A method was developed to quantify lead in bronze coins that utilized hot plate digestion of the samples followed by flame atomic absorption spectrophotometric measurements. Hot plate temperatures can be difficult to set and are often uneven across the apparatus surface, however. As well, a quantitative transfer of the digest solution is required. A second, more efficient method was then developed that utilized a programmable temperature feedback digestion block with volumetric vessels. Data were collected using both digestion methods as provided below. Four coins (numbered 1–4) were divided in half with a sample from each half (designated as a and b) being digested and analyzed. Also, a standard reference material, NIST SRM 872, was analyzed using both methods; the certified lead content in this material is 4.13 ± 0.03%. Evaluate the data. Comment on accuracy and precision. How could the homogeneity of lead in the bronze coins affect precision? Is one method “better” or “different” than the other? How could this be shown quantitatively? Coin
Hot Plate Method (%)
Digestion Block Method (%)
1a 1b 2a 2b 3a 3b 4a 4b NIST SRM 872
14.44 8.41 23.77 27.23 6.34 8.04 16.16 19.07 4.21
15.37 7.24 24.14 24.87 6.77 7.34 17.20 18.26 4.12
An Excel file for the solution is in the website supplement.
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37. Potassium dichromate is an oxidizing agent that is used for the volumetric determination of iron by titrating iron(II). Although potassium dichromate is a high-purity material that can be used for the direct preparation of a standard solution of known concentration, the solution is frequently standardized by titrating a known amount of iron(II) prepared from high-purity iron wire or electrolytic iron, using the same procedure as for the sample. This is because the color of the iron(III) product of the titration tends to mask the indicator color (used to detect the end of the titration), causing a slight error. A solution prepared to be 0.1012 M was standardized with the following results: 0.1017, 0.1019, 0.1016, 0.1015 M. Is the supposition that the titration values are statistically different from the actual prepared concentration valid? 38. In the nuclear industry, detailed records are kept of the quantity of plutonium received, transported, or used. Each shipment of plutonium pellets received is carefully analyzed to check that the purity and hence the total quantity is as the supplier claims. A particular shipment is analyzed with the following results: 99.93, 99.87, 99.91, and 99.86%. The listed purity as received from the supplier is 99.95%. Is the shipment acceptable?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Amanda Grannas, Villanova University 39. You have developed a new method to measure cholesterol levels in blood that would be cheap, quick and patients could do tests at home (much like glucose tests for diabetics). You need to validate your method, so that you can patent it! Use the information given below and the various statistical methods of data validation you have learned to evaluate the effectiveness of your new testing method. (a) NIST makes a cholesterol in human serum standard that is 182.15 mg/dL. Your method reports values of 181.83, 182.12, 182.32 and 182.20 when taking 4 replicate measurements of this standard. Is your value the same? (b) To be comprehensive you tested the same sample (not the NIST standard) numerous times using your method and the “accepted” method for measuring cholesterol. (c) You do not want to give critics an opportunity; there are many at the FDA. You compared the results you get to the accepted method when measuring many different samples. Using the data obtained below, compare your method of analysis to the accepted method for measuring cholesterol. Do your results agree with the accepted method? Sample #
Your Method (mg/dL)
Accepted Method (mg/dL)
1 2 3 4 5
174.60 142.32 210.67 188.32 112.41
174.93 142.81 209.06 187.92 112.37
For an Excel-based Problem related to the t-test, courtesy of Professor Steven Goates, Brigham Young University, see the website supplement.
Q TEST 40. The following replicate molarities were obtained when standardizing a solution: 0.1067, 0.1071, 0.1066, and 0.1050. Can one of the results be discarded as due to accidental error at the 95% confidence level? 41. Can any of the data in Problem 14 be rejected at the 95% confidence level? 42. The precision of a method is being established, and the following data are obtained: 22.23. 22.18, 22.25, 22.09, and 22.17%. Is 22.09% a valid measurement at the 95% confidence level?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, University of Texas at El Paso 43. Police have a hit-and-run case and need to identify the brand of red auto paint. The percentages of iron oxide, which gives paint its red color, found in paint from the victim’s car are as follows: 43.15, 43.81, 45.71, 43.23, 41.99, and 43.56%.
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(a) Can we include all of the numbers above to calculate the average percentage of the iron oxide in the paint sample at 90% confidence level? (b) Based on your finding from subquestion (a), take only reliable numbers to calculate the average with the uncertainty, that is, average standard deviation. (c) The police found a suspect and collected some red paint samples from the front bumper of the suspect’s car. The percentage of iron oxide was found to be (42.60 ± 0.44) % (average standard deviation) with five measurements. Do you think the police got the right person who should be charged with “hit-and-run” (at 95% confidence level)?
STATISTICS FOR SMALL SETS OF DATA 44. For Problem 15, estimate the standard deviation from the range. Compare with the standard deviation calculated in the problem. 45. For Problem 25, use the range to estimate the confidence limit at the 95% confidence level, and compare with the value calculated in the problem using the standard deviation. 46. For Problem 29, use the range to estimate the confidence limits at the 95 and 99% confidence levels, and compare with the values calculated in the problem using standard deviation.
LEAST SQUARES 47. Calculate the slope of the line in Example 3.21, using Equation 3.22. Compare with the value calculated using Equation 3.23. 48. A calibration curve for the colorimetric determination of phosphorous in urine is prepared by reacting standard solutions of phosphate with molybdenum(VI) and reducing the phosphomolybdic acid complex to produce the characteristic blue color. The measured absorbance A is plotted against the concentration of phosphorous. From the following data, determine the linear least-squares line and calculate the phosphorous concentration in the urine sample: ppm P
A
1.00 2.00 3.00 4.00 Urine sample
0.205 0.410 0.615 0.820 0.625
49. Calculate the uncertainties in the slope and intercept of the least-squares line in Problem 48, and the uncertainty in the phosphorus concentration in the urine sample. For an Excel-based Problem related to least squares, courtesy of Professor Steven Goates, Brigham Young University, see the website supplement. For an Excel-based least-squares plotting problem, courtesy of Professor Wen-Yee Lee, The University of Texas at El Paso, see the website supplement.
PROFESSOR’S FAVORITE CHALLENGE Contributed by Professor Michael D. Morris, University of Michigan 50. Explain situations when a least-squares linear fit is not appropriate and should not be used. There are at least two common important cases.
CORRELATION COEFFICIENT 51. From the data given below, determine the correlation coefficient between the amount of toxin produced by a fungus and the percent of yeast extract in the growth medium. Sample (a) (b) (c) (d) (e)
% Yeast Extract
Toxin (mg)
1.000 0.200 0.100 0.010 0.001
0.487 0.260 0.195 0.007 0.002
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52. The cultures described in Problem 51 had the following fungal dry weights: sample (a) 116 mg, (b) 53 mg, (c) 37 mg, (d) 8 mg, and (e) 1 mg. Determine the correlation coefficient between the dry weight and the amount of toxin produced. 53. A new method for the determination of cholesterol in serum is being developed in which the rate of depletion of oxygen is measured with an oxygen electrode upon reaction of the cholesterol with oxygen, when catalyzed by the enzyme cholesterol oxidase. The results for several samples are compared with those of the standard Lieberman colorimetric method. From the following data, determine by the t-test if there is a statistically significant difference between the two methods and calculate the correlation coefficient. Assume the two methods have similar precisions. Sample 1 2 3 4 5 6 7 8 9 10
Enzyme Method (mg/dL)
Colorimetric Method (mg/dL)
305 385 193 162 478 455 238 298 408 323
300 392 185 152 480 461 232 290 401 315
DETECTION LIMIT 54. You are determining aluminum in plants by a fluorometric procedure. Seven prepared blanks give fluorescence readings of 0.12, 0.18, 0.25, 0.11, 0.16, 0.26, and 0.16 units. A 1.0 aluminum standard solution gave a reading of 1.25. What is the detection limit? What would be the total reading at this level?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, University of Texas at El Paso 55. In spectrophotometry, the concentration of analyte is measured by its absorbance of light. Nine reagent blanks were also measured and gave values of 0.0006, 0.0012, 0.0022, 0.0005, 0.0016, 0.0008, 0.0017, 0.0010, and 0.0009. (a) Find the minimum detectable signal. (b) A calibration curve was conducted using a series of standard solutions. The concentrations and the absorbance of the standards are listed in the following table. Absorbance is a dimensionless quantity. Find the slope of the calibration curve (including the unit). (c) Find the concentration detection limit. conc. (ppb) 0.01 0.10 0.50 1.00 2.50
Absorbance 0.0078 0.0880 0.4467 0.8980 1.8770
SAMPLING STATISTICS 56. Four-tenth gram samples of paint from a bridge, analyzed for the lead content by a precise method (< 1% rsd), gives a relative sampling precision, R, of 5%. What weight sample should be taken to improve this to 2.5%?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, University of Texas at El Paso 57. (a) You and your friends are visiting the M&M candy factory at Christmas time. Just as you walk in the door, there is a terrible spill and 200,000 red M&Ms along with 50,000 green
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M&Ms have just spilled on the floor. You madly start grabbing M&Ms and have managed to collect 1000 of them before management catches you. How many red M&Ms have you probably picked up? (b) Continued from question above, if you repeat this scenario many times, what will be the absolute standard deviation of green M&Ms retrieved? 58. Copper in an ore sample is at a concentration of about 3% (wt/wt). How many samples should be analyzed to obtain a percent relative standard deviation of 5% in the analytical result at the 95% confidence level, if the sampling precision is 0.15% (wt/wt)?
REAL-LIFE SAMPLING CONSIDERATIONS, BEYOND STATISTICS PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Christine Blaine, Carthage College 59. Many, if not most, environmental sampling problems go beyond mere statistical considerations. As stated, typically the greatest error occurs in the process of obtaining a representative sample of the analyte of interest. Obtaining a representative sample becomes challenging when it involves an environmental sample or a large sampling site. Let’s examine the following two scenarios that focus specifically on collecting samples for analysis. (a) State authorities in Wisconsin recently asked you to analyze the salmon in Lake Michigan for mercury contamination. Lake Michigan has a surface area of over 22,000 square miles. What factors relating to the salmon do you need to consider when collecting the samples? (b) Poultry farmers often mix small amounts of an arsenic-containing compound, roxarsone, into chicken feed to combat parasites. As the saying goes, “what goes in, comes out.” Research shows that almost all of the roxarsone is excreted by the poultry (Environ. Sci. Technol. 2003, 37, 1509–1514). Over time, the roxarsone degrades to arsenate, AsO4 3− , so the concentrations of both species need to be included in your laboratory analysis. You were recently appointed to sample a 1-acre area adjacent to the coop where chickens roam free for arsenic compounds. How would you identify sites to sample? When would you sample these sites? What other types of sampling might you perform? Obviously there are no unique solutions to these problems. The suggested solutions in the Solutions Manual provide guidelines. Consult the text website for more thought-provoking sample scenarios by Professor Blaine for group discussion.
Recommended References STATISTICS 1. P. C. Meier and R. E. Zund, Statistical Methods in Analytical Chemistry, 2nd ed. New York: Wiley, 2000. 2. J. C. Miller and J. N. Miller, Statistics and Chemometrics for Analytical Chemistry, 4th ed. Englewood Cliffs, NJ: Prentice Hall, 2000. 3. J. C. Miller and J. N. Miller, “Basic Statistical Methods for Analytical Chemistry. A Review. Part 1. Statistics of Repeated Measurements.” “Part 2. Calibration and Regression Methods,” Analyst, 113 (1988) 1351; 116 (1991) 3. 4. A series of articles on Statistics in Analytical Chemistry by D. Coleman and L. Vanatta, Am. Lab.: Part 24—Glossary (November/December (2006) 25; Part 26—Detection Limits: Editorial Comments and Introduction, June/July (2007) 24; Part 30—Statistically Derived Detection Limits (concluded), June/July (2008) 34; Part 32—Detection Limits via 3-Sigma, November/December (2008) 60; Part 34—Detection Limit Summary, May (2009) 50; Part 35—Reporting Data and Significant Figures, August (2009) 34; Part 40—Blanks. Go to their website at www.americanlaboratory.com, click on article/archives, and you can search by title.
Q TEST 5. R. B. Dean and W. J. Dixon, “Simplified Statistics for Small Numbers of Observations,” Anal. Chem., 23 (1951) 636. 6. W. J. Blaedel, V. W. Meloche, and J. A. Ramsay, “A Comparison of Critiera for the Rejection of Measurements,” J. Chem. Educ., 28 (1951) 643.
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7. D. B. Rorabacher, “Statistical Treatment for Rejection of Deviant Values; Critical Values of Dixon’s ‘Q’ Parameter and Related Subrange Ratios at the 95% Confidence Level,” Anal. Chem., 63 (1991) 139. 8. C. E. Efstathiou, “A Test for the Simultaneous Detection of Two Outliers Among Extreme Values of Small Data Sets,” Anal. Lett., 26 (1993) 379.
QUALITY CONTROL 9. J. K. Taylor, “Quality Assurance of Chemical Measurements,” Anal. Chem., 53 (1981) 1588A. 10. J. K. Taylor, Quality Assurance of Chemical Measurements. Boca Raton, FL: CRC Press/Lewis, 1987. 11. J. K. Taylor, “Validation of Analytical Methods,” Anal. Chem., 55 (1983) 600A. 12. J. O. Westgard, P. L. Barry, and M. R. Hunt, “A Multi-Rule Shewhart Chart for Quality Control in Clinical Chemistry,” Clin. Chem., 27 (1981) 493.
LEAST SQUARES 13. P. Galadi and B. R. Kowalski, “Partial Least Squares Regression (PLS): A Tutorial,” Anal. Chim. Acta, 185 (1986) 1.
DETECTION LIMITS 14. G. L. Long and J. D. Winefordner, “Limit of Detection. A Closer Look at the IUPAC Definition,” Anal. Chem., 55 (1983) 712A. 15. J. P. Foley and J. G. Dorsey, “Clarification of the Limit of Detection in Chromatography,” Chromatographia, 18 (1984) 503. 16. J. E. Knoll, “Estimation of the Limit of Detection in Chromatography,” J. Chromatogr. Sci., 23 (1985) 422. 17. Analytical Methods Committee, “Recommendations for the Definition, Estimation and Use of the Detection Limit,” Analyst, 112 (1987) 199. 18. ICH-Q2B Validation of Analytical Procedures: Methodology (International Conference on Harmonization of Technical Requirements for Registration of Pharmaceuticals for Human Use, Geneva, Switzerland, November 1996).
SAMPLING STATISTICS 19. B. Kratochvil and J. K. Taylor, “Sampling for Chemical Analysis,” Anal. Chem., 53 (1981) 924A. 20. M. H. Ramsey, “Sampling as a Source of Measurement Uncertainty: Techniques for Quantification and Comparison with Analytical Sources,” J. Anal. Atomic Spectrosc., 13 (1998) 97. 21. G. Brands, “Theory of Sampling. I. Uniform Inhomogeneous Material,” Fresenius’ Z. Anal. Chem., 314 (1983) 6; II. “Sampling from Segregated Material,” Z. Anal. Chem., 314 (1983) 646. 22. N. T. Crosby and I. Patel, eds., General Principles of Good Sampling Practice. Cambridge, UK: Royal Society of Chemistry, 1995. 23. S. K. Thompson, Sampling, New York: Wiley, 1992.
SPREADSHEETS 24. D. Diamond and V. C. A. Hanratty, Spreadsheet Applications in Chemistry Using Microsoft Excel. New York: Wiley, 1997. 25. H. Freiser, Concepts and Calculations in Analytical Chemistry: A Spreadsheet Approach. Boca Rato, FL: CRC Press, 1992. 26. R. De Levie, Advanced Excel For Scientific Data Analysis. 2nd ed. Oxford University Press, 2008. 27. J. Workman and H. Mark, “Statistics and Chemometrics for Clinical Data Reporting, Part II: Using Excel for Computations,” Spectroscopy, October, 20 (2009) Chapter 2. www.spectroscopyonline.com. 28. E. J. Billo, Excel for Scientists and Engineers: Numerical Methods. Hoboken, NY: Wiley, 2007.
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Chapter Four GOOD LABORATORY PRACTICE: QUALITY ASSURANCE AND METHOD VALIDATION “We can lick gravity, but sometimes the paperwork is overwhelming.” —Werner von Braun
Chapter 4 URLs
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Good laboratory—what it is, how to apply it, p. 133
●
Electronic records, p. 145
●
How to validate a method: selectivity, linearity, accuracy, precision, sensitivity, range, LOD, LOQ, ruggedness, p. 135
●
Official organizations that provide GLP information, p. 146
●
Quality assurance: control charts, documenting, proficiency testing, p. 143
We described in Chapter 1 the general principles of performing quantitative analyses, and in Chapters 2 and 3 we discussed aspects of sampling methodology and statistics and proper data handling and analysis. When you, as an analyst, adhere to these general guidelines, you will generally perform measurements properly, and if well-established methods are used, chances are good you will achieve acceptable (accurate) results. But, depending on what the results are to be used for, this may not be enough to satisfy the client. This is especially so if the measurements are for regulatory purposes or forensic analyses, all of which may have to be defended in court. As a result, the concepts of good laboratory practice (GLP), method validation, and quality assurance for testing laboratories have evolved as an approach to assure, to the extent possible, that reported analyzed results are correct within prescribed or documented limits. Various government agencies [the Environmental Protection Agency (EPA), the Food and Drug Administration (FDA)] and private agencies (e.g., AOAC International, ASTM) have promulgated their own specific guidelines for GLP or method validation and quality assurance. We will briefly describe some of these below, but they all have elements in common. We will first describe the basic elements for GLP. The bottom line is that the lab management and analysts should use common sense in judging what quality assurance procedure should be implemented, based on the goal of the analysis, experience, available methods, time and cost constraints, and the like. But the closer you can adhere to accepted guidelines, the more confident you (and others) will be in your results. Remember, a proper analysis is more than simply receiving a sample and performing a one-shot analysis. If it is not carried out by a validated method and properly documented, the analysis effort, time, and cost will likely be wasted. 132
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Why Have Good Laboratory Practice? The answer to this question is probably obvious. But it can be illustrated by the embarrassment faced by one of the premier analytical laboratories in the world, the Federal Bureau of Investigation (FBI) laboratory. In 1995, it was involved in a high-profile case, the bombing of the Alfred P. Murrah federal building in Oklahoma City, which partially leveled the building, killed 168 people, and injured hundreds of others. The FBI lab had performed analyses for explosives at the scene and provided key evidence at trial. The jury found Timothy McVeigh guilty on all counts of conspiracy, bombing, and first-degree murder. But McVeigh’s legal defense team, looking for weaknesses in the prosecution’s case, introduced a 157-page Justice Department report on the FBI lab that had recently been released that listed a number of purported shoddy policies and practices (only 3 pages were admitted into evidence). The report was the result of an 18-month investigation that was triggered by a whistleblower in the lab who filed hundreds of complaints claiming contamination in the explosives unit lab, among hundreds of other accusations. The whistleblower even testified for the defense at the trial! While the Justice Department team found no evidence of contamination, and most of the whistleblower’s allegations were not substantiated, the team did find evidence of insufficient documentation of test results, improper preparation of lab reports, and an inadequate record management and record retention system. The Justice Department concluded that management had failed to establish and enforce validated procedures and protocols. The outcome of the investigation was some 40 systemic suggestions for correcting or improving lab practices and procedures, including pursuing accreditation by the American Society of Crime Laboratory Directors/Laboratory Accreditation Board (ASCLD/LAB). Some practices the lab was to institute include: ●
●
● ●
Each examiner who analyzes evidence should prepare and sign a separate report. All case files should contain notes, printouts, charts, and other data records used to reach conclusions. The lab must develop a record retention and retrieval system. Written procedures for handling evidence and for avoiding contamination should be refined.
Now, many of these concerns relevant to the FBI lab are not applicable to many other laboratories. But they illustrate the importance of instituting good laboratory practices. Had the FBI lab been more diligent in the care of its practices, it may have avoided the turmoil of this investigation.
4.1 What Is Good Laboratory Practice? The exact definition of good laboratory practice depends on who is defining it and for what purpose. A broad definition encompasses such issues as organization of the laboratory, management, personnel, facilities, equipment, operations, method validation, quality assurance, and record keeping. The goal is to certify that every step of the analysis is valid. The aspects that need to be particularly addressed will vary by laboratory.
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GLP ensures correct results are reported.
The laboratory should have SOPs for every method.
The QAU is responsible for assuring good laboratory practices are implemented. Everyone in the lab is responsible for following them.
Good laboratory practices have been established by worldwide bodies such as the Organization for Economic Cooperation and Development (OECD) and the International Organization for Standardization (ISO). Government agencies have adopted them for their purposes as rules that must be followed for laboratories involved in analyzing substances that require regulation. Examples are pharmaceutical formulations, foods, and environmentally important samples. GLPs can be defined as “a body of rules, operating procedures, and practices established by a given organization that are considered to be mandatory with a view to ensuring quality and correctness in the results produced by a laboratory” (M. Valcarcel, Principles of Analytical Chemistry, Berlin: Springer, 2000, p. 323). They all contain two common elements: standard operating procedures (SOPs) and a quality assurance unit (QAU). Standard operating procedures provide detailed descriptions of activities performed by the laboratory. Examples are sample custody chain, sample handling and preparation, the analytical method, instrument maintenance, archiving (record keeping), and the like. Detailed sample analysis procedures are provided for laboratory analysts or technicians to follow. These are generally more detailed than given in scientific publications of developed methods since the level of training and experience of different laboratory personnel will vary, even though highly trained analytical chemists may need less direction. The quality assurance unit is generally independent from the laboratory and answers to the manager of the organization with which the laboratory is affiliated. The QAU is responsible for implementing quality procedures and assessing them on a continuing basis; this will include audits of the laboratory from time to time.
4.2 Validation of Analytical Methods First identify the problem and the requirements, then select the method to meet those requirements.
Method validation is the process of documenting or proving that an analytical method provides analytical data acceptable for the intended use. The basic concept of the validation process encompasses two aspects: ● ●
The problem and the data requirements The method and its performance characteristics
As mentioned in Chapter 1, the analytical process benefits when the analyst can be involved in defining the problem, that is, in making sure the proper questions are posed. When data requirements are poorly conceived or unrealistic, analytical measurements can be unnecessarily expensive if the method selected is more accurate than needed. Or, it may be inadequate if the method is less accurate than required, or of questionable value if the accuracy of the method is unknown. The first step in method development and validation is setting minimum requirements, which essentially are the specifications of the method for the intended purpose. How accurate and precise does it have to be? What is the target concentration? HIERARCHY OF METHODOLOGY We described in Chapter 1 the general procedure for establishing how an analysis will proceed. The hierarchy of methodology (Table 4.1) may be considered as follows: Technique → method → procedure → protocol These are critical steps in developing a method for a specific purpose that eventually leads to a validated method and address the list of validation characteristics above. The level of the hierarchy reached or used will depend on the need.
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135
Table 4.1
Hierarchy of Analytical Methodologya Definition Technique
Method Procedure
Protocol
a Reprinted
Scientific principle useful for providing compositional information Distinct adaptation of a technique for a selected measurement purpose Written directions necessary to use a method
Set of definitive directions that must be followed, without exception, if the analytical results are to be accepted for a given purpose
Example Spectrophotometry
Pararosaniline method for measurement of sulfur dioxide ASTM D2914—Standard Test Method for the Sulfur Dioxide Content of the Atmosphere (West-Gaeke Method) EPA Reference Method for the Determination of Sulfur Dioxide in the Atmosphere (Pararosaniline Method)
from J. K. Taylor, Anal. Chem., 55 (1983) 600A. Published 1983 by the American Chemical Society.
A technique is the scientific principle selected for providing compositional information. Spectrophotometry gives information about concentration, from the amount of light absorbed by the prepared sample solution. A method is the adaptation of the technique (using the appropriate chemistry), so it is selective for a given analyte. A procedure consists of the written directions necessary to utilize the method. (This is where we enter the broader area of GLP.) It does not necessarily reach the status of a standard method. Finally, a protocol is a set of specifically prescribed directions that must be followed, without exception, if the results are to be accepted for a given purpose, for example, for EPA regulations or action; the method has been validated to provide accurate results for the specified analyte in the specified matrix, and it is then a reference method. It is analogous to a pilot’s checklist for each stage of a flight. Such lists reduce the likelihood of landing with the landing gear retracted. Similarly, an analytical protocol/checklist reduces the likelihood that pH will be inappropriate or a step omitted. If the procedure has been validated, then going straight through according the instructions, making all the measurements in order should give reliable results. Further, if there’s a problem, the specifics of what caused the problem will be evident. VALIDATION PROCESS The need to validate a method and the procedure to be followed are matters of professional judgment; fairly well-prescribed procedures and guidelines are now available that aid in decision making. Government and international agencies have issued guidelines for appropriate method validation, particularly for methods for regulatory submission. Generally, they include studies on: ● ● ● ●
Selectivity Linearity Accuracy Precision
● ● ●
Sensitivity Range Limit of detection
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●
●
Limit of quantitation Ruggedness or robustness
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Spike/ Surrogate
Independent Method
Candidate Method
Quality Control
Replicate Measurements
Precision
SRM
Bias
Validated/ Evaluated Method
Method of Known Accuracy
Collaborative Test
Fig. 4.1.
General process for evaluation/validation of methodology. SRM denotes standard reference material. [Reprinted from J. K. Taylor, Anal. Chem., 55 (1983) 600A. Published 1983 by the American Chemical Society.]
These are best done during the development of a method. (If a method does not possess the required sensitivity, why proceed?) Figure 4.1 gives an overall view of the validation process. Different aspects are discussed in the following paragraphs. SELECTIVITY The matrix is everything in the sample except the actual analyte of interest. Components in the matrix can interfere with determination of the analyte (i.e., induce a matrix effect).
Selectivity is the extent that the method can measure the analyte of interest in the matrices of the samples being analyzed without interference from the matrix (including other analytes). Matrix effects may be either positive or negative. The analytical response of the analyte in the presence of potential sample components is compared with the response of a solution containing only the analyte. The selection of an appropriate measurement methodology is a key consideration. Methods, even previously validated in general terms, may not be assured to be valid for a particular sample matrix. LINEARITY A linearity study verifies that the response is linearly proportional to the analyte concentration in the concentration range of sample solutions. The study should be performed using standard solutions at five concentration levels, in the range of 50 to 150% of the target analyte concentration. Five concentration levels should allow detection of curvature on the calibration curve. Each standard should be measured at least three times. Linearity data are often judged from the coefficient of determination (r2 ) and the y intercept of the linear regression line. An r2 value of > 0.998 is considered as evidence of acceptable fit of the data to the regression line. The y intercept should be a small percentage of the analyte target concentration, for example, < 2%. While these statistical evaluations are a practical way to assess linearity, they do not guarantee it. You should always do a visual inspection of the calibration curve. The linearity will often deviate somewhat at high and low values. (This is the reason weighted least-squares plots may be preferred. In one mode of weighted least squares, the
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Response factor (y−0.595)/C
70 y = −1.48x + 54.98
60 50 40 30
Fig. 4.2.
Response factor plot for Figure 3.5. The y-intercept value is subtracted from the corresponding fluorescence intensity for each concentration (0.1, 0.2, 0.4, and 0.8 ppm) and divided by each concentration.
20 10 0 0
0.2
0.4 0.6 Concentration, ppm
0.8
1
points with the least relative deviation are given more weight in the regression line.) One way of evaluating the range of linearity is to plot a response factor (RF) versus concentration. Such a plot has been referred to as a Cassidy Plot (after R. M. Cassidy and M. Janoski, LC.GC Mag. 10 (1992) 692). Response factor = (signal − y intercept)/concentration
(4.1)
If a plot with zero slope is obtained, this indicates that a linear response is obtained over this concentration range. A response factor change over the calibration concentration range within, for example, 2 to 3% of the target-level response factor or the average RF may be considered acceptable linearity. In Figure 3.5 the regression line is y = mx + b. The y intercept is 0.595. A plot of the response factor versus concentration is shown in Figure 4.2. The slope of the line is −1.48 RF/1 ppm. This corresponds to −1.0 over the concentration range 0.1 to 0.8 ppm, which is 1.8% of the average RF value of 54.4. This is acceptable linearity. If a calibration curve deviates from linearity over the 50 to 150% target-level range, selection of a narrower range of, for example, 80 to 120% may provide the desired linearity. Current use of embedded computers/microprocessors as integral components of instruments often includes on-board raw signal manipulations to extend linear range, often by power transformation. This leads to some unusual effects, see Anal. Chem. 82 (2010) 10143. ACCURACY Accuracy of a method is the closeness of the obtained value to the true value for the sample. This is probably the most difficult parameter to validate. One should consider the sampling and sample treatment, in addition to the measurement method accuracy. Accuracy of the method can be determined in one of three ways. The most preferred first, in order, these are: (a) Analysis of a reference material (b) Comparison with results using another method known to be accurate (c) Recovery studies (a) is preferred; if one can’t do (a) then (b) is second choice and (c) third choice. Recovery studies are performed by spiking (adding) a known amount of the analyte either to a blank matrix (a sample that has an unmeasurable level of the test analyte)
The response per unit concentration should be nearly constant for good linearity.
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Spike recovery can be calculated as the percentage of the amount spiked into the sample, which is measured to be present by the method (i.e., the percentage of a known amount that is recovered).
Accuracy is best determined by analysis of a standard reference material.
Perform at least seven measurements for statistical validation.
or by spiking a sample in which the background analyte is measured by the same procedure and subtracting from the total (sample + spike) value to obtain the recovery. The spiked samples should be prepared at three levels, the extremes and the midrange. They should be prepared at least in triplicate. Good spike recovery cannot, however, ensure lack of positive interferences. A better validation method is to perform the analysis by two independent methods, in which the second method is an accepted procedure known to be accurate for the sample matrix of interest. Ideally, even the sample treatment should be different. You can often find in the scientific literature (journals, reference books, standard methods books) a method that is applicable to your sample (but that may not be appropriate to use because of expense, unavailability of equipment, etc.). If none can be found that has been applied to your sample matrix, but one is known to be generally applicable and accurate, then use this. If results by your method and the second method agree, that is good evidence they both work for your sample. If there is disagreement, then it is not possible to draw any conclusions since either may give erroneous results with your particular sample, although it is probably more likely your new method is the culprit. The ideal way to validate a method is to analyze a reference material identical in composition to your sample. The National Institute of Standards (NIST) has the goal of ensuring accurate and compatible measurements through the development, certification, and distribution of standard reference materials (SRMs). The SRM program has over 1000 SRMs available for use in (1) basic measurements in science and metrology (which is the science of measurement), (2) environmental analysis, (3) health measurements, and (4) industrial materials and production. The NIST has standards for chemical composition, physical properties, engineering materials, and the like (http://www.nist.gov/srm/). It serves as the main contact point for interfacing with similar efforts in the private sector, other federal agencies, and internationally. The website for the Canadian equivalent of the NIST SRM program is www.nrc-cnrc.gc.ca/eng/solutions/advisory/crm_index.html. Other programs include the American Society for Testing and Materials (ASTM), the American Association for Clinical Chemistry (AACC), the International Union of Pure and Applied Materials (IUPAC), the International Organization for Standardization (ISO), and the European Union (EU). You can obtain information about each of these from their websites. Chemical composition standards are certified for given concentrations, with a statistical (standard deviation) range given. If your method falls two standard deviations from the certified value, there is a 95% chance there is a significant (nonrandom) difference between the results. Depending on the concentration levels being measured, you may establish that the measurement should be within, for example, ±2% of the certified value, or perhaps ±10% if it is a trace analysis, and so forth. One way to detect interferences is to run an SRM, then spike the SRM with the potential interferent. If the result is unchanged, one is confident that the interferent isn’t going to cause trouble. There may not be reference materials available that are identical in composition to your sample, but similar. These will still provide a high level of confidence in the validation. When performing measurements on reference materials or by comparison with another method, statistical considerations suggest that at least six degrees of freedom (seven measurements) be made for proper validation. PRECISION
Repeatability is intralaboratory precision while reproducibility is interlaboratory precision.
The precision of an analytical method is obtained from multiple analyses of a homogeneous sample. You can determine overall precision of the method, including sample preparation. Such precision data are obtained by one laboratory on one day,
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using aliquots of the homogeneous sample that have been independently prepared. Such intralaboratory precision is called repeatability. Interlaboratory precision, if appropriate, is also determined as part of a measurement of reproducibility or robustness of the method (see below). You can also determine the precision of different steps of the analysis, for example, the precision of injecting a sample into a gas chromatograph determined from multiple injections of the same sample solution. Again, statistical considerations dictate that at least seven measurements should be made for each evaluation step. SENSITIVITY The sensitivity is determined by the slope of the calibration curve and generally reflects the ability to distinguish two different concentrations. You can measure the slope or measure samples of closely related concentrations at high, intermediate, and low concentrations. The sensitivity and precision will govern how many significant figures should be reported in a measurement. Do not report 11.25% when the method can at most distinguish a 0.1% difference.
Sensitivity is often confused with limit of detection.
RANGE The working range of a method is the concentration range over which acceptable accuracy and precision are obtained. Usually it also includes linearity. The acceptable accuracy and precision are generally specified in establishing a method. The precision will, of course, vary with the concentration, becoming poorer at low concentrations (Figure 4.3), as well as sometimes at high concentrations, as in spectrophotometric measurements. LIMIT OF DETECTION (LOD) The limit of detection should be determined using a definition given in Chapter 3. Typically, replicate blanks of the sample matrix are analyzed to determine the mean blank value and its standard deviation. Then a matrix is spiked with analyte near the quantitation limit (e.g., to give a signal 10 times the standard deviation above the blank mean signal). The limit of detection is the concentration calculated to give a response equal to the blank signal plus three standard deviations.
When someone asks, “How sensitive is your method?”, most of the time they really want to know the detection limit of your method.
LIMIT OF QUANTITATION (LOQ) This is the lowest concentration of analyte that can be measured in the sample matrix at an acceptable level of precision and accuracy. An acceptable precision is typically 10 to 20% relative standard deviation, depending on the concentration levels measured. Absent a specified precision, then the concentration that gives a signal 10 standard deviations above the blank is used.
4
%RSD
3 2 1 0 0
500 1000 Iron concentration, ppm
1500
Fig. 4.3.
Dependence of relative standard deviation on concentration.
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RUGGEDNESS/ROBUSTNESS
Ruggedness and robustness are often confused. Ruggedness refers to intermediate (day-to-day intralaboratory) precision, whereas robustness refers to the effect of deliberate small changes to the method on its performance.
Interlaboratory variability is about double intralaboratory variability.
We have defined the precision of a method. Repeatability is the long-term precision over several weeks for an analysis conducted in the same laboratory. Ruggedness refers to the precision of one lab over multiple days, which may include multiple analysts, multiple instruments, different sources of reagents, different chromatographic columns, and the like. A ruggedness study will identify those factors that will contribute to variability of the results and should not be changed. This is related to robustness or reliability of the method, which refers to how sensitive it is to deliberate or uncontrolled small changes in parameters, such as the size of the sample, the temperature, pH of the solution, reagent concentration, time of reaction, and so forth. It includes an evaluation of the stability of reagents, standards, and samples with time. Each parameter should be tested separately, unless statistically more sophisticated factorial analysis experiments for varying several parameters at once are designed, which we will not go into here. Reproducibility (or transferability) is the analysis of the same sample between labs, in which a homogeneous sample is analyzed by multiple labs with one lab serving as the primary comparison lab. A reproducibility study generally focuses on bias between labs, besides precision. One strives for a bias that is within defined acceptable limits.
Reality Check on Interlaboratory Variability Is interlaboratory variability significant? Is it different than intralaboratory variability? The answer on both accounts is yes. And it varies with concentration. William Horwitz and collaborators documented interlaboratory variability over two decades by examining over 10,000 interlaboratory data sets (see R. H. Albert, Chemical & Engineering News, September 13, 1999, p. 2). They developed an expression relating interlaboratory standard deviations of results, sR , to concentration, C (expressed as decimal fraction, e.g., 1 mg/kg = 10−6 ). They found sR = 0.02C0.85 , or as a relative standard deviation among laboratories, rsd(%) = 2C−0.15 . These expressions indicate that starting with pure materials (C = 1) with an sR of 2%, the interlaboratory precision increases by a factor of 2 for every one hundredfold decrease in concentration. This holds, irrespective of the analyte, method, matrix, or date. The precision of different types of analyses, whether agricultural, geological, or pharmaceutical, did not change for a half-century, even with the advent of modern instrumentation. Thus, the relative standard deviation for pesticide residue levels of 1 ppm (1 mg/kg; 10−6 ) is 16%. (Do the calculation using either formula. Putting the equation in Excel makes it easy.) The fact that this empirical function follows collaborative study statistics was substantiated by a separate study [M. Thompson and P. J. Lowthian, J. AOAC Int., 80 (1997) 6786] in which it was shown that interlaboratory variability is about double that of intralaboratory variability. The variability for EPA-tested analytes with extensive quality assurance built into the procedure was somewhat better than predicted by the above [see J. AOC Int., 79 (1996) 589], but the analyses cost approximately $1000 each, besides being slow. There is an obvious trade-off between quality assurance efforts and cost and time.
Professor Wes Steiner (Eastern Washington University) suggests that the following figures of merit normally need to be met in a laboratory adopting GLP: Precision: ±15% in Relative Standard Deviation (RSD); Accuracy: ±20% in Relative
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Error; LOD defined as by the IUPAC, three standard deviations of the blank over the blank, LOQ defined as ten standard deviations of the sample over the blank (see Section 3.17, on LODs and LOQs), Linear range: the range spanning 50 to 150% of the target analyte value must have a coefficient of determination (r2 ) ≥ 0.995; Specificity: no interfering signal from any substance other than the desired analyte that is 25% of response of the LOQ. Review Figure 4.1, which places in context most of the validation concepts and steps for a candidate method that we have discussed. We will discuss quality control in the context of quality assurance below.
Professor’s Favorite Examples Contributed by Professor Wes Steiner, Eastern Washington University
Example 4.1 One definition of analytical method precision in a GLP laboratory is ±15 percent relative standard deviation, %rsd, of the mean sample concentration where s is the sample standard deviation and x is the sample mean s %rsd = × 100 x A single quality control (QC) sample was run in replicates of eight to produce the following concentrations values: 12.1, 11.9, 11.6, 13.3, 12.8, 12.4, 13.1, and 12.6 μg/L. Is this level of analytical method precision acceptable for this group of QC replicates? Solution
0.58 × 100 = 4.72 12.47 Yes, this level of analytical method precision is acceptable for this group of QC replicates because the %rsd of 4.72 % was within the ±15 %rsd threshold of the mean sample concentration. %rsd =
Example 4.2 One definition of analytical method accuracy in a GLP laboratory is ±20 percent relative error (RE) of the measured mean sample concentration to that of the true concentration such that %re =
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Measured Mean Concentration − True Concentration × 100 True Concentration
A single quality control (QC) sample was prepared at a concentration of 11.6 μg/L and run in replicates of eight to produce the following concentrations values: 13.5, 11.6, 11.2, 12.4, 14.2, 12.2, 13.7, and 14.1 μg/L. Is this level of analytical method accuracy acceptable for this QC sample?
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Solution
12.86 − 11.6 × 100 = 10.86 11.6 Yes, this level of analytical method precision is acceptable for this group of QC replicates because the %RE of 10.86 % was within the ±20 %RE threshold of the true concentration. %re =
Example 4.3 One definition of analytical method limit of quantitation, LOD, in a GLP laboratory is a signal that is 3 times as great as the standard deviation s of a blank or a low level sample. If m is the slope of the calibration curve: 3s LOD = m A single quality control (QC) sample was run in replicates of eight to produce the following mass spectral peak area signals: 2.2, 1.7, 1.9, 2.3, 2.1, 1.8, 2.7, and 2.3 peak area. The slope of the calibration curve is m = 0.456 peak area/μM. Find the minimum detectable concentration. Solution
The standard deviation of the above set of results s = 0.32 LOD =
3(0.32 ) = 2.1 μM 0.456
Example 4.4 One definition of analytical method limit of quantitation, LOQ, in a GLP laboratory is a signal that is 10 times as great as the standard deviation s of a blank or a low level sample. If m is the slope of the calibration curve: 10s LOQ = m What is the LOQ for the results in the previous example? Solution
LOQ =
10(0.32 ) = 7.0 μM 0.456
Example 4.5 Linear Range One definition of analytical method linear range is if the target concentration of the analyte is 1.0, and a 6-point calibration curve spanning the values 0.5:1.0:1.5 is constructed, the linear R2 value must be ≥ 0.995. The target analyte concentration was 10.20 μg/L and exhibited a R2 value of 0.996. The 6 point calibration curve standards were: 5.10, 6.12, 8.16, 12.24, 14.28, and 15.30 μg/L. (a) Was the linearity of the calibration curve acceptable? (b) Did the calibration curve properly span the target concentration?
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Solution
(a) Yes, the correlation coefficient or measure of linearity, R2 , was ≥ 0.995. (b) Yes, the calibration curve properly spanned the target concentration with 5.10, 6.12, and 8.16 μg/L standards spanning from 0.5 to 1.0 in concentration and 12.24, 14.28, and 15.30 μg/L standards spanning from 1.0 to 1.5 in concentration.
4.3 Quality Assurance—Does the Method Still Work? Once a method has been validated, an important aspect of applying it is to assure that it is working properly. Quality assurance (QA) is the implementation of procedures to ensure and document that the method continues to perform as required and is part of the responsibility of the quality assurance unit. It includes written documentation of validation of the method, procedures followed, and the sample custody chain. A number of quality control procedures are implemented, based on quantitative measurements. Typical quality control activities are listed below. CONTROL CHARTS The laboratory should maintain a continuing quality control chart (Figure 3.3) for each method. A reference material of known analyte content is blindly and randomly run each day, or preferably with each batch of samples. If measured values fall outside prescribed standard deviation limits, then you should check for some systematic error such as reagent deterioration or instrument drift (needs recalibration). DOCUMENTING AND ARCHIVING This is a tedious and time-consuming, but critical, part of quality assurance. All activities performed by the laboratory dealing with quality assurance should be documented in written form. This includes recording the chain of custody of the sample, the calibration and performance of instruments, standard operating procedures, original measurement data, results, and reports. Documents should be traced to individuals, meaning they should be signed and dated by the individual creating or responsible for them. PROFICIENCY TESTING One way of documenting performance of the laboratory is to participate in collaborative interlaboratory studies. An official body provides aliquots of the same homogeneous material to laboratories for analysis. The goal is to compare results among laboratories and the uncertainties in the results. The mean of the results of the participating laboratories can be used as the reference, if the actual concentration is not known. Or better, a certified reference material whose concentration and certainty is known (not to the participating laboratories) is used. The latter is particularly more informative if the laboratories use different methods. One way of expressing the results of a collaborative testing exercise is to report the laboratory’s z score, a measure of its deviation from the standard deviation of the known concentration: X − X z= i (4.2) s X is the accepted where X i is the mean of i replicate measurements by the laboratory, concentration, and s is the standard deviation of the accepted concentration X.
QA is an ongoing checking procedure to assure proper performance of a method.
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3 2
z-Value
1 0 −1
Fig. 4.4.
Representative z-value distributions for proficiency tests with a series of laboratories.
−2 −3 Laboratory
Example 4.6 You agree to participate in a collaborative study for the determination of calcium in serum. A sample containing 5.2 meq/dL Ca, with a specified standard deviation of ±0.2 meq/dL, is sent to 10 laboratories for analysis using atomic absorption spectroscopy. You obtain triplicate results of 5.0, 4.7, and 4.8 meq/dL Ca. What is the z value for your laboratory? What do the results imply? Solution
A z-score greater than 2.0 or less than –2.0 can indicate a difference that is due to nonrandom error. The Z -test is one of the statistical functions available in Excel.
The mean is 4.8 meq/dL with a standard deviation of ±0.15 meq/dL. The z value is 4.8 − 5.2 z= = −2.0 0.2 This means your reported mean is low by two standard deviations away from the accepted value. There is a 95% chance this difference is due to a nonrandom systematic error. Also, the range for one standard deviation of your measurements is 4.6 to 5.0 meq/dL. The accepted range is 5.0 to 5.4 for one standard deviation. There is just a 68% chance that your value overlaps the accepted range. You need to look into your method. It has a low enough standard deviation; perhaps you need to recalibrate your instrument with new standards. Figure 4.4 shows the results of a representative proficiency collaborative test. It is not uncommon for several laboratories to be outside acceptable ranges.
4.4 Laboratory Accreditation Another form of external evaluation is laboratory accreditation by a formal organization or government agency. This is generally voluntary but may be required for laboratories dealing with regulatory measurements. Accreditation is a procedure by which an authoritative body gives formal recognition that the laboratory is competent to carry out specific tasks. The accreditation procedure may take the form of qualitative inspection of the laboratory operations, to verify that good laboratory practice policies are followed, that is, proper documentation and record keeping, validation, proficiency testing, and the like. Or it may include measurement of submitted reference materials. In any event, certification will involve periodic laboratory audits, which may be unannounced.
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4.5 ELECTRONIC RECORDS AND ELECTRONIC SIGNATURES: 21 CFR, PART 11
4.5 Electronic Records and Electronic Signatures: 21 CFR, Part 11 Most laboratory tasks are dependent on computers, from sample log-in to reporting. The traditional way of maintaining records for audits, regulatory actions, and the like has been by printing hardcopy for signature, submission, and archiving. This process is time-consuming, requires storage facilities, records can be lost or misplaced, and in part it defeats the purpose of computers. If acceptable records and signatures could be achieved electronically, this would improve efficiency. It would provide faster access to documents, the ability to search databases and view information from multiple perspectives, to determine trends or patterns. The Food and Drug Administration (FDA) worked for 6 years with the pharmaceutical industry to develop procedures to accommodate paperless record systems under the current good manufacturing practices (GMP) regulations. In 1997, the FDA issued the Final Rule on electronic records, signatures, and submissions, known as 21 Code of Federal Regulations (CFR), Part 11 [“Electronic Records; Electronic Signatures,” Fed. Reg., 62 (1997) 1000, 13,230; 64 (1999) 41442]. You can find it at http://www.fda.gov/RegulatoryInformation/Guidances/ucm125067.htm. The main concern and challenge is that electronic records can be too easily changed or falsified, either accidentally or intentionally. The Final Rule provides criteria under which the FDA will consider electronic records to be equivalent to paper records, and electronic signatures equivalent to handwritten signatures, to ensure the integrity, accuracy, and authenticity of information stored in the systems. ELECTRONIC RECORDS Electronic validation will need to be done to document data integrity, backup and recovery, archiving and restoring, and how electronic signatures are used. A validated system must be for the life cycle of the software. If it is changed or updated, the data must be transferable. One problem is that electronic records are comprised of databases, which are dynamic, that is, the content changes as new information is added. Worse, data can be changed or deleted, with no evidence and in a manner that destroys the original data. System access must be limited to authorized individuals. There must be regular system checks. There must be time- and date-stamped audit trails. If changes are made in the database, the audit trail must show who made the changes, when, what the old and new values are, and why the data were modified. ELECTRONIC SIGNATURES Access to the system must be limited to authorized persons. The type of security will depend on whether the system is open or closed. Electronic signature technologies include identification codes (usernames, passwords) or more sophisticated biometric systems (based on measurement of physical features such as palm prints, fingerprints, or iris or retinal pattern scanners). The latter is expensive and less likely to be implemented, especially for multiple users. User names must be unique and never reassigned. Passwords should be unique and changed periodically. 21 CRF, Part 11, permits but does not require the use of electronic records and signatures. As more systems become validated and accepted, and as more instrument manufacturers incorporate validated systems, this will become more commonplace. It is likely other agencies will adopt similar standards. EPA: CROMERR The EPA’s Office of Environmental Information (OEI) has defined a Cross Media Electronic Reporting and Record-Keeping Rule (CROMERR) to remove existing
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regulatory obstacles to electronic reporting and record keeping across a broad spectrum of EPA programs. Information can be found at the URL at http://www.epa.gov/ cromerr/index.html. CROMERR requires criteria for electronic records that are consistent with 21 CFR, Part 11.
What about Cost?
QA costs are about a quarter of laboratory costs. Do it well!
With quality assurance, it is necessary to have reasonable documentation of what the accuracy is and to identify areas where significant contributors to inaccuracies may occur and actions taken to minimize them. This, of course, does not come without cost. Implementing a quality assurance program will involve substantial initial investment, both in expense and time. It has been estimated that ongoing quality assurance costs amount to 20 to 30% of the laboratory budget. So it is important that the system be properly set up, as efficiently as possible (which will require an understanding by management of what is needed— —could that be you?), and that it be taken seriously by all laboratory personnel (which certainly includes you). Quality assurance programs also do not guarantee accurate results, see the box Reality Check on Interlaboratory Variability.
4.6 Some Official Organizations A number of government agencies and national and international organizations have established their own guidelines for method validation and good laboratory practice. Most are based on principles espoused by multinational organizations. Some of the major ones are listed below. Detailed information on each is given on the text website. Do take a look at these. You can obtain more (a lot more!) information on each by browsing their websites. It gives a flavor of the real world of standardization and regulation. International Organization for Standardization (ISO): www.iso.ch International Conference on Harmonization (ICH): www.ich.org Organization for Economic Cooperation and Development (OECD): www. oecd.org Food and Drug Administration (FDA): www.fda.gov/cder/guidance Environmental Protection Agency (EPA): www.epa.gov Quality systems: http://www.epa.gov/quality/bestlabs.html Office of Solid Waste: www.epa.gov/osw US-EPA, Region 4, Science and Ecosystem Support Division: www.epa .gov/region4/sesd American Association for Clinical Chemistry (AACC): www.aacc.org American Association of Cereal Chemistry (AACC): www.aaccnet.org/ American Oil Chemists Society (AOCS): www.aocs.org The Society of Quality Assurance (SQA): www.sqa.org American Society for Testing and Materials (ASTM): www.astm.org Association of Official Analytical Chemists International (AOAC International): www.aoac.org National Institute of Standards and Technology (NIST): www.nist.gov
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PROBLEMS
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PRACTICING GLP PROCEDURES On the companion website for the book, some experiments are available. Experiment 45 provides practice in method validation and quality control, and Experiment 46 is an exercise in proficiency testing. These are class team experiments. Read these, even if they are not part of your assigned laboratory exercises.
Questions GOOD LABORATORY PRACTICE 1. What is good laboratory practice? 2. What are the common elements of GLP implementation? 3. What are SOPs? 4. What are the characteristics of a quality assurance unit?
METHOD VALIDATION 5. What are the two aspects of the validation process? 6. What is the first step in a method development? 7. Distinguish a technique, a method, a procedure, and a protocol. 8. What are the essential features of most method validation processes? 9. What is the response factor? 10. What are ways of assessing calibration linearity? 11. What are the main ways of assessing accuracy of a method? 12. How many measurements should you make to obtain reasonable statistical validation? 13. Distinguish among repeatability, ruggedness, robustness, and reproducibility of a method. 14. What are the main requirements for validation of electronic records and signatures?
QUALITY ASSURANCE 15. What is quality assurance? Quality control? 16. What are some typical quality control procedures? 17. What is a z score? 18. What is laboratory accreditation?
Problems VALIDATION 19. You prepare a calibration curve for the measurements of blood ethanol by gas chromatography. The recorded peak area as a function of concentration is: Concentration, % (wt/vol) 0 0.020 0.040 0.080 0.120 0.160 0.200
Peak Area (arbitrary units) 0.0 43 80 155 253 302 425
Plot the calibration curve using Excel and determine the least-squares line, indicating the y intercept and slope. Calculate the response factor and the slope of its plot versus concentration. What percent of the average response factor is the RF change over the calibration range? See the text website for spreadsheets.
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20. Calculate the 16% relative standard deviation interlaboratory variation listed for 1 ppm pesticide residue levels in the box Reality Check on Interlaboratory Variability. Perform the calculation using both formulas given. Place each formula in an Excel spreadsheet cell to perform the calculations.
QUALITY ASSURANCE 21. You participate in a collaborative study for measuring lead in leaves. A homogeneous standard reference material of ground leaves, certified to contain 10.3 ± 0.5 ppm lead, is given to the participating labs. You analyze the sample, using acid digestion and atomic absorption spectrometry. You report 9.8 ± 0.3 ppm for seven analyzed aliquots. What is the z value for your laboratory?
WEB EXERCISE 22. Look up the websites of at least three of the government agencies and professional societies listed in Section 4.6, and link to their pages dealing with method validation. Document the similarities and any differences between them.
Recommended References WEBSITES 1. http://21cfrpart11.com. This commercial site dealing with compliance-related issues includes useful links. 2. www.PDA.org. The Parenteral Drug Association features an online conference on computer validation issues and 21CFR, Part 11.
GOOD LABORATORY PRACTICES 3. J. M. Miller and J. B. Crowther, eds., Analytical Chemistry in a GMP Environment: A Practical Guide. New York: Wiley, 2000. 4. M. P. Balogh, “How to Build a GLP Bioanalytical Lab,” LCGC North America, 24 (10), October (2006) 1088. www.chromatographyonline.com. 5. M. Swartz and I. Krull, “Glossary of Terms Related to Chromatographic Method Validation,” LCGC North America, 25 (8) August (2007) 718. www.chromatographyonline.com. 6. P. Konieczka and J. Namiesnik, Quality Assurance and Quality Control in the Analytical Chemical Laboratory. Boca Raton, FL: CRC Press, 2009.
QUALITY ASSURANCE/QUALITY CONTROL 7. H. Y. Aboul-Enein, R-I. Stefan, and G-E. Baiulescu, Quality and Reliability in Analytical Chemistry. Boca Raton, FL: CRC Press, 2000. 8. E. Prichard and V. Barwick, Quality Assurance in Analytical Chemistry. New York: Wiley, 2007. 9. C. C. Chan, H. Lam, and X-M Zhang, Eds., Practical Approaches to Method Validation and Essential Instrument Qualification. Hoboken, NJ: Wiley, 2010. 10. M. Swartz and I. Krull, “21 CFR Part 11 and Risk Assessment: Adapting Fundamental Methodologies to a Current Rule,” LCGC North America, 25 (1), January (2007) 48. 11. M. Swartz and I. S. Krull, Analytical Method Development and Validation. New York: Marcel Dekker, 1997. 12. H. Marchandise, “Quality and Accuracy in Analytical Chemistry,” Fresenius’ J. Anal. Chem., 345 (1993) 82. 13. J. M. Green, “A Practical Guide to Analytical Method Validation,” Anal. Chem., 68 (1996) 305A. 14. M. Stoeppler, W. R. Wolf, and P. S. Jenks, eds., Reference Materials for Chemical Analysis: Certification, Availability and Proper Usage. New York: Wiley, 2001. 15. D. G. Rhoads, Lab Statistics Fun and Easy: A Practical Approach to Method Validation. Washington, DC: AACC (American Association for Clinical Chemistry), 1999. 16. M. E. Schwartz and I. S. Krull, Handbook of Analytical Validation. Boca Raton, FL: CRC Press, 2012.
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Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
How to calculate molarities and moles (key equations: 5.4, 5.5), p. 152
●
How to express analytical results, p. 159
●
How to calculate weight and percent analyted from molarities, volumes, and reaction ratios (key equations: 5.5, 5.17–5.20, 5.25), pp. 152, 169, 171
●
Weight relationships for gravimetric analysis (key equation: 5.28), p. 181
Analytical chemistry deals with measurements of analytes in solids and concentrations in solution, from which we calculate masses. Thus, we prepare solutions of known concentrations that can be used to calibrate instruments or to titrate sample solutions. We calculate the mass of an analyte in a solution from its concentration and the volume. We calculate the mass of product expected from the mass of reactants. All of these calculations require a knowledge of stoichiometry, that is, the ratios in which chemicals react, from which we apply appropriate conversion factors to arrive at the desired calculated results. In this chapter we review the fundamental concepts of mass, moles, and equivalents; the ways in which analytical results may be expressed for solids and liquids; and the principles of volumetric analysis and how stoichiometric relationships are used in titrations to calculate the mass of analyte.
Stoichiometry deals with the ratios in which chemicals react.
5.1 Review of the Fundamentals Quantitative analysis is based on a few fundamental atomic and molecular concepts, which we review below. You have undoubtedly been introduced to these in your general chemistry course, but we briefly review them here since they are so fundamental to quantitative calculations. THE BASICS: ATOMIC, MOLECULAR, AND FORMULA WEIGHTS The atomic weight for any element is the weight of a specified number of atoms of that element, and that number is the same from one element to another. A gram-atomic weight of any element contains exactly the same number of atoms of that element 149
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We will use formula weight (fw) to express grams per mole.
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as there are carbon atoms in exactly 12 g of carbon 12. This number is Avogadro’s number, 6.022 × 1023 , the number of atoms present in 1 g-at wt of any element.1 Since naturally occurring elements consist of mixtures of isotopes, the chemical atomic weights will be an average of the isotope weights of each element, taking into account their relative naturally occurring abundances. For example, bromine has two isotopes: 79 Br with atomic weight 78.981338 at a 50.69% relative abundance, and 81 Br with atomic weight 80.9162921 with a 49.31% relative abundance. These average to 79.904, the natural atomic weight we use in chemical calculations. Another measurement used by chemists is molecular weight (mw), defined as the sum of the atomic weights of the atoms that make up a compound. The term formula weight (fw) is a more accurate description for substances that don’t exist as molecules but exist as ionic compounds (strong electrolytes—acids, bases, salts). The term molar mass is sometimes used in place of formula weight.
WHAT IS A DALTON?
Theodore W. Richards, the first American Nobel Laureate in Chemistry, won his accolades to a large extent for his exact measurements. In particular, he is credited with the measurement of the exact atomic weight of chlorine, which he measured through the formation of AgCl.
There are 6.022 × 1023 atoms in a mole of atoms.
Biologists and biochemists sometimes use the unit dalton (Da) to report masses of large biomolecules and small biological entities such as chromosomes, ribosomes, viruses, and mitochondria, where the term molecular weight would be inappropriate. The mass of a single carbon-12 atom is equivalent to 12 daltons, and 1 dalton is therefore 1.661 × 10−24 g, the reciprocal of Avogadro’s number. The number of daltons in a single molecule is numerically equivalent to the molecular weight (g/mol). Strictly speaking, it is not correct to use the dalton as a unit of molecular weight, and it should be reserved for the types of substances mentioned above. For example, the mass of an Escherichia coli bacterium cell is about 1 × 10−12 g, or 6 × 1011 daltons.
MOLES: THE BASIC UNIT FOR EQUATING THINGS The chemist knows that atoms and molecules react in definite proportions. Unfortunately, he or she cannot conveniently count the number of atoms or molecules that participate in a reaction. But since the chemist has determined their relative masses, he or she can describe their reactions on the basis of the relative masses of atoms and molecules reacting, instead of the number of atoms and molecules reacting. For example, in the reaction Ag+ + Cl− → AgCl we know that one silver ion will combine with one chloride ion. We know further, since the atomic weight of silver is 107.870 and the atomic weight of chlorine is 35.453, that 107.870 mass units of the silver will combine with 35.453 mass units of chlorine. To simplify calculations, chemists have developed the concept of the mole, which is 1 There is a proposal to change the definition of Avogadro’s number in redefining the kilogram to be an invariant
unit. The kilogram is the only base unit in the International System of Units (SI) that is defined by a physical artifact rather than an unvarying physical property of nature, the others being the meter (length), second (time), ampere (electric current), kelvin (temperature), mole (amount of substance), and candela (light intensity). It is currently equal to the mass of a small cylinder of platinum-iridium alloy, known as the international prototype, that was ratified as the official kilogram in 1889, and is kept in a vault at the International Bureau of Weights & Measures near Paris. But it has inexplicably lost about 50 μg over time compared with copies. There are two proposals for an invariant definition of the kilogram, one based on Plank’s constant and one based on Avogadro’s constant, either of which would marginally change the value or definition of Avogadro’s number and hence the definition of the mole. But the change would be insignificant for the units we use. Details of the proposals may be found in “Avogadro’s Number Is Up . . . ,” P. J. Karol, Chem. & Eng. News, March 17 (2008) 48, “Redefining the Kilogram,” S. K. Ritter, Chem. & Eng. News, May 26 (2008) 43, and “Redefining the Kilogram and Mole,” P. F. Rusch, Chem. & Eng. News, May 30 (2011) 58: http://pubs.acs.org/isubscribe/journals/cen/89/i22/html/8922acscomment.html.
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Avogadro’s number (6.022 × 1023 ) of atoms, molecules, ions, or other species. Numerically, it is the atomic, molecular, or formula weight of a substance expressed in grams.2 Now, since a mole of any substance contains the same number of atoms or molecules as a mole of any other substance, atoms will react in the same mole ratio as their atom ratio in the reaction. In the above example, one silver ion reacts with one chloride ion, and so each mole of silver ion will react with one mole of chloride ion. (Each 107.87 g of silver will react with 35.453 g of chlorine.)
Example 5.1 Calculate the weight of one mole of CaSO4 · 7H2 O. Solution
One mole is the formula weight expressed in grams. The formula weight is Ca S 11 O 14 H
40.08 32.06 176.00 14.11 262.25 g/mol
The number of moles of a substance is calculated from Moles =
grams formula weight (g/mol)
(5.1)
where formula weight represents the atomic or molecular weight of the substance. Thus, g g = Moles Na2 SO4 = fw 142.04 g/mol g g Moles Ag+ = = fw 107.870 g/mol Since many experiments deal with very small quantities, a more convenient form of measurement is the millimole. The formula for calculating millimoles is Millimoles =
milligrams formula weight (mg/mmol)
(5.2)
Just as we can calculate the number of moles from the grams of material, we can likewise calculate the grams of material from the number of moles: g Na2 SO4 = moles × fw = moles × 142.04 g/mol g Ag = moles × fw = moles × 107.870 g/mol Again, we usually work with millimole quantities, so Milligrams = millimoles × formula weight (mg/mmol)
(5.3)
2 Actually, the term gram-atomic weight is more correct for atoms, gram-formula weight for ionic substances,
and gram-molecular weight for molecules, but we will use moles in a broad sense to include all substances. In place of gram-formula weight we will simply use formula weight (fw).
g/mol = mg/mmol = formula weight; g/L = mg/mL; mol/L = mmol/mL = molarity.
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Note that g/mol is the same as mg/mmol, g/L the same as mg/mL, and mol/L the same as mmol/mL.
Example 5.2 Calculate the number of moles in 500 mg Na2 WO4 (sodium tungstate). Solution
500 mg × 0.001 mol/mmol = 0.00170 mol 293.8 mg/mmol
Example 5.3 What is the weight, in milligrams, of 0.250 mmol Fe2 O3 (ferric oxide)? Solution
0.250 mmol × 159.7 mg/mmol = 39.9 mg
5.2 How Do We Express Concentrations of Solutions? Chemists express solution concentrations in a number of ways. Some are more useful than others in quantitative calculations. We will review here the common concentration units that chemists use. Their use in quantitative volumetric calculations is treated in more detail below. MOLARITY——THE MOST WIDELY USED The mole concept is useful in expressing concentrations of solutions, especially in analytical chemistry, where we need to know the volume ratios in which solutions of different materials will react. A one-molar solution is defined as one that contains one mole of substance in each liter of a solution. It is prepared by dissolving one mole of the substance in the solvent and diluting to a final volume of one liter in a volumetric flask; or a fraction or multiple of the mole may be dissolved and diluted to the corresponding fraction or multiple of a liter (e.g., 0.01 mol in 10 mL). More generally, the molarity of a solution is expressed as moles per liter or as millimoles per milliliter. Molar is abbreviated as M, and we talk of the molarity of a solution when we speak of its concentration. A one-molar solution of silver nitrate and a one-molar solution of sodium chloride will react on an equal-volume basis, since they react in a 1:1 ratio: Ag+ + Cl− → AgCl. We can be more general and calculate the moles of substance in any volume of the solution. Moles = (moles/liter) × liters
(5.4)
= molarity × liters We often work with millimoles in analytical chemistry. Remember this formula!
The liter is an impractical unit for the relatively small quantities encountered in titrations, and we normally work with milliliters. This is what your buret reads. So, Millimoles = molarity × milliliters (or mmol = M × mL)
(5.5)
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Example 5.4 A solution is prepared by dissolving 1.26 g AgNO3 in a 250-mL volumetric flask and diluting to volume. Calculate the molarity of the silver nitrate solution. How many millimoles AgNO3 were dissolved? Solution
M= Then,
1.26 g/169.9 g/mol = 0.0297 mol/L (or 0.0297 mmol/mL) 0.250 L
Millimoles = (0.0297 mmol/mL)(250 mL) = 7.42 mmol
Always remember that the units in a calculation must combine to give the proper units in the answer. Thus, in this example, grams cancel to leave the proper unit, moles/liter, or molarity. Using units in the calculation to check if the final units are proper is called dimensional analysis. Accurate use of dimensional analysis is essential to properly setting up computations.
Example 5.5 How many grams per milliliter of NaCl are contained in a 0.250 M solution? Solution
0.250 mol/L = 0.250 mmol/mL 0.250 mmol/mL × 58.4 mg/mmol × 0.001 g/mg = 0.0146 g/mL
Example 5.6 How many grams Na2 SO4 should be weighed out to prepare 500 mL of a 0.100 M solution? Solution
500 mL × 0.100 mmol/mL = 50.0 mmol 50.0 mmol × 142 mg/mmol × 0.001 g/mg = 7.10 g
Example 5.7 Calculate the concentration of potassium ion in grams per liter after mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2 SO4 . Solution
mmol K+ = mmol KCl + 2 × mmol K2 SO4 = 100 mL × 0.250 mmol/mL +2 × 200 mL × 0.100 mmol/mL
Always use dimensional analysis to set up a calculation properly. Don’t just memorize a formula.
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= 65.0 mmol in 300 mL 65.0 mmol × 39.1 mg/mmol × 0.001 g/mg × 1000 mL/L = 8.47 g/L 300 mL
NORMALITY The equivalent weight (or the number of reacting units) depends on the chemical reaction. It may vary most often in redox reactions, when different products are obtained.
Although molarity is widely used in chemistry, some chemists use a unit of concentration in quantitative analysis called normality (N). A one-normal solution contains one equivalent per liter. An equivalent represents the mass of material providing Avogadro’s number of reacting units. A reacting unit is a proton or an electron. The number of equivalents is given by the number of moles multiplied by the number of reacting units per molecule or atom; the equivalent weight is the formula weight divided by the number of reacting units. Table 5.1 lists the reacting units used for different types of reactions. For acids and bases, the number of reacting units is based on the number of protons (i.e., hydrogen ions) an acid will furnish or a base will react with. For oxidation–reduction reactions it is based on the number of electrons an oxidizing or reducing agent will take on or supply. Thus, for example, sulfuric acid, H2 SO4 , has two reacting units of protons; that is, there are two equivalents of protons in each mole. Therefore, Equivalent weight =
Equivalent weight g/eq = mg/meq; eq/L = meq/mL = normality.
98.08 g/mol = 49.04 g/eq 2 eq/mol
So, the normality of a sulfuric acid solution is twice its molarity, that is, N = (g/eq wt)/L. The number of equivalents is given by Number of equivalents (eq) =
wt (g) = normality (eq/L) × volume (L) eq wt (g/eq) (5.6)
Just as we ordinarilly use millimoles (mmol) instead of moles, we typically use milliequivalents (meq) instead of equivalents meq =
There is no ambiguity in a molar concentration. Instead of normality, eq/L is often presently used.
mg = normality (meq/mL) × mL eq wt (mg/meq)
(5.7)
In clinical chemistry, equivalents are frequently defined in terms of the number of charges on an ion rather than on the number of reacting units. Thus, for example, the equivalent weight of Ca2+ is one-half its atomic weight, and the number of equivalents is twice the number of moles. This use is convenient for electroneutrality calculations. We discuss equivalents in more detail in Section 5.3. While normality has been used extensively in the past and is found in the scientific literature, it is not as widely used today as molarity. We discuss normality in
Table 5.1
Reacting Units in Different Reactions Reaction Type Acid–base Oxidation–reduction
Reacting Unit H+ Electron
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the book’s website for those who do make use of it. We will use moles and molarity throughout most of this text so there will be no ambiguity about what the concentration represents. Molarity calculations require a knowledge of the stoichiometry of reactions, that is, the ratio in which substances react. The journal Analytical Chemistry does not allow normality in articles it publishes, but other publications do. The unit eq/L is the same as normality and is accepted by most publications. FORMALITY— —INSTEAD OF MOLARITY Chemists sometimes use the term formality for solutions of ionic salts that do not exist as molecules in the solid or in solution. The concentration is given as formal (F). Operationally, formality is identical to molarity: The former is sometimes reserved for describing makeup concentrations of solutions (i.e., total analytical concentration), and the latter for equilibrium concentrations. For convenience, we shall use molarity exclusively, a common practice.
Formality is numerically the same as molarity.
MOLALITY——THE TEMPERATURE-INDEPENDENT CONCENTRATION In addition to molarity and normality, another useful concentration unit is molality, m. A one-molal solution contains one mole per 1000 g of solvent. The molal concentration is convenient in physicochemical measurements of the colligative properties of substances, such as freezing point depression, vapor pressure lowering, and osmotic pressure because colligative properties depend solely on the number of solute particles present in solution per mole of solvent. Molal concentrations are not temperature dependent as molar and normal concentrations are (since the solution volume in molar and normal concentrations is temperature dependent). DENSITY CALCULATIONS——HOW DO WE CONVERT TO MOLARITY? The concentrations of many fairly concentrated commercial acids and bases are usually given in terms of percent by weight. It is frequently necessary to prepare solutions of a given approximate molarity from these substances. In order to do so, we must know the density in order to calculate the molarity. Density is the weight per unit volume at the specified temperature, usually g/mL or g/cm3 at 20◦ C. (One milliliter is the volume occupied by 1 cm3 .) Sometimes substances list specific gravity rather than density. Specific gravity is defined as the ratio of the mass of a body (e.g., a solution), usually at 20◦ C, to the mass of an equal volume of water at 4◦ C (or sometimes 20◦ C). That is, specific gravity is the ratio of the densities of the two substances; it is a dimensionless quantity. Since the density of water at 4◦ C is 1.00000 g/mL, density and specific gravity are equal when referred to water at 4◦ C. But normally specific gravity is referred to water at 20◦ C; density is equal to specific gravity × 0.99821 (the density of water is 0.99821 g/mL at 20◦ C).
◦
Density of solution at 20 C = Specific gravity of solution × 0.99821 g/mL
Note that the density of the solution at a temperature other than 20◦ C cannot be precisely computed from the specific gravity specified for 20◦ C without knowing the volumetric expansion behavior of the solution, which is not the same as that of water.
Molality does not change with temperature.
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Example 5.8 How many milliliters of concentrated sulfuric acid, 94.0% (g/100 g solution), density 1.831 g/cm3 , are required to prepare 1 liter of a 0.100 M solution? Solution
Mass is conserved: M1 V1 = M2 V2 If a solution of molarity M1 and volume V1 is diluted to V2 , the molarity M2 will obey this relationship. The general equation C1 V1 = C2 V2 will hold as long as C1 and C2 are in the same units, regardless of the specific unit. Memorize this equation.
Consider 1 cm3 = 1 mL. The concentrated acid contains 0.940 g H2 SO4 per gram of solution, and the solution weighs 1.831 g/mL. The product of these two numbers, then, gives the gram H2 SO4 per milliliter of solution: M=
(0.940 g H2 SO4 /g solution)(1.831 g/mL) × 1000 mL/L 98.1 g/mol
= 17.5 mol H2 SO4 /L solution We must dilute this solution to prepare 1 liter of a 0.100 M solution. The same number of millimoles of H2 SO4 must be taken as will be contained in the final solution. Since mmol = M × mL and mmol dilute acid = mmol concentrated acid, 0.100 M × 1000 mL = 17.5 M × mL x = 5.71 mL concentrated acid to be diluted to 1000 mL
See Sections 5.5 and the text website for volumetric calculations using molarity (or normality).
Molarity and normality are the most useful concentrations in quantitative analysis. Calculations using these for volumetric analysis are discussed in more detail below.
The analytical concentration represents the concentration of total dissolved substance, i.e., the sum of all species of the substance in solution = CX .
Analytical chemists prepare solutions of known analytical concentrations, but the dissolved substances may partially or totally dissociate to give equilibrium concentrations of different species. Acetic acid, for example, is a weak acid that dissociates a few percent depending on the concentration,
An equilibrium concentration is that of a given dissolved form of the substance = [X].
to give equilibrium amounts of the proton and the acetate ion. The more dilute the solutions, the greater the dissociation. We often use these equilibrium concentrations in calculations involving equilibrium constants (Chapter 6), usually using molarity concentrations. The analytical molarity is given by the notation CX , while equilibrium molarity is given by [X]. A solution of 1 M CaCl2 (analytical molarity) is completely ionized into constituent ions in solution and gives at equilibrium, 0 M CaCl2 , 1 M Ca2+ , and 2 M Cl− (equilibrium molarities). Hence, we say the solution is 1 M in Ca2+ and 2 M in Cl− .
ANALYTICAL AND EQUILIBRIUM CONCENTRATIONS— —THEY ARE NOT THE SAME
HOAc H+ + OAc−
DILUTIONS——PREPARING THE RIGHT CONCENTRATION The millimoles taken for dilution will be the same as the millimoles in the diluted solution, i.e., Mstock × mLstock = Mdiluted × mLdiluted
We often must prepare dilute solutions from more concentrated stock solutions. For example, we may prepare a dilute HCl solution from concentrated HCl to be used for titrations (following standardization). Or, we may have a stock standard solution from which we wish to prepare a series of more dilute standards. The millimoles of stock solution taken for dilution will be identical to the millimoles in the final diluted solution, remember, C1 V1 = C2 V2 .
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Example 5.9 You wish to prepare a calibration curve for the spectrophotometric determination of permanganate. You have a stock 0.100 M solution of KMnO4 and a series of 100-mL volumetric flasks. What volumes of the stock solution will you have to pipet into the flasks to prepare standards of 1.00, 2.00, 5.00, and 10.0 × 10−3 M KMnO4 solutions? Solution
x mL of the stock solution of 0.100 M concentration will be diluted to 100 mL of some specified concentration, say C2 . Remembering C1 V1 = C2 V2 Let us do this for the first concentration, C2 = 1.00 × 10−3 . Here V1 is x, C1 = 0.100 M and V2 = 100 mL 0.100 M × x mL = 1.00 × 10−3 M × 100 mL x = 1.00 mL Similarly, for the other solutions we will need 2.00, 5.00, and 10.0 mL of the stock solution, which will be diluted to 100 mL.
Example 5.10 You are analyzing for the manganese content in an ore sample by dissolving it and oxidizing the manganese to permanganate for spectrophotometric measurement. The ore contains about 5% Mn. A 5-g sample is dissolved and diluted to 100 mL, following the oxidation step. By how much must the solution be diluted to be in the range of the calibration curve prepared in Example 5.9, that is, about 3 × 10−3 M permanganate? Solution
The solution contains 0.05 × 5-g sample = 0.25 g Mn. This corresponds to [0.25 g/ (55 g Mn/mol)]/100 mL = 4.5 × 10−3 mol MnO4 − /100 mL = 4.5 × 10−2 M. For 3 × 10−3 M, we must dilute it by 4.5 × 10−2 /3 × 10−3 = 15-fold. If we have a 100-mL volumetric flask, using C1 V1 = C2 V2 , 4.5 × 10−2 M × x mL = 3 × 10−3 M × 100 mL x = 6.7 mL needed for dilution to 100 mL Since we need to pipet accurately, we could probably take an accurate 10-mL aliquot, which would give about 4.5 × 10−3 M permanganate for measurement. MORE DILUTION CALCULATIONS We can use the relationship C1 V1 = C2 V2 to calculate the dilution required to prepare a certain concentration of a solution from a more concentrated solution. For example, if we wish to prepare 500 mL of a 0.100 M solution by diluting a more concentrated solution, we can calculate it from this relationship.
Example 5.11 You wish to prepare 500 mL of a 0.100 M K2 Cr2 O7 solution from a 0.250 M solution. What volume of the 0.250 M solution must be diluted to 500 mL?
Remember, the millimoles before and after dilution are the same. See Section 5.5 (and the text’s website) for volumetric calculations using molarity (and normality).
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Solution
Mfinal × mLfinal = Moriginal × mLoriginal 0.100 mmol/mL × 500 mL = 0.250 mmol/mL × mLoriginal mLoriginal = 200 mL
Example 5.12 What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 M NaOH to give a solution 0.50 M in OH− ? Solution
Volumes of dilute aqueous solutions can be assumed to be additive, i.e., if x mL of Ba(OH)2 is added to 50 mL NaOH, the total volume is going to be 50 + x mL. Wex can use a modified form of C1 V1 = C2 V2 where all the initial solution components are added in this manner and these sum up to the final solution components: Cin Vin = Cfin Vfin In the present case, MNaOH VNaOH + 2 × MBa(OH)2 VBa(OH)2 = MOH− × Vfin , note that 1 M Ba(OH)2 is 2 M in OH− . Thus 0.30 M × 50 mL + 2 × 0.40 M × xmL = 0.50 M × (50 + x) mL. Solving, x = 33 mL. Alternatively, Let x = mL Ba(OH)2 . The final volume is (50 + x) mL. mmol OH− = mmol NaOH + 2 × mmol Ba(OH)2 0.50 M × (50 + x)mL = 0.30 M NaOH × 50 mL + 2 × 0.40 M Ba(OH)2 × x mL x = 33 mL Ba(OH)2
Often, the analyst is confronted with serial dilutions of a sample or standard solution. Again, obtaining the final concentration simply requires keeping track of the number of millimoles and the volumes.
Example 5.13 You are to determine the concentration of iron in a sample by spectrophotometry by reacting Fe2+ with 1,10-phenanthroline to form an orange-colored complex. This requires preparation of a series of standards against which to compare absorbances or color intensities (i.e., to prepare a calibration curve). A stock standard solution of 1.000 × 10−3 M iron is prepared from ferrous ammonium sulfate. Working standards A and B are prepared by adding with pipets 2.000 and 1.000 mL, respectively, of this solution to 100-mL volumetric flasks and diluting to volume. Working standards C, D, and E are prepared by adding 20.00, 10.00, and 5.000 mL of working standard A to 100-mL volumetric flasks and diluting to volume. What are the concentrations of the prepared working solutions?
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Solution
Solution A: Mstock × mLstock = MA × mLA (1.000 × 10−3 M)(2.000 mL) = MA × 100.0 mL MA = 2.000 × 10−5 M Solution B: (1.000 × 10−3 M)(1.000 mL) = MB × 100.0 mL MB = 1.000 × 10−5 M Solution C: MA × mLA = MC × mLC (2.000 × 10−5 M)(20.00 mL) = MC × 100.0 mL MC = 4.000 × 10−6 M Solution D: (2.000 × 10−5 M)(10.00 mL) = MD × 100.0 mL MD = 2.000 × 10−6 M Solution E: (2.000 × 10−5 M)(5.000 mL) = ME × 100.0 mL ME = 1.000 × 10−6 M
The above calculations apply to all types of reactions, including acid–base, redox, precipitation, and complexometric reactions. The primary requirement before making calculations is to know the ratio in which the substances react, that is, start with a balanced reaction. Solution preparation procedures in the chemical literature often call for the dilution of concentrated stock solutions, and authors may use different terms. For example, a procedure may call for 1 + 9 dilution (solute + solvent) of sulfuric acid. In some cases, a 1:10 dilution (original volume:final volume) may be indicated. The first procedure calls for diluting a concentrated solution to 1/10th of its original concentration by adding 1 part to 9 parts of solvent; the second procedure by diluting to 10 times the original volume. The first procedure does not give an exact 10-fold dilution because volumes are not completely additive, except when all components are dilute aqueous solutions, whereas the second procedure does (e.g., adding 10 mL with a pipet to a 100-mL volumetric flask and diluting to volume—fill the flask partially with water before adding sulfuric acid!). The solute + solvent approach is fine for reagents whose concentrations need not be known accurately.
The solute + solvent method of dilution should not be used for quantitative dilutions. Added volumes are not completely additive, especially mixed solvents. Water and ethanol always have negative excess volumes when mixed, indicating the partial molar volume of each component is less when mixed than its molar volume when pure. That is, volume of pure alcohol plus volume of water does not equal the volume of vodka! Y = yotta = 1024 Z = zetta = 1021 E = exa = 1018 P = peta = 1015 T = tera = 1012
5.3 Expressions of Analytical Results—So Many Ways We can report the results of analysis in many ways, and the beginning analytical chemist should be familiar with some of the common expressions and units of measure employed. Results will nearly always be reported as concentration, on either a weight or a volume basis: the quantity of analyte per unit weight or per volume of sample. The units used for the analyte will vary. We shall first review the common units of weight and volume in the metric system and then describe methods of expressing results. The gram (g) is the basic unit of mass and is the unit employed most often in macro analyses. For small samples or trace constituents, chemists use smaller units. The milligram (mg) is 10−3 g, the microgram (μg) is 10−6 g, and the nanogram (ng) is 10−9 g. The basic unit of volume is the liter (L). The milliliter (mL) is 10−3 L and is used commonly in volumetric analysis. The microliter (μL) is 10−6 L (10−3 mL), and the nanoliter (nL) is 10−9 L (10−6 mL). (Prefixes for even smaller quantities include pico for 10−12 and femto for 10−15 .)
G = giga = 109 M = mega = 106 k = kilo = 103 d = deci = 10−1 c = centi = 10−2 m = milli = 10−3 μ = micro = 10−6 n = nano = 10−9 p = pico = 10−12 f = femto = 10−15 a = atto = 10−18 z = zepto = 10−21 y = yocto = 10−24
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SOLID SAMPLES
Mass and weight are really different. See Chapter 2. We deal with masses but will use mass and weight interchangeably.
1 ppt (thousand) = 1000 ppm = 1, 000, 000 ppb; 1 ppm = 1000 ppb = 1, 000, 000 ppt (trillion). Usually ppt refers to parts per trillion, but in some cases it could be used as parts per thousand. Take note of the units when you see this!
ppt = mg/g = g/kg ppm = μg/g = mg/kg ppb = ng/g = μg/kg
Calculations for solid samples are based on weight.3 The most common way of expressing the results of macro determinations is to give the weight of analyte as a percent of the weight of sample (weight/weight basis). The weight units of analyte and sample are the same. For example, a limestone sample weighing 1.267 g and containing 0.3684 g iron would contain 0.3684 g × 100% = 29.08% Fe 1.267 g The general formula for calculating percent on a weight/weight basis, which is the same as parts per hundred, then is wt solute (g) × 102 (%/g solute/g sample) %(wt/wt) = (5.8) wt sample (g) It is important to note that in such calculations, grams of solute do not cancel with grams of sample solution; the fraction represents grams of solute per gram of sample. Multiplication of the above by 102 converts to grams of solute per 100 g of sample. Since the conversion factors for converting weight of solute and weight of sample (weights expressed in any units) to grams of solute and grams of sample are always the same, the conversion factors will always cancel. Thus, we can use any weight in the definition. Trace concentrations are usually given in smaller units, such as parts per thousand (ppt, ‰), parts per million (ppm), or parts per billion (ppb). These are calculated in a manner similar to parts per hundred (%): wt solute (g) × 103 (ppt/g solute/g sample) ppt (wt/wt) = wt sample (g) wt solute (g) × 106 (ppm/g solute/g sample) ppm (wt/wt) = wt sample (g) wt solute (g) × 109 (ppb/g solute/g sample) ppb (wt/wt) = wt sample (g)
(5.9) (5.10) (5.11)
You can use any weight units in your calculations so long as both analyte and sample weights are in the same units. Parts per trillion (parts per 1012 parts) is also abbreviated ppt, so be careful to define which one you mean. Some authors like to use ppth to denote parts per thousand and pptr to denote parts per trillion. In the above example, we have 29.08 parts per hundred of iron in the sample, or 290.8 parts per thousand and 290,800 parts per million (290,800 g of iron per 1 million grams of sample, 290,800 lb of iron per 1 million pounds of sample, etc.). Working backward, 1 ppm corresponds to 0.0001 part per hundred, or 10−4 %. Table 5.2 summarizes the concentration relationships for ppm and ppb. Note that ppm is simply mg/kg or μg/g and that ppb is μg/kg, or ng/g. Trace gas concentrations are also expressed in ppb, ppm, and so forth. In this case the ratio refers not to mass ratios, but to volume ratios (which for gases is the same as mole ratios). Thus, present atmospheric CO2 concentration of 390 ppm means that each liter of air (this is a million microliters) contains 390 microliters of CO2 . 3 They are really based on mass, but the term weight is commonly used. See Chapter 2 for a description and determination of mass and weight.
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Table 5.2
Common Units for Expressing Trace Concentrations Unit Parts per million (1 ppm = 10−4 %) Parts per billion (1 ppb = 10−7 % = 10−3 ppm) Milligram percent a pL
Abbreviation
wt/wt
wt/vol
vol/vol
ppm
mg/kg μg/g μg/kg ng/g mg/100 g
mg/L μg/mL μg/L ng/mL mg/100 mL
μL/L nL/mL nL/L pL/mLa
ppb mg%
= picoliter = 10−12 L.
Sometimes for this reason, these concentrations are written as ppmv or ppbv, and so on, indicating “by volume”.
Example 5.14 A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 μg zinc. What is the concentration of zinc in the plant in ppm? In ppb? Solution
3.6 μg = 1.4 μg/g ≡ 1.4 ppm 2.6 g 3.6 × 103 ng = 1.4 × 103 ng/g ≡ 1400 ppb 2.6 g One ppm is equal to 1000 ppb. One ppb is equal to 10−7 %.
Clinical chemists sometimes prefer to use the unit milligram percent (mg%) rather than ppm for small concentrations. This is defined as milligrams of analyte per 100 g of sample. The sample in Example 5.14 would then contain (3.6 × 10−3 mg/2.6 g) × 100 mg% = 0.14 mg% zinc.
Example 5.15 Concentrations of Gases and Particles in Air The current National Ambient Air Quality Standards for the seven criteria pollutants listed by the U.S. Environmental Protection Agency is listed below from http://www.epa.gov/air/criteria.html. Other than lead and particulate matter, all others are gases. For particulate matter, wt/vol (μg/m3 ) units are used; m3 (equal to 1000 L) rather than μg/L. The gas concentrations are expressed by ppm(v) or ppb(v). Concentrations of CO are also given in mg/m3 as CO largely comes from automotive exhaust and mass of CO emitted per vehicle-mile driven is often of interest. (a) Show how 35 ppm CO is 40 mg/m3 . (b) What is 75 ppb SO2 at 25◦ C in μg/m3 ?
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National Ambient Air Quality Standards Primary Standards Secondary Standards Level Averaging Time Level Averaging Time 9 ppm (10 mg/m3 ) 8-hour (1) None 35 ppm (40 mg/m3 ) 1-hour (1) None 3 (2) Lead 0.15 μg/m Rolling 3-Month Average Same as Primary 1.5 μg/m3 Quarterly Average Same as Primary Nitrogen Dioxide 53 ppb (3) Annual (Arithmetic Average) Same as Primary (4) 100 ppb 1-hour None Particulate Matter (PM10 ) 150 μg/m3 24-hour (5) Same as Primary Particulate Matter (PM2.5 ) 15.0 μg/m3 Annual (6) (Arithmetic Average) Same as Primary 35 μg/m3 24-hour (7) Same as Primary (8) Ozone 0.075 ppm (2008 std) 8-hour Same as Primary 0.08 ppm (1997 std) 8-hour (9) Same as Primary 0.12 ppm 1-hour (10) Same as Primary Sulfur Dioxide 0.03 ppm Annual (Arithmetic Average) 0.5 ppm 3-hour (1) (1) 0.14 ppm 24-hour None 75 ppb (11) 1-hour None Pollutant Carbon Monoxide
Solution
Because gas volumes change as a function of temperature and pressure, we must refer to some temperature and pressure. When this is not specified, we assume a temperature of 25◦ C and a pressure of 1 atm. The ideal gas laws (PV = RT, where R is the universal gas constant, 0.0821 L-atm/(mole K)) dictate that the volume of 1 mole of any gas at 1 atm pressure and 25◦ C (298.15 K) is 24.5 L (at 0 ◦ C, this is 22.4 L). (a) 35 ppm is 35 μmol CO per 1 mole air. Since FW of CO is 28, we can write this as 35 μmole × 28 μg/μmole = 980 μg CO in 24.5 L air. 1 mg 1000 L 980 μg × × = 40 mg/m3 24.5 L 1000 μg 1 m3 (b) 75 ppb SO2 is 75 nmol SO2 in 24.5 L air. The FW of SO2 is 64, so 1 μmol 64 μg 1000 L 75 nmol × × × = 196 μg/m3 ; round to 0.20 mg/m3 . 24.5 L 1000 nmol μmol 1 m3
LIQUID SAMPLES A deciliter is 0.1 L or 100 mL.
You can report results for liquid samples on a weight/weight basis, as above, or they may be reported on a weight/volume basis. The latter is more common, at least in the clinical laboratory. The calculations are similar to those above. Percent on a weight/volume basis is equal to grams of analyte per 100 mL of sample, while mg% is equal to milligrams of analyte per 100 mL of sample. This latter unit is often used by clinical chemists for biological fluids, and their accepted terminology is milligrams per deciliter (mg/dL) to distinguish from mg% on a weight/weight basis. Whenever a concentration is expressed as a percentage, it should be clearly specified whether this is wt/vol or wt/wt. In all but dilute aqueous solutions, this distinction is important. In dilute aqueous solutions, wt/vol and wt/wt ratios are numerically the same because the density of water is unity (1 mL = 1 g) for all practical purposes. Parts per million, parts per billion, and parts per trillion can also be expressed on a weight/volume basis; ppm is calculated from mg/L or μg/mL; ppb is calculated from μg/L or ng/mL;
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and ppt is calculated from pg/mL or ng/L. Alternatively, the following fundamental calculations may be used: wt solute (g) × 102 (%/g solute/mL sample) %(wt/vol) = (5.12) vol sample (mL) wt solute (g) × 106 (ppm/g solute/mL sample) (5.13) ppm (wt/vol) = vol sample (mL) wt solute (g) × 109 (ppb/g solute/mL sample) (5.14) ppb (wt/vol) = vol sample (mL) wt solute (g) × 1012 (ppt/g solute/mL sample) ppt (wt/vol) = (5.15) vol sample (mL)
In dilute aqueous solution ppm = μg/mL = mg/L ppb = ng/mL = μg/L ppt = pg/mL = ng/L
Note that % (wt/vol) is not pounds/100 gal of solution; the units must be expressed in grams of solute and milliliters of solution. To avoid ambiguities, increasingly it is recommended that ppm, ppb, and ppt units are not used to describe solution phase concentrations; most journals require that μg/mL or ng/mL units should be used instead.
Example 5.16 A 25.0-μL serum sample was analyzed for glucose content and found to contain 26.7 μg. Calculate the concentration of glucose in μg/mL and in mg/dL. Solution
1 mL = 2.50 × 10−2 mL 1000 μL 1g μg 26.7 = 2.67 × 10−5 g × 5 10 μg
× 25.0 μL
Glucose Concentration =
2.67 × 10−5 g glucose × 106 μg/g = 1.07 × 103 μg/mL 2.50 × 10−2 mL serum
This is numerically the same in ppm units. Also, 0.001 mg 100 mL μg × × Glucose Concentration =1.07 × 103 mL 1 μg 1 dL =107 mg/dL [Note the relationship: 10 ppm (wt/vol) = 1 mg/dL]
What does 1 μg/mL, often called 1 ppm, represent in terms of moles per liter? It depends on the formula weight. Let’s do some actual conversions using real formula weights. We begin with a solution that contains 2.5 μg/mL benzene. The formula weight (C6 H6 ) is 78.1. The concentration in moles per liter is (2.5 × 10−3 g/L)/(78 g/mol) = 3.8 × 10−5 M Another solution contains 5.8 × 10−8 M lead. The concentration in parts per billion is (5.8 × 10−8 mol/L)(207 g/mol) = 1.20 × 10−5 g/L. For parts per billion (μg/L), then (1.20 × 10−5 g/L) × (106 μg/g) = 1.20 × 101 μg/L or 12 ppb. A drinking water sample that contains 350 pg/L of carbon tetrachloride has a concentration in ng/L
The relationship between μg/mL and molarity (M) units depends on the formula weight.
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of (350 × 10−12 g/L)/(109 ng/g) = 350 × 10−3 ng/L = 0.35 ng/L or 0.35 ppt. The molar concentration is (350 × 10−12 g/L)/(154 g/mol) = 2.3 × 10−12 M. (Chlorinetreated water may contain traces of chlorinated hydrocarbons—this is very low.) A key point to remember is that solutions that have the same numerical concentrations on a weight/weight or weight/volume basis do not have the same number of molecules4 , but solutions of the same molarity do.
Example 5.17 (a) Calculate the molar concentrations of 1 mg/L (1.00 ppm) solutions each of Li+ and Pb2+ . (b) What weight of Pb(NO3 )2 will have to be dissolved in 1 liter of water to prepare a 100 mg/L (100 ppm) Pb2+ solution? Solution
(a) Li concentration = 1.00 mg/L Pb concentration = 1.00 mg/L MLi =
1.00 mg Li/L × 10−3 g/mg = 1.44 × 10−4 mol/L Li 6.94 g Li/mol
MPb =
1.00 mg Pb/L × 10−3 g/mg = 4.83 × 10−6 mol/L Pb 207 g Pb/mol
Because lead is much heavier than lithium, a given weight contains a smaller number of moles and its molar concentration is less. (b) If 1 ppm Pb = 4.83 × 10−6 mol/L Pb 100 ppm Pb = 4.83 × 10−4 mol/L Pb Therefore, we need 4.83 × 10−4 mol Pb(NO3 )2 . 4.83 × 10−4 mol × 283.2 g Pb(NO3 )2 /mol = 0.137 g Pb(NO3 )2 For dilute aqueous solutions, wt/wt ≈ wt/vol because the density of water is near 1.000 mg/ml.
Alcohol in wine and liquor is expressed as vol/vol (200 proof = 100%vol/vol). Since the specific gravity of alcohol is 0.8, wt/vol concentration = 0.8 × (vol/vol) = 0.4 × proof
The concentration units wt/wt and wt/vol are related through the density. They are numerically the same for dilute aqueous solutions as the density is 1 g/mL. If the analyte is a liquid dissolved in another liquid, the results may be expressed on a volume:volume basis, but you will likely encounter this only in rare situations. One exception is specification of eluents in liquid chromatography, 40:60 methanol:water connotes that 40 volumes of methanol added to 60 volumes of water; the final volume is generally not available in such specifications. On the other hand, in a vol/vol designation, the first volume refers to that of the solute and the second to that of the solution; this unit is commonly used in the alcoholic beverage industry to specify ethanol content. You would handle the calculations in the same manner as those above, using the same volume units for solute and solution. As illustrated in Example 5.15, gas concentrations may be reported on a weight/volume, volume/ volume, and rarely, on a weight/weight basis. It is always best to specify clearly what is meant. In the absence of clear labels, it is best to assume that solids are being reported wt/wt, gases vol/vol, and liquids may be reported wt/wt (concentrated acid and base reagents), wt/vol (most dilute aqueous solutions), or vol/vol (the U.S. alcoholic beverage industry). Clinical chemists frequently prefer to use a unit other than weight for expressing the amount of major electrolytes in biological fluids (Na+ , K+ , Ca2+ , Mg2+ , 4 Unless they have the same formula weight.
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Table 5.3
Major Electrolyte Composition of Normal Human Plasmaa Cations
meq/L
Anions
meq/L
Na+ K+ Ca2+ Mg2+
143 4.5 5 2.5
Total
155
Cl− HCO3 − Protein HPO4 − SO4 2− Organic acids Total
104 29 16 2 1 3 155
a Reproduced
At blood pH of 7.4, the phosphate actually exists primarily as a mixture of mono- and dihydrogen phosphate.
from Joseph S. Annino, Clinical Chemistry, 3rd ed., by Boston: Little,
Brown, 1964.
Cl− , H2 PO4 − , etc.). This is the unit milliequivalent (meq). In this context, milliequivalent is defined as the number of millimoles of analyte multiplied by the charge on the analyte ion. Results are generally reported as meq/L. This concept gives an overall view of the electrolyte balance. The physician can tell at a glance if total electrolyte concentration has increased or decreased markedly. Obviously, the milliequivalents of cations will be equal to the milliequivalents of anions. One mole of a monovalent (+1) cation (1 eq) and half a mole of a divalent (−2) anion (1 eq) have the same number of positive and negative charges (one mole each). As an example of electrolyte or charge balance, Table 5.3 summarizes the averages of major electrolyte compositions normally present in human blood plasma and urine. Chapter 25 discusses the ranges and physiological significant of some chemical constituents of the human body. We can calculate the milliequivalents of a substance from its weight in milligrams simply as follows (similar to how we calculate millimoles): meq =
mg mg = eq wt (mg/meq) fw (mg/mmol)/n (meq/mmol)
(5.16)
n = charge on ion The equivalent weight of Na+ is 23.0 (mg/mmol)/1 (meq/mmol) = 23.0 mg/meq. The equivalent weight of Ca2+ is 40.1 (mg/mmol)/2 (meq/mmol) = 20.0 mg/meq.
Example 5.18 The concentration of zinc ion in blood serum is about 1 mg/L. Express this as meq/L. Solution
The equivalent weight of Zn2+ is 65.4 (mg/mmol)/2 (meq/mmol) = 32.7 mg/meq. Therefore, 1 mg Zn/L = 3.06 × 10−2 meq/L Zn 32.7 mg/meq This unit is more often used for the major electrolyte constituents as in Table 5.3 rather than the trace constituents, as in the example here.
The equivalents of cations and anions in any solution must be equal.
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REPORTING CONCENTRATIONS AS DIFFERENT CHEMICAL SPECIES We may express results in any form of the analyte. This is often done to facilitate the interpretation by other professionals.
Water hardness due to calcium ion is expressed as ppm CaCO3 . Conveniently, the fw of CaCO3 is 100. Converting from ppm CaCO3 to molar units is very simple!
Thus far, we have implied that the analyte is determined in the form it exists or for which we want to report the results. However, this is often not true. In the determination of the iron content of an ore, for example, we may measure the iron in the form of Fe2 O3 and then report it as % Fe. Or we may determine the iron in the form of Fe2+ (e.g., by titration) and report it as % Fe2 O3 . This is perfectly proper so long as we know the relationship between what is really being measured and the form it is to be reported in. We may actually determine the calcium content of water, for example, but we may wish to report it as parts per million (mg/L) of CaCO3 (this is the typical way of expressing water hardness). We know that each gram of Ca2+ is equivalent to (or could be converted to) grams of CaCO3 by multiplying grams Ca by fw CaCO3 /fw Ca2+ . That is, multiplying the milligrams of Ca2+ determined by 100.09/40.08 will give us the equivalent number of milligrams of CaCO3 . The calcium does not have to exist in this form (we may not even know in what form it actually exists); we simply have calculated the weight that could exist and will report the result as if it did. Specific operations necessary for calculating the weight of the desired constituent will be described below. At this point we should mention some of the different weight criteria used for expressing results with biological tissues and solids. The sample may be weighed in one of three physical forms; wet, dry, or ashed. This can apply also to fluids, although fluid volume taken is usually used for the analysis. The wet weight is taken on the fresh, untreated sample. The dry weight is taken after the sample has been dried by heating, desiccation, or freeze-drying. If the test substance is unstable to heat, the sample should not be dried by heating. The weight of the ash residue after the organic matter has been burned off is sometimes used as the weight. This can obviously be used only for mineral (inorganic) analysis.
5.4 Volumetric Analysis: How Do We Make Stoichiometric Calculations? Volumetric or titrimetric analyses are among the most useful and accurate analytical techniques, especially for millimole amounts of analyte. They are rapid and can be automated, and they can be applied to smaller amounts of analyte when combined with a sensitive instrumental technique for detecting the completion of the titration reaction, for example, pH measurement. Other than pedagogic purposes, manual titrations nowadays are generally used only in situations that require high accuracy for relatively small numbers of samples. They are used, for example, to calibrate or validate more routine instrumental methods. Automated titrations are useful when large numbers of samples must be processed. (A titration may be automated, for instance, by means of a color change or a pH change that activates a motor-driven buret to stop delivery. The volume delivered may be electronically registered. (Automatic titrators are discussed in Chapter 14.) Below, we describe the types of titrations that can be performed and the applicable principles, including the requirements of a titration and of standard solutions. The volumetric relationship described earlier in this chapter may be used for calculating quantitative information about the titrated analyte. Volumetric calculations are given in Section 5.5. TITRATION— —WHAT ARE THE REQUIREMENTS? We calculate the moles of analyte titrated from the moles of titrant added and the ratio in which they react.
In a titration, the test substance (analyte) reacts with an added reagent of known concentration, generally instantaneously. The reagent of known concentration is referred to as a standard solution. It is typically delivered from a buret; the solution
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delivered by the buret is called the titrant. (In some instances, the reverse may also be carried out where a known volume of the standard solution is taken and it is titrated with the analyte of unknown concentration as the titrant.) The volume of titrant required to just completely react with the analyte is measured. Since we know the reagent concentration as well as the reaction stoichiometry between the analyte and the reagent, we can calculate the amount of analyte. The requirements of a titration are as follows:
1. The reaction must be stoichiometric. That is, there must be a well-defined and known reaction between the analyte and the titrant. In the titration of acetic acid in vinegar with sodium hydroxide, for example, a well-defined reaction takes place: CH3 COOH + NaOH → CH3 COONa + H2 O 2. The reaction should be rapid. Most ionic reactions, as above, are very rapid. 3. There should be no side reactions; the reaction should be specific. If there are interfering substances, these must be removed or independently determined and their influence subtracted from the overall signal (Canalyte = Ctotal − Cinterference ). In the above example, there should be no other acids present. 4. There should be a marked change in some property of the solution when the reaction is complete. This may be a change in color of the solution or in some electrical or other physical property of the solution. In the titration of acetic acid with sodium hydroxide, there is a marked increase in the pH of the solution when the reaction is complete. A color change is usually brought about by addition of an indicator, whose color is dependent on the properties of the solution, for example, the pH. 5. The point at which an equivalent or stoichiometric amount of titrant is added is called the equivalence point. The point at which the reaction is observed to be complete is called the end point, that is, when a change in some property of the solution is detected. The end point should coincide with the equivalence point or be at a reproducible interval from it. 6. The reaction should be quantitative. That is, the equilibrium of the reaction should be far to the right so that a sufficiently sharp change will occur at the end point to obtain the desired accuracy. If the equilibrium does not lie far to the right, then there will be gradual change in the property marking the end point (e.g., pH) and this will be difficult to detect precisely.
The equivalence point is the theoretical end of the titration where the number equivalents of the analyte exactly equals the number of equivalents of the titrant added. The end point is the observed end of the titration. The difference is the titration error.
STANDARD SOLUTIONS— —THERE ARE DIFFERENT KINDS A standard solution is prepared by dissolving an accurately weighed quantity of a highly pure material called a primary standard and diluting to an accurately known volume in a volumetric flask. Alternatively, if the material is not sufficiently pure, a solution is prepared to give approximately the desired concentration, and this is standardized by titrating a weighed quantity of a primary standard. For example, sodium hydroxide is not sufficiently pure to prepare a standard solution directly. It is therefore standardized by titrating a primary standard acid, such as potassium acid phthalate (KHP). Potassium acid phthalate is a solid that can be weighed accurately. Standardization calculations are treated below.
A solution standardized by titrating a primary standard is itself a secondary standard. It will be less accurate than a primary standard solution due to the errors of titrations.
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A primary standard should fulfill these requirements: 1. It should be 100.00% pure, although 0.01 to 0.02% impurity is tolerable if it is accurately known. 2. It should be stable to drying temperatures, and it should be stable indefinitely at room temperature. The primary standard is always dried before weighing.5 3. It should be readily and relatively inexpensively available. 4. Although not essential, it should have a high formula weight. This is so that a relatively large amount of it will have to be weighed. The relative error in weighing a greater amount of material will be smaller than that for a small amount. 5. If it is to be used in titration, it should possess the properties required for a titration listed above. In particular, the equilibrium of the reaction should be far to the right so that a sharp end point will be obtained.
A high formula weight means a larger weight must be taken for a given molar concentration of titrant to be made. This reduces the relative error in weighing.
CLASSIFICATION OF TITRATION METHODS——WHAT KINDS ARE THERE? There are four general classes of volumetric or titrimetric methods. 1. Acid–Base. Many compounds, both inorganic and organic, are either acids or bases and can be titrated, respectively, with a standard solution of a strong base or a strong acid. The end points of these titrations are easy to detect, either by means of an indicator or by following the change in pH with a pH meter. The acidity and basicity of many organic acids and bases can be enhanced by titrating in a nonaqueous solvent. The result is a sharper end point, and weaker acids and bases can be titrated in this manner. 2. Precipitation. In the case of precipitation, the titrant forms an insoluble product with the analyte. An example is the titration of chloride ion with silver nitrate solution to form silver chloride precipitate. Again, indicators can be used to detect the end point, or the potential of the solution can be monitored electrically. 3. Complexometric. In complexometric titrations, the titrant is a reagent that forms a water-soluble complex with the analyte, a metal ion. The titrant is often a chelating agent.6 Ethylenediaminetetraacetic acid (EDTA) is one of the most useful chelating agents used for titration. It will react with a large number of metal ions, and the reactions can be controlled by adjustment of pH. Indicators can be used to form a highly colored complex with the metal ion. 4. Reduction–Oxidation. These “redox” titrations involve the titration of an oxidizing agent with a reducing agent, or vice versa. An oxidizing agent gains electrons and a reducing agent loses electrons in a reaction between them. There must be a sufficiently large difference between the oxidizing and reducing capabilities of these agents for the reaction to go to completion and give a sharp end point; that is, one should be a fairly strong oxidizing agent (strong tendency to gain electrons) and the other a fairly strong reducing agent (strong tendency to lose electrons). Appropriate indicators for these titrations are available; various electrometric means to detect the end point may also be used.
5
There are a few exceptions when the primary standard is a hydrate.
6A
chelating agent (the term is derived from the Greek word for clawlike) is a type of complexing agent that contains two or more groups capable of complexing with a metal ion. EDTA has six such groups.
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These different types of titrations and the means of detecting their end points will be treated separately in succeeding chapters.
5.5 Volumetric Calculations—Let’s Use Molarity We shall use molarity throughout the majority of the text for volumetric calculations. Some instructors prefer to introduce the concept of normality, and students are likely to encounter it in reference books. A section on normality-based calculations can be found in the text website. In Equations 5.1–5.5, we previously discussed the ways of expressing in molar and millimolar units. By rearranging these equations, we obtain the expressions for calculating other quantities. M (mol/L) × L = mol
M (mmol/mL) × mL = mmol
(5.17)
g = mol × fw (g/mol)
mg = mmol × fw (mg/mmol)
(5.18)
g = M (mol/L) × L × fw (g/mol)
(5.19)
mg = M (mmol/mL) × mL × fw (mg/mmol)
Learn these relationships well. They are the basis of all volumetric calculations, solution preparation, and dilutions. Think units!
We usually work with millimole (mmol) and milliliter (mL) quantities in titrations; therefore, the right-hand equations are more useful. Note that the expression for formula weight contains the same numerical value whether it be in g/mol or mg/mmol. Note also that care must be taken in utilizing “milli” quantities (millimoles, milligrams, milliliters). Incorrect use could result in calculations errors of 1000-fold. Assume 25.0 mL of 0.100 M AgNO3 is required to titrate a sample containing sodium chloride. The reaction is Cl− (aq) + Ag+ (aq) → AgCl(s) Since Ag+ and Cl− react on a 1:1 molar basis, the number of millimoles of Cl− is equal to the number of millimoles of Ag+ needed for titration. We can calculate the milligrams of NaCl as follows: mmolNaCl = mLAgNO3 × MAgNO3 = 25.0 mL × 0.100 (mmol/mL) = 2.50 mmol mgNaCl = mmol × fwNaCl = 2.50 mmol × 58.44 mg/mmol = 146 mg We can calculate the percentage of analyte A that reacts on a 1:1 mole basis with the titrant using the following general formula: %Analyte = fractionanalyte × 100% = = =
mganalyte mgsample
× 100%
mmol analyte × fwanalyte (mg/mol) mgsample
× 100%
Mtitrant (mmol/mL) × mLtitrant × fwanalyte (mg/mmol) mgsample
× 100% (5.20)
For 1:1 reactions, mmolanalyte = mmoltitrant .
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Note that this computation is a summary of the individual calculation steps taken to arrive at the fraction of analyte in the sample using proper dimensional analysis. You should use it in that sense rather than simply memorizing a formula.
Example 5.19 A 0.4671-g sample containing sodium bicarbonate was dissolved and titrated with standard 0.1067 M hydrochloric acid solution, requiring 40.72 mL. The reaction is HCO3 − + H+ → H2 O + CO2 Calculate the percent sodium bicarbonate in the sample. Solution
The millimoles of sodium bicarbonate are equal to the millimoles of acid used to titrate it, since they react in a 1:1 ratio. mmolHCl = 0.1067 mmol/mL × 40.72 mL = 4.3448 mmolHCl ≡ mmol NaHCO3 (Extra figures are carried so an identical answer is obtained when all steps are done together below.) mgNaHCO3 = 4.3448 mmol × 84.01 mg/mmol = 365.01 mg NaHCO3 %NaHCO3 =
365.01 mg NaHCO3 × 100% = 78.14% NaHCO3 467.1 mgsample
Or, combining all the steps, % NaHCO3 = =
MHCl × mLHCl × fwNaHCO3 mgsample
× 100%
0.1067 mmol HCl/mL × 40.72 mL HCl × 84.01 mg NaHCO3 /mmol × 100% 467.1 mg
= 78.14% NaHCO3
SOME USEFUL THINGS TO KNOW FOR MOLARITY CALCULATIONS When the reaction is not 1:1, a stoichiometric factor must be used to equate the moles of analyte and titrant.
Many substances do not react on a 1:1 mole basis, and so the simple calculation in the above example cannot be applied to all reactions. It is possible, however, to write a generalized formula for calculations applicable to all reactions based on the balanced equation for reactions. Consider the general reaction aA × tT → P
(5.21)
where A is the analyte, T is the titrant, and they react in the ratio a/t to give products P. Then, noting the units and using dimensional analysis, Still think units! We have added mmolanalyte /mmoltitrant .
a (mmol A/mmol T) t a mmolA = MT (mmol/mL) × mLT × (mmol A/mmol T) t
mmolA = mmolT ×
(5.22) (5.23)
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mgA = mmolA × fwA (mg/mmol)
(5.24)
a mgA = MT (mmol/mL) × mLT × (mmol A/mmol T) t × fwA (mg/mmol)
(5.25)
Note that the a/t factor serves to equate the analyte and titrant. To avoid a mistake in setting up the factor, it is helpful to remember that when you calculate the amount of analyte, you must multiply the amount of titrant by the a/t ratio (a comes first). Conversely, if you are calculating the amount of titrant (e.g., molarity) from a known amount of analyte titrated, you must multiply the amount of analyte by the t/a ratio (t comes first). The best way, of course, to ascertain the correct ratio is to always do a dimensional analysis to obtain the correct units. In a manner similar to that used to derive Equation 5.22, we can list the steps in arriving at a general expression for calculating the percent analyte A in a sample determined by titrating a known weight of sample with a standard solution of titrant T: %Analyte = fractionanalyte ×100% = = =
mganalyte mgsample
×100%
mmoltitrant ×(a/t)(mmolanalyte /mmoltitrant )×fwanalyte (mg/mmol) mgsample
×100%
Mtitrant (mmol/mL)×mLtitrant ×(a/t)(mmolanalyte /mmoltitrant )×fwanalyte (mg/mmol) mgsample
×100% (5.26)
Again, note that we simply use dimensional analysis, that is, we perform stepwise calculations in which units cancel to give the desired units. In this general procedure, the dimensional analysis includes the stoichiometric factor a/t that converts millimoles of titrant to an equivalent number of millimoles of titrated analyte.
Example 5.20 A 0.2638-g soda ash sample is analyzed by titrating the sodium carbonate with the standard 0.1288 M hydrochloride solution, requiring 38.27 mL. The reaction is CO3 2− + 2H+ → H2 O + CO2 Calculate the percent sodium carbonate in the sample. Solution
The millimoles of sodium carbonate is equal to one-half the millimoles of acid used to titrate it, since they react in a 1:2 ratio a/t = 12 . mmolHCl = 0.1288 mmol/mL×38.27 mL = 4.929 mmol HCl 1 mmolNaCO3 = 4.929 mmol HCl× (mmol Na2 CO3 /mmol HCl) = 2.4645 mmol Na2 CO3 2 mgNa2 CO3 = 2.4645 mmol×105.99 mg Na2 CO3 /mmol = 261.21 mg Na2 CO3 %Na2 CO3 =
261.21 mg Na2 CO3 ×100% = 99.02% Na2 CO3 263.8 mgsample
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Or, combining all the steps at once, %Na2 CO3 = =
MHCl × mLHCl × 12 (mmol Na2 CO3 /mmol HCl) × fwNa2 CO3 mgsample
× 100%
0.1288 mmol HCl×38.27 mL HCl× 12 (mmol Na2 CO3 /mmol HCl)×105.99 (mg Na2 CO3 /mmol) ×100% 263.8 mgsample
= 99.02% Na2 CO3
Example 5.21 How many milliliters of 0.25 M solution of H2 SO4 will react with 10 mL of a 0.25 M solution of NaOH? Solution
The reaction is H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O One-half as many millimoles of H2 SO4 as of NaOH will react, or MH2 SO4 × mLH2 SO4 = MNaOH × mLNaOH ×
1 (mmol H2 SO4 /mmol NaOH) 2
Therefore, mLH2 SO4 =
0.25 mmol NaOH/mL × 10 mL NaOH × 12 (mmol H2 SO4 /mmol NaOH) 0.25 mmol H2 SO4 /mL
= 5.0 mL H2 SO4 Note that, in this case, we multiplied the amount of titrant by the a/t ratio (mmol analyte/mmol titrant).
Example 5.22 A sample of impure salicylic acid, C6 H4 (OH)COOH (one titratable proton), is analyzed by titration. What size sample should be taken so that the percent purity is equal to five times the milliliters of 0.0500 M NaOH used to titrate it? Solution
Let x = mL NaOH; % salicylic acid (HA) = 5x: % HA = 5x% =
MNaOH × mLNaOH × 1 (mmol HA/mmol NaOH) × fwHA (mg/mmol) × 100% mgsample 0.0500 M × x mL NaOH × 1 × 138 mg HA/mmol × 100% mgsample
mgsample = 138 mg
You can apply the above examples of acid–base calculations to the titrations described in Chapter 8.
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STANDARDIZATION AND TITRATION CALCULATIONS——THEY ARE THE REVERSE OF ONE ANOTHER When a titrant material of high or known purity is not available, the concentration of the approximately prepared titrant solution must be accurately determined by standardization; that is, by titrating an accurately weighed quantity (a known number of millimoles) of a primary standard. From the volume of titrant used to titrate the primary standard, we can calculate the molar concentration of the titrant. Taking the analyte A in Equation 5.21 to be the primary standard, mgstandard mmolstandard = fwstandard (mg/mmol)
In standardization, generally it is the concentration of the titrant that is unknown and the moles of analyte (primary standard) are known.
mmoltitrant = Mtitrant (mmol/mL) × mLtitrant = mmolstandard × t/a (mmoltitrant /mmolstandard ) Mtitrant (mmol/mL) =
mmolstandard × t/a (mmoltitrant /mmolstandard ) mLtitrant
Or, combining all steps at once, Mtitrant (mmol/mL) =
mgstandard /fwstandard (mg/mmol) × t/a (mmoltitrant /mmolstandard ) mLtitrant (5.27)
Note once again that dimensional analysis (cancellation of units) results in the desired units of mmol/mL.
Example 5.23 An approximate 0.1 M hydrochloric acid solution is prepared by 120-fold dilution of concentrated hydrochloric acid. It is standardized by titrating 0.1876 g of dried primary standard sodium carbonate: CO3 2− + 2H+ → H2 O + CO2 The titration required 35.86 mL acid. Calculate the molar concentration of the hydrochloric acid. Solution
The millimoles of hydrochloric acid are equal to twice the millimoles of sodium carbonate titrated. mmolNa2 CO3 = 187.6 mg Na2 CO3 /105.99 (mg Na2 CO3 /mmol) = 1.7700 mmol Na2 CO3 mmolHCl = MHCl (mmol/mL) × 35.86 mL HCl = 1.7700 mmol Na2 CO3 ×2 (mmol HCl/mmol Na2 CO3 ) MHCl =
1.7700 mmol Na2 CO3 × 2 (mmol HCl/mmol Na2 CO3 ) = 0.09872 M 35.86 mL HCl
Units!
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Or, combining all steps at once, (mgNa2 CO3 /fwNa2 CO3 ) × (2/1)(mmol HCl/mmol Na2 CO3 ) MHCl = mLHCl [187.6 mg/105.99 (mg/mmol)] × 2 (mmol HCl/mmol Na2 CO3 ) 35.86 mL = 0.09872 mmol/mL =
Note that we multiplied the amount of analyte, Na2 CO3 , by the t/a ratio (mmol titrant/mmol analyte). Note also that although all measurements were to four significant figures, we computed the formula weight of Na2 CO3 to five figures. This is because with four figures, it would have had an uncertainty of about one part per thousand compared to 187.6 with an uncertainty of about half that. It is not bad practice, as a matter of routine, to carry the formula weight to one additional figure. The following examples illustrate titration calculations for different types of reactions and stoichiometry.
Example 5.24 This is a “redox” titration (see Chapter 14).
The iron(II) in an acidified solution is titrated with a 0.0206 M solution of potassium permanganate: 5Fe2+ + MnO4 − + 8H+ → 5Fe3+ + Mn2+ + 4H2 O If the titration required 40.2 mL, how many milligrams iron are in the solution? Solution
There are five times as many millimoles of iron as there are of permanganate that react with it, so mgFe 5 mmolFe = = MKMnO4 × mLKMnO4 × (mmol Fe/mmol KMnO4 ) fwFe 1 mgFe = 0.0206 mmol KMnO4 /mL × 40.2 mL KMnO4 × 5 (mmol Fe/mmol MnO4 − ) ×55.8 mg Fe/mmol = 231 mg Fe
Calculations of this type are used for the redox titrations described in Chapter 14. Following is a list of typical precipitation and complexometric titration reactions and the factors for calculating the milligrams of analyte from millimoles of titrant.7 Cl− + Ag+ → AgCl 2Cl− + Pb2+ → PbCl2
mgCl− = MAg+ × mLAg+ × 1 (mmol Cl− /mmol Ag+ ) × fwCl− mgCl− = MPb2+ × mLPb2+ × 2 (mmol Cl− /mmol Pb2+ ) × fwCl− 1 (mmol PO4 3− /mmol Ag+ ) × fwPO4 3− 3
PO4 3− + 3Ag+ → Ag3 PO4
mgPO4 3− = MAg+ × mLAg+ ×
2CN− + Ag+ → Ag(CN)2 −
mgCN− = MAg+ × mLAg+ × 2 (mmol CN− /mmol Ag+ ) × fwCN−
2CN− + 2Ag+ → Ag[Ag(CN)2 ] mgCN− = MAg+ × mLAg+ × 1 (mmol CN− /mmol Ag+ ) × fwCN− Ba2+ + SO4 2− → BaSO4
mgBa2+ = MSO4 2− × 1 (mmol Ba2+ /mmol SO4 2− ) × fwBa2+
Ca2+ + H2 Y2− → CaY2− + 2H+ mgCa2+ = MEDTA × 1 (mmol Ca2+ /mmol EDTA) × fwCa2+ 7 H Y = EDTA in the last equation. 4
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These formulas are useful calculations involving the precipitation and complexometric titrations described in Chapters 8 and 11.
Example 5.25 Aluminum is determined by titrating with EDTA: Al
3+
+ H2 Y
2−
−
+
→ AlY + 2H
A 1.00-g sample requires 20.5 mL EDTA for titration. The EDTA was standardized by titrating 25.0 mL of a 0.100 M CaCl2 solution, requiring 30.0 mL EDTA. Calculate the percent Al2 O3 in the sample. Solution
Since Ca2+ and EDTA react on a 1:1 mole ratio, 0.100 mmol CaCl2 /mL × 25.0 mL CaCl2 = 0.0833 mmol/mL 30.0 mL EDTA The millimoles Al3+ are equal to the millimoles EDTA used in the sample titration, but there are one-half this number of millimoles of Al2 O3 (since 1Al3+ → 12 Al2 O3 ). Therefore, MEDTA =
% Al2 O3 = % Al2 O3 =
MEDTA × mLEDTA ×
1 2
(mmol Al2 O3 /mmol EDTA) × fwAl2 O3 mgsample
0.0833 mmol EDTA/mL × 20.5 mL EDTA × 1000-mg sample
1 2
× 100%
× 101.96 mg Al2 O3 /mmol
× 100% = 8.71% Al2 O3
WHAT IF THE ANALYTE AND TITRANT CAN REACT IN DIFFERENT RATIOS? As you might be aware from your introductory chemistry course, some substances can undergo reaction to different products. The factor used in calculating millimoles of such a substance from the millimoles of titrant reacted with it will depend on the specific reaction. Sodium carbonate, for example, can react as a diprotic or a monoprotic base: CO3 2− + 2H+ → H2 O + CO2 or
CO3 2− + H+ → HCO3 −
In the first case, mmol Na2 CO3 = mmol acid × 12 (mmol CO3 2− /mmol H+ ). In the second case, mmol Na2 CO3 = mmol acid. Similarly, phosphoric acid can be titrated as a monoprotic or a diprotic acid: H3 PO4 + OH− → H2 PO4 − + H2 O or
H3 PO4 + 2OH− → HPO4 2− + 2H2 O
This is a “complexometric” titration (see Chapter 9). EDTA has four protons, and in this titration, two are dissociated at the pH of the titration.
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Example 5.26 In acid solution, potassium permanganate reacts with H2 O2 to form Mn2+ : 5H2 O2 + 2MnO4 − + 6H+ → 5O2 + 2Mn2+ + 8H2 O In neutral solution, it reacts with MnSO4 to form MnO2 : 3Mn2+ + 2MnO4 − + 4OH− → 5MnO2 + 2H2 O Calculate the number of milliliters of 0.100 M KMnO4 that will react with 50.0 mL of 0.200 M H2 O2 and with 50.0 mL of 0.200 M MnSO4 . Solution Keep track of millimoles!
The number of millimoles of MnO4 − will be equal to two-fifths of the number of millimoles of H2 O2 reacted: MMnO4 − × mLMnO4 − = MH2 O2 × mLH2 O2 × mLMnO4 − =
2 (mmol MnO4 − /mmol H2 O2 ) 5
0.200 mmol H2 O2 /mL × 50.0 mL H2 O2 × 0.100 mmol MnO4 − /mL
2 5
= 40.0 mL KMnO4
The number of millimoles of MnO4 − reacting with Mn2+ will be equal to two-thirds of the number of millimoles of Mn2+ : MMnO4 − × mLMnO4 − = MMn2+ × mLMnO4 − =
2 (mmol MnO4 − /mmol Mn2+ ) 3
0.200 mmol Mn2+ /mL × 50.0 mL Mn2+ × 0.100 mmol MnO4 − /mL
2 3
= 66.7 mL KMnO4
Example 5.27 Oxalic acid, H2 C2 O4 , is a reducing agent that reacts with KMnO4 as follows: 5H2 C2 O4 + 2MnO4 − + 6H+ → 10CO2 + 2Mn2+ + 8H2 O Its two protons are also titratable with a base. How many milliliters of 0.100 M NaOH and 0.100 M KMnO4 will react with 500 mg H2 C2 O4 ? Solution mmol NaOH = 2 × mmol H2 C2 O4 0.100 mmol/mL × x mL NaOH =
500 mgH2 C2 O4 × 2 (mmol OH− /mmol H2 C2 O4 ) 90.0 mg/mmol
x = 111 mL NaOH 2 mmol KMnO4 = × mmol H2 C2 O4 5 500 mg H2 C2 O4 2 0.100 mmol/mL × x mL KMnO4 = × (mmol KMnO4 /mmol H2 C2 O4 ) 90.0 mg/mmol 5 x = 22.2 mL KMnO4
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Example 5.28 Pure Na2 C2 O4 plus KHC2 O4 · H2 C2 O4 (three replaceable protons, KH3 A2 ) are mixed in such a proportion that each gram of the mixture will react with equal volumes of 0.100 M KMnO4 and 0.100 M NaOH. What is the proportion? Solution
Assume 10.0-mL titrant, so it will react with 1.00 mmol NaOH or KMnO4 . The acidity is due to KHC2 O4 · H2 C2 O4 denoted in the following as KH3 A2 : 1 mmol KH3 A2 = mmol NaOH × (mmol KH3 A2 /mmol OH− ) 3 1 1.00 mmol NaOH × = 0.333 mmol KH3 A2 3 From Example 5.27, each mmol Na2 C2 O4 (Na2 A) reacts with
2 5
mmol KMnO4 .
2 (mmol MnO4 − /mmol Na2 A) + mmol KH3 A2 5 4 × (mmol MnO4 − /mmol KH3 A2 ) 5 2 4 1.00 mmol KMnO4 = mmol Na2 A × + 0.333 mmol KH3 A2 × 5 5 mmol Na2 A = 1.83 mmol
mmol KMnO4 = mmol Na2 A ×
The ratio is 1.83 mmol Na2 A/0.333 mmol KH3 A2 = 5.50 mmol Na2 A/mmol KH3 A2 . The weight ratio is 5.50 mmol Na2 A × 134 mg/mmol = 3.38 g Na2 A/g KH3 A2 218 mg KH3 A2 /mmol
IF THE REACTION IS SLOW, DO A BACK-TITRATION Sometimes a reaction is slow to go to completion, and a sharp end point cannot be obtained. One example is the titration of antacid tablets with a strong acid such as HCl. In these cases, a back-titration will often yield useful results. In this technique, a measured amount of the reagent, which would normally be the titrant, is added to the sample so that there is a slight excess. After the reaction with the analyte is allowed to go to completion, the amount of excess (unreacted) reagent is determined by titration with another standard solution; the kinetics of the analyte reaction may be increased in the presence of excess reagent. So by knowing the number of millimoles of reagent taken and by measuring the number of millimoles remaining unreacted, we can calculate the number of millimoles of sample that reacted with the reagent: mmol reagent reacted = mmol taken − mmol back-titrated mg analyte = mmol reagent reacted × factor (mmol analyte/mmol reagent) × fw analyte (mg/mmol)
In back-titrations, a known number of millimoles of reactant is taken, in excess of the analyte. The unreacted portion is titrated.
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Example 5.29 Chromium(III) is slow to react with EDTA (H4 Y) and is therefore determined by back-titration. Chromium(III) picolinate, Cr(C6 H4 NO2 )3 , is sold as a nutritional supplement for athletes with the claim that it aids muscle building. A nutraceutical preparation containing chromium(III) is analyzed by treating a 2.63-g sample with 5.00 mL of 0.0103 M EDTA. Following reaction, the unreacted EDTA is back-titrated with 1.32 mL of 0.0122 M zinc solution. What is the percent chromium picolinate in the pharmaceutical preparation? Solution
Both Cr3+ and Zn2+ react in a 1:1 ratio with EDTA: Cr3+ + H4 Y → CrY− + 4H+ Zn2+ + H4 Y → ZnY2 + 4H+ The millimoles of EDTA taken is 0.0103 mmol EDTA/mL × 5.00 mL EDTA = 0.0515 mmol EDTA The millimoles of unreacted EDTA is 0.0112 mmol Zn2+ /mL × 1.32 mL Zn2+ = 0.0148 mmol unreacted EDTA The millimoles of reacted EDTA is 0.0515 mmol taken − 0.0148 mmol left = 0.0367 mmol EDTA ≡ mmol Cr3+ The milligrams of Cr(C6 H4 NO2 )3 titrated is 0.0367 mmol Cr(C6 H4 NO2 )3 × 418.3 mg/mmol = 15.35 mg Cr(C6 H4 NO2 )3 %Cr(C6 H4 NO2 )3 =
15.35 mg Cr(C6 H4 NO2 )3 × 100% = 0.584% Cr(C6 H4 NO2 )3 2630 mg sample
Or, combining all steps, %Cr(C6 H4 NO2 )3 = =
(MEDTA × mLEDTA − MZn × mLZn2+ ) × 1(mmol Cr(C6 H4 NO2 )3 /mmol EDTA) × fwCr(C6 H4 NO2 )3 mgsample
× 100%
(0.0103 mmol EDTA/mL×5.00 mL EDTA−0.0112 mmol Zn2+/mL×1.32 mL Zn2+)×1×418.3 mg Cr(C6 H4 NO2 )3 /mmol ×100% 2630 mg sample
= 0.584% Cr(C6 H4 NO2 )3
Example 5.30 A 0.200-g sample of pyrolusite is analyzed for manganese content as follows. Add 50.0 mL of a 0.100 M solution of ferrous ammonium sulfate to reduce the MnO2 to Mn2+ . After reduction is complete, the excess ferrous ion is titrated in acid solution with 0.0200 M KMnO4 , requiring 15.0 mL. Calculate the percentage of manganese in the sample as Mn3 O4 (the manganese may or may not exist in this form, but we can make the calculations on the assumption that it does).
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Solution
The reaction between Fe2+ and MnO4 − is 5Fe2+ + MnO4 − + 8H+ → 5Fe3+ + Mn2+ + 4H2 O and so there are five times as many millimoles of excess Fe2+ as of MnO4 − that reacted with it. The reaction between Fe2+ and MnO2 is
The reactant may react in different ratios with the analyte and titrant.
MnO2 + 2Fe2+ + 4H+ → Mn2+ + 2Fe3+ + 2H2 O and there are one-half as many millimoles of MnO2 as millimoles of Fe2+ that react with it. There are one-third as many millimoles of Mn3 O4 as of MnO2 (1MnO2 → 1 3 Mn3 O4 ). Therefore, mmol Fe2+ reacted = 0.100 mmol Fe2+ /mL × 50.0 mL Fe2+ − 0.0200 mmol MnO4 − /mL ×15.0 mL MnO4 − × 5 mmol Fe2+ /mmol MnO4 − = 3.5 mmol Fe2+ reacted 1 mmol MnO2 = 3.5 mmol Fe2+ × (mmol MnO2 /mmol Fe2+ ) = 1.75 mmol MnO2 2 1 mmol Mn3 O4 = 1.75 mmol MnO2 × (mmol Mn3 O4 )/mmol MnO2 ) 3 = 0.583 mmol Mn3 O4 % Mn3 O4 =
0.583 mmol Mn3 O4 × 228.8 (mg Mn3 O4 /mmol) × 100% 200 mg sample = 66.7% Mn3 O4
Or, combining all steps at once, % Mn3 O4 = {[MFe2+ × mLFe2− − MMnO4 − × mLMnO4 − × 5(mmol Fe2+ /mmol MnO4 − ) ×
=
1 1 (mmol MnO2 /mmol Fe2+ ) × (mmol Mn3 O4 /mmol MnO2 ) 2 3 ×fwMn3 O4 ]/mgsample } × 100% (0.100 × 50.0 − 0.0200 × 15.0 × 5) × 12 × 200 mg sample
1 3
× 228.8 mg/mmol
= 66.7% Mn3 O4
× 100%
5.6 Titer—How to Make Rapid Routine Calculations For routine titrations, it is often convenient to calculate the titer of the titrant. The titer is the weight of analyte that is chemically equivalent to 1 mL of the titrant, usually expressed in milligrams. For example, if a potassium dichromate solution has a titer of 1.267 mg Fe, each milliliter potassium dichromate will react with 1.267 mg iron, and the weight of iron titrated is obtained by simply multiplying the volume of titrant used by the titer. The titer can be expressed in terms of any form of the analyte desired, for example, milligrams FeO or Fe2 O3 .
Titer = milligrams analyte that react with 1 mL of titrant.
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Example 5.31 A standard solution of potassium dichromate contains 5.442 g/L. What is its titer in terms of milligrams Fe3 O4 ? Solution
The iron is titrated as Fe2+ and each Cr2 O7 2− will react with 6Fe2+ (or the iron from 2Fe3 O4 ): 6Fe2+ + Cr2 O7 2− + 14H+ → 6Fe3+ + 2Cr3+ + 7H2 O The molarity of the K2 Cr2 O7 , solution is MCr2 O7 2− =
g/L 5.442 g/L = 0.01850 mol/L = fwK2 Cr2 O7 294.19 g/mol
Therefore the titer is mmol K2 Cr2 O7 2 mmol Fe3 O4 mg Fe3 O4 0.01850 × × 231.54 mL 1 mmol K2 Cr2 O7 mmol Fe3 O4 = 8.567 mg Fe3 O4 /mL K2 Cr2 O7
5.7 Weight Relationships—You Need These for Gravimetric Calculations In the technique of gravimetric analysis (Chapter 10), the analyte is converted to an insoluble form, which is weighed. From the weight of the precipitate formed and the weight relationship between the analyte and the precipitate, we can calculate the weight of analyte. We review here some of the calculation concepts. The analyte is almost always weighed in a form different from what we wish to report. We must, therefore, calculate the weight of the desired substance from the weight of the gravimetric precipitate. We can do this by using a direct proportion. For example, if we are analyzing for the percentage of chloride in a sample by weighing it as AgCl, we can write precipitating reagent
Cl− −−−−−−−−−−−→ AgCl(s) We derive one mole AgCl from one mole Cl− , so at wt Cl g Cl− = g AgCl fw AgCl or g Cl− = g AgCl × In gravimetric analysis, the moles of analyte is a multiple of the moles of precipitate formed (the moles of analyte contained in each mole of precipitate).
at wt Cl fw AgCl
Note that when we specify at wt or fw of a substance x, it implicitly has the units g x/mol x. In other words, the weight of Cl contained in or used to create AgCl is equal to the weight of AgCl times the fraction of Cl in it. Calculation of the corresponding weight of Cl2 that would be contained in the sample would proceed thus: precipitating reagent
Cl2 −−−−−−−−−−−→ 2AgCl(s)
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181
We derive two moles of AgCl from each mole of Cl2 , so g Cl2 fw Cl2 = g AgCl 2 × fw AgCl and g Cl2 = g AgCl ×
fw Cl2 2 (fw AgCl)
g Cl2 = g AgCl ×
70.906 2 × 143.32
or
We may also write g AgCl ×
Remember to keep track of the units!
1 mol Cl2 70.906 g Cl2 1 mol AgCl × × = g Cl2 143.32 g AgCl 2 mol AgCl 1 mol Cl2
The gravimetric factor (GF) is the appropriate ratio of the formula weight of the substance sought to that of the substance weighed: GF = gravimetric factor =
The gravimetric factor is the weight of analyte per unit weight of precipitate.
a fw of substance sought × (mol sought/mol weighed) fw of substance weighed b (5.28)
where a and b are integers that make the formula weights in the numerator and denominator chemically equivalent. In the above examples, the gravimetric factors were (fw Cl/fw AgCl) × 1/1, and (fw Cl2 /fw AgCl) × 12 . Note that one or both of the formula weights may be multiplied by an integer in order to keep the same number of atoms of the key element in the numerator and denominator. The weight of the substance sought is obtained by multiplying the weight of the precipitate by the gravimetric factor: weight (g) ×
fw of substance sought a × = sought (g) fw of substance weighed b
(5.29)
Note that the species and the units of the equation can be checked by dimensional analysis (canceling of like species and units). Note also that we have calculated the amount of Cl2 gas derivable from the sample instead of the amount of Cl− ion, the form in which it probably exists in the sample and the form in which it is weighed. If we precipitate the chloride as PbCl2 , precipitating agent
2Cl− −−−−−−−−−−→ PbCl2 and Cl2 → PbCl2 then, g Cl− = g PbCl2 ×
2(fw Cl) = g PbCl2 × GF fw PbCl2
g Cl2 = g PbCl2 ×
fw Cl2 = g PbCl2 × GF fw PbCl2
or
Conversion from weight of one substance to the derived there from weight of another is done using dimensional analysis of the units to arrive at the desired weight. The gravimetric factor is one step of that calculation and is useful for routine calculations. That is, if we know the gravimetric factor, we simply multiply the weight of the precipitate by the gravimetric factor to arrive at the weight of the analyte.
The grams of analyte = grams precipitate × GF.
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Example 5.32 Calculate the weight of barium and the weight of Cl present in 25.0 g BaCl2 . Solution
25.0 g BaCl2 ×
fw Ba = fw BaCl2
137.3 = 16.5 g Ba 208.2 2 × fw Cl = 25.0 g BaCl2 × fw BaCl2 25.0 g ×
25.0 g ×
2 × 35.45 = 8.51 g Cl 208.2
Example 5.33 Aluminum in an ore sample is determined by dissolving it and then precipitating with base as Al(OH)3 and igniting to Al2 O3 , which is weighed. What weight of aluminum was in the sample if the ignited precipitate weighed 0.2385 g? Solution
g Al = g Al2 O3 × = 0.2385 g ×
2 × fw Al fw Al2 O3 2 × 26.982 = 0.12623 g Al 101.96
The gravimetric factor is
or
2 × 26.982 2 × fw Al = = 0.52927 (g Al/g Al2 O3 ) fw Al2 O3 101.96 0.2385 g Al2 O3 × 0.52927 (g Al/g Al2 O3 ) = 0.12623 g Al
Following are some other examples of gravimetric factors: Sought
The operations of gravimetric analyses are described in detail in Chapter 10.
Weighed
Gravimetric Factor
SO3
BaSO4
fw SO3 fw BaSO4
Fe3 O4
Fe2 O3
2 × fw Fe3 O4 3 × fw Fe2 O3
Fe
Fe2 O3
2 × fw Fe fw Fe2 O3
MgO
Mg2 P2 O7
2 × fw MgO fw Mg2 P2 O7
P2 O5
Mg2 P2 O7
fw P2 O5 fw Mg2 P2 O7
More examples of gravimetric calculations are given in Chapter 10.
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PROBLEMS
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Questions 1. Distinguish between the expression of concentration on weight/weight, weight/volume, and volume/volume bases. 2. Express ppm and ppb on weight/weight, weight/volume, and volume/volume bases. 3. Define the term “equivalent weight,” used for electrolytes in clinical chemistry. Why is this used? 4. List the requirements for a titration. What are the four classes of titrations? 5. What is the equivalence point of a titration? The end point? 6. What is a standard solution? How is it prepared? 7. What are the requirements of a primary standard? 8. Why should a primary standard have a high formula weight?
Problems WEIGHT/MOLE CALCULATIONS 9. Calculate the grams of substance required to prepare the following solutions: (a) 250 mL of 5.00% (wt/vol) NaNO3 ; (b) 500 mL of 1.00% (wt/vol) NH4 NO3 , (c) 1000 mL of 10.0% (wt/vol) AgNO3 . 10. What is the wt/vol % of the solute in each of the following solutions? (a) 52.3 g Na2 SO4 /L, (b) 275 g KBr in 500 mL, (c) 3.65 g SO2 in 200 mL. 11. Calculate the formula weights of the following substances: (a) BaCl2 · 2H2 O, (b) KHC2 O4 · H2 C2 O4 , (c) Ag2 Cr2 O7 , (d) Ca3 (PO4 )2 . 12. Calculate the number of millimoles contained in 500 mg of each of the following substances: (a) BaCrO4 , (b) CHCl3 , (c) KIO3 · HIO3 , (d) MgNH4 PO4 , (e) Mg2 P2 O7 , (f) FeSO4 · C2 H4 (NH3 )2 SO4 · 4H2 O. 13. Calculate the number of grams of each of the substances in Problem 12 that would have to be dissolved and diluted to 100 mL to prepare a 0.200 M solution. 14. Calculate the number of milligrams of each of the following substances you would have to weigh out in order to prepare the listed solutions: (a) 1.00 L of 1.00 M NaCl, (b) 0.500 L of 0.200 M sucrose (C12 H22 O11 ), (c) 10.0 mL of 0.500 M sucrose, (d) 0.0100 L of 0.200 M Na2 SO4 , (e) 250 mL of 0.500 M KOH, (f) 250 mL of 0.900% NaCl (g/100 mL solution). 15. The chemical stockroom is supplied with the following stock solutions: 0.100 M HCl, 0.0200 M NaOH, 0.0500 M KOH, 10.0% HBr (wt/vol), and 5.00% Na2 CO3 (wt/vol). What volume of stock solution would be needed to obtain the following amounts of solutes? (a) 0.0500 mol HCl, (b) 0.0100 mol NaOH, (c) 0.100 mol KOH, (d) 5.00 g HBr, (e) 4.00 g Na2 CO3 , (f) 1.00 mol HBr, (g) 0.500 mol Na2 CO3 .
MOLARITY CALCULATIONS 16. Calculate the molar concentrations of all the cations and anions in a solution prepared by mixing 10.0 mL each of the following solutions: 0.100 M Mn(NO3 )2 , 0.100 M KNO3 , and 0.100 M K2 SO4 . 17. A solution containing 10.0 mmol CaCl2 is diluted to 1 L. Calculate the number of grams of CaCl2 · 2H2 O per milliliter of the final solution. 18. Calculate the molarity of each of the following solutions: (a) 10.0 g H2 SO4 in 250 mL of solution, (b) 6.00 g NaOH in 500 mL of solution, (c) 25.0 g AgNO3 in 1.00 L of solution. 19. Calculate the number of grams in 500 mL of each of the following solutions: (a) 0.100 M Na2 SO4 , (b) 0.250 M Fe(NH4 )2 (SO4 )2 · 6H2 O, (c) 0.667 M Ca(C9 H6 ON)2 . 20. Calculate the grams of each substance required to prepare the following solutions: (a) 250 mL of 0.100 M KOH, (b) 1.00 L of 0.0275 M K2 Cr2 O7 , (c) 500 mL of 0.0500 M CuSO4 . 21. How many milliliters of concentrated hydrochloric acid, 38.0% (wt/wt), specific gravity 1.19, are required to prepare 1 L of a 0.100 M solution? (Assume density and specific gravity are equal within three significant figures.)
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22. Calculate the molarity of each of the following commercial acid or base solutions: (a) 70.0% HClO4 , specific gravity 1.668, (b) 69.0% HNO3 , specific gravity 1.409, (c) 85.0% H3 PO4 , specific gravity 1.689, (d) 99.5% CH3 COOH (acetic acid), specific gravity 1.051, (e) 28.0% NH3 , specific gravity 0.898. (Assume density and specific gravity are equal within three significant figures.)
mg/L (PPM) CALCULATIONS 23. A solution contains 6.0 μmol Na2 SO4 in 250 mL. How many mg/L sodium does it contain? Of sulfate? 24. A solution (100 mL) containing 325 mg/L K+ is analyzed by precipitating it as the tetraphenyl borate, K(C6 H5 )4 B, dissolving the precipitate in acetone solution, and measuring the concentration of tetraphenyl borate ion, (C6 H5 )4 B− , in the solution. If the acetone solution volume is 250 mL, what is the concentration of the tetraphenyl borate in mg/L? 25. Calculate the molar concentrations of 1.00-mg/L solutions of each of the following. (a) AgNO3 , (b) Al2 (SO4 )3 , (c) CO2 , (d) (NH4 )4 Ce(SO4 )4 · 2H2 O, (e) HCl, (f) HClO4 . 26. Calculate the mg/L concentrations of 2.50 × 10−4 M solutions of each of the following. (a) Ca2+ , (b) CaCl2 , (c) HNO3 , (d) KCN, (e) Mn2+ , (f) MnO4 − . 27. You want to prepare 1 L of a solution containing 1.00 mg/L Fe2+ . How many grams ferrous ammonium sulfate, FeSO4 · (NH4 )2 SO4 · 6H2 O, must be dissolved and diluted in 1 L? What would be the molarity of this solution? 28. A 0.456-g sample of an ore is analyzed for chromium and found to contain 0.560 mg Cr2 O3 . Express the concentration of Cr2 O3 in the sample as (a) percent, (b) parts per thousand, and (c) parts per million. 29. How many grams NaCl should be weighed out to prepare 1 L of a 100-mg/L solution of (a) Na+ and (b) Cl− ? 30. You have a 250-mg/L solution of K+ as KCl. You wish to prepare from this a 0.00100 M solution of Cl− . How many milliliters must be diluted to 1 L? 31. One liter of a 500-mg/L solution of KClO3 contains how many grams K+ ?
DILUTION CALCULATIONS 32. A 12.5-mL portion of a solution is diluted to 500 mL, and its molarity is determined to be 0.125. What is the molarity of the original solution? 33. What volume of 0.50 M H2 SO4 must be added to 65 mL of 0.20 M H2 SO4 to give a final solution of 0.35 M? Assume volumes are additive. 34. How many milliliters of 0.10 M H2 SO4 must be added to 50 mL of 0.10 M NaOH to give a solution that is 0.050 M in H2 SO4 ? Assume volumes are additive. 35. You are required to prepare working standard solutions of 1.00 × 10−5 , 2.00 × 10−5 , 5.00 × 10−5 , and 1.00 × 10−4 M glucose from a 0.100 M stock solution. You have available 100-mL volumetric flasks and pipets of 1.00-, 2.00-, 5.00-, and 10.00-mL volume. Outline a procedure for preparing the working standards. 36. A 0.500-g sample is analyzed spectrophotometrically for manganese by dissolving it in acid and transferring to a 250-mL flask and diluting to volume. Three aliquots are analyzed by transferring 50-mL portions with a pipet to 500-mL Erlenmeyer flasks and reacting with an oxidizing agent, potassium peroxydisulfate, to convert the manganese to permanganate. After reaction, these are quantitatively transferred to 250-mL volumetric flasks, diluted to volume, and measured spectrophotometrically. By comparison with standards, the average concentration in the final solution is determined to be 1.25 × 10−5 M. What is the percent manganese in the sample? 37. A stock solution of analyte is made by dissolving 34.83 mg of copper (II) acetate hexahydrate (fw = 289.73 g/mol) in 25.00 mL of water. A second stock solution of internal standard is made by dissolving 28.43 mg of germanium (I) acetate (fw = 190.74 g/mol) into 25.00 mL of water. These solutions are used to make a series of standards for flame atomic absorption analysis calibration. The standard solutions (each 10.00 mL total volume) should have the following concentrations of copper: 10.00; 25.00; 50.00; 100.0; and 200.0 μM. Each calibration solution should also contain 50.00 μM of germanium.
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185
What is the analyte stock solution concentration? What is the internal standard stock solution concentration? Complete this table. Concentration Cu (μM)
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Volume of analyte stock (mL)
Volume of internal standard stock (mL)
10.00 25.00 50.00 100.0 200.0
Volume of diluent (mL)
Total Volume (mL) 10.00 10.00 10.00 10.00 10.00
PROFESSOR’s FAVORITE PROBLEM Contributed by Professor Wen-Yee Lee, The University of Texas at El Paso 38. Tom, an analytical chemist, bought a bag of decaffeinated coffee from a grocery store. However, Tom suspected that he might have received regular coffee and therefore decided to analyze his coffee for caffeine. In the lab, he took 0.5 mL of the brewed coffee and diluted it in water to make a 100.0 mL solution. He performed four analyses and found the concentrations to be 4.69, 3.99, 4.12, and 4.50 mg/L, respectively. (Assume the density of all solutions is 1.000 g/mL, 1 oz = 28.35 mL). (a) Report the concentration (in mg/L) of caffeine in the brewed coffee using the format as average ± standard deviation. (Note this is not the concentration in the diluted solution.) (b) Look up the caffeine content of regular vs. decaffeinated coffee. Do you think that Tom was given the wrong type of coffee? (c) Caffeine intake of 300 mg per day reportedly has no adverse effects in the vast majority of the adult population. If Tom drinks 3 cups (8 oz/cup) of this coffee daily, is his intake within this known safe zone?
STANDARDIZATION CALCULATIONS 39. A preparation of soda ash is known to contain 98.6% Na2 CO3 . If a 0.678-g sample requires 36.8 mL of a sulfuric acid solution for complete neutralization, what is the molarity of the sulfuric acid solution? 40. A 0.1 M sodium hydroxide solution is to be standardized by titrating primary standard sulfamic acid (NH2 SO3 H). What weight of sulfamic acid should be taken so that the volume of NaOH delivered from the buret is about 40 mL?
ANALYSIS CALCULATIONS 41. A sample of USP-grade citric acid (H3 C6 H5 O7 , three titratable protons) is analyzed by titrating with 0.1087 M NaOH. If a 0.2678-g sample requires 38.31 mL for titration, what is the purity of the preparation? USP specification requires 99.5%. 42. Calcium in a 200-μL serum sample is titrated with 1.87 × 10−4 M EDTA solution, requiring 2.47 mL. What is the calcium concentration in the blood in mg/dL? 43. A 0.372-g sample of impure BaCl2 ·2H2 O is titrated with 0.100 M AgNO3 , requiring 27.2 mL. Calculate (a) the percent Cl in the sample and (b) the percent purity of the compound. 44. An iron ore is analyzed for iron content by dissolving in acid, converting the iron to Fe2+ , and then titrating with standard 0.0150 M K2 Cr2 O7 solution. If 35.6 mL is required to titrate the iron in a 1.68-g ore sample, how much iron is in the sample, expressed as percent Fe2 O3 ? (See Example 5.31 for the titration reaction.) 45. Calcium in a 2.00-g sample is determined by precipitating CaC2 O4 , dissolving this in acid, and titrating the oxalate with 0.0200 M KMnO4 . What percent of CaO is in the sample if 35.6 mL KMnO4 is required for titration? (The reaction is 5H2 C2 O4 + 2MnO4 − + 6H+ → 10CO2 + 2Mn2+ + 8H2 O.)
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46. A potassium permanganate solution is prepared by dissolving 4.68 g KMnO4 in water and diluting to 500 mL. How many milliliters of this will react with the iron in 0.500 g of an ore containing 35.6% Fe2 O3 ? (See Example 5.30 for the titration reaction.) 47. A sample contains BaCl2 plus inert matter. What weight must be taken so that when the solution is titrated with 0.100 AgNO3 , the milliliters of titrant will be equal to the percent BaCl2 in the sample? 48. A 0.250-g sample of impure AlCl3 is titrated with 0.100 M AgNO3 , requiring 48.6 mL. What volume of 0.100 M EDTA would react with a 0.350-g sample? (EDTA reacts with Al3+ in a 1:1 ratio.) 49. A 425.2-mg sample of a purified monoprotic organic acid is titrated with 0.1027 M NaOH, requiring 28.78 mL. What is the formula weight of the acid? 50. The purity of a 0.287-g sample of Zn(OH)2 is determined by titrating with a standard HCl solution, requiring 37.8 mL. The HCl solution was standardized by precipitating AgCl in a 25.0-mL aliquot and weighing (0.462 g AgCl obtained). What is the purity of the Zn(OH)2 ? 51. A sample of pure KHC2 O4 ·H2 C2 O4 ·2H2 O (three replaceable hydrogens) requires 46.2 mL of 0.100 M NaOH for titration. How many milliliters of 0.100 M KMnO4 will the same-size sample react with? (See Problem 45 for reaction with KMnO4 .)
BACK-TITRATIONS 52. A 0.500-g sample containing Na2 CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to remove CO2 , and then back-titrating the excess acid with 0.100 M NaOH. If 5.6 mL NaOH is required for the back-titration, what is the percent Na2 CO3 in the sample? 53. A hydrogen peroxide solution is analyzed by adding a slight excess of standard KMnO4 solution and back-titrating the unreacted KMnO4 with standard Fe2+ solution. A 0.587-g sample of the H2 O2 solution is taken, 25.0 mL of 0.0215 M KMnO4 is added, and the back-titration requires 5.10 mL of 0.112 M Fe2+ solution. What is the percent H2 O2 in the sample? (See Examples 5.26 and 5.30 for the reactions.) 54. The sulfur content of an iron pyrite ore sample is determined by converting it to H2 S gas, absorbing the H2 S in 10.0 mL of 0.00500 M I2 , and then back-titrating the excess I2 with 0.00200 M Na2 S2 O3 . If 2.6 mL Na2 S2 O3 is required for the titration, how many milligrams of sulfur are contained in the sample? Reactions: H2 S + I2 → S + 2I− + 2H+ I2 + 2S2 O3 2− → 2I− + S4 O6 2−
TITER 55. Express the titer of a 0.100 M EDTA solution in mg BaO/mL. 56. Express the titer of a 0.0500 M KMnO4 solution in mg Fe2 O3 /mL. 57. The titer of a silver nitrate solution is 22.7 mg Cl/mL. What is its titer in mg Br/mL?
EQUIVALENT WEIGHT CALCULATIONS 58. Calculate the equivalent weights of the following substances as acids or bases: (a) HCl, (b) Ba(OH)2 , (c) KH(IO3 )2 , (d) H2 SO3 , (e) CH3 COOH. 59. Calculate the molarity of a 0.250 eq/L solution of each of the acids or bases in Problem 58.
EQUIVALENT WEIGHT 60. Calculate the equivalent weight of KHC2 O4 (a) as an acid and (b) as a reducing agent in reaction with MnO4 − (5HC2 O4 − + 2MnO4 − + 11H+ → 10CO2 + 2Mn2+ + 8H2 O). 61. Mercuric oxide, HgO, can be analyzed by reaction with iodide and then titration with an acid: HgO + 4I− → Hgl4 2− + 2OH− . What is its equivalent weight? 62. Calculate the grams of one equivalent each of the following for the indicated reaction: (a) FeSO4 (Fe2+ → Fe3+ ), (b) H2 S (→ S0 ), (c) H2 O2 (→ O2 ), (d) H2 O2 (→ H2 O).
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63. BaCl2 · 2H2 O is to be used to titrate Ag+ to yield AgCl. How many milliequivalents are contained in 0.5000 g BaCl2 ·2H2 O?
EQUIVALENTS/L (eq/L) 64. A solution is prepared by dissolving 7.82 g NaOH and 9.26 g Ba(OH)2 in water and diluting to 500 mL. What is the concentration of the solution as a base in eq/L? 65. What weight of arsenic trioxide, As2 O3 , is required to prepare 1 L of 0.1000 eq/L arsenic(III) solution (arsenic As3+ is oxidized to As5+ in redox reactions)? 66. If 2.73 g KHC2 O4 ·H2 C2 O4 (three ionizable protons) having 2.0% inert impurities and 1.68 g KHC8 H4 O4 (one ionizable proton) are dissolved in water and diluted to 250 mL, what is the concentration of the solution as an acid in eq/L, assuming complete ionization? 67. A solution of KHC2 O4 ·H2 C2 O4 ·2H2 O (three replaceable hydrogens) is 0.200 eq/L as an acid. What is its concentration in eq/L as reducing agent? (See Problem 45 for its reaction as a reducing agent.) 68. Na2 C2 O4 and KHC2 O4 ·H2 C2 O4 are mixed in such a proportion by weight that the concentration of the resulting solution as a reducing agent in eq/L is 3.62 times the concentration as an acid in eq/L. What is the proportion? (See Problem 45 for its reaction as a reducing agent.) 69. What weight of K2 Cr2 O7 is required to prepare 1.000 L of 0.1000 eq/L solution? (It reacts as: Cr2 O7 2− + 14H+ + 6e− 2Cr3 + 7H2 O.)
CHARGE EQUIVALENT CALCULATIONS 70. A chloride concentration is reported as 300 mg/dL. What is the concentration in meq/L? 71. A calcium concentration is reported as 5.00 meq/L. What is the concentration in mg/dL? 72. A urine specimen has a chloride concentration of 150 meq/L. If we assume that the chloride is present in urine as sodium chloride, what is the concentration of NaCl in g/L?
GRAVIMETRIC CALCULATIONS 73. What weight of manganese is present in 2.58 g of Mn3 O4 ? 74. Zinc is determined by precipitating and weighing as Zn2 Fe(CN)6 . (a) What weight of zinc is contained in a sample that gives 0.348 g precipitate? (b) What weight of precipitate would be formed from 0.500 g of zinc? 75. Calculate the gravimetric factors for: Substance Sought
Substance Weighed
Mn Mn2 O3 Ag2 S CuCl2 MgI2
Mn3 O4 Mn3 O4 BaSO4 AgCl PbI2
PROFESSOR’s FAVORITE PROBLEM Contributed by Professor Thomas L. Isenhour, Old Dominion University 76. A 10.00 g sample contains only NaCl and KCl. The sample is dissolved and AgNO3 is added to precipitate AgCl. After the precipitate is washed and dried, it weighs 21.62 g. What is the weight percent of NaCl in the original sample?
Recommended References 1. T. P. Hadjiioannou, G. D. Christian, C. E. Efstathiou, and D. Nikolelis, Problem Solving in Analytical Chemistry. Oxford: Pergamon, 1988. 2. Q. Fernando and M. D. Ryan, Calculations in Analytical Chemistry. New York: Harcourt Brace Jovanovich, 1982.
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Chapter Six GENERAL CONCEPTS OF CHEMICAL EQUILIBRIUM “The worst form of inequality is to try to make unequal things equal.” —Aristotle
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
The equilibrium constant (key equations: 6.12, 6.15), pp. 194
●
Activity and activity coefficients (key equation: 6.19), p. 211
●
Calculation of equilibrium concentrations, p. 195
●
●
Using Excel Goal Seek to solve one-variable equations, p. 197
Thermodynamic equilibrium constants (key equation: 6.23), p. 217
●
The systematic approach to equilibrium calculations: mass balance and charge balance equations, p. 204
Even though in a chemical reaction the reactants may almost quantitatively react to form the products, reactions never go in only one direction. In fact, reactions reach an equilibrium in which the rates of reactions in both directions are equal. In this chapter we review the equilibrium concept and the equilibrium constant and describe general approaches for calculations using equilibrium constants. We discuss the activity of ionic species along with the calculation of activity coefficients. These values are required for calculations using thermodynamic equilibrium constants, that is, for the diverse ion effect, described at the end of the chapter. They are also used in potentiometric calculations (Chapter 13).
6.1 Chemical Reactions: The Rate Concept In 1863 Guldberg and Waage described what we now call the law of mass action, which states that the rate of a chemical reaction is proportional to the “active masses” of the reacting substances present at any time. The active masses may be concentrations or pressures. Guldberg and Waage derived an equilibrium constant by defining equilibrium as the condition when the rates of the forward and reverse reactions are equal. Consider the chemical reaction aA + bB cC + dD
(6.1)
According to Guldberg and Waage, the rate of the forward reaction is equal to a constant times the concentration of each species raised to the power of the number of molecules participating in the reaction: that is,1 Ratef wd = kf wd [A]a [B]b
(6.2)
where ratefwd is the rate of the forward reaction and kfwd is the rate constant, which is dependent on such factors as the temperature and the presence of catalysts. [A] and 1 [ ] represents moles/liter and here represents the effective concentration. The effective concentration will be
discussed under the diverse ion effect, when we talk about activities.
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[B] represent the molar concentrations of A and B. Similarly, for the reverse reaction, Guldberg and Waage wrote Raterev = krev [C]c [D]d
(6.3)
and for a system at equilibrium, the forward and reverse rates are equal: kfwd [A]a [B]b = krev [C]c [D]d
(6.4)
At equilibrium, the rate of the reverse reaction equals the rate of the forward reaction.
Rearranging these equations gives the molar equilibrium constant (which holds for dilute solutions) for the reaction, K: kfwd [C]c [D]d = =K [A]a [B]b krev
(6.5)
The expression obtained here is the correct expression for the equilibrium constant, but the method of derivation has no general validity. This is because reaction rates actually depend on the mechanism of the reaction, determined by the number of colliding species, whereas the equilibrium constant expression depends only on the stoichiometry of the chemical reaction. The sum of the exponents in the rate constant gives the order of the reaction, and this may be entirely different from the stoichiometry of the reaction (see Chapter 22). An example is the rate of reduction of S2 O8 2− with I− : S2 O8 2− + 3I− → 2SO4 2− + I3 −
Concentrations
The rate is actually given by kfwd [S2 O8 2− ][I− ] (a second-order reaction) and not kfwd [S2 O8 2− ][I− ]3 , as might be expected from the balanced chemical reaction (a fourth-order reaction would be predicted). The only sound theoretical basis for the equilibrium constant comes from thermodynamic arguments. See Gibbs free energy in Section 6.3 for the thermodynamic computation of equilibrium constant values. The value of K can be calculated empirically by measuring the concentrations of A, B, C, and D at equilibrium. Note that the more favorable the rate constant of the forward reaction relative to the backward reaction, the larger will be the equilibrium constant and the farther to the right the reaction will be at equilibrium. When the reaction between A and B is initiated, the rate of the forward reaction is large because the concentrations of A and B are large, whereas the backward reaction is slow because the concentrations of C and D are small (that rate is initially zero). As the reaction progresses, concentrations of A and B decrease and concentrations of C and D increase, so that the rate of the forward reaction diminishes while that for the backward reaction increases (Figure 6.1). Eventually, the two rates become equal, and the system is in a state of equilibrium. At this point, the individual concentrations of A, B, C, and D remain constant (the relative values will depend on the reaction stoichiometry, the initial concentrations, and how far the equilibrium lies to the right). However, the system remains in dynamic equilibrium, with the forward and backward reactions continuing at equal rates.
The larger the equilibrium constant, the farther to the right is the reaction at equilibrium.
A and B disappearing Large K
Equilibrium concentrations C and D appearing 0 Time 0
Initial Change state
Equilibrium
Small K
Fig. 6.1. reaction.
Progress of a chemical
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E
Fig. 6.2.
D
Extent of Reaction Completion
C B
A
Time
A large equilibrium constant does not assure the reaction will proceed at an appreciable rate.
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For the same equilibrium constant, the approach to equilibrium is controlled by the kinetics of the reaction, this cannot be predicted a priori. Ionic reactions are often instantaneous. Besides these and reactions involving combustion, most other reactions take measurable time. From A–E, at each step, the reaction rate increases by a factor of 3, but the position of equilibrium is not altered. An accelerated reaction rate can be brought about by an appropriate catalyst. Heating will also typically increase the reaction rate, but it may also affect the position of the equilibrium.
You will notice that the equilibrium constant expression is the ratio in which the concentrations of the products appear in the numerator and the concentrations of the reactants appear in the denominator. This is quite arbitrary, but it is the accepted convention. Hence, a large equilibrium constant indicates the equilibrium lies far to the right. We should point out that although a particular reaction may have a rather large equilibrium constant, the reaction may proceed from right to left if sufficiently large concentrations of the products are initially present. Also, the equilibrium constant tells us nothing about how fast a reaction will proceed toward equilibrium. Some reactions, in fact, may be so slow as to be unmeasurable. The equilibrium constant merely tells us the tendency of a reaction to occur and in what direction, not whether it is fast enough to be feasible in practice. (See Chapter 22 on kinetic methods of analysis for the measurement of reaction rates and their application to analyses.) For the reaction depicted in Equation 6.1, the rate at which equilibrium is approached will likely be different for either the forward or the reverse reaction. That is, if we start with a mixture of C and D, the rate at which equilibrium is approached may be much slower or faster than for the converse reaction. Figure 6.2 illustrates the approach to equilibrium at different reaction rates.
6.2 Types of Equilibria Equilibrium constants may be written for dissociations, associations, reactions, or distributions.
We can write equilibrium constants for virtually any type of chemical process. Some common equilibria are listed in Table 6.1. The equilibria may represent dissociation (acid/base, solubility), formation of products (complexes), reactions
Table 6.1
Types of Equilibria Equilibrium
Reaction
Equilibrium Constant
Acid–base dissociation Solubility Complex formation Reduction–oxidation Phase distribution
HA + H2 O H3 O+ + A− MA Mn+ + An− Mn+ + aLb− ML(n−ab)+ a Ared + Box Aox + Bred AH2 O Aorganic
Ka , acid dissociation constant Ksp , solubility product Kf , formation constant Keq , reaction equilibrium constant KD , distribution coefficient
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(redox), a distribution between two phases (water and nonaqueous solvent—solvent extraction; adsorption from water onto a surface, as in chromatography, etc.). We will describe some of these equilibria below and in later chapters.
6.3 Gibbs Free Energy and the Equilibrium Constant The tendency for a reaction to occur is defined thermodynamically from its change in enthalpy (H) and entropy (S). Enthalpy is the heat absorbed when an endothermic reaction occurs under constant pressure. When heat is given off (exothermic reaction), H is negative. Entropy is a measure of the disorder, or randomness, of a substance or system. A system will always tend toward lower energy and increased randomness, that is, lower enthalpy and higher entropy. For example, a stone on a hill will tend to roll spontaneously down the hill (lower energy state), and a box of marbles ordered by color will tend to become randomly ordered when shaken. The combined effect of enthalpy and entropy is given by the Gibbs free energy, G: G = H − TS
(6.6)
where T is the absolute temperature in kelvin; G is a measure of the energy of the system, and a system spontaneously tends toward lower energy states. The change in energy of a system at a constant temperature is G = H − TS
(6.7)
So a process will be spontaneous when G is negative, will be spontaneous in the reverse direction when G is positive, and will be at equilibrium when G is zero. Hence, a reaction is favored by heat given off (negative H), as in exothermic reactions, and by increased entropy (positive S). Both H and S can be either positive or negative, and the relative magnitudes of each and the temperature will determine whether G will be negative so that the reaction will be spontaneous. Enthalpy or entropy change alone cannot decide if a process will be spontaneous. Many salts, NH4 Cl for example, spontaneously dissolve in water in an endothermic process (heat is absorbed, the solution gets cold). In such cases, the large positive entropy of dissolution exceeds the positive enthalpy change. Standard enthalpy H ◦ , standard entropy S◦ , and standard free energy G◦ represent the thermodynamic quantities for one mole of a substance at standard state (P = 1 atm, T = 298 K, unit concentration). Then, ◦
◦
G = H − TS
◦
(6.8)
◦
G is related to the equilibrium constant of a reaction by ◦
K = e−G or
◦
/RT
G = −RT ln K = −2.303RT log K
(6.9)
J. Willard Gibbs (1839–1903) was the founder of chemical thermodynamics. He introduced the concept of free energy, now universally called Gibbs free energy in his honor. The Gibbs free energy relates the tendency of a physical or chemical system to simultaneously lower its energy and increase its disorder, or entropy, in a spontaneous natural process. Gibbs was awarded the first Ph.D. degree in engineering in the United States in 1863 by Yale University (his thesis was on the design of gearing). Albert Einstein, who relied on Gibb’s studies of thermodynamics and discoveries in statistical mechanics for his own work, called Gibbs “the greatest mind in American history”.
Everything in the universe tends toward increased disorder (increased entropy) and lower energy (lower enthalpy).
A spontaneous reaction results in energy given off and a lower free energy. At equilibrium, the free energy does not change.
(6.10)
where R is the gas constant (8.314 J K−1 mol−1 ). Hence, if we know the standard free energy of a reaction, we can calculate the equilibrium constant. Obviously, the larger G◦ (when negative), the larger will be K. Note that while G◦ and G give information about the spontaneity of a reaction, they say nothing about the rate at which it will occur. The reaction between hydrogen and oxygen to form water has a very large negative free energy change associated with it. But at room temperature, in the absence of a catalyst (or a spark!), these gases may coexist together for years without observable reaction.
A large equilibrium constant results from a large negative free energy change for the reaction in question.
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We can shift an unfavorable equilibrium by increasing the reactant concentration. All equilibrium constants are temperature dependent, as are the rates of reactions.
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6.4 Le Chˆatelier’s Principle The equilibrium concentrations of reactants and products can be altered by applying stress to the system, for example, by changing the temperature, the pressure, or the concentration of one of the reactants. The effects of such changes can be predicted from Le Chˆatelier’s principle, which states that when stress is applied to a system at chemical equilibrium, the equilibrium will shift in a direction that tends to relieve or counteract that stress. The effects of temperature, pressure, and concentrations on chemical equilibria are considered below.
6.5 Temperature Effects on Equilibrium Constants
French chemist Henry-Louise Le Chˆatelier (1850–1936), while working with thermodynamics, devised what became known as Le Chˆatelier’s principle in 1884. This principle states that if a system is in a state of equilibrium and one of the conditions is changed, such as the pressure or temperature, the equilibrium will shift in such a way as to try to restore the original equilibrium condition.
As we have mentioned, temperature influences the individual rate constants for the forward and backward reactions and therefore the equilibrium constant (more correctly, temperature affects the free energy—see Equation 6.10). An increase in temperature will displace the equilibrium in the direction that results in absorbing heat, since this removes the source of the stress. So an endothermic forward reaction (which absorbs heat) will be displaced to the right, with an increase in the equilibrium constant. The reverse will be true for an exothermic forward reaction, which releases heat. An exothermic reaction needs to release heat to proceed, a process that will be hindered at higher temperature. The extent of the displacement will depend on the magnitude of the heat of reaction for the system. Strictly speaking, enthalpy and entropy changes are not temperature independent. But in most cases it is a reasonable approximation that these are constant over a modest change in temperature, and the change in the equilibrium constant (K1 at temperature T1 and K2 at temperature T2 ) can be predicted by the Clausius-Clapeyron equation: K1 H 1 1 ln = − K2 R T2 T1 In addition to influencing the position of equilibrium, temperature has a pronounced effect on the rates of the forward and backward reactions involved in the equilibrium, and so it influences the rate at which equilibrium is approached. This is because the number and the energy of collisions among the reacting species increase with increasing temperature. The rates of many endothermic reactions increase about two- to threefold for every 10◦ C rise in temperature. See Figure 6.2.
6.6 Pressure Effects on Equilibria Rudolf Julius Emanuel Clausius (1822–1888), German physicist and mathematician, was one of the pioneers of thermodynamics. He coined the word entropy.
For solutions, pressure effects are usually negligible.
Pressure can have a large influence on the position of chemical equilibrium for reactions occurring in the gaseous phase. An increase in pressure favors a shift in the direction that results in a reduction in the volume of the system, for example, when one volume of nitrogen reacts with three volumes of hydrogen to produce two volumes of ammonia. But for reactions occurring in solutions, normal pressure changes have a negligible effect on the equilibrium because liquids cannot be compressed the way gases can. Note, however, that above a few atmospheres pressure, even “incompressible” fluids compress somewhat and one can learn about the electronic and mechanical changes to molecules by studying their spectroscopy under such conditions, for example, using high-pressure diamond cells.
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6.7 Concentration Effects on Equilibria The value of an equilibrium constant is independent of the concentrations of the reactants and products. However, the position of equilibrium is very definitely influenced by the concentrations. The direction of change is readily predictable from Le Chˆatelier’s principle. Consider the reaction of iron(III) with iodide:
Changes in concentration do not affect the equilibrium constant. They do affect the position of the equilibrium.
3I− + 2Fe3+ I3 − + 2Fe2+ If the four components are in a state of equilibrium, as determined by the equilibrium constant, addition or removal of one of the components would cause the equilibrium to reestablish itself. For example, suppose we add more iron(II) to the solution. According to Le Chˆatelier’s principle, the reaction will shift to the left to relieve the stress. Equilibrium will eventually be reestablished, and its position will still be defined by the same equilibrium constant.
6.8 Catalysts Catalysts either speed up or retard2 the rate at which an equilibrium is attained by affecting the rates of both the forward and the backward reactions. But catalysts affect both rates to the same extent and thus have no effect on the value of an equilibrium constant. See Figure 6.2. Catalysts are very important to the analytical chemist in a number of reactions that are normally too slow to be analytically useful. An example is the use of an osmium tetroxide catalyst to speed up the titration reaction between arsenic(III) and cerium(IV), whose equilibrium is very favorable but whose rate is normally too slow to be useful for titrations. The measurement of the change in the rate of a kinetically slow reaction in the presence of a catalyst can actually be used for determining the catalyst concentration. The same reaction between arsenic(III) and cerium(IV) is catalyzed also by iodide and the measurement of this reaction rate (through the disappearance of the yellow cerium(IV) color) constitutes the basis for what used to be the most widely used (and is still frequently used) method for measuring iodine, also called the Sandell-Kolthoff method. Modern methods used now for iodide include ion chromatography, ICP-MS, and ion-selective electrodes—see later chapters.
Catalysts do not affect the equilibrium constant or the position at equilibrium. See Chapter 22 for analytical uses of enzyme catalysts.
6.9 Completeness of Reactions If the equilibrium of a reaction lies sufficiently to the right that the remaining amount of the substance being determined (reacted) is too small to be measured by the measurement technique, we say the reaction has gone to completion. If the equilibrium is not so favorable, then Le Chˆatelier’s principle may be applied to make it so. We may either increase the concentration of a reactant or decrease the concentration of a product. Production of more product may be achieved by (1) allowing a gaseous product to escape, (2) precipitating the product, (3) forming a stable ionic complex of the product in solution, or (4) preferential extraction. It is apparent from the above discussion that Le Chˆatelier’s principle is the dominant concept behind most chemical reactions in the real world. It is particularly important in biochemical reactions, and external factors such as temperature can have a significant effect on biological equilibria. Catalysts (enzymes) are also key players in many biological and physiological reactions, as we shall see in Chapter 22. 2 Such “negative catalysts” are generally called inhibitors.
For quantitative analysis, equilibria should be at least 99.9% to the right for precise measurements. A reaction that is 75% to the right is still a “complete” reaction.
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6.10 Equilibrium Constants for Dissociating or Combining Species—Weak Electrolytes and Precipitates Equilibrium constants are finite when dissociations are less than 100%.
A weak electrolyte is only partially dissociated. Many slightly soluble substances are strong electrolytes because the portion that dissolves is totally ionized.
When a substance dissolves in water, it will often partially or completely dissociate or ionize. Electrolytes that tend to dissociate only partially are called weak electrolytes, and those that tend to dissociate completely are strong electrolytes. For example, acetic acid only partially ionizes in water and is therefore a weak electrolyte. But hydrochloric acid is completely ionized and is therefore a strong electrolyte. (Acid dissociations in water are really proton transfer reactions: HOAc + H2 O H3 O+ + OAc− .) Some substances completely ionize in water but have limited solubility; we call these slightly soluble substances. Substances may combine in solution to form a dissociable product, for example, a complex. An example is the reaction of copper(II) with ammonia to form the Cu(NH3 )4 2+ species. The dissociation of weak electrolytes or the solubility of slightly soluble substances can be quantitatively described by equilibrium constants. Equilibrium constants for completely dissolved and dissociated electrolytes are effectively infinite. Consider the dissociating species AB: AB A + B (6.11) The equilibrium constant for such a dissociation can be written generally as [A][B] = Keq [AB]
(6.12)
The larger Keq , the greater will be the dissociation. For example, the larger the equilibrium constant of an acid, the stronger will be the acid. Some species dissociate stepwise, and an equilibrium constant can be written for each dissociation step. A compound A2 B, for example, may dissociate as follows:
Successive stepwise dissociation constants become smaller.
A2 B A + AB
K1 =
[A][AB] [A2 B]
(6.13)
AB A + B
K2 =
[A][B] [AB]
(6.14)
The overall dissociation process of the compound is the sum of these two equilibria: A2 B 2A + B
Keq =
[A]2 [B] [A2 B]
(6.15)
If we multiply Equations 6.13 and 6.14 together, we arrive at the overall equilibrium constant: [A][AB] [A][B] · Keq = K1 K2 = [A2 B] [AB] =
[A]2 [B] [A2 B] (6.16)
When chemical species dissociate in a stepwise manner like this, the successive equilibrium constants generally become progressively smaller. For a diprotic acid (e.g., HOOCCOOH), the dissociation of the second proton is inhibited relative to the
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first (K2 < K1 ), because the negative charge on the monoanion makes it more difficult for the second proton to ionize. This effect is more pronounced the closer the proton sites are. Note that in equilibrium calculations we always use mol/L for solution concentrations. If a reaction is written in the reverse, the same equilibria apply, but the equilibrium constant is inverted. Thus, in the above example, for A + B AB, Keq(reverse) = [AB]/([A][B]) = 1/Keq(forward) . If Keq for the forward reaction is 105 , Kforward = 1/Kbackward then Keq for the reverse reaction is 10−5 . Similar concepts apply for combining species, except, generally, the equilibrium constant will be greater than unity rather than smaller, since the reaction is favorable for forming the product (e.g., complex). We will discuss equilibrium constants for acids, complexes, and precipitates in later chapters.
195
Kforward = 1/Kbackward
6.11 Calculations Using Equilibrium Constants—Composition at Equilibrium? Equilibrium constants are useful for calculating the concentrations of the various species in equilibrium, for example, the hydrogen ion concentration from the dissociation of a weak acid. In this section we present the general approach for calculations using equilibrium constants. The applications to specific equilibria are treated in later chapters dealing with these equilibria. CHEMICAL REACTIONS It is sometimes useful to know the concentrations of reactants and products in equilibrium in a chemical reaction. For example, we may need to know the amount of reactant for the construction of a titration curve or for calculating the potential of an electrode in the solution. These are, in fact, applications we consider in later chapters. Some example calculations will illustrate the general approach to solving equilibrium problems.
Example 6.1 The chemicals A and B react as follows to produce C and D: [C][D] A+BC+D K= [A][B] The equilibrium constant K has a value of 0.30. Assume 0.20 mol of A and 0.50 mol of B are dissolved in 1.00 L, and the reaction proceeds. Calculate the concentrations of reactants and products at equilibrium. Solution
The initial concentration of A is 0.20 M and that of B is 0.50 M, while C and D are initially 0 M. After the reaction has reached equilibrium, the concentrations of A and B will be decreased and those of C and D will be increased. Let x represent the equilibrium concentration of C or the moles/liter of A and B reacting. Since we get one mole of D with each mole of C, the concentration of D will also be x. We may represent the initial concentration of A and B as the analytical concentrations, CA and CB . The equilibrium concentrations are [A] and [B]. The concentrations of A and B will each be diminished by x, that is, [A] = CA − x and [B] = CB − x. So the equilibrium concentrations are
The equilibrium concentration is the initial (analytical) concentration minus the amount reacted.
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This approach to to solving such problems is often called creating an “ICE” diagram. Initial, Change, and Equilibrium conditions are charted to help construct the equilibrium expression to be solved. See for example http://www.youtube .com/user/genchemconcepts#p/a/u /5/LZtVQnILdrE.
Initial Change (x = mmol/mL reacting) Equilibrium
[A] 0.20 −x 0.20 − x
[B] 0.50 −x 0.50 − x
[C] 0 +x x
[D] 0 +x x
We can substitute these values in the equilibrium constant expression and solve for x: (x)(x) = 0.30 (0.20 − x)(0.50 − x) x2 = (0.10 − 0.70x + x2 )0.30 0.70x2 + 0.21x − 0.030 = 0
ICE diagram for Example 6.1 Concentration, M Initial Change Equilibrium
[A]
[B]
[C]
[D]
0.20 0.50 0.00 0.00 –x –x +x +x 0.20–x 0.50–x x x
This is a quadratic equation and can be solved algebraically for x using the quadratic formula given in Appendix B (see also Example 6.1 quadratic equation solution .xlsx on the website supplement for a quadratic equation solution calculator, also the Chapter 3 Solver video for solution of the quadratice equation): √ −b ± b2 − 4ac x= 2a
−0.21 ± (0.21)2 − 4(0.70)(−0.030) = 2(0.70) √ −0.21 ± 0.044 + 0.084 = 0.11 M = 1.40 [A] = 0.20 − x = 0.09 M [B] = 0.50 − x = 0.39 M [C] = [D] = x = 0.11 M
In successive approximations, we begin by taking the analytical concentration as the equilibrium concentration, to calculate the amount reacted. Then we repeat the calculation after subtracting the calculated reacted amount, until it is constant.
Instead of using the quadratic equation, we could also use the method of successive approximations. In this procedure, we will first neglect x compared to the initial concentrations to simplify calculations, and calculate an initial value of x. Then we can use this first estimate of x to subtract from CA and CB to give an initial estimate of the equilibrium concentration of A and B, and calculate a new x. The process is repeated until x is essentially constant. (x)(x) = 0.30 First calculation (0.20)(0.50) x = 0.173 The calculations converge more quickly if we keep an extra digit throughout. (x)(x) = 0.30 Second calculation (0.20 − 0.173)(0.50 − 0.173) x = 0.051 Third calculation
(x)(x) = 0.30 (0.20 − 0.051)(0.50 − 0.051) x = 0.142
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Fourth calculation
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(x)(x) = 0.30 (0.20 − 0.142)(0.50 − 0.142) x = 0.079
Fifth calculation
(x)(x) = 0.30 (0.20 − 0.79)(0.50 − 0.079) x = 0.124
Sixth calculation
(x)(x) = 0.30 (0.20 − 0.124)(0.50 − 0.124) x = 0.093
Seventh calculation
(x)(x) = 0.30 (0.20 − 0.093)(0.50 − 0.093) x = 0.114
Eighth calculation
(x)(x) = 0.30 (0.20 − 0.114)(0.50 − 0.114) x = 0.104
Ninth calculation
(x)(x) = 0.30 (0.20 − 0.104)(0.50 − 0.104) x = 0.107
We will take 0.11 as the equilibrium value of x since it essentially repeated the value of the seventh calculation. Note that in these iterations, x oscillates above and below the equilibrium value. The larger x solved for in a particular problem is compared to C, the larger will be the oscillations and the more iterations that will be required to reach an equilibrium value (as in this example—not the best for this approach). There is a more efficient way of completing the iteration. Take the average of the first and second for the third iteration, which should be close to the final value (in this case, 0.112 ). One or two more iterations will tell us we have reached the equilibrium value. Try it! In Example 6.1, appreciable amounts of A remained, even though B was in excess, because the equilibrium constant was not very large. In fact, the equilibrium was only about halfway to the right since C and D were about the same concentration as A. In most reactions of analytical interest, the equilibrium constants are large, and the equilibrium lies far to the right. In these cases, the equilibrium concentrations of reactants that are not in excess are generally very small compared to the other concentrations. This simplifies our calculations.
EXCEL GOAL SEEK FOR ITERATIVE PROBLEM SOLVING Microsoft ExcelTM provides several powerful tools to perform iterative or “trial-anderror” solutions. Frequently only one parameter needs to be solved for in a problem; this is the case for most equilibrium calculations. There may be more than one reactant or product but a single reaction parameter is multiplied by stoichiometric coefficients to calculate change in concentrations of all the reactants and products. In this case, the “Goal Seek” function in Excel is ideal to use; it is already built-in to your Excel
Shorten the number of iterations by taking the average of the first two for the next.
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software and nothing needs to be installed. Consider a slightly more challenging version of Example 6.1 where the reaction stoichiometry is A + 3B = C + 2D 0.2–x 0.5–x x 2x 2
[C][D] the equilibrium constant being [A] = 3.00 and the initial concentrations of A and [B]3 B being 0.20 and 0.50 M, while no C and D are initially present. We construct an ICE diagram on an Excel sheet. A screenshot below shows the formulas in the cells, rather than the values. (We have formatted the numbers in columns B to E to 2 decimals.) Note that a “reaction parameter,” equivalent to x in the previous ICE diagram for Example 6.1, is set up in cell G2 with an initial value of 0. (Don’t confuse [A] with the column B, etc., in setting formulas.) The changes in cells B4 to E4 are thus noted in terms of G2, and the equilibrium values in cells B5 to E5 are similarly noted in terms of initial values and G2. The equilibrium constant in cell C7 is accordingly expressed in terms of the equilibrium values.
The actual spreadsheet will not show the formulas and will look like:
Now put your cursor on cell C7 (the equilibrium constant expression), which is known to be 3.00 but presently reads zero. Click on the data tab. [This and following instructions pertain to Excel 2010 and will be slightly different for other versions. However, the Goal Seek function exists in all versions of Excel and you can go to Excel Help (key F1) and type in Goal Seek; it will give you guidance on how to access this function on your version of Excel. In older versions, it may be under Tools.] Once the Data menu bar opens, locate the “What-If Analysis” submenu and click on it, a drop-down menu will open. Click on Goal Seek. There are three data entry boxes in the Goal Seek operation. These are: “Set Cell,” “To Value,” and “By changing Cell.” Because you had your cursor on C7, the Goal Seek operation will already open with C7 in the “Set Cell” box. If this is not so, type in C7 or the location of your equilibrium constant expression, if you have decided to type it in some other cell. In the “To Value” box, type in the value of the equilibrium constant 3 and type in G2 in the “By changing Cell” box (or when you have clicked on this box, put your cursor on cell G2 and click). Click OK on the Goal Seek box. Voila! You have a solution with all the equilibrium values calculated.
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You can access the spreadsheets “Practice Goal Seek spreadsheet, Section 6.11, setup.xlsx” and “Practice Goal Seek spreadsheet, Section 6.11, answer.xlsx” in Chapter 6 of the text website (Goal Seek problems). USING GOAL SEEK TO SOLVE AN EQUATION Goal Seek can solve a single variable polynomial equation if there is a real unique solution to it. Solutions to a quadratic equation will be such a case. Consider that in Example 6.1 we were trying to solve the equation 0.70x2 + 0.21x = 0.03 Consider an Excel spreadsheet in which in cell A2, we have the variable x that we wish to vary (presently just zero is entered) and cell B2 the expression of the polynomial we wish to solve is entered in terms of A2. First the formula version is illustrated; then what the spreadsheet will actually look like is shown.
Next you click on B2, access Goal Seek, set cell B2 to the value of 0.03 and vary cell A2 to get to this goal. You will immediately be rewarded with the solution that x = 0.105 as shown below.
Do note that a nth degree polynomial ultimately has n possible solutions, except that for a real problem only one of them is logically possible. It will be up to you decide if the solution being produced is logically tenable. For example, negative values of x or values greater than 0.2 are not logically possible in the present illustration. The answer given here is close to the true value. But note that the target value did not quite converge to 0.03. This is a limitation of Goal Seek, and there is a way to get around it, as illustrated and explained below (Shortcomings of Excel-Based Approaches and How to Get Around Them), by multiplying the target equation by a large number, e.g., 10,000, and setting the target value 10,000 larger. Try it. If you do, the target value comes to 0.0300, and the x value is 0.01565, which agrees with
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the Solver calculation given in the Chapter 3 video and the Example 6.1 Quadratic Equation solution given in the text website.
Example 6.2 Assume that in Example 6.1 the equilibrium constant was 2.0 × 1016 instead of 0.30. Calculate the equilibrium concentrations of A, B, C, and D (starting with 0.20 mol A and 0.50 mol B in 1.00 L). Solution If the equilibrium constant for a reaction is very large, x is very small compared to the analytical concentration, which simplifies calculations.
Since K is very large, the reaction of A with B will be virtually complete to the right, leaving only traces of A at equilibrium. Let x represent the equilibrium concentration of A. An amount of B equal to A will have reacted to form an equivalent amount of C and D (about 0.20 M for each). We can summarize the equilibrium concentrations as follows: [A] = x [B] = (0.50 − 0.20) + x = 0.30 + x [C] = 0.20 − x [D] = 0.20 − x Or, looking at the equilibrium, A + x
B C 0.30 + x 0.20 − x
+
D 0.20 − x
Basically, we have said that all of A is converted to a like amount of C and D, except for a small amount x. Now x will be very small compared to 0.20 and 0.30 and can be neglected. So we can say [A] = x [B] ≈ 0.30 [C] ≈ 0.20 [D] ≈ 0.20 The only unknown concentration is [A]. Substituting these numbers in the equilibrium constant expression, we have (0.20)(0.20) = 2.0 × 1016 (x)(0.30) x = [A] = 6.7 × 10−18 M (usually analytically undetectable)
Neglect x compared to C (product) if C ≤ 0.01Keq in a reaction.
In this case the calculation was considerably simplified by neglecting x in comparison to other concentrations. If x should turn out to be significant compared to these concentrations, then the solution should be reworked using the quadratic formula, or Goal Seek. Generally, if the value of x is less than about 5% of the assumed concentration, it can be neglected. In this case, the error in x itself is usually 5% or less. This simplification will generally hold if the product concentration is less than 1% at K eq , that is ≤ 0.01 K eq . However, once you have mastered using Excel, especially its Goal Seek and Solver functions discussed later in the book, (Solver-based
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equilibrium calculations are also discussed in reference 8), you may find that it is just as simple or simpler to solve a problem without any approximation using Excel. This is because no judgments on what can or cannot be neglected is needed. A video illustrating the use of Goal Seek to solve an equilibrium problem is on the text website.
Example 6.3
Video: Goal Seek Equilibrium
A and B react as follows: [C]2 [A][B]2 Assume 0.10 mol of A is reacted with 0.20 mol of B in a volume of 1000 mL; K = 1.0 × 1010 . What are the equilibrium concentrations of A, B, and C? A + 2B 2C
K=
Solution
We have stoichiometrically equal amounts of A and B, so both are virtually all reacted, with trace amounts remaining. Let x represent the equilibrium concentration of A. At equilibrium, we have A + x
2B 2C 2x 0.20 − 2x ≈ 0.20
For each mole of A that either reacts (or is produced), we produce (or remove) two moles of C, and consume (or produce) two moles of B. Substituting into the equilibrium constant expression,
x = [A] =
3
(0.20)2 = 1.0 × 1010 (x)(2x)2 0.040 = 1.0 × 1010 4x3
4.0 × 10−2 3 = 1.0 × 10−12 = 1.0 × 10−4 M 10 4.0 × 10 B = 2x = 2.0 × 10−4 M
(analytically detectable, but not appreciable compared to the starting}?> concentration)
DISSOCIATION EQUILIBRIA Calculations involving dissociating species are not much different from the example just given for chemical reactions.
Example 6.4 Calculate the equilibrium concentrations of A and B in a 0.10 M solution of a weak electrolyte AB with an equilibrium constant of 3.0 × 10−6 . Solution
AB A + B
Keq =
[A][B] [AB]
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Both [A] and [B] are unknown and equal. Let x represent their equilibrium concentrations. The concentration of AB at equilibrium is equal to its initial analytical concentration minus x. AB A + B 0.10 − x x x In a dissociation, neglect x compared to the initial concentration C if C ≥ 100Keq in a dissociation.
The value of Keq is quite small, so we are probably justified in neglecting x compared to 0.10. Otherwise, we will have to use a quadratic equation. Substituting into the Keq expression, (x)(x) = 3.0 × 10−6 0.10
x = [A] = [B] = 3.0 × 10−7 = 5.5 × 10−4 M After the calculation is done, check if the approximation you made was valid. Here the calculated value of x can indeed be neglected compared to 0.10.
SHORTCOMINGS OF EXCEL-BASED APPROACHES AND HOW TO GET AROUND THEM Goal Seek and Solver functions in Excel attempt to achieve convergence to some specified value by varying the value(s) of some other parameters. Because of quirks built into the software, convergence to the specified value is judged in terms of absolute difference rather than relative difference. For example, if you are using Goal Seek to converge to a value of 3 × 10−6 , the software may regard that reaching zero is close enough. (If you were an accountant and counting pennies, this may indeed be close enough, but this is not the case in most scientific problems.) Now let us consider Example 6.4. Without any approximation, the equation that you need to solve is: x2 = 3.0 × 10−6 0.10 − x One advantage of solving such a problem in Excel is that we do not really need to make a nice equation in one line out of this; we can put in the variable x as being in the cell A2 and then write the expression on the left-hand side of the above equation in cell B2 as shown below.
If you now go through the Goal Seek exercise by setting cell B2 to 3E-6 (a shorthand notation for writing 3 × 10−6 in Excel) by changing cell A2, you will find that it says it found a solution but in fact it did not change anything! That is, the answer in cell A2 is 0.00E+00, still zero. This is because the software mistakenly believes that for your purposes 0 is close enough to 3 × 10−6 . Consider that we multiply both sides of the above equation with the same large number, say 1010 , so that the equation now is x2 = 3.0 × 104 0.10 − x In the Excel spreadsheet in cell B2 we have to put in a 1E10 multiplier before the rest of the expression. Of course, mathematically there is no difference between the two equations and solutions to either will produce the same correct value of x. From the software’s point of view, however, you will now want the convergence value to be 1010 ×
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3 × 104 . If you go through the Goal Seek exercise again (setting the new expression in cell B2 to attain a value of 3E4 by varying A2), you will see that this time it does work, giving you the correct solution for x as 5.5E-4.
You can see the spreadsheet setup and the answer in the text website (Goal Seek problems).
6.12 The Common Ion Effect—Shifting the Equilibrium Equilibria can be markedly affected by adding one or more of the species present, as is predicted from Le Chˆatelier’s principle. Example 6.5 illustrates this principle.
Example 6.5 Assume that A and B are an ion pair, which can dissociate into A (a cation) and B (an anion). Recalculate the concentration of A in Example 6.4, assuming that the solution also contains 0.20 M B. Solution
We can represent the equilibrium concentration as follows:
Initial Change (x = mmol/mL of AB dissociated) Equilibrium
[AB] 0.10 −x 0.10 − x ≈ 0.10
[A] 0 +x x
[B] 0.20 +x 0.20 + x ≈ 0.20
The value of x will be smaller now than before because of the common ion effect of B, so we can certainly neglect it compared to the initial concentrations. Substituting in the equilibrium constant expression, (x)(0.20) = 3.0 × 10−6 (0.10) x = 1.5 × 10−6 M The concentration of A was decreased nearly 400-fold.
The common ion effect can be used to make analytical reactions more favorable or quantitative. The adjustment of acidity, for example, is frequently used to shift equilibria. Titrations with potassium dichromate, for example, are favored in acid solution, since protons are consumed in the reaction. Titrations with iodine, a weak oxidizing agent, are usually done in slightly alkaline solution to shift the equilibrium toward completion of the reaction, for example, in titrating arsenic(III): H3 AsO3 + I2 + H2 O H3 AsO4 + 2I− + 2H+
Adjusting the pH is a common way of shifting the equilibrium.
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6.13 SYSTEMATIC APPROACH TO EQUILIBRIUM CALCULATIONS—HOW TO SOLVE ANY EQUILIBRIUM PROBLEM Now that some familiarity has been gained with equilibrium problems, we will consider a systematic approach for calculating equilibrium concentrations that will work with all equilibria, no matter how complex. It consists of identifying the unknown concentrations involved and writing a set of simultaneous equations equal to the number of unknowns. Simplifying assumptions are made with respect to relative concentrations of species (not unlike the approach we have already taken) to shorten the solving of the equations. This approach involves writing expressions for mass balance of species and one for charge balance of species as part of our equations. We will first describe how to arrive at these expressions. MASS BALANCE EQUATIONS The principle of mass balance is based on the law of mass conservation, and it states that the number of atoms of an element remains constant in chemical reactions because no atoms are produced or destroyed. The principle is expressed mathematically by equating the concentrations, usually in molarities. The equations for all the pertinent chemical equilibria are written, from which appropriate relations between species concentrations are written.
Example 6.6 Write the equation of mass balance for a 0.100 M solution of acetic acid. Solution
The equilibria are HOAc H+ + OAc− H2 O H+ + OH− In a mass balance, the analytical concentration is equal to the sum of the concentrations of the equilibrium species derived from the parent compound (or an appropriate multiple).
We know that the analytical concentration of acetic acid is equal to the sum of the equilibrium concentrations of all its species: CHOAc = [HOAc] + [OAc− ] = 0.100 M A second mass balance expression may be written for the equilibrium concentration of H+ , which is derived from both HOAc and H2 O. We obtain one H+ for each OAc− and one for each OH− : [H+ ] = [OAc− ] + [OH− ]
Example 6.7 Write the equations of mass balance for a 1.00 × 10−5 M[Ag(NH3 )2 ]Cl solution. Solution
The equilibria are [Ag(NH3 )2 ]Cl → Ag(NH3 )2 + + Cl− Ag(NH3 )2 + Ag(NH3 )+ + NH3
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Ag(NH3 )+ Ag+ + NH3 NH3 + H2 O NH4 + + OH− H2 O H+ + OH− The Cl− concentration is equal to the concentration of the salt that dissociated, that is, 1.00 × 10−5 M. Likewise, the sum of the concentrations of all silver species is equal to the concentration of Ag in the original salt that dissociated: CAg = [Ag+ ] + [Ag(NH3 )+ ] + [Ag(NH3 )2 + ] = [Cl− ] = 1.00 × 10−5 M We have the following nitrogen-containing species: NH4 + NH3 Ag(NH3 )+ Ag(NH3 )2 + The concentration of N from the last species is twice the concentration of Ag(NH3 )2 + . For this species, the concentration of the nitrogen is twice the concentration of the Ag(NH3 )2 + , since there are two NH3 per molecule. Hence, we can write CNH3 = [NH4 + ] + [NH3 ] + [Ag(NH3 )+ ] + 2[Ag(NH3 )2 + ] = 2.00 × 10−5 M Finally, we can write
[OH− ] = [NH4 + ] + [H+ ]
Some of the equilibria and the concentrations derived from them may be insignificant compared to others and may not be needed in subsequent calculations. We have seen that several mass balance expressions may be written. Some may not be needed for calculations (we may have more equations than unknowns), or some may be simplified or ignored due to the small concentrations involved compared to others. This will become apparent in the equilibrium calculations below. CHARGE BALANCE EQUATIONS According to the principle of electroneutrality, all solutions are electrically neutral; that is, there is no solution containing a detectable excess of positive or negative charge because the sum of the positive charges equals the sum of negative charges. We may write a single charge balance equation for a given set of equilibria.
Example 6.8 Write a charge balance equation for a solution of H2 CO3 . Solution
The equilibria are +
H2 CO3 H + HCO3
−
HCO3 − H+ + CO3 2− H2 O H+ + OH− Dissociation of H2 CO3 gives H+ and two anionic species, HCO3 − and CO3 2− , and that of water gives H+ and OH− . The amount of H+ from that portion of completely dissociated H2 CO3 is equal to twice the amount of CO3 2− formed, and from partial (first step) dissociation is equal to the amount of HCO3 − formed. That is, for each CO3 2− formed, there are 2 H+ ; for each HCO3 − formed, there is 1 H+ ; and for each OH− formed, there is 1 H+ . Now, for the singly charged species, the charge
In a charge balance, the sum of the charge concentration of cationic species equals the sum of charge concentration of the anionic species in equilibrium.
The charge concentration is equal to the molar concentration times the charge of a species.
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concentration is identical to the concentration of the species. But for CO3 2− , the charge concentration is twice the concentration of the species, so we must multiply the CO3 2− concentration by 2 to arrive at the charge concentration from it. According to the principle of electroneutrality, positive charge concentration must equal the negative charge concentration. Hence, [H+ ] = 2[CO3 2− ] + [HCO3 − ] + [OH− ] Note that while there may be more than one source for a given species (H+ in this case), the total charge concentrations from all sources is always equal to the net equilibrium concentration of the species multiplied by its charge.
Example 6.9 Here, we neglect the dissociation of water. However, to be comprehensive, you can always include the formation of H+ and OH− from water in the charge balance expression for an aqueous system. Incorporating ions derived from the dissociation of water becomes more important as the concentration of other ions in the solutions decrease, i.e., it is more important in very dilute solutions. The pH of 10−3 M HCl can be correctly calculated, for example without taking into account the dissociation of water but the pH of 10−6 M HCl cannot.
Write a charge balance expression for a solution containing KCl, Al2 (SO4 )3 , and KNO3 . Neglect the dissociation of water. Solution
[K+ ] + 3[Al3+ ] = [Cl− ] + 2[SO4 2− ] + [NO3 − ]
Example 6.10 Write a charge balance equation for a saturated solution of CdCO3 . Solution
The equilibria are CdCO3 Cd2+ + CO3 2− CO3 2− + H2 O HCO3 − + OH− HCO3 − + H2 O H2 CO3 + OH− H2 O H+ + OH− Again, the charge concentration for the singly charged species (H+ , OH− , HCO3 − ) will be equal to the concentrations of the species. But for Cd2+ and CO3 2− the charge concentration will be twice their concentrations. We must again equate the positive and negative charge concentrations. 2[Cd2+ ] + [H+ ] = 2[CO3 2− ] + [HCO3 − ] + [OH− ]
Example 6.11 Write a charge balance equation for Example 6.7. Solution
[Ag+ ] + [Ag(NH3 )+ ] + [Ag(NH3 )2 + ] + [NH4 + ] + [H+ ] = [Cl− ] + [OH− ] Since all are singly charged species, the charge concentrations are equal to the molar concentrations.
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EQUILIBRIUM CALCULATIONS USING THE SYSTEMATIC APPROACH——THE STEPS We may now describe the systematic approach for calculating equilibrium concentrations in problems involving several equilibria. The basic steps can be summarized as follows:
In the systematic approach, a series of equations equal in number to the number of unknown species is written. These are simultaneously solved, using approximations to simplify.
1. 2. 3. 4. 5.
Write the chemical reactions appropriate for the system. Write the equilibrium constant expressions for these reactions. Write all the mass balance expressions. Write the charge balance expression. Count the number of chemical species involved and the number of independent equations (from steps 2, 3, and 4). If the number of equations is greater than or equal to the number of chemical species, then a solution is possible. At this point, it is possible to proceed to an answer. 6. Make simplifying assumptions concerning the relative concentrations of chemical species. At this point you need to think like a chemist so that the math will be simplified. 7. Calculate the answer. 8. Check the validity of your assumptions!
Let us examine one of the examples worked before, but using this approach.
Example 6.12 Repeat the problem stated in Example 6.4 using the systematic approach outlined above. Chemical reactions AB = A + B Equilibrium constant expressions [A][B] = 3.0 × 10−6 [AB]
(1)
CAB = [AB] + [A] = 0.10 M
(2)
[A] = [B]
(3)
Keq = Mass balance expressions
Remember that C represents the total analytical concentration of AB. Charge balance expression There is none because none of the species is charged. Number of expressions versus number of unknowns There are three unknowns ([AB], [A], and [B]) and three expressions (one equilibrium and two mass balance).
Use equilibrium constant expressions plus mass and charge balance expressions to write the equations.
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Use the same rules as before for simplifying assumptions (CA ≥ 100Keq for dissociations, C ≤ 0.01Keq for reactions).
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Simplifying assumptions: We want the equilibrium concentrations of A, B, and AB. Because K is small, very little AB will dissociate, so from (2): [AB] = CAB − [A] = 0.10 − [A] ≈ 0.10 M Calculate [AB] was found above. [A] can be found from (1) and (3). [A][B] = 3.0 × 10−6 0.10
[A] = 3.0 × 10−7 = 5.5 × 10−4 M [B] can be found from (3): [B] = [A] = 5.5 × 10−4 M Check [AB] = 0.10 − 5.5 × 10−4 = 0.10 M (within significant figures)
The systematic approach is applicable to multiple equilibria.
You see that the same answer was obtained as when the problem was worked intuitively as in Example 6.4. You may think that the systematic approach is excessively complicated and formal. For this extremely simple problem that may be a justified opinion. However, you should realize that the systematic approach will be applicable to all equilibrium calculations, regardless of the difficulty of the problem. You may find problems involving multiple equilibria and/or many species to be hopelessly complicated if you use only an intuitive approach. Nevertheless, you should also realize that a good intuitive “feel” for equilibrium problems is an extremely valuable asset. You should attempt to improve your intuition concerning equilibrium problems. Such intuition comes from experience gained by working a number of problems of different varieties. As you gain experience you will be able to shorten some of the formalism of the systematic approach, and you will find it easier to make appropriate simplifying assumptions. Although need for making approximations may no longer exist, the ability to set up the charge balance and mass balance equations is needed even if you use an Excel-based approach to solve the problem.
6.14 Some Hints for Applying the Systematic Approach for Equilibrium Calculations Mass balances: 1. One will be for the total analytical concentration of the main parent species. 2. Others will be for species of interest, e.g., H+ and other (dissociated) species in equilibrium. Charge balances: 1. Charge balance equations are simply adding all cationic species on one side and all anionic species on the other, each multiplied by the respective charges. 2. Both mass and charge balance equations are rarely needed for solving the equilibrium calculations; in the case of ionic equilibria, charge balance equations are often easier to write.
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Solving the equilibrium concentrations: 1. Using simplifying assumptions, at least one of the equilibrium species concentrations can be estimated. 2. From (substituting) this, the other species can be calculated. Follow the rules given after Example 6.11.
Example 6.13 Repeat the problem outlined in Example 6.5 using the systematic approach. Assume the charge on A is +1, the charge on B is −1, and that the extra B (0.20 M) comes from MB; MB is completely dissociated. Solution
Chemical reactions AB = A+ + B− MB → M+ + B− Equilibrium expressions Keq =
[A+ ][B− ] = 3.0 × 10−6 [AB]
Mass balance expressions CAB = [AB] + [A+ ] = 0.10 M [B− ] = [A+ ] + [M+ ] = [A+ ] + 0.20 M
(1)
(2) (3)
Charge balance expression [A+ ] + [M+ ] = [B− ]
(4)
Number of expressions versus number of unknowns There are three unknowns ([AB], [A+ ], and [B− ]; the concentration of M+ is known to be 0.20 M) and three independent expressions (one equilibrium and two mass balance; the charge balance is the same as the second mass balance). Simplifying assumptions (i) Because Keq is small, very little AB will dissociate, so from (2). [AB] = 0.10 − [A+ ] ≈ 0.10M (ii) [A] [M] so from (3) or (4): [B− ] = 0.20 + [A+ ] ≈ 0.20 M Calculate [A] is now found from (1): [A+ ](0.20) = 3.0 × 10−6 0.10 [A+ ] = 1.5 × 10−6 M
Both mass and charge balance equations are often not needed. Both are needed, however, to derive the shape of titration curve.
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Check (i) [AB] = 0.10 − 1.5 × 10−6 = 0.10 M (ii) [B] = 0.20 + 1.5 × 10−6 = 0.20 M
SOLVING EXAMPLE 6.13 USING GOAL SEEK Set up the spreadsheet using individual columns for [A+ ], [B− ], [M+ ], [AB], and Keq . The “formula view” of the spreadsheet then would look like:
Note the 1E10 multiplier so as to avoid the aforementioned pitfall in Excel. Now Goal Seek, setting E2 to 3E4 by varying A2 (which is equal to x), you will find that A2 will readily converge to 1.50E-6, see the text website. We will in general use the approximation approaches given in Sections 6.10 and 6.11, which actually incorporate many of the equilibria and assumptions used in the systematic approach. The use of the systematic approach for problems involving multiple equilibria is discussed in Chapter 8. We can now write some general rules for solving chemical equilibrium problems, using the approximation approach. These rules should be applicable to acid–base dissociation, complex formation, oxidation–reduction reactions, and others. That is, all equilibria can be treated similarly.
1. Write down the equilibria involved. 2. Write the equilibrium constant expressions and the numerical values. 3. From a knowledge of the chemistry involved, let x represent the equilibrium concentration of the species that will be unknown and small compared to other equilibrium concentrations; other species of unknown and small concentrations will be multiples of this. 4. List the equilibrium concentrations of all species, adding or subtracting the appropriate multiple of x from the analytical concentration where needed. 5. Make suitable approximations by neglecting x compared to finite equilibrium concentrations. This is generally valid if the finite concentration is about 100 × Keq or more. Also, if the calculated x is less than approximately 5% of the finite concentration, the assumption is valid. 6. Substitute the approximate representation of individual concentration into the equilibrium constant expression and solve for x. 7. If the approximations in step 5 are invalid, use the quadratic formula to solve for x or use an Excel-based approach.
The application of these rules will become more apparent in subsequent chapters when we deal with specific equilibria in detail.
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6.15 Heterogeneous Equilibria—Solids Don’t Count Equilibria in which all the components are in solution (a “homogeneous” medium) generally occur quite rapidly. If an equilibrium involves two phases (“heterogeneous”), the rate of approach to equilibrium will generally be substantially slower than in the case of solutions. An example is the dissolution of a poorly soluble solid or the formation of a precipitate, neither of these processes will be instantaneous. Another way in which heterogeneous equilibria differ from homogeneous equilibria is the manner in which the different constituents offset the equilibrium. Guldberg and Waage showed that when a solid is a component of a reversible chemical process, its active mass can be considered constant, regardless of how much of the solid is present. That is, when any amount of the solid is already present, adding more solid does not bring about a shift in the equilibrium. So the expression for the equilibrium constant need not contain any concentration terms for substances present as solids. That is, the standard state of a solid is taken as that of the solid itself, or unity. Thus, for the equilibrium CaF2 Ca2+ + 2F− we write that
Keq = [Ca2+ ][F− ]2
The same is true for pure liquids (undissolved) in equilibrium, such as mercury. The standard state of water is taken as unity in dilute aqueous solutions, and water does not appear in equilibrium constant expressions.
6.16 Activity and Activity Coefficients—Concentration Is Not the Whole Story Generally, the presence of diverse salts (not containing ions common to the equilibrium involved) will cause an increase in dissociation of a weak electrolyte or in the solubility of a precipitate. Cations attract anions, and vice versa, and so the cations of the analyte
Ca2+ H2O
+
H2O
Na
H2O
Cl
H2O
H2O
H2O
H2O H2O
H2O H2O
H2O
(a)
K+
NO3−
SO42−
Ca2+ H2O
H2O
H2O
(a) In solution, Na+ and Cl from salt form an ion atmosphere where each ion type has more of the oppositely charged ions as nearest neighbors. The structure is very dynamic in nature (all species are rapidly shifting). (b) Addition of an inert salt, such as KNO3 , decreases the attraction between ion pairs by shielding and reducing the effective charge and thus increases the solubility of sparingly soluble salts such as CaSO4 (see Section 6.16). −
Na
Cl − H2O
Fig. 6.3.
KNO3
+
H2O
The “effective concentration” of an ion is decreased by shielding it with other “inert” ions, and it represents the activity of the ion.
H2O
H2O
Na +
Consider a saturated solution of sugar with undissolved sugar remaining at the bottom. The relevant equilibrium constant is Sugarsolution /Sugarsolid . We know well that adding more solid sugar to a saturated solution will not increase the solution concentration further. If the equilibrium constant is indeed a constant, obviously adding more solid sugar to the system does not change the “concentration” of the solid sugar. Any amount of undissolved solid in the saturated solution system represents the same “concentration” of the solid.
m= 0
H2O
Cl−
The “concentration” of a pure solid or liquid is unity.
SO42−
H2O
H2O
H2O
−
H2O
Heterogeneous equilibria are slower than solution equilibria.
NO3−
NO3−
K+
H2O
H2O
Diverse ion effect
(b)
K+
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Activities are important in potentiometric measurements. See Chapter 13.
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attract anions of the diverse electrolyte and the anions of the analyte are surrounded by the cations of the diverse electrolyte. The attraction of the ions participating in the equilibrium of interest by the dissolved electrolyte effectively shields them, decreasing their effective concentration and shifting the equilibrium. We say that an “ion atmosphere” is formed about the cation and anion of interest. As the charge on either the diverse salt or the ions of the equilibrium reaction is increased, the diverse salt effect generally increases. This effect on the equilibrium is not predicted by Le Chˆatelier’s principle; but it is readily understood if you think in terms of the effective concentrations being changed. This “effective concentration” of an ion in the presence of an electrolyte is called the activity of the ion. To quantitatively describe the effects of salts on equilibrium constants, one must use activities, not concentrations (see the diverse salt effect below). In potentiometric measurements, it is activity that is measured, not concentration (see Chapter 13). In this section we describe how to estimate activity. THE ACTIVITY COEFFICIENT
G. N. Lewis introduced the thermodynamic concept of activity in 1908 in a paper entitled “The Osmotic Pressure of Concentrated Solutions and the Laws of Perfect Solutions.”
The activity of an ion ai is defined by ai = Ci fi
(6.17)
where Ci is the concentration of the ion i and fi is its activity coefficient. The concentration is usually expressed as molarity, and the activity has the same units as the concentration. The activity coefficient is dimensionless, but numerical values for activity coefficients do depend on the choice of standard state. The activity coefficient varies with the total number of ions in the solution and with their charge, and it is a correction for interionic attraction. In dilute solution, less than 10−4 M, the activity coefficient of a simple electrolyte is near unity, and activity is approximately equal to the concentration. As the concentration of an electrolyte increases, or as an extraneous salt is added, the activity coefficients of ions decrease, and the activity becomes less than the concentration. Note, however, that at much higher concentrations a different effect comes into play. Ions, especially cations, are hydrated in aqueous solution and the associated water of solvation becomes unavailable to function as solvent. This causes the activity coefficient to reach a minimum as a function of concentration and at very high concentrations; it has a value greater than unity. IONIC STRENGTH
For ionic strengths less than 10−4 , activity coefficients are near unity. In 1921 Lewis and Randall first introduced the empirical concept of ionic strength and showed that in dilute solution, the logarithm of the activity coefficient is proportional to the square root of the ionic strength.
From the above discussion, we can see that the activity coefficient is a function of the total electrolyte concentration of the solution. The ionic strength is a measure of total electrolyte concentration and is defined by μ=
1 Ci Zi2 2
(6.18)
where μ is the ionic strength and Zi is the charge on each individual ion. All cations and anions present in solution are included in the calculation. Obviously, for each positive charge there will be a negative charge.
Example 6.14 Calculate the ionic strength of a 0.2 M solution of KNO3 and a 0.2 M solution of K2 SO4 .
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Solution
For KNO3 , μ=
2 CK+ ZK2 + + CNO3 − ZNO
[K+ ] = 0.2 M μ=
3
−
2 [NO3 − ] = 0.2 M Higher charged ions contribute more to the ionic strength.
0.2 × (1)2 + 0.2 × (1)2 = 0.2 2
For K2 SO4 , μ=
CK+ ZK2 + + CSO4 2− Z 2
−
SO4 2
2
+
[K ] = 0.4 M
[SO4 2− ] = 0.2 M
So, 0.4 × (1)2 + 0.2 × (2)2 = 0.6 2 Note that due to the doubly charged SO4 2− , the ionic strength of the same molar concentration of K2 SO4 is three times that of the KNO3 . μ=
In 1923, Dutch physicist Petrus (Peter) Debye (1884–1966), together with his assistant Erich H¨uckel (1896–1980), developed the Debye-H¨uckel theory of electrolyte solutions, an improvement of Svante Arrhenius’s theory of electrical conductivity in electrolytic solutions.
If more than one salt is present, then the ionic strength is calculated from the total concentration and charges of all the different ions. For any given electrolyte, the ionic strength will be proportional to the concentration. Strong acids that are completely ionized are treated in the same manner as salts. If the acids are partially ionized, then the concentration of the ionized species must be estimated from the ionization constant before the ionic strength is computed. Very weak acids can usually be considered to be nonionized and do not contribute to the ionic strength.
Example 6.15 Calculate the ionic strength of a solution consisting of 0.30 M NaCl and 0.20 M Na2 SO4 . μ=
2 CNa+ ZNa +
+
2 CCl− ZCl −
+
Peter J. W. Debye
2 CSO4 2− ZSO 2− 4
2 0.70 × (1) + 0.30 × (1)2 + 0.20 × (2)2 = 2 = 0.90 2
CALCULATION OF ACTIVITY COEFFICIENTS In 1923, Debye and H¨uckel derived a theoretical expression for calculating activity ¨ coefficients. The original Debye-Huckel equation is given as Equation 6.19a below but it is of limited use as it can be used only in extremely dilute solutions: √ −log fi = AZi2 μ (6.19a) They later provided a more useful equation, known as the Extended Debye– ¨ Huckel equation: √ AZi2 μ − log fi = (6.19b) √ 1 + Bai μ
Erich A. A. J. H¨uckel
This equation applies for ionic strengths up to 0.2.
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The estimation of the ion size parameter places a limit on the accuracy of the calculated activity coefficient.
A and B are constants; the values are, respectively, 0.51 and 0.33 for water at 25◦ C. At other temperatures, the values can be computed from A = 1.82 × 106 (DT)−3/2 and B = 50.3(DT)−1/2 where D and T are the dielectric constant and the absolute temperature, respectively; ai is the ion size parameter, which is the effective diameter ˚ An angstrom is 100 picometers (pm, 10−10 of the hydrated ion in angstrom units, A. meter). A limitation of the Debye–H¨uckel equation is the accuracy to which ai can be ˚ and for practical evaluated. For many singly charged ions, ai is generally about 3 A, purposes Equation 6.19b simplifies to √ 0.51Zi2 μ − log fi = √ 1+ μ
This equation may be used for ionic strengths less than 0.01. See Reference 10 for a tabulation of ai values.
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(6.20)
˚ But at ionic For common multiply charged ions, ai may become as large as 11 A. strengths less than 0.01, the second term of the denominator becomes small with respect to 1, so uncertainties in ai become relatively unimportant, and Equation 6.20 can be applied at ionic strengths of 0.01 or less. Equation 6.19b can be applied up to ionic strengths of about 0.2. Reference 10 at the end of the chapter lists values for ai for different ions and also includes a table of calculated activity coefficients, using Equation 6.19b, at ionic strengths ranging from 0.0005 to 0.1. This paper, with the complete list of ion size parameters, is available on the text website. Excel answers using Equations 6.19b and 6.20 for the following two problems are on the website (Spreadsheets). Table 6.2 above contains a list of ion size parameters for some common ions taken from this reference.
Table 6.2
Ion Size Parameters for Common Ions ˚ (Angstroms) Ion size parameter A
Ion H+ (C3 H7 )4 N+ (C3 H7 )3 NH+ , {OC6 H2 (NO3 )3 }− Li+ , C6 H5 COO− , (C2 H5 )4 N+ CHCl2 COO− , (C2 H5 )3 NH+ Na+ , IO3 − , HSO3 − , (CH3 )3 NH+ , C2 H5 NH3 + K+ , Cl− , Br− , I− , CN− , NO2 − , NO3 − Rb+ , Cs+ , NH4 + , Tl+ , Ag+ Mg2+ , Be2+
8
Ca2+ , Cu2+ , Zn2+ , Mn2+ , Ni2+ , Co2+ 2+
2+
9 8 7 6 5 4–4.5 3 2.5
2+
Sr , Ba , Cd , H2 C(COO)2 Hg2 2+ , SO4 2− , CrO4 2−
2−
Al3+ , Fe3+ , Cr3+ , La3+ 3−
6 5 4 9
Citrate PO4 3− , Fe(CN)6 3− , {CO(NH3 )6 }3+
5 4
Th4+ , Zr4+ , Ce4+ Fe(CN)6 4−
11 5
Taken from Kielland, Reference 10 (See the text website for an arrangement of inorganic and organic ions.)
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6.16 ACTIVITY AND ACTIVITY COEFFICIENTS— —CONCENTRATION IS NOT THE WHOLE STORY
Example 6.16 Calculate the activity coefficients for K+ and SO4 2− in a 0.0020 M solution of potassium sulfate. Solution
The ionic strength is 0.0060, so we can apply Equation 6.20: √ 0.51(1)2 0.0060 − log fK + = = 0.037 √ 1 + 0.0060 fK + = 10−0.037 = 10−1 × 100.963 = 0.918 √ 0.51(2)2 0.0060 = 0.147 − log fSO4 2− = √ 1 + 0.0060 fSO4 2− = 10−0.147 = 10−1 × 100.853 = 0.713
Example 6.17 Calculate the activity coefficients for K+ and SO4 2− in a 0.020 M solution of potassium sulfate. Solution
The ionic strength is 0.060, so we would use Equation 6.19b. From Table 6.2, we find ˚ and aSO 2− = 4.0 A. ˚ For K+ , we can use Equation 6.20: that aK+ = 3 A 4 √ 0.51(1)2 0.060 − log fK+ = = 0.101 √ 1 + 0.060 fK+ = 10−0.101 = 10−1 × 100.899 = 0.794 For SO4 2− , use Equation 6.19b: − log fSO4 2−
√ 0.51(2)2 0.060 = 0.378 = √ 1 + 0.33 × 4.0 0.060
fSO4 2− = 10−1 × 100.622 = 0.419 This latter compares with a calculated value of 0.396 using Equation 6.20. Note the decrease in the activity coefficients compared to 0.002 M K2 SO4 , especially for the SO4 2− ion. Spreadsheets for calculating activity coefficients using Equations 6.19b and 6.20 are given in the textbook’s website for Chapter 6. For higher ionic strengths, a number of empirical equations have been developed. One of the more useful is the Davies modification (see Reference 9): √ μ 2 − log fi = 0.51Zi (6.21) √ − 0.3 μ 1+ μ It is valid up to ionic strengths of about 0.5.
Use this equation for ionic strengths of 0.2–0.5. It gives higher activity coefficients compared to the Extended Debye–H¨uckel equation.
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Fig. 6.4.
Mean ionic activity coefficient of calcium nitrate as a function of concentration. Data taken from R. H. Stokes and R. A. Robinson, J. Am. Chem. Soc. 70 (1948) 1870.
A 0.01 M solution of HCl prepared in 8 M NaCl has an activity about 100 times that in water! Its pH is actually 0.0. See F. E. Critchfield and J. B. Johnson, Anal. Chem., 30, (1958) 1247 and G. D. Christian, CRC Crit. Rev. in Anal. Chem., 5(2) (1975) 119–153.
Mean Activity Coefficient of Calcium Nitrate
2.4
2
1.6
1.2
0.8
0.4
0 0
1
2
3
4
5
Square Root of Solution Molarity, M
At very high electrolyte concentrations, activity coefficients increase and become greater than unity. Note that Equation 6.21 does predict this; the last term causes an increase in the activity coefficient as μ increases. This is because the activity of the solvent, water, decreases and the water that is tied in the primary solvation shell of the ions (cations are especially hydrated) are not available to function as solvent. Consider that if the solvation number of Na+ is 4 (meaning that 4 molecules of water are attached to each sodium ion), an 8 molal NaCl solution has 32 of the 55.5 moles of the water already tied up with the sodium, the effective amount of water remaining to function as solvent is only 43% of that in a very dilute solution. The effective concentration is thus 2.4× higher. This change in concentration is ultimately reflected in an increased activity coefficient. Unfortunately, the real situation is more complex than this because water becomes so scarce at very high concentrations, solvation numbers are not constant and they also begin to decrease from their values in more dilute solutions. For more detailed discussion of activity coefficients in concentrated solutions, see the paper by Stokes and Robinson cited in Figure 6.4. The Stokes-Robinson equation, which works for AB and AB2 type electrolytes up to several molar in concentration, is given by: √ 0.51ZA ZB μ n − log f± = (6.22) √ + log aw + log(1 − 0.018)(n − ν)m 1 + 0.33at μ ν where f± is the mean activity coefficient of the positive and the negative ion (this is a geometric mean), ZA and ZB are respectively the charge on ion A and ion B, aw is the activity of water (the ratio of vapor pressure of the solution to that of pure water), m is the molality of the solution, n is the hydration number per solute molecule, and ν is the number of species formed from each solute molecule, e.g., for Ca(NO3 )2 , this is 3. We can draw some general conclusions about the estimation of activity coefficients.
The greater the charge on diverse ions, the greater their effect on the activity. The activity of nonelectrolytes is the same as the concentration, up to ionic strengths of 1.
1. The activity coefficients of ions of a given charge magnitude are approximately the same in solutions of a given ionic strength, and this activity coefficient is the same regardless of their individual concentrations.
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2. The behavior of ions become less ideal as the charge magnitude increases, resulting in less confidence in calculated activity coefficients. 3. The calculated activity coefficient of an ion in a mixed electrolyte solution will be less accurate than in a single-electrolyte solution. 4. The activity coefficients of nonelectrolytes (uncharged molecules) can generally be considered equal to unity in ionic strengths up to 0.1, and deviations from this approximation are only moderate in ionic strengths as high as 1. Undissociated acids, HA, are nonelectrolytes whose activity coefficients can be taken as unity. However, in highly concentrated electrolytes, the activity coefficients of nonelectrolytes do exceed unity, again because solvent becomes unavailable. This is the basis of “salting out” a nonelectrolyte from solution, often used in organic synthesis.
A final comment about activity coefficients: Kenneth S. Pitzer recast activity coefficient corrections using quantum mechanics, and provides rigorous treatment of concentrated solutions. See Reference 11.
Salting out can be useful in analytical chemistry as well. It is possible to add sufficient CaCl2 to a mixture of water and acetone containing an organic chelate of cobalt that the water will all be taken up and separate as a highly concentrated CaCl2 solution layer, distinct from the acetone layer bearing the Co chelate. See C.E. Matkovich and G. D. Christian Anal. Chem. 45 (1973) 1915.
6.17 The Diverse Ion Effect: The Thermodynamic Equilibrium Constant and Activity Coefficients We mentioned at the beginning of the last section on activity that the presence of diverse salts will generally increase the dissociation of weak electrolytes due to a shielding (or decrease in the activity) of the ionic species produced upon dissociation. We can quantitatively predict the extent of the effect on the equilibrium by taking into account the activities of the species in the equilibrium. In our consideration of equilibrium constants thus far, we have assumed no diverse ion effect, that is, an ionic strength of zero and an activity coefficient of 1. Equilibrium constants should more exactly be expressed in terms of activities rather than concentrations. Consider the dissociation of AB. The thermodynamic equilibrium constant (i.e., the equilibrium constant extrapolated to the case of infinite ◦ dilution) Keq is ◦
Keq =
aA · aB [A]fA · [B]fB = aAB [AB]fAB
(6.23)
Thermodynamic equilibrium constants hold at all ionic strengths. For a qualitative picture see panel (b), Figure 6.3
Since the concentration equilibrium constant Keq = [A][B]/[AB], then ◦
Keq = Keq
fA · fB fAB
(6.24)
fAB fA · fB
(6.25)
or ◦
Keq = Keq
◦ ◦ holds for all activities. Keq = Keq at zero ionic strength, but The numerical value of Keq at appreciable ionic strengths, a value for Keq must be calculated for each ionic strength using Equation 6.25. The equilibrium constants listed in Appendix C are for zero ionic strength; that is, they are really thermodynamic equilibrium constants. (For some reaction systems, experimental Keq values are available at different ionic strengths
Concentration equilibrium constants must be corrected for ionic strength.
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and can be used for equilibrium calculations at the listed ionic strength, using molar concentrations without having to calculate activity coefficients. For example, chemical oceanography is of sufficient importance that all relevant equilibrium constants uniquely applicable to a seawater matrix are available.)
Example 6.18 The weak electrolyte AB dissociates to A+ and B− , with a thermodynamic equilibrium ◦ constant Keq of 2 × 10−8 in the presence of a diverse salt of ionic strength 0.1. If the activity coefficients of A+ and B− are 0.6 and 0.7, respectively, at μ = 0.1: (a) Calculate the molar equilibrium constant Keq in terms of concentration. (b) Calculate the percent dissociation of a 1.0 × 10−4 M solution of AB in water. Solution
(a)
AB A+ + B− [A+ ][B− ] Keq = [AB] aA+ · aB− [A+ ] fA+ · [B− ]fB− = aAB [AB]fAB The activity coefficient of a neutral species is unity, so [A+ ][B− ] ◦ Keq = · fA+ · fB− = Keq fA+ · fB− [AB] ◦
Keq =
Keq =
◦ Keq
fA+ · fB−
=
2 × 10−8 = 5 × 10−8 (0.6)(0.7)
AB A+ −4 x 1 × 10 − x
(b)
+
B− x
In water, fA+ = fB− ≈ 1 (since μ < 10−4 ), x 10−4 [A+ ][B− ] = 2 × 10−8 [AB] (x)(x) = 2 × 10−8 1.0 × 10−4 x = 1.4 × 10−6 M % dissociated =
1.4 × 10−6 M × 100% = 1.4 % 1.0 × 10−4 M
For 0.1 M salt, [A+ ][B− ] = 5 × 10−8 [AB] (x)(x) = 5 × 10−8 1.0 × 10−4 x = 2.2 × 10−6 2.2 × 10−6 × 100% = 2.2 % 1.0 × 10−4 which represents a 57% increase in dissociation. % dissociated =
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PROBLEMS
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Calculations using the diverse ion effect are illustrated in Chapter 7 for acid dissociation and in Chapter 10 for precipitate solubilities. For illustrative purposes throughout this book, we will in general neglect the diverse ion effects on equilibria. In most cases, we are interested in relative changes in equilibrium concentrations, and the neglect of activities will not change our arguments. Electrostatic effects are important in the behavior of charged molecules such as proteins, DNA and other charged biopolymers, how ions move in capillary electrophoresis, and the behavior of glass and ion selective electrodes, to name a few areas. Read Professor’s Favorite Musings: Where Do Activity Coefficients Come From? The thoughts of Professor Michael D. Morris, University of Michigan, on this topic are on the text website.
We will generally ignore diverse salt effects.
Problems EQUILIBRIUM CALCULATIONS 1. A and B react as follows: A + B C + D. The equilibrium constant is 2.0 × 103 . If 0.30 mol of A and 0.80 mol of B are mixed in 1 L, what are the concentrations of A, B, C, and D after reaction? 2. A and B react as follows: A + B 2C. The equilibrium constant is 5.0 × 106 . If 0.40 mol of A and 0.70 mol of B are mixed in 1 L, what are the concentrations of A, B, and C after reaction? (See the text website for a video using Goal Seek to solve this problem.) 3. The dissociation constant for salicylic acid, C6 H4 (OH)COOH, is 1.0 × 10−3 . Calculate the percent dissociation of a 1.0 × 10−3 M solution. There is one dissociable proton. (See also Excel Problem 26 below.) 4. The dissociation constant for hydrocyanic acid, HCN, is 7.2 × 10−10 . Calculate the percent dissociation of a 1.0 × 10−3 M solution. 5. Calculate the percent dissociation of the salicylic acid in Problem 3 if the solution also contained 1.0 × 10−2 M sodium salicylate (the salt of salicylic acid). −8
6. Hydrogen sulfide, H2 S, dissociates stepwise, with dissociation constants of 9.1 × 10 and 1.2 × 10−15 , respectively. Write the overall dissociation reaction and the overall equilibrium constant. 7. Fe2+ and Cr2 O7 2− react as follows: 6Fe2+ + Cr2 O7 2− + 14H+ 6 Fe3+ + 2Cr3+ + 7H2 O. The equilibrium constant for the reaction is 1 × 1057 . Calculate the equilibrium concentrations of the iron and chromium species if 10 mL each of 0.02 M K2 Cr2 O7 in 1.14 M HCl and 0.12 M FeSO4 in 1.14 M HCl are reacted.
SYSTEMATIC APPROACH TO EQUILIBRIUM CALCULATIONS 8. Write charge balance expressions for (a) a saturated solution of Bi2 S3 ; (b) a solution of Na2 S. 9. Write the equations of mass balance and electroneutrality for a 0.100 M [Cd(NH3 )4 ]Cl2 solution. 10. Prove the following relations using the principles of electroneutrality and mass balance: (a) [NO2 − ] = [H+ ] − [OH− ] for 0.2 M HNO2 solution (b) [CH3 COOH] = 0.2 − [H+ ] + [OH− ] for 0.2 M CH3 COOH solution (c) [H2 C2 O4 ] = 0.1 − [H+ ] + [OH− ] − [C2 O4 2− ] for 0.1 M H2 C2 O4 solution (d) [HCN] = [OH− ] − [H+ ] for 0.1 M KCN solution [OH− ] − [H+ ] − [HPO4 2− ] − 3[H3 PO4 ] for 0.1 M Na3 PO4 solution (e) [H2 PO4 − ] = 2 (f) [HSO4 − ] = 0.2 − [H+ ] − [OH− ] for 0.1 M H2 SO4 solution (assume that the dissociation of H2 SO4 to H+ and HSO4 − is quantitative) 11. Write equations of mass balance for an aqueous saturated solution of BaF2 containing the species F− , HF, HF2 − , and Ba2+ .
Video: Goal Seek Problem 6.2
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12. Write an equation of mass balance for an aqueous solution of Ba3 (PO4 )2 . 13. Calculate the pH of a 0.100 M solution of acetic acid using the charge/mass balance approach.
IONIC STRENGTH 14. Calculate the ionic strengths of the following solutions: (a) 0.30 M NaCl; (b) 0.30 M Na2 SO4 ; (c) 0.30 M NaCl and 0.20 M K2 SO4 ; (d) 0.20 M Al2 (SO4 )3 and 0.10 M Na2 SO4 . 15. Calculate the ionic strengths of the following solutions: (a) 0.20 M ZnSO4 ; (b) 0.40 M MgCl2 ; (c) 0.50 M LaCl3 ; (d) 1.0 M K2 Cr2 O7 ; (e) 1.0 M Tl(NO3 )3 + 1.0 M Pb(NO3 )2 .
ACTIVITY See the text website, Spreadsheet Problems, for Excel answers to problems 16–19. 16. Calculate the activity coefficients of the sodium and chloride ions for a 0.00100 M solution of NaCl. 17. Calculate the activity coefficients of each ion in a solution containing 0.0020 M Na2 SO4 and 0.0010 M Al2 (SO4 )3 . 18. Calculate the activity of the NO3 − ion in a solution of 0.0020 M KNO3 . 19. Calculate the activity of the CrO4 2− ion in a 0.020 M solution of Na2 CrO4 . 20. 2.5 M sulfuric acid (H2 SO4 ) has a density of 1.15. The relative humidity over such a solution is 88.8%. If you assume that each proton is solvated by 4 molecules of water, what will be the mean activity coefficient according to Equation 6.22?
THERMODYNAMIC EQUILIBRIUM CONSTANTS 21. Write thermodynamic equilibrium constant expressions for the following: (a)HCN H+ + CN− (b)NH3 + H2 O NH4 + + OH− 22. Calculate the pH of a solution of 5.0 × 10−3 M benzoic acid (a) in water and (b) in the presence of 0.05 M K2 SO4 .
EXCEL EXERCISES See the text website, Spreadsheet Problems, for Excel solutions to these problems. 23. Write a spreadsheet program for calculating activity coefficients using Equation 6.20. Then compare it with the one on the text’s website. Do a calculation with both to check the accuracy. 24. Calculate the activity coefficients for K+ and SO4 2− in Example 6.16 using the website spreadsheet for Equation 6.20. Compare your results with the manually calculated values in the example. 25. Calculate the activity coefficients in Example 6.17 using the website spreadsheets for Equations 6.19b and 6.20. Compare your results with the manually calculated values in the example. 26. Use Excel Goal Seek to calculate the concentration, x, in Problem 3 above. (The problem requires use of the quadratic equation.) 27. Solve Problems 16 to 19 above using Excel.
Recommended References EQUILIBRIA 1. A. J. Bard, Chemical Equilibrium. New York: Harper & Row, 1966. 2. T. R. Blackburn, Equilibrium: A Chemistry of Solutions. New York: Holt, Rinehart and Winston, 1969. 3. J. N. Butler. Ionic Equilibrium. A Mathematical Approach. Reading, MA: Addison-Wesley, 1964.
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RECOMMENDED REFERENCES
4. G. M. Fleck, Ionic Equilibria in Solution. New York: Holt, Rinehart and Winston, 1966. 5. H. Freiser and Q. Fernando, Ionic Equilibria in Analytical Chemistry. New York: Wiley, 1963. 6. A. E. Martell and R. J. Motekaitis, The Determination and Use of Stability Constants. New York: VCH, 1989.
METHOD OF SUCCESSIVE APPROXIMATIONS 7. S. Brewer, Solving Problems in Analytical Chemistry. New York: Wiley, 1980. 8. J. J. Baeza-Baeza and M. C. Garcia-Alvarez-Coque, “Systematic Approach to Calculate the Concentration of Chemical Species in Multi-Equilibrium Problems,” J. Chem. Educ. 88 (2011) 169. This article demonstrates the solution of multiple simultaneous equlibria using Excel Solver.
ACTIVITY 9. C. W. Davies, Ion Association. London: Butterworth, 1962. 10. J. Kielland, “Individual Activity Coefficients of Ions in Aqueous Solutions,” J. Am. Chem. Soc., 59 (1937) 1675. 11. K. S. Pitzer. Activity Coefficients in Electrolyte Solutions, 2nd ed. Boca Raton, FL: CRC Press, 1991. 12. P. C. Meier, “Two-Parameter Debye-Huckel Approximation for the Evaluation of Mean Activity Coefficients of 109 Electrolytes,” Anal. Chim. Acta, 136 (1982) 363.
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Chapter Seven ACID–BASE EQUILIBRIA Police arrested two kids yesterday, one was drinking battery acid, the other was eating fireworks. They charged one and let the other one off. —Tommy Cooper
Chapter 7 URLs
Saying sulfates do not cause acid rain is the same as saying that smoking does not cause lung cancer. —Drew Lewis
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Acid–base theories, p. 223
●
●
Acid–base equilibria in water, p. 225 (key equations: 7.11, 7.13, 7.19)
●
●
Weak acids and bases, p. 232
●
●
Salts of weak acids and bases, p. 234 (key equations: 7.27, 7.29, 7.32, 7.36, 7.39)
Salts of polyprotic acids, p. 255 (key equations: 7.96, 7.97, 7.99, 7.100)
●
Logarithmic concentration diagrams, p. 266
Buffers, p. 238 (key equations: 7.45, 7.58)
●
pH calculator programs, p. 269
●
Polyprotic acids—α values, pp. 245, 248 (key equations: 7.74–7.77) Using spreadsheets to prepare α vs. pH plots, p. 251
The acidity or basicity of a solution is frequently an important factor in chemical reactions. The use of buffers to maintain the solution pH at a desired level is very important. In addition, fundamental acid–base equilibria are important in understanding acid–base titrations and the effects of acids on chemical species and reactions, for example, the effects of complexation or precipitation. In Chapter 6, we described the fundamental concept of equilibrium constants. In this chapter, we consider in more detail various acid–base equilibrium calculations, including weak acids and bases, hydrolysis of salts of weak acids and bases, buffers, polyprotic acids and their salts, and physiological buffers. Acid–base theories and the basic pH concept are reviewed first.
Sir Humphry Davy (1778–1829). Best known for his discovery of several alkali and alkaline earth metals and for establishing chlorine and iodine as elements. Berzelius called Davy’s 1806 Bakerian lecture On Some Chemical Agencies of Electricity as one of the highpoints ever to have enriched the theories of chemistry.
222
7.1 The Early History of Acid—Base Concepts The word acid derives from Latin acere, meaning sour. Bases were referred to as alkali in early history and that word derives from Arabic al-qili, the ashes of the plant saltwort, rich in sodium carbonate. In the mid seventeenth century it was recognized that acids and bases (called alkali in early history) tend to neutralize each other (known as the Silvio-Tachenio theory) but the concepts were vague. Acids, for example, were thought to be substances that would cause limestone to effervesce and alkalis as those that would effervesce with acids. In 1664 Robert Boyle published in The
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7.2 ACID–BASE THEORIES— —NOT ALL ARE CREATED EQUAL
Experimental History of Colours that extracts of certain plants such as red roses and Brazil wood changed color reversibly as the solution was made alternately acidic and basic. Many other plant and flower extracts were shown subsequently to behave in a similar fashion. In 1675, Boyle objected to the vagueness of the Silvio–Tachenio theory and largely because of his efforts, a set of definitions emerged about acids that sought to incorporate their known properties: acids taste sour, cause limestone to effervesce, turn blue plant dyes to red, and precipitate sulfur from alkaline solutions. Alkalies are substances that are slippery to the touch and can reverse the effect of acids. Almost a hundred years elapsed before Antoine-Laurent Lavoisier formed his own opinion of how acids come to be. Based primarily on his observations on combustion and respiration, in which carbon is converted to carbon dioxide (the acidic nature of carbon dioxide dissolved in water was already obvious), he named the gas recently (1774) discovered by Joseph Priestley, so essential for combustion or respiration, as oxygen (from Greek, meaning acid former), since he surmised it was what created the acidic product. Alessandro Volta announced the electric pile—an early type of battery—in 1800. Humphry Davy started playing with electricity immediately thereafter. Through electrolysis he discovered several new elements. In 1807 he electrolyzed fused potash and then soda—substances that many thought to be elements—and isolated potassium and sodium. He also similarly isolated magnesium, calcium, strontium, and barium. Davy recognized that these alkali and alkaline earth metals combine with oxygen and form already known oxides that are highly basic, which challenged Lavoisier’s theory that oxygen was the acidifying element. He went on to establish that hydrochloric acid, not “oxymuriatic acid” as Lavoisier called it, was acidifying; by electrolysis he isolated hydrogen and one other element, chlorine (that he so named in 1810), which until then was believed to be a compound containing oxygen. Rather than oxygen, Davy suggested in 1815 that hydrogen may be the acidifying element. All substances that contain hydrogen, however, are not acids. It would wait for Justus von Liebig, to identify an acid in 1838 as a compound of hydrogen where the hydrogen can be replaced by a metal.
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Justus Von Liebig (1803–1873) was a German chemist who profoundly enriched agricultural and biological chemistry. Liebig first recognized nitrogen as an essential plant nutrient and is hence often called the father of the fertilizer industry. Liebig first advocated experimentally centered teaching of chemistry. He was also the first to attempt to systematize organic chemistry.
7.2 Acid–Base Theories—Not All Are Created Equal Several acid–base theories have been proposed to explain or classify acidic and basic properties of substances. You are probably most familiar with the Arrhenius theory, which is applicable only to water. Other theories are more general and are applicable to other solvents or even the gas phase. We describe the common acid–base theories here. ARRHENIUS THEORY——H+ AND OH− Arrhenius, as a graduate student, introduced a dramatically new theory that an acid is any substance that ionizes (partially or completely) in water to give hydrogen ions (which associate with the solvent to give hydronium ions, H3 O+ ): HA + H2 O H3 O+ + A− A base ionizes in water to give hydroxide ions. Weak (partially ionized) bases generally ionize as follows: B + H2 O BH+ + OH− while strong bases such as metal hydroxides (e.g., NaOH) dissociate as M(OH)n → Mn+ + nOH− This theory is obviously restricted to water as the solvent.
Svante Arrhenius (1859–1927) submitted a 150-page dissertation in 1884 on electrolytic conductivity to Uppsala for the doctorate. It did not impress the professors, and he received a fourth class degree, but upon his defense it was reclassified as third class. Nineteen years later, extensions of this very work would earn him the 1903 Nobel Prize in Chemistry.
The Arrhenius theory is restricted to aqueous solutions. See J. Am. Chem. Soc., 34 (1912) 353 for his personal observations of the difficulty Arrhenius had in the acceptance of his theory.
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THEORY OF SOLVENT SYSTEMS——SOLVENT CATIONS AND ANIONS
Franklin and Germann’s theory is similar to the Arrhenius theory but is applicable also to other ionizable solvents.
The Brønsted–Lowry theory assumes a transfer of protons from an acid to a base, i.e., conjugate pairs.
In 1905, Franklin was working in liquid NH3 as solvent and noticed the similarity with acid-base behavior in water. In 1925, Germann, working with liquid COCl2 as solvent observed the similarities as well and formulated a general solvent system concept of acids and bases. This theory recognizes the ionization of a solvent to give a cation and an anion; for example, 2H2 O H3 O+ + OH− or 2NH3 NH4 + + NH2 − . An acid is defined as a solute that yields the characteristic cation of the solvent while a base is a solute that yields the characteristic anion of the solvent. Thus, NH4 Cl (which produces ammoniated NH4 + , i.e., [NH4 (NH3 )+ ], and Cl− ) is a strong acid in liquid ammonia (similar to HCl in water: HCl + H2 O → H3 O+ + Cl− ) while NaNH2 is a strong base in ammonia (similar to NaOH in water); both of these compounds ionize to give the characteristic solvent cation and anion, respectively. Ethanol ionizes as follows: 2C2 H5 OH C2 H5 OH2 + + C2 H5 O− . Hence, sodium ethoxide, NaOC2 H5 , is a strong base in this solvent. BRØNSTED—LOWRY THEORY——TAKING AND GIVING PROTONS The theory of solvent systems is suitable for ionizable solvents, but it is not applicable to acid–base reactions in nonionizable solvents such as benzene or dioxane. In 1923, Brønsted and Lowry independently described what is now known as the Brønsted–Lowry theory. This theory states that an acid is any substance that can donate a proton, and a base is any substance that can accept a proton. Thus, we can write a “half-reaction”a acid = H+ + base
Thomas M. Lowry (1874–1936). In 1923, Lowry and Brønsted independently described the theory that is named after them.
(7.1)
The acid and base of a half-reaction are called conjugate pairs. Free protons do not exist in solution, and there must be a proton acceptor (base) before a proton donor (acid) will release its proton. That is, there must be a combination of two half-reactions. Another way to look at it is that an acid is an acid because it can lose a proton. However, it cannot exhibit its acidic behavior unless there is a base present to accept the proton. It is like being wealthy on a deserted island with no one to accept your money. Some acid–base reactions in different solvents are illustrated in Table 7.1. In the first example, acetate ion is the conjugate base of acetic acid and ammonium ion is the conjugate acid of ammonia. The first four examples represent ionization of an acid or a base in a solvent, while the others represent a neutralization reaction between an acid and a base in the solvent. It is apparent from the above definition that a substance cannot act as an acid unless a base is present to accept the protons. Thus, acids will undergo complete
Table 7.1
Brønsted Acid–Base Reactions: Conjugate acid base pairs are denoted in the same color Solvent
Acid1
NH3 (liq.) H2 O H2 O H2 O H2 O C2 H5 OH C6 H6
HOAc HCl NH4 + H2 O HCO3 − NH4 + H picrate
+
Base2 NH3 H2 O H2 O OAc− OH− C2 H5 O− C6 H5 NH2
→
Acid2 NH4 + H3 O+ H3 O+ HOAc H2 O C2 H5 OH C6 H5 NH3 +
+
Base1 OAc− Cl− NH3 OH− CO3 2− NH3 picrate –
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or partial ionization in basic solvents such as water, liquid ammonia, or ethanol, depending on the basicity of the solvent and the strength of the acid. But in neutral or “inert” solvents, ionization is insignificant. However, ionization in the solvent is not a prerequisite for an acid–base reaction, as in the last example in the table, where picric acid reacts with aniline. LEWIS THEORY——TAKING AND GIVING ELECTRONS Also in 1923, G. N. Lewis introduced the electronic theory of acids and bases. In the Lewis theory, an acid is a substance that can accept an electron pair and a base is a substance that can donate an electron pair. The latter frequently contains an oxygen or a nitrogen as the electron donor. Thus, nonhydrogen-containing substances are included as acids. Examples of acid–base reactions in the Lewis theory are as follows:
H (solvated)
:NH3
The Lewis theory assumes a donation (sharing) of electrons from a base to an acid.
H:NH3
R AlCl3
:O
Cl3Al:OR2 R
H O:
H
H2O:H
:OH
H:OH
H H
Gilbert N. Lewis (1875–1946) developed theories of covalent bonding, leading to the Lewis electronic theory of acids and bases.
In the second example, aluminum chloride is an acid and ether is a base.
7.3 Acid–Base Equilibria in Water We see from the above that when an acid or base is dissolved in water, it will dissociate, or ionize, the amount of ionization being dependent on the strength of the acid or the base. A “strong” electrolyte is completely dissociated, while a “weak” electrolyte is partially dissociated. Table 7.2 lists some common electrolytes, some strong and some weak. Other weak acids and bases are listed in Appendix C. Hydrochloric acid is a strong acid, and in water, its ionization is complete: HCl + H2 O → H3 O+ + Cl−
(7.2)
An equilibrium constant for Equation 7.2 would have a value of infinity. The proton H+ exists in water as a hydrated ion, the hydronium ion, H3 O+ . Higher hydrates probably exist, particularly H9 O4 + . The hydronium ion is written as H3 O+ for convenience and to emphasize Brønsted behavior. Acetic acid1 is a weak acid, which ionizes only partially in water (a few percent): HOAc + H2 O H3 O+ + OAc− We can write an equilibrium constant for this reaction: aH O+ · aOAc− ◦ Ka = 3 aHOAc · aH2 O O 1 We shall use the symbol OAc− to represent the acetate ion CH3
C
O .
(7.3)
(7.4)
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Table 7.2
Some Strong Electrolytes and Some Weak Electrolytes Strong
Weak
HCl HClO4 H2 SO4a HNO3 NaOH
CH3 COOH (acetic acid) NH3 C6 H5 OH (phenol) HCHO2 (formic acid) C6 H5 NH2 (aniline) CH3 COONa
a The
Autoprotolysis is the self-ionization of a solvent to give a characteristic cation and anion, e.g., 2CH3 OH CH3 OH+ + CH3 O− .
first proton is completely ionized in dilute solution, but the second proton is partially ionized (K2 = 10−2 ).
where Ka◦ is the thermodynamic acidity constant (see Section 6.16) and a is the activity of the indicated species. Salt cations or anions may also partially react with water after they are dissociated. For example, acetate ion is formed from dissociated acetate salts, to give HOAc. The activity can be thought of as representing the effective concentration of an ion (described in Chapter 6). The effects of protons in reactions are often governed by their activities, and it is the activity that is measured by the widely used pH meter (Chapter 13). Methods for predicting numerical values of activity coefficients were described in Chapter 6. In dilute solutions, the activity of water remains essentially constant, and is taken as unity at standard state. Therefore, Equation 7.4 can be written as aH O+ · aOAc− ◦ Ka = 3 (7.5) aHOAc Pure water ionizes slightly, or undergoes autoprotolysis: 2H2 O H3 O+ + OH− The equilibrium constant for this is ◦
Kw =
aH3 O+ · aOH− aH2 O2
(7.6)
(7.7)
Again, the activity of water is constant in dilute solutions (its concentration is essentially constant at ∼55.5 M), so ◦
Kw = aH3 O+ · aOH− We will use H+ in place of H3 O+ , for simplicity. Also, molar concentrations will generally be used instead of activities.
(7.8)
Kw◦
is the thermodynamic autoprotolysis, or self-ionization, constant. where Calculations are simplified if we neglect activity coefficients. This simplification results in only slight errors for dilute solutions, and we shall use molar concentrations in all our calculations. This will satisfactorily illustrate the equilibria involved. Most of the solutions we will be concerned with are rather dilute, and we will frequently be interested in relative changes in pH (and large ones) in which case small errors are insignificant. We will simplify our expressions by using H+ in place of H3 O+ . This is not inconsistent since the waters of solvation associated with other ions or molecules (e.g., metal ions) are not generally written and H3 O+ is not an accurate representation of the actual species present; typically the proton in dilute aqueous solution has at least four water molecules in its solvation shell. Molar concentration will be represented by square brackets [ ] around the species. Simplified equations for the above reactions are HCl → H+ + Cl−
(7.9)
HOAc H+ + OAc−
(7.10)
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[H+ ] [OAc− ] [HOAc]
(7.11)
H2 O H+ + OH−
(7.12)
Kw = [H+ ] [OH− ]
(7.13)
Ka =
Ka and Kw are the molar equilibrium constants. Kw is exactly 1.00 × 10−14 at 24◦ C and even at 25◦ C, to a smaller number of significant figures, it is still accurately represented as 1.0 × 10−14 . The product of the hydrogen ion concentration and the hydroxide ion concentration in aqueous solution is always equal to 1.0 × 10−14 at room temperature: [H+ ] [OH− ] = 1.0 × 10−14
(7.14)
In pure water, then, the concentrations of these two species are equal since there are no other sources of H+ or OH− except H2 O dissociation:
Chemists (and especially students!) are lucky that nature made Kw an even unit number at room temperature. Imagine doing pH calculations with a Kw like 2.39 × 10−13 . However, see Section 7.5 where you must indeed do this for other temperatures.
[H+ ] = [OH− ] Therefore, [H+ ] [H+ ] = 1.0 × 10−14 [H+ ] = 1.0 × 10−7 M ≡ [OH− ] If an acid is added to water, we can calculate the hydroxide ion concentration if we know the hydrogen ion concentration from the acid. But when the hydrogen ion concentration from the acid is very small, 10−6 M or less, the contribution to [H + ] from the ionization of water cannot be neglected.
Example 7.1 A 1.0 × 10−3 M solution of hydrochloric acid is prepared. What is the hydroxide ion concentration? Solution
Since hydrochloric acid is a strong electrolyte and is completely ionized, the H+ concentration is 1.0 × 10−3 M. Thus, (1.0 × 10−3 )[OH− ] = 1.0 × 10−14 [OH− ] = 1.0 × 10−11 M
7.4 The pH Scale The concentration of H+ or OH− in aqueous solution can vary over extremely wide ranges, from 1 M or greater to 10−14 M or less. To construct a plot of H+ concentration against some variable would be very difficult if the concentration changed from, say, 10−1 M to 10−13 M. This range is common in a titration. It is more convenient to
pScales are used to compress and more conveniently express a range of numbers that span several decades in magnitude.
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pH is really –log aH+ . This is what a pH meter (glass electrode) measures—see Chapter 13.
compress the acidity scale by placing it on a logarithm basis. The pH of a solution was defined by Sørenson as pH = −log[H+ ]
(7.15)
The minus sign is used because most of the concentrations encountered are less than 1 M, and so this designation gives a positive number. (More strictly, pH is now defined as − log aH+ , but we will use the simpler definition of Equation 7.15.) In general, pAnything = − log Anything, and this method of notation will be used later for other numbers that can vary by large amounts, or are very large or small (e.g., equilibrium constants).
Example 7.2 Calculate the pH of a 2.0 × 10−3 M solution of HCl. Solution
HCl is completely ionized, so Carlsberg Laboratory archives In 1909, Søren Sørenson, head of the chemistry department at Carlsberg Laboratory (Carlsberg Brewery) invented the term pH to describe this effect and defined it as −log[H+ ]. The term pH refers simply to “the power of hydrogen.” In 1924, he realized that the pH of a solution is a function of the “activity” of the H+ ion, and published a second paper on the subject, defining it as pH = − log aH+ .
A 1 M HCl solution has a pH of 0 and pOH of 14. A 1 M NaOH solution has a pH of 14 and a pOH of 0.
[H+ ] = 2.0 × 10−3 M pH = − log(2.0 × 10−3 ) = 3 − log 2.0 = 3 − 0.30 = 2.70
A similar definition is made for the hydroxide ion concentration: pOH = −log[OH− ]
(7.16)
Equation 7.13 can be used to calculate the hydroxyl ion concentration if the hydrogen ion concentration is known, and vice versa. The equation in logarithm form for a more direct calculation of pH or pOH is − log Kw = −log[H+ ][OH− ] = −log[H+ ] − log [OH− ]
(7.17)
pKw = pH + pOH
(7.18)
14.00 = pH + pOH
(7.19)
At 25◦ C,
Example 7.3 Calculate the pOH and the pH of a 5.0 × 10−2 M solution of NaOH at 25◦ C. Solution
[OH− ] = 5.0 × 10−2 M pOH = − log(5.0 × 10−2 ) = 2 − log 5.0 = 2 − 0.70 = 1.30 pH + 1.30 = 14.00 pH = 12.70
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or [H+ ] =
229
1.0 × 10−14 = 2.0 × 10−13 M 5.0 × 10−2
pH = − log(2.0 × 10−13 ) = 13 − log 2.0 = 13 − 0.30 = 12.70
Example 7.4 Calculate the pH of a solution prepared by mixing 2.0 mL of a strong acid solution of pH 3.00 and 3.0 mL of a strong base of pH 10.00.
Keep track of millimoles!
Solution
[H+ ] of acid solution = 1.0 × 10−3 M mmol H+ = 1.0 × 10−3 M × 2.0 mL = 2.0 × 10−3 mmol pOH of base solution = 14.00 − 10.00 = 4.00 [OH− ] = 1.0 × 10−4 M mmol OH− = 1.0 × 10−4 M × 3.0 mL = 3.0 × 10−4 mmol There is an excess of acid. mmol H+ = 0.0020 − 0.0003 = 0.0017 mmol Total Volume = (2.0 + 3.0) mL = 5.0 mL [H+ ] = 0.0017 mmol/5.0 mL = 3.4 × 10−4 M pH = − log 3.4 × 10−4 = 4 − 0.53 = 3.47
Example 7.5 The pH of a solution is 9.67. Calculate the hydrogen ion concentration in the solution. Solution
−log[H+ ] = 9.67 [H+ ] = 10−9.67 = 10−10 × 100.33 [H+ ] = 2.1 × 10−10 M
Remember, this answer is reported to two significant figures (2.1 × 10−10 M) because the mantissa of the pH value (9.67) has two significant figures.
When [H+ ] = [OH− ], then a solution is said to be neutral. If [H+ ] > [OH− ], then the solution is acidic. And if [H+ ] < [OH− ], the solution is alkaline. The hydrogen ion and hydroxide ion concentrations in pure water at 25◦ C are each 10−7 M, and the pH of water is 7. A pH of 7 is therefore neutral. Values of pH that are greater than this are alkaline, and pH values less than this are acidic. The reverse is true of pOH values. A pOH of 7 is also neutral. Note that the product of [H+ ] and [OH− ] is always 10−14 at 25◦ C, and the sum of pH and pOH is always 14. If the temperature is other than 25◦ C, then Kw is different from 1.0 × 10−14 , and a neutral solution will have other than 10−7 M H+ and OH− (see below). Some mistakenly believe that it is impossible to have a negative pH. There is no theoretical basis for this. A negative pH only means that the hydrogen ion
[H+ ] = 10−pH .
A 10 M HCl solution should have a pH of −1 and pOH of 15.
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The pH of 10−9 M HCl is not 9!
concentration is greater than 1 M. In actual practice, a negative pH is uncommon for two reasons. First, even strong acids may become partially undissociated at high concentrations. For example, 100% H2 SO4 is so weakly dissociated that it can be stored in iron containers; more dilute H2 SO4 solutions would contain sufficient protons from dissociation to attack and dissolve the iron. The second reason has to do with the activity, which we have chosen to neglect for dilute solutions. Since pH is really − log aH+ (this is what a pH meter reading is a measure of), a solution that is 1.1 M in H+ may actually have a positive pH because the activity of the H+ is less than 1.0 M.2 This is because at these high concentrations, the activity coefficient is less than unity (although at still higher concentrations the activity coefficient may become greater than unity—see Chapter 6). Nevertheless, there is mathematically no basis for not having a negative pH (or a negative pOH), although it may be rarely encountered in situations relevant to analytical chemistry. If the concentration of an acid or base is much less than 10−7 M, then its contribution to the acidity or basicity will be negligible compared with the contribution from water. The pH of a 10−8 M sodium hydroxide solution would therefore not differ significantly from 7. If the concentration of the acid or base is around 10−7 M, then its contribution is not negligible and neither is that from water; hence the sum of the two contributions must be taken.
Example 7.6 Calculate the pH and pOH of a 1.0 × 10−7 M solution of HCl. Solution
Equilibria: HCl → H+ + Cl− H2 O H+ + OH− [H+ ][OH− ] = 1.0 × 10−14 [H+ ]H2 Odiss. = [OH− ]H2 Odiss. = x Since the hydrogen ions contributed from the ionization of water are not negligible compared to the HCl added, [H+ ] = CHCl + [H+ ]H2 Odiss. Then, ([H+ ]HCl + x)(x) = 1.0 × 10−14 (1.00 × 10−7 + x)(x) = 1.0 × 10−14 x2 + 1.00 × 10−7 x − 1.0 × 10−14 = 0 Using the quadratic equation to solve [see Appendix B] or the use of Excel Goal Seek (Section 6.11),
−1.00 × 10−7 ± 1.0 × 10−14 + 4(1.0 × 10−14 ) x= = 6.2 × 10−8 M 2 2 As will be seen in Chapter 13, it is also difficult to measure the pH of a solution having a negative pH or
pOH because high concentrations of acids or bases tend to introduce an error in the measurement by adding a significant and unknown liquid-junction potential in the measurements.
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7.5 PH AT ELEVATED TEMPERATURES: BLOOD PH
total
Therefore, the 1.62 × 10−7 M:
H+
231
concentration = (1.00 × 10−7 + 6.2 × 10−8 ) =
pH = − log 1.62 × 10−7 = 7 − 0.21 = 6.79 pOH = 14.00 − 6.79 = 7.21 −
or, since [OH ] = x, pOH = − log(6.2 × 10−8 ) = 8 − 0.79 = 7.21 Note that, owing to the presence of the added H+ , the ionization of water is suppressed by 38% by the common ion effect (Le Chˆatelier’s principle). At higher acid (or base) concentrations, the suppression is even greater and the contribution from the water becomes negligible. The contribution from the autoionization of water can be considered negligible if the concentration of protons or hydroxyl ions from an acid or base is 10−6 M or greater. The calculation in this example is more academic than practical because carbon dioxide from the air dissolved in water substantially exceeds these concentrations, being about 1.2 × 10−5 M carbonic acid. Since carbon dioxide in water forms an acid, extreme care would have to be taken to remove and keep this from the water, to have a solution of 10−7 M acid.
We usually neglect the contribution of water to the acidity in the presence of an acid since its ionization is suppressed in the presence of the acid.
7.5 pH at Elevated Temperatures: Blood pH It is a convenient fact of nature for students and chemists who deal with acidity calculations and pH scales in aqueous solutions at room temperature that pKw is an integer number. At 100◦ C, for example, Kw = 5.5 × 10−13 , and a neutral solution has
[H+ ] = [OH− ] = 5.5 × 10−13 = 7.4 × 10−7 M pH = pOH = 6.13 pKw = 12.26 = pH + pOH Not all measurements or interpretations are done at room temperature, however, and the temperature dependence of Kw must be taken into account (recall from Chapter 6 that equilibrium constants are temperature dependent). An important example is the pH of the body. The pH of blood at body temperature (37◦ C) is 7.35 to 7.45. This value represents a slightly more alkaline solution relative to neutral water than the same value would be at room temperature. At 37◦ C, Kw = 2.5 × 10−14 and pKw = 13.60. The pH (and pOH) of a neutral
solution is 13.60/2 = 6.80. The hydrogen ion (and hydroxide ion) concentration is 2.5 × 10−14 = 1.6 × 10−7 M. Since a neutral solution at 37◦ C would have pH 6.8, a blood pH of 7.4 is more alkaline at 37◦ C by 0.2 pH units than it would be at 25◦ C. This is important when one considers that a change of 0.3 pH units in the body is extreme. The hydrochloric acid concentration in the stomach is about 0.1 to 0.02 M. Since pH = − log[H+ ], the pH at 0.02 M would be 1.7. It will be the same regardless of the temperature since the hydrogen ion concentration is the same (neglecting solvent volume changes), and the same pH would be measured at either temperature. But, while the pOH would be 14.0 − 1.7 = 12.3 at 25◦ C, it is 13.6 − 1.7 = 11.9 at 37◦ C. Not only does the temperature affect the ionization of water in the body and therefore change the pH of neutrality, it also affects the ionization constants of the acids and bases from which the buffer systems in the body are derived. As we shall see later in the chapter, this influences the pH of the buffers, and so a blood pH of 7.4
A neutral solution has pH < 7 above room temperature.
The pH of blood must be measured at body temperature to accurately reflect the status of blood buffers.
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measured at 37◦ C will not be the same when measured at room temperature, in contrast to the stomach pH, whose value was determined by the concentration of a strong acid. For this reason, measurement of blood pH for diagnostic purposes is generally done at 37◦ C (see Chapter 13).
7.6 Weak Acids and Bases—What Is the pH? We have limited our calculations so far to strong acids and bases in which ionization is assumed to be complete. Since the concentration of H+ or OH− is determined readily from the concentration of the acid or base, the calculations are straightforward. As seen in Equation 7.3, weak acids (or bases) are only partially ionized. While mineral (inorganic) acids and bases such as HCl, HClO4 , HNO3 , and NaOH are strong electrolytes that are totally ionized in water; most organic acids and bases, as found in clinical applications, are weak. The ionization constant can be used to calculate the amount ionized and, from this, the pH. The acidity constant for acetic acid at 25◦ C is 1.75 × 10−5 : [H+ ][OAc− ] = 1.75 × 10−5 [HOAc]
(7.20)
When acetic acid ionizes, it dissociates to equal portions of H+ and OAc− by such an amount that the computation on the left side of Equation 7.20 will always be equal to 1.75 × 10−5 : (7.21) HOAc H+ + OAc− If the original concentration of acetic acid is C and the concentration of ionized acetic acid species (H+ and OAc− ) is x, then the final concentration for each species at equilibrium is given by HOAc H+ + OAc− (7.22) (C − x) x x
Example 7.7 Calculate the pH and pOH of a 1.00 × 10−3 M solution of acetic acid. Solution
HOAc H+ + OAc− The concentrations of the various species in the form of an ICE table are as follows:
Initial Change (x = mmol/mL HOAc ionized) Equilibrium From Equation 7.20
[HOAc] 1.00 × 10−3
[H+] 0
[OAc−] 0
−x 1.00 × 10−3 − x
+x x
+x x
(x)(x) = 1.75 × 10−5 1.00 × 10−3 − x
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233
The solution is that of a quadratic equation. If less than about 10 or 15% of the acid is ionized, the expression may be simplified by neglecting x compared with C (10−3 M in this case). This is an arbitrary (and not very demanding) criterion. The simplification applies if Ka is smaller than about 0.01C, that is, smaller than 10−4 at C = 0.01 M, 10−3 at C = 0.1 M, and so forth. Under these conditions, the error in calculation is 5% or less (results come out too high), and within the probable accuracy of the equilibrium constant. Our calculation simplifies to
If CHA > 100Ka , x can be neglected compared to CHA .
x2 = 1.75 × 10−5 −3 1.00 × 10 x = 1.32 × 10−4 M ≡ [H+ ] Therefore, pH = − log(1.32 × 10−4 ) = 4 − log 1.32 = 4 − 0.12 = 3.88 pOH = 14.00 − 3.88 = 10.12
The simplification in the calculation does not lead to serious errors, particularly since equilibrium constants are often not known to a high degree of accuracy (frequently no better than ±10%). In the above example, solution of the quadratic equation results in [H+ ] = 1.26 × 10−4 M (5% less) and pH = 3.91. This pH is within 0.03 unit of that calculated using the simplification, which is near the limit of accuracy to which pH measurements can be made. It is almost certainly as close a calculation as is justified in view of the experimental errors in Ka or Kb values and the fact that we are using concentrations rather than activities in the calculations. In our calculations, we also neglected the contribution of hydrogen ions from the ionization of water (which was obviously justified); this is generally permissible except for very dilute ( 100Kb , which will generally be the case for such weakly ionized bases. We can solve for the OH− concentration using Equation 7.30: [OH− ][OH− ] K = w = Kb CA− Ka Compare this with the algebraic setup in Example 7.8. They are identical:
Kw − [OH ] = · CA− = Kb · CA− Ka
(7.31)
(7.32)
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This equation holds only if CA− > 100Kb , and x can be neglected compared to CA− . If this is not the case, then the quadratic formula must be solved as for other bases in this situation.
Example 7.9 Calculate the pH of a 0.10 M solution of sodium acetate. Solution Compare this base “ionization” with that of NH3 , Example 7.8.
Write the equilibria NaOAc → Na+ + OAc− (ionization) OAc− + H2 O HOAc + OH− (hydrolysis) Write the equilibrium constant K 1.0 × 10−14 [HOAc][OH− ] = Kb = w = = 5.7 × 10−10 − Ka [OAc ] 1.75 × 10−5 Let x represent the concentration of HOAc and OH− at equilibrium. Then, at equilibrium, [HOAc] = [OH− ] = x [OAc− ] = COAc− − x = 0.10 − x Since COAc− Kb , neglect x compared to COAc− . Then, (x)(x) = 5.7 × 10−10 0.10
x = 5.7 × 10−10 × 0.10 = 7.6 × 10−6 M Compare this last step with Equation 7.32. Also, compare the entire setup and solution with those in Example 7.8. The HOAc formed is undissociated and does not contribute to the pH: [OH− ] = 7.6 × 10−6 M [H+ ] =
1.0 × 10−14 = 1.3 × 10−9 M 7.6 × 10−6
pH = − log 1.3 × 10−9 = 9 − 0.11 = 8.89 For an Excel Goal Seek solution of Example 7.9 without approximation, see the text website. Similar equations can be derived for the cations of salts of weak bases (the salts are completely dissociated). These are Brønsted acids and ionize (hydrolyze) in water: (7.33) BH+ + H2 O B + H3 O+ The B is undissociated and does not contribute to the pH. The acidity constant is KH = Ka =
[B][H3 O+ ] [BH+ ]
(7.34)
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The acidity constant (hydrolysis constant) can be derived by multiplying the numerator and denominator by [OH− ]:
Ka
[B] [H3O ]
[OH ]
[BH ]
[OH ]
(7.35)
Again, the quantity inside the dashed lines is Kw , while the remainder is 1/Kb . Therefore, K [B][H3 O+ ] = w = Ka + Kb [BH ] and for NH4 + , Ka =
Kw 1.0 × 10−14 = = 5.7 × 10−10 Kb 1.75 × 10−5
(7.36)
(7.37)
We could, of course, have derived Ka from Equation 7.27. It is again coincidence that the numerical value of Ka for NH4 + equals Kb for OAc− . The salt of a weak base ionizes to form equal amounts of B and H3 O+ (H+ if we disregard hydronium ion formation as was done previously). We can therefore solve for the hydrogen ion concentration (by assuming CBH+ > 100Ka ): [H+ ][H+ ] K = w = Ka CBH+ Kb +
[H ] =
(7.38)
Kw · CBH+ = Ka · CBH+ Kb
(7.39)
Again, this equation only holds if CBH+ > 100Ka . Otherwise, the quadratic formula must be solved. Note: One can obtain Ka directly from a list of acidity constants, as in Table C2.b in Appendix C for the acid equilibrium as given in Equation 7.33; substituting in Equation 7.34 and solving for the hydrogen ion concentration gives Equation 7.39.
Example 7.10 Calculate the pH of a 0.25 M solution of ammonium chloride.
Solution
Write the equilibria NH4 Cl → NH4 + + Cl− +
+
NH4 + H2 O NH4 OH + H
(ionization) (hydrolysis)
(NH4 + + H2 O NH3 + H3 O+ ) Write the equilibrium constant Kw 1.0 × 10−14 [NH4 OH][H+ ] = K = = = 5.7 × 10−10 a −5 [NH4 + ] Kb 1.75 × 10
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Let x represent the concentration of [NH4 OH] and [H+ ] at equilibrium. Then, at equilibrium, [NH4 OH] = [H+ ] = x [NH4 + ] = CNH4 + − x = 0.25 − x Since CNH4 + Ka , neglect x compared to CNH4 + . Then, (x)(x) = 5.7 × 10−10 0.25
x = 5.7 × 10−10 × 0.25 = 1.2 × 10−5 M Compare this last step with Equation 7.39. Also, compare the entire setup and solution with those in Example 7.7. The NH4 OH formed is undissociated and does not contribute to the pH: [H+ ] = 1.2 × 10−5 M pH = − log(1.2 × 10−5 ) = 5 − 0.08 = 4.92 For an Excel Goal Seek solution of Example 7.8 without approximation, see the chapter’s website. See also the video illustrating the use of Goal Seek to calculate the pH of an NH4 F solution.3 Video: Goal Seek pH NH4 F
7.8 Buffers—Keeping the pH Constant (or Nearly So) A buffer is defined as a solution that resists change in pH when a small amount of an acid or base is added or when the solution is diluted. While carrying out a reaction, this is very useful for maintaining the pH within an optimum range. A buffer solution consists of a mixture of a weak acid and its conjugate base, or a weak base, and its conjugate acid at predetermined concentrations or ratios. That is, we have a mixture of a weak acid and its salt or a weak base and its salt. Consider an acetic acid–acetate buffer. The equilibrium that governs this system is HOAc H+ + OAc− But now, since we have added a supply of acetate ions to the system (e.g., from sodium acetate), the hydrogen ion concentration is no longer equal to the acetate ion concentration. The hydrogen ion concentration is [HOAc] (7.40) [H+ ] = Ka [OAc− ] Taking the negative logarithm of each side of this equation, we have [HOAc] −log[H+ ] = − log Ka − log (7.41) [OAc− ] [HOAc] (7.42) pH = pKa − log [OAc− ] Upon inverting the last log term, it becomes positive: [OAc− ] (7.43) [HOAc] This form of the ionization constant equation is called the Henderson–Hasselbalch pH = pKa + log
The pH of a buffer is determined by the ratio of the conjugate acid–base pair concentrations.
3 This is an unedited student video and contains some errors of statement, e,g., it talks about K of NH whereas a 3
it should really refer to it as Ka of NH4 + ; it mistakenly states that bases like to take up electrons whereas bases of course like to take up hydrogen ions, etc. But despite these errors of statement, it is a nicely set up example of a correctly solved problem that shows the use of Excel Goal Seek!
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equation. It is useful for calculating the pH of a weak acid solution containing its salt. A general form can be written for a weak acid HA that ionizes to its salt, A− , and H+ : HA H+ + A− pH = pKa + log
[A− ] [HA]
(7.44) (7.45)
[conjugate base] [acid] [proton acceptor] pH = pKa + log [proton donor]
pH = pKa + log
(7.46) (7.47)
Example 7.11 Calculate the pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.10 M sodium acetate. Solution
We need to calculate the concentration of the acid and salt in the solution. The final volume is 30 mL: M1 × mL1 = M2 × mL2 For HOAc,
0.10 mmol/mL × 10 mL = MHOAc × 30 mL MHOAc = 0.033 mmol/mL
−
For OAc , 0.10 mmol/mL × 20 mL = MOAc− × 30 mL MOAc− = 0.067 mmol/mL Some of the HOAc dissociates to H+ + OAc− , and the equilibrium concentration of HOAc would be the amount added (0.033 M) minus the amount dissociated, while that of OAc− would be the amount added (0.067 M) plus the amount of HOAc dissociated. However, the amount of acid dissociated is very small, particularly in the presence of the added salt (ionization suppressed by the common ion effect), and can be neglected. Hence, we can assume the added concentrations to be the equilibrium concentrations: [proton acceptor] pH = − log Ka + log [proton donor] pH = − log(1.75 × 10−5 ) + log
The ionization of the acid is suppressed by the salt and can be neglected.
0.067 mmol/mL 0.033 mmol/mL
= 4.76 + log 2.0 = 5.06 We could have shortened the calculation by recognizing that in the log term the volumes cancel. So we can take the ratio of millimoles only: mmolHOAc = 0.10 mmol/mL × 10 mL = 1.0 mmol mmolOAc− = 0.10 mmol/mL × 20 mL = 2.0 mmol H = 4.76 + log
2.0 mmol = 5.06 1.0 mmol
We can use millimoles of acid and salt in place of molarity. Because the terms appear in a ratio, as long as the units are the same, they will cancel out. But it has to relate to moles or molarity, not mass.
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The mixture of a weak acid and its salt may also be obtained by mixing an excess of weak acid with some strong base to produce the salt by neutralization, or by mixing an excess of salt with strong acid to produce the weak acid component of the buffer.
Example 7.12 Calculate the pH of a solution prepared by adding 25 mL of 0.10 M sodium hydroxide to 30 mL of 0.20 M acetic acid (this would actually be a step in a typical titration). Keep track of millimoles of reactants!
Solution
mmol HOAc = 0.20 M × 30 mL = 6.0 mmol mmol NaOH = 0.10 M × 25 mL = 2.5 mmol These react as follows: HOAc + NaOH NaOAc + H2 O After reaction, mmol NaOAc = 2.5 mmol mmol HOAc = 6.0 − 2.5 = 3.5 mmol 2.5 = 4.61 pH = 4.76 + log 3.5
Dilution does not change the ratio of the buffering species.
The buffering capacity increases with the concentrations of the buffering species.
The buffering mechanism for a mixture of a weak acid and its salt can be explained as follows. The pH is governed by the logarithm of the ratio of the salt and acid: [A− ] (7.48) pH = constant + log [HA] If the solution is diluted, the ratio remains constant, and so the pH of the solution does not change.4 If a small amount of a strong acid is added, it will combine with an equal amount of the A− to convert it to HA. That is, in the equilibrium HA H+ + A− , Le Chˆatelier’s principle dictates added H+ will combine with A− to form HA, with the equilibrium lying far to the left if there is an excess of A− . The change in the ratio [A− ]/[HA] is small and hence the change in pH is small. If the acid had been added to an unbuffered solution (e.g., a solution of NaCl), the pH would have decreased markedly. If a small amount of a strong base is added, it will combine with part of the HA to form an equivalent amount of A− . Again, the change in the ratio is small. The amount of acid or base that can be added without causing a large change in pH is governed by the buffering capacity of the solution. This is determined by the concentrations of HA and A− . The higher their concentrations, the more acid or base the solution can tolerate. The buffer intensity or buffer index of a solution is defined as β = dCB /dpH = −dCHA /dpH
(7.49)
where dCB and dCHA represent the number of moles per liter of strong base or acid, respectively, needed to bring about a pH change of dpH. Although the terms buffer intensity and buffer capacity are often used interchangeably, the buffer capacity is the integrated form of buffer intensity (e.g., the amount of strong acid/base needed to 4 In actuality, the pH will increase slightly because the activity coefficient of the salt has been increased by decreasing the ionic strength. The activity of an uncharged molecule (i.e., undissociated acid) is equal to its molarity (see Chapter 6), and so the ratio increases, causing a slight increase in pH. See the end of the chapter.
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change the pH by a certain finite amount) and is always a positive number. The larger it is, the more resistant the solution is to pH change. For a simple monoprotic weak acid/conjugate base buffer solutions of concentration greater than 0.001 M, the buffer intensity is approximated by: CHA CA− (7.50) β = 2.303 CHA + CA− where CHA and CA− represent the analytical concentrations of the acid and its salt, respectively. Thus, if we have a mixture of 0.10 mol/L acetic acid and 0.10 mol/L sodium acetate, the buffer intensity is 0.10 × 0.10 = 0.050 mol/L per pH β = 2.303 0.10 + 0.10 If we add 0.0050 mol/L solid sodium hydroxide, the change in pH is
241
See Chapter 8, Section 8.11 for a derivation of buffer intensity.
dpH = dCB /β = 0.0050/0.050 = 0.10 = pH In addition to concentration, the buffer intensity is governed by the ratio of HA to A− . It is maximum when the ratio is unity, that is, when the pH = pKa : 1 (7.51) pH = pKa + log = pKa 1 This corresponds to the midpoint of a titration of a weak acid. In general, provided the concentration is not too dilute, the buffering capacity is satisfactory over a pH range of pKa ± 1. We will discuss the buffering capacity in more detail in Chapter 8, when the titration curves of weak acids are discussed.
Example 7.13: Professor’s Favorite Example Contributed by Professor Kris Varazo, Francis Marion University, Florence, South Carolina Calculating the pH of a buffer when strong acid or base is added As an example, suppose you have 100 mL of a buffer containing 0.100 M acetic acid and 0.0500 M sodium acetate. Calculate the pH of the buffer when 3.00 mL of 1.00 M HCl is added to it. As a first step, calculate the pH of the buffer before adding the strong acid using the Henderson–Hasselbalch equation: [A− ] pH = pKa + log [HA] All we need to determine the pH of this buffer is the pKa of acetic acid, which is 4.76: 0.0500 = 4.46 pH = 4.76 + log 0.100 Remember that adding acid to a solution will necessarily lower the pH, so we should expect the pH of the buffer to be lower than 4.46. The best way to solve this problem is to calculate the moles of acetic acid, moles of sodium acetate, and added moles of HCl: 0.100 moles = 0.0100 moles acetic acid 100 mL × 1000 mL 0.0500 moles 100 mL × = 0.00500 moles sodium acetate 1000 mL 1.00 moles = 0.00300 moles hydrochloric acid 3.00 mL × 1000 mL
The buffer intensity is maximum at pH = pKa .
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Note that we expressed molarity in moles per 1000 mL instead of liters. Even though the Henderson–Hasselbalch equation uses molar concentrations, there is only one volume of buffer solution, so we can simply use the mole values we just calculated. We also need to know the chemical reaction occurring when strong acid is added to the buffer: A− + H+ → HA The reaction says that the acetate ion in the buffer will react with the added strong acid, and the moles of acetate will decrease and the moles of acetic acid will increase. How much will the decrease and increase be? It is equal to the amount of strong acid added. We can now write the Henderson–Hasselbalch equation and account for the decrease in moles of acetate and increase in moles of acetic acid: (moles A− − moles H+ added) (7.52) pH = pKa + log (moles HA + moles H+ added) (0.00500 − 0.00300) pH = 4.76 + log (0.0100 + 0.00300) pH = 4.22 The new pH of the buffer is lower than the original value, and it makes sense because adding strong acid to a solution, even a buffer, will cause the pH to decrease. This approach also works in reverse, when you add a strong base to a buffer. In this case, the pH of the buffer will increase, and the relevant chemical reaction is: HA + OH− → H2 O + A− This time the moles of acetic acid will decrease and the moles of acetate will increase. The Henderson–Hasselbalch equation can be written in the following form to solve for the pH: (moles A− + moles OH− added) pH = pKa + log (7.53) (moles HA − moles OH− added)
Note that a buffer can resist a pH change, even when there is added an amount of strong acid or base greater (in moles) than the equilibrium amount of H+ or OH− (in moles) in the buffer. For example, in Example 7.13, the pH of the buffer is 4.46 and [H+ ] = 3.5 × 10−5 M, and millimoles H+ = (3.5 × 10−5 mmol/mL)(100 mL) = 3.5 × 10−3 mmol (in equilibrium with the buffer components). We added 3.00 mmol H+ , far in excess of this. However, due to the reserve of buffer components (OAc− to react with H+ in this case), the added H+ is consumed so that the pH remains relatively constant, so long as we do not exceed the amount of buffer reserve. Similar calculations apply for mixtures of a weak base and its salt. We can consider the equilibrium between the base B and its conjugate acid BH+ and write a Ka for the conjugate (Brønsted) acid: BH+ B + H+ Ka =
[B][H+ ] +
[BH ]
(7.54) =
Kw Kb
(7.55)
The logarithmic Henderson–Hasselbalch form is derived exactly as above: [H+ ] = Ka ·
K [BH+ ] [BH+ ] = w· [B] Kb [B]
−log[H+ ] = − log Ka − log
K [BH+ ] [BH+ ] = − log w − log [B] Kb [B]
(7.56) (7.57)
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[B] [B] + =(pKw −pKb )+log [BH ] [BH+ ] [proton acceptor] [proton acceptor] =(pKw −pKb )+log pH=pKa +log [proton donor] [proton donor] pH = pKa +log
243
(7.58) (7.59)
Since pOH = pKw − pH, we can also write, by subtracting either Equation 7.58 or Equation 7.59 from pKw , pOH = pKb + log
[BH+ ] [proton donor] = pKb + log [B] [proton acceptor]
(7.60)
A mixture of a weak base and its salt acts as a buffer in the same manner as a weak acid and its salt. When a strong acid is added, it combines with some of the base B to form the salt BH+ . Conversely, a base combines with BH+ to form B. Since the change in the ratio will be small, the change in pH will be small. Again, the buffering capacity is maximum at a pH equal to pKa = 14 − pKb (or at pOH = pKb ), with a useful range of pKa ± 1. Although we show the calculations in terms of pKb as well, we recommend that you do all calculations using the pKa of the conjugate acid; consistency keeps you in the comfort zone. When a buffer is diluted, the pH will not change appreciably because the ratio [proton donor] / [proton acceptor] will remain the same.5
pKa = 14 − pKb for a weak base. The alkaline buffering capacity is maximum at pOH = pKb (pH = pKa ).
Example 7.14 Calculate the volume of concentrated ammonia and the weight of ammonium chloride you would have to take to prepare 100 mL of a buffer at pH 10.00 if the final concentration of salt is to be 0.200 M. Solution
We want 100 mL of 0.200 M NH4 Cl. Therefore, mmol NH4 Cl = 0.200 mmol/mL × 100 mL = 20.0 mmol mg NH4 Cl = 20.0 mmol × 53.5 mg/mmol = 1.07 × 103 mg Therefore, we need 1.07 g NH4 Cl. We calculate the concentration of NH3 by [proton acceptor] pH = pKa + log [proton donor] = (14.00 − pKb ) + log 10.0 = (14.00 − 4.76) + log log
[NH3 ] [NH4 + ] [NH3 ] 0.200 mmol/mL
[NH3 ] = 0.76 0.200 mmol/mL [NH3 ] = 100.76 = 5.8 0.200 mmol/mL [NH3 ] = (0.200)(5.8) = 1.16 mmol/mL
5 Buying one bottle of buffer and keeping on diluting and reusing it is not a good business plan, however. As you
dilute, you lose buffering capacity. If you get it really dilute, dissolution of atmospheric CO2 and autoionization of water will affect the buffer pH.
pKa can be obtained directly from Ka given in Table C.2b in Appendix C.
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The molarity of concentrated ammonia is 14.8 M. Therefore, remember, C1 V1 = C2 V2 , 100 mL × 1.16 mmol/mL = 14.8 mmol/mL × mL NH3
mL NH3 = 7.8 mL
Example 7.15 How many grams ammonium chloride and how many milliliters 3.0 M sodium hydroxide should be added to 200 mL water and diluted to 500 mL to prepare a buffer of pH 9.50 with a salt concentration of 0.10 M? Solution
We need the ratio of [NH3 ]/[NH4 + ]. From Example 7.14. pH = pKa + log
[NH3 ] [NH3 ] = 9.24 + log + [NH4 ] [NH4 + ]
9.50 = 9.24 + log log
[NH3 ] [NH4 + ]
[NH3 ] = 0.26 [NH4 + ] [NH3 ] = 100.26 = 1.8 [NH4 + ]
The final concentration of NH4 + is 0.10 M, so [NH3 ] = (1.8)(0.10) = 0.18 M +
mmol NH4 in final solution = 0.10 M × 500 mL = 50 mmol mmol NH3 in final solution = 0.18 M × 500 mL = 90 mmol The NH3 is formed by reacting an equal number of millimoles of NH4 Cl with NaOH. Therefore, a total of 50 + 90 = 140 mmol NH4 Cl must be taken: mg NH4 Cl = 140 mmol × 53.5 mg/mmol = 7.49 × 103 mg = 7.49 g The volume of NaOH needed to react with NH4 + to give 90 millimoles of NH3 is: 3.0 M × x mL = 90 mmol Video: Goal Seek pH mixture
Select a buffer with a pKa value near the desired pH. Buffer salts do not hydrolyze appreciably. See Chapter 13 for a list of NIST standard buffers.
x = 30 mL NaOH
We see that a buffer solution for a given pH is prepared by choosing a weak acid (or a weak base) and its salt, with a pKa value near the pH that we want. There are a number of such acids and bases, and any pH region can be buffered by a proper choice of these. A weak acid and its salt give the best buffering in acid solution, and a weak base and its salt give the best buffering in alkaline solution. Some useful buffers for measurements in physiological solutions are described below. National Institute of Standards and Technology (NIST) buffers used for calibrating pH electrodes are described in Chapter 13. You may have wondered why, in buffer mixtures, the salt does not react with water to hydrolyze as an acid or base. This is because the reaction is suppressed by the presence of the acid or base. In Equation 7.28, the presence of appreciable amounts of
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245
either HA or OH− will suppress the ionization almost completely. In Equation 7.33, the presence of either B or H3 O+ will suppress the ionization. Goal Seek may be used to calculate the pH of mixtures of acids and bases, as in preparing buffers illustrated above. See the text website for an example video for calculating the pH of a mixture of carbonic acid and sodium hydroxide: Goal Seek pH mixture.
7.9 Polyprotic Acids and Their Salts Many acids or bases are polyfunctional, that is, have more than one ionizable proton or hydroxide ion. These substances ionize stepwise, and an equilibrium constant can be written for each step. Consider, for example, the ionization of phosphoric acid: H3 PO4 H+ + H2 PO4 −
Ka1 = 1.1 × 10−2 =
[H+ ] [H2 PO4 − ] [H3 PO4 ]
(7.61)
H2 PO4 − H+ + HPO4 2−
Ka2 = 7.5 × 10−8 =
[H+ ] [HPO4 2− ] [H2 PO4 − ]
(7.62)
HPO4 2− H+ + PO4 3−
Ka3 = 4.8 × 10−13 =
[H+ ] [PO4 3− ] [HPO4 2− ]
(7.63)
The stepwise Ka values of polyprotic acids get progressively smaller as the increased negative charge makes dissociation of the next proton more difficult.
Recall from Chapter 6 that the overall ionization is the sum of these individual steps and the overall ionization constant is the product of the individual ionization constants: H3 PO4 3H+ + PO4 3− Ka = Ka1 Ka2 Ka3 = 4.0 × 10−22 =
[H+ ]3 [PO4 3− ] [H3 PO4 ]
(7.64)
The individual pKa values are 1.96, 7.12, and 12.32, respectively, for pKa1 , pKa2 , and pKa3 . In order to make precise pH calculations, the contributions of protons from each ionization step must be taken into account. Exact calculation is difficult and requires a tedious iterative procedure since [H+ ] is unknown in addition to the various phosphoric acid species. See, for example, References 8 and 11 for calculations. Excel or other spreadsheet-based calculations can be simple. This is illustrated later. In most cases, approximations can be made so that each ionization step can be considered individually. If the difference between successive ionization constants is more than 103 , each proton can be differentiated in a titration, that is, each is titrated separately to give stepwise pH breaks in the titration curve. (If an ionization constant is less than about 10−9 , then the ionization is too small for a pH break to be exhibited in the titration curve—for example, the third proton for H3 PO4 .) When the individual pKa ’s are separated by three units or more, calculations are simplified because the system can be considered as simply a mixture of three weak acids of equal concentration that largely do not interact with each other.
We can titrate the first two protons of H3 PO4 separately. The third is too weak to titrate.
BUFFER CALCULATIONS FOR POLYPROTIC ACIDS The anion on the right side in each ionization step can be considered the salt (conjugate base) of the acid from which it is derived. That is, in Equation 7.61, H2 PO4 − is the salt of the acid H3 PO4 . In Equation 7.62, HPO4 2− is the salt of the acid H2 PO4 − , and in Equation 7.63, PO4 3− is the salt of the acid HPO4 2− . So each of these pairs constitutes a buffer system, and orthophosphate buffers can be prepared over a wide pH range.
We can prepare phosphate buffers with pH centered around 1.96 (pKa1 ), 7.12 (pKa2 ), and 12.32 (pKa3 ).
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The optimum buffering capacity of each pair occurs at a pH corresponding to its pKa . The HPO4 2− /H2 PO4 − couple is an effective buffer system in the blood (see below).
Example 7.16 The pH of blood is 7.40. What is the ratio of [HPO4 2− ]/[H2 PO4 − ] in the blood (assume 25◦ C)? [proton acceptor] Solution pH = pKa + log [proton donor] pKa2 = 7.12 Therefore, pH = 7.12 + log
[HPO4 2− ] [H2 PO4 − ]
7.40 = 7.12 + log
[HPO4 2− ] [H2 PO4 − ]
[HPO4 2− ] = 10(7.40−7.12) = 100.28 = 1.9 [H2 PO4 − ]
DISSOCIATION CALCULATIONS FOR POLYPROTIC ACIDS Because the individual ionization constants are sufficiently different, the pH of a solution of H3 PO4 can be calculated by treating it just as we would any weak acid. The H+ from the first ionization step effectively suppresses the other two ionization steps, so that the H+ contribution from them is negligible compared to the first ionization. The quadratic equation must, however, be solved because Ka1 is relatively large.
Example 7.17 Calculate the pH of a 0.100 M H3 PO4 solution. Solution
H3 PO4 ≈ H+ 0.100 − x x
+ H2 PO4 − x
From Equation 7.61,
(x)(x) = 1.1 × 10−2 0.100 − x In order to neglect x, C should be ≥ 100Ka . Here, it is only 10 times as large. Therefore, use the quadratic equation to solve:
Treat H3 PO4 as a monoprotic acid. But x can’t be neglected compared to C.
x2 + 0.011x − 1.1 × 10−3 = 0
−0.011 ± (0.011)2 − 4(−1.1 × 10−3 ) x= 2 + x = [H ] = 0.028 M The acid is 28% ionized: pH = − log 2.8 × 10−2 = 2 − 0.45 = 1.55
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We can determine if our assumption that the only important source of protons is H3 PO4 was a realistic one. H2 PO4 − would be the next most likely source of protons. From Equation 7.62, [HPO4 2− ] = Ka2 [H2 PO4 − ]/[H+ ]. Assuming the concentrations of H2 PO4 − and H+ as a first approximation are 0.028 M as calculated, then [HPO4 2− ] ≈ Ka2 = 7.5 × 10−8 M. This is very small compared to 0.028 M H2 PO4 − , and so further dissociation is indeed insignificant. We were justified in our approach.
7.10 Ladder Diagrams Professor’s Favorite Example Contributed by Professor Galena Talanova, Howard University What is the dominant species at a given pH? A simple way to visualize what is the dominant species at a given pH is to construct what David Harvey of Depauw University (http://acad.depauw.edu/ ˜harvey/ASDL2008/introduction.html) calls Ladder Diagrams. We draw a vertical axis in pH with horizontal bars for the different pKa ’s present in the system, as illustrated in Figure 7.1 below. Referring to Figure 7.1(a), the unionized acid HOAc dominates below a pH of 4.76 (the pKa value of HOAc) while above this value the OAc− anion is dominant. Referring to Figure 7.1(b), one can consider that all three acid–base systems can be simultaneously or individually present. For the HF–F− system depicted in blue, HF dominates at pH values below its pKa of 3.17 and above this F− dominates. Similarly H2 S and NH4 + dominate at pH values below 6.88 and 9.25, respectively, while HS− and NH3 dominates at respective pH values above that. Such diagrams also indicate what species will be dominant in a mixed system at a given pH: at a pH of 6 for
12
F− 9.25
OAc−
pK6 = 10.24
H2S pKNH4+
pH
HY3−
8
8
8
pKH2S
6.88
pH
Basic
12
Y4 −
12
NH3
pK5 = 6.16
NH4
+
Fig. 7.1.
pKHOAc
4.76
Acidic
pKHF HS− pK = 2.66 4
3.17
HOAc
H5Y+
(a)
H3Y− pK3 = 2 H4Y
pK2 = 1.5
HF 0
H2Y
4
4
4
2−
pK1 = 0
H6Y2+ (b)
(c)
Ladder diagrams depicting systems containing (a) acetic acid and acetate, (b) ammonium and ammonia, hydrofluoric acid and fluoride, (di)hydrogen sulfide and hydrosulfide, and (c) diprotonated ethylenediaminetetraacetic acid (EDTA) (H6 Y2+ ) and the other six species obtained by successive deprotonation.
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example, we will expect F− , NH4 + , and H2 S to be the dominant species relative to their respective conjugate acid/base. Finally, Figure 7.1(c) shows the case for EDTA, this represents a hexaprotic acid system; the free acid is actually the diprotonated form H6 Y2+ and it exists in dominant form only in extremely strongly acid solutions (pH < 0). The respective zones in which the other individual species dominate (e.g., Y4− is dominant at pH > 10). More details and example problems are given in a PowerPoint file (ch7 7.10 ladder diagrams.ppt) available in the text website.
7.11 FRACTIONS OF DISSOCIATING SPECIES AT A GIVEN pH: α VALUES—HOW MUCH OF EACH SPECIES? Often, it is of interest to know the distribution of the different species of a polyprotic acid as a function of pH, that is, at known hydrogen ion concentration as in a buffered solution. Consider, for example, the dissociation of phosphoric acid. The equilibria are given in Equations 7.61 to 7.63. All the four phosphoric acid species coexist in equilibrium with one another, although the concentrations of some may be very small at a given pH. By changing the pH, the equilibria shift; the relative concentrations change. It is possible to derive general equations for calculating the fraction of the acid that exists in a given form, from the given hydrogen ion concentration. For a given total analytical concentration of phosphoric acid, CH3 PO4 , we can write H3 PO4 , H2 PO4 − , HPO4 2− , and PO4 3− all exist together in equilibrium. The pH determines the fraction of each.
CH3 PO4 = [PO4 3− ] + [HPO4 2− ] + [H2 PO4 − ] + [H3 PO4 ]
(7.65)
where the terms on the right-hand side of the equation represent the equilibrium concentrations of the individual species. We presumably know the initial total concentration CH3 PO4 and wish to find the fractions or concentrations of the individual species at equilibrium. We define [H2 PO4 − ] CH3 PO4
α0 =
[H3 PO4 ] CH3 PO4
α1 =
α3 =
[PO4 3− ] CH3 PO4
α0 + α1 + α2 + α3 = 1
α2 =
[HPO4 2− ] CH3 PO4
where the α’s are the fractions of each species present at equilibrium. Note that the subscripts denote the number of dissociated protons or the charge on the species. We can use Equation 7.65 and the equilibrium constant expressions 7.61 through 7.63 to obtain an expression for CH3 PO4 in terms of the desired species. This is substituted into the appropriate equation to obtain α in terms of [H+ ] and the equilibrium constants. In order to calculate α0 , for example, we can rearrange Equations 7.61 through 7.63 to solve for all the species except [H3 PO4 ] and substitute into Equation 7.65: [PO4 3− ] =
Ka3 [HPO4 2− ] [H+ ]
(7.66)
[HPO4 2− ] =
Ka2 [H2 PO4 − ] [H+ ]
(7.67)
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[H2 PO4 − ] =
Ka1 [H3 PO4 ] [H+ ]
249
(7.68)
We want all these to contain only [H3 PO4 ] (and [H+ ], the variable). We can substitute Equation 7.68 for [H2 PO4 − ] in Equation 7.67: [HPO4 2− ] =
Ka1 Ka2 [H3 PO4 ] [H+ ]2
(7.69)
And we can substitute Equation 7.69 into Equation 7.66 for [HPO4 2− ]: [PO4 3− ] =
Ka1 Ka2 Ka3 [H3 PO4 ] [H+ ]3
(7.70)
Finally, we can substitute 7.68 through 7.70 in Equation 7.65: Ka1 Ka2 Ka3 [H3 PO4 ] Ka1 Ka2 [H3 PO4 ] Ka1 [H3 PO4 ] + + + [H3 PO4 ] [H+ ]3 [H+ ]2 [H+ ] (7.71) We can divide each side of this expression by [H3 PO4 ] to obtain 1/α0 : CH3 PO4 =
CH3 PO4 [H3 PO4 ]
=
1 K K K K K K +1 = a1 +a2 3 a3 + a1+ a2 + a1 2 α0 [H ] [H ] [H+ ]
(7.72)
Taking the reciprocal of both sides α0 =
(Ka1 Ka2 Ka3
/[H+ ]3 )
1 + (Ka1 Ka2 /[H+ ]2 ) + (Ka1 /[H+ ]) + 1
(7.73)
multiplying both the numerator and denominator on the right with [H+ ]3 , we have: α0 =
[H+ ]3 [H+ ]3 + Ka1 [H+ ]2 + Ka1 Ka2 [H+ ] + Ka1 Ka2 Ka3
(7.74)
Use this equation to calculate the fraction of H3 PO4 in solution.
Similar approaches can be taken to obtain expressions for the other αs. For α1 , for example, the equilibrium constant expressions would be solved for all species in terms of [H2 PO4 − ] and substituted into Equation 7.65 to obtain an expression for CH3 PO4 containing only [H2 PO4 − ] and [H+ ], from which α1 is calculated. The results for the other αs are α1 =
α2 =
α3 =
Ka1 [H+ ]2 [H+ ]3 + Ka1 [H+ ]2 + Ka1 Ka2 [H+ ] + Ka1 Ka2 Ka3 Ka1 Ka2 [H+ ] [H+ ]3 + Ka1 [H+ ]2 + Ka1 Ka2 [H+ ] + Ka1 Ka2 Ka3 [H+ ]3
+ Ka1
Ka1 Ka2 Ka3 + Ka1 Ka2 [H+ ] + Ka1 Ka2 Ka3
[H+ ]2
(7.75)
(7.76) (7.77)
Note that all have the same denominator and that the sum of the numerators equals the denominator. For α0 , the first term in the denominator becomes the numerator; for α1 , the second term in the denominator becomes the numerator; for α2 , the third term becomes the numerator, and so on. See Problem 63 for a more detailed derivation of the other α’s. In general, an n-protic acid (n = 1, 2, 3, respectively, for HOAc, H2 C2 O4 , H3 PO4 , etc.) will have n + 1 species other than H+ derived from the acid (e.g., H2 C2 O4 ,
Derive these equations in Problem 62.
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HC2 O4 − , and C2 O4 2− ) and thus, n + 1 α-values. The denominator in the α-values, Qn , will consist in each case of n + 1 terms: Qn =
i=n
[H + ]n−i Ka0 . . . Kai
(7.78)
i=0
where Ka0 is taken to be 1. Thus Q1 = [H+ ] + Ka
(7.79)
Q2 = [H+ ]2 + Ka1 [H+ ] + Ka1 Ka2
(7.80)
Q3 = [H+ ]3 + Ka1 [H+ ]2 + Ka1 Ka2 [H+ ] + Ka1 Ka2 Ka3
(7.81)
Q4 = [H+ ]4 + Ka1 [H+ ]3 + Ka1 Ka2 [H+ ]2 + Ka1 Ka2 Ka3 [H+ ] + Ka1 Ka2 Ka3 Ka4 (7.82) While Equation 7.78 may look complicated, it is easy to remember the pattern represented by Equations 7.79 through 7.82. You start with [H+ ]n where n is the number of dissociable proton and then replace an H+ with a Ka term, beginning with Ka1 until you run out of both. For α0 to αn , remember that the first term in the Q expression becomes the numerator for α0 , the second term for α1 and so on. Thus, α1 for a monoprotic acid α1,m is: α1,m = Ka /Q1 =
Ka + Ka
[H+ ]
(7.83)
and α3 for a tetraprotic acid α3,te is: α3,te = Ka1 Ka2 Ka3 [H+ ]/Q4 = Ka1 Ka2 Ka3 [H+ ]/([H+ ]4 + Ka1 [H+ ]3 + Ka1 Ka2 [H+ ]2 +Ka1 Ka2 Ka3 [H+ ] + Ka1 Ka2 Ka3 Ka4 )
(7.84)
Example 7.18 Calculate the equilibrium concentration of the different species in a 0.10 M phosphoric acid solution at pH 3.00 ([H+ ] = 1.0 × 10−3 M). Solution
Substituting into Equation 7.79, α0 =
=
(1.0 × 10−3 )3 (1.0 × 10−3 )3 + (1.1 × 10−2 )(1.0 × 10−3 )2 + (1.1 × 10−2 ) (7.5 × 10−8 )(1.0 × 10−3 ) + (1.1 × 10−2 )(7.5 × 10−8 )(4.8 × 10−13 ) 1.0 × 10−9 = 8.3 × 10−2 1.2 × 10−8
[H3 PO4 ] = CH3 PO4 α0 = 0.10 × 8.3 × 10−2 = 8.3 × 10−3 M Similarly, α1 = 0.92 [H2 PO4 − ] = CH3 PO4 α1 = 0.10 × 0.92 = 9.2 × 10−2 M
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α2 = 6.9 × 10−5 [HPO4 2− ] = CH3 PO4 α2 = 0.10 × 6.9 × 10−5 = 6.9 × 10−6 M α3 = 3.3 × 10−14 [PO4 3− ] = CH3 PO4 α3 = 0.10 × 3.3 × 10−14 = 3.3 × 10−15 M We see that at pH 3, the majority (92%) of the phosphoric acid exists as H2 PO4 − and 8.3% exists as H3 PO4 . Only 3.3 × 10−12 % exists as PO4 3− ! We can prepare a spreadsheet to calculate the fraction of each species as a function of pH. Formulas and calculations are shown in the spreadsheet6 , and Figure 7.2 shows the corresponding α versus pH plot. The Ka values are entered in cells B4, D4, and F4. The pH values are entered in column A. All the formulas needed for each cell are listed at the bottom of the spreadsheet, and they are initially entered in the boldfaced cells. The formula for calculating the corresponding hydrogen ion concentration (used in the α calculation) is entered in cell B6. The formula for the denominator used for each α calculation is entered in cell C6. Note that the constants are entered as absolute values, while the hydrogen ion concentration is computed from the specific pH values entered in column A. The formulas for the three α calculations are entered in cells D6, E6, and F6. All formulas are copied down to row 34. Referring to Figure 7.2.xlsx that is available to you from the website (the screenshot is reproduced as Figure 7.3), we want to plot α0 , α1 , α2 , and α3 (which appear in the columns D:G, titled a0 , a1 , a2 , and a3 ) as a function of pH (column A).The most expedient way to accomplish this is to leave the x-data (pH) where it is in column A and move the three sets of y-data we want to plot in contiguous columns. We therefore select the [H+ ] and denominator (Q) data (B5:C34) that we do not wish to plot and cut and paste (Ctrl-X, Ctrl V) beginning in cell H5. We select B5:C34 again and delete the empty space (Alt-E D, Shift cells left). Now highlight all the data
The overlapping curves represent buffer regions. The pH values where α1 and α2 are 1 represent the end points in titrating H3 PO4 . See Section 7.16 for a way of representing these plots as straight lines (log–log plots).
By far the best way to understand Excel-related text is to have the relevant Excel file open on your computer while you read this text.
H3PO4 alpha values 1.0 0.9 𝛂1: H2PO4−
0.8
𝛂2: HPO42−
0.7 𝛂0: H3PO4
Alpha
0.6 0.5 0.4
𝛂3: PO43−
0.3 0.2 1st end point
0.1 Buffer region
0.0 0.0
1.0
2.0
3.0
Buffer region
4.0
5.0
Fig. 7.2.
2nd end point
6.0
7.0 pH
8.0
Buffer region
9.0 10.0 11.0 12.0 13.0 14.0
6 The “standard view” of any Excel spreadsheet consists of the numbers entered in a cell or the results of the
formula entered into a cell. It does not normally show the formula contained in the cell until your cursor is on the cell and then it displays the formula in the formula bar. At any time, you can change the “standard view” into “formula view” by pressing the Ctrl and ‘ (accent key) together.
Fractions of H3 PO4 species as a function of pH. (Buffer region designation suggestions courtesy of Professor Galina Talanova, Howard University)
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A B C 1 Calculation of alpha values for H3PO4 vs. pH.
D
E
F
G
2 Alpha (ai) denominator = [H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3
Fig. 7.3.
Screenshot of the file Figure 7.2.xlsx (the spreadsheet itself is available in the text website).
3 Numerators: α0 = [H+]3; α1 = Ka1[H+]2 ; α2 = Ka1Ka2[H+]; α3 = Ka1Ka2Ka3 Ka1= 1.10E-02 Ka2= Ka3= 4.80E-13 4 7.50E-08 [H+] α0 α1 α2 Denominator pH 5 1 1.01E+00 9.89E−01 1.09E−02 8.16E−10 0.0 6 7 0.5 0.316228 3.27E-02 9.66E-01 3.36E-02 7.97E-09 8 1.0 0.1 1.11E-03 9.01E-01 9.91E-02 7.43E-08 9 1.5 0.031623 4.26E-05 7.42E-01 2.58E-01 6.12E-07 10 2.0 0.01 2.10E-06 4.76E-01 5.24E-01 3.93E-06 11 2.5 0.003162 1.42E-07 2.23E-01 7.77E-01 1.84E-05 3.0 0.001 1.20E-08 8.33E-02 9.17E-01 6.87E-05 12 13 3.5 0.000316 1.13E-09 2.79E-02 9.72E-01 2.30E-04 14 4.0 0.0001 1.11E-10 9.00E-03 9.90E-01 7.43E-04 15 4.5 3.16E-05 1.11E-11 2.86E-03 9.95E-01 2.36E-03 16 5.0 0.00001 1.11E-12 9.02E-04 9.92E-01 7.44E-03 17 5.5 3.16E-06 1.13E-13 2.81E-04 9.77E-01 2.32E-02 18 6.0 0.000001 1.18E-14 8.46E-05 9.30E-01 6.98E-02 19 6.5 3.16E-07 1.36E-15 2.32E-05 8.08E-01 1.92E-01 20 7.0 1E-07 1.93E-16 5.19E-06 5.71E-01 4.29E-01 21 7.5 3.16E-08 3.71E-17 8.53E-07 2.97E-01 7.03E-01 22 8.0 1E-08 9.35E-18 1.07E-07 1.18E-01 8.82E-01 23 8.5 3.16E-09 2.72E-18 1.16E-08 4.05E-02 9.59E-01 24 9.0 1E-09 8.36E-19 1.20E-09 1.32E-02 9.86E-01 9.5 3.16E-10 2.62E-19 1.21E-10 4.19E-03 9.94E-01 25 26 10.0 1E-10 8.30E-20 1.20E-11 1.33E-03 9.94E-01 27 10.5 3.16E-11 2.65E-20 1.19E-12 4.15E-04 9.85E-01 28 11.0 1E-11 8.65E-21 1.16E-13 1.27E-04 9.54E-01 29 11.5 3.16E-12 3.00E-21 1.05E-14 3.66E-05 8.68E-01 12.0 1E-12 1.22E-21 8.19E-16 9.01E-06 6.76E-01 30 12.5 3.16E-13 6.57E-22 4.81E-17 1.67E-06 3.97E-01 31 13.0 1E-13 4.79E-22 2.09E-18 2.30E-07 1.72E-01 32 33 13.5 3.16E-14 4.22E-22 7.49E-20 2.61E-08 6.18E-02 34 14.0 1E-14 4.04E-22 2.47E-21 2.72E-09 2.04E-02 35 Formulas for cells in boldface: 10^-A6 36 Cell B6 = [H+] = 37 Cell C6=denom.= B6^3+$B$4*B6^2+$B$4*$D$4*B6+$B$4*$D$4*$F$4 B6^3/C6 38 Cell D6 = α0 = ($B$4*B6^2)/C6 39 Cell E6 = α1 = ($B$4*$D$4*B6)/C6 40 Cell F6 = α2 = ($B$4*$D$4*$F$4)/C6 41 Cell G6 = α3 = 42 Copy each formula down through Cell 34 43 Plot A6:A34 vs. D6:D34, E6:E34, F6:F34, and G6:G34 (series 1, 2, 3, and 4)
α3 3.92E−22 1.21E-20 3.57E-19 9.29E-18 1.89E-16 2.80E-15 3.30E-14 3.50E-13 3.56E-12 3.58E-11 3.57E-10 3.52E-09 3.35E-08 2.91E-07 2.06E-06 1.07E-05 4.24E-05 1.46E-04 4.73E-04 1.51E-03 4.77E-03 1.49E-02 4.58E-02 1.32E-01 3.24E-01 6.03E-01 8.28E-01 9.38E-01 9.80E-01
columns we wish to plot (A5:E34) and click on Insert—Charts- Scatter. You can choose the scatter plot shown in column 1, row 2 in the drop-down menu if you just want to see the line plot. Otherwise the scatter plot shown in column 2, row 1 if you want to see the specific points plotted as in Figure 7.2. The plot appears. You can move it to a separate sheet (by right-clicking on the frame of the chart, selecting Move Chart, New Sheet, and OK). Play with the Chart layout on the menu bar: You can click on Layout 3 (the grid line and linear fit layout), the linear fits are irrelevant here can be subsequently deleted by clicking on the fit lines and deleting them, leaving the gridlines. You can also click on the axes and chart titles, and change them to what you want them to read, etc. The plot generated by the procedure above is given in Figure 7.2. This figure illustrates how the ratios of the four phosphoric acid species change as the pH is adjusted, for example, in titrating H3 PO4 with NaOH. While some appear to go to zero concentration above or below certain pH values, they are not really zero,
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but diminishingly small. For example, we saw in Example 7.18 that at pH 3.00, the concentration of the PO4 3− ion for 0.1 M H3 PO4 is only 3.3 × 10−15 M, but it is indeed present in equilibrium. The pH regions where two curves overlap (with appreciable concentrations) represent regions in which buffers may be prepared using those two species. For example, mixtures of H3 PO4 and H2 PO4 − can be used to prepare buffers around pH 2.0 ± 1, mixtures of H2 PO4 − and HPO4 2− around pH 7.1 ± 1, and mixtures of HPO4 2− and PO4 3− around pH 12.3 ± 1. The pH values at which the fraction of a species is essentially 1.0 correspond to the end points in the titration of phosphoric acid with a strong base, that is, H2 PO4 2− at the first end point (pH 4.5), HPO4 2− at the second end point (pH 9.7). Equation 7.71 could be used for a rigorous calculation of the hydrogen ion concentration from dissociation of a phosphoric acid solution at a given H3 PO4 concentration (no other added H+ ), but this involves tedious iterations. As a first approximation, [H+ ] could be calculated from Ka1 as in Example 7.17, assuming that only the first dissociation step of phosphoric acid was significant. (This is, in fact, what we did in that example.) The first calculated [H+ ] could then be substituted in Equation 7.71 to calculate a second approximation of [H3 PO4 ], which would be used for a second iterative calculation of [H+ ] using Ka1 , and so forth, until the concentration was constant. A simpler way is to use Excel and Goal Seek as illustrated in Example 7.19 below. A useful applet developed by Professor Constantinos Efstathiou at the University of Athens allows easy plotting of distribution diagrams of mono- to tetraprotic acids: http://www.chem.uoa.gr/applets/AppletAcid/Appl_Distr2.html. Check out the phosphoric acid one and compare with Figure 7.1, and the EDTA (H4 A) one with Figure 9.1 in Chapter 9. You can change the pKa values to see how the plots change. The applet also plots log distribution diagrams for the acids (Section 7.16 below). You can also change the Ka values in the spreadsheet Figure 7.2.xlsx to see how the distribution will change.
Example 7.19 The Method of Charge Balance. Calculation of pH in a Phosphoric Acid System Calculate the pH of 0.050 M H3 PO4 using Excel and Goal Seek. What will be the pH if you add 0.11 mole of NaOAc and 0.02 mole of K2 HPO4 to 1 L of this of this solution? The charge balance method invokes that the sum of the positive charges in any solution equals the sum of negative charges. So for a solution containing only phosphoric acid, the relevant charge balance equation is: [H+ ] = [OH− ] + [H2 PO4 − ] + 2[HPO4 2− ] + 3[PO4 3− ]
(7.85)
Note that the multipliers of 2 and 3 need, respectively, to be applied to the concentrations of HPO4 2− and PO4 3− because these ions, respectively, carry 2 and 3 units of charge. Putting all terms on one side, expressing [OH− ] as Kw /[H+ ] and expressing the various phosphate species concentrations in terms of their α-values, we have: [H+ ] − Kw /[H+ ] − Cp (α1 + 2α2 + 3α3 ) = 0
(7.86)
where Cp is the concentration of the total phosphate species, in this case the concentration of H3 PO4 taken. The Example 7.19.xlsx spreadsheet is available in your text website. In cells B1:B5, we have, respectively, written down the values of Ka1 , Ka2 , Ka3 , Cp , and Kw . It is unfortunate that many of the symbols we traditionally use in chemical problems aren’t allowed by Excel as KA1, etc. It actually refers to the cell
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in column KA and row 1. Similarly C and R refer to columns and rows and are not allowed to be used for any other meaning. Thus, for our purposes, we have named Ka1 , Ka2 , Ka3 , and Cp as KAA, KAB, KAC, and CP and written these names in column A next to the numeric values in column bar. Next we want to permanently ascribe these names to the specific numbers, so that every time we write KAA, Excel will know that we are referring to the value of KAA, 1.1 × 10−2 . To do this, we put our cursor on cell B2. In the top right corner on the formula bar normally it would say B2, when we put our cursor on cell B2. But notice that it says KAA. This is because we have given the number in cell B2 the name KAA. We did this (Excel 2010 only—previous versions have a different procedure) by putting the cursor on cell B2, clicking on the name box (top left corner of formula bar), typing KAA and hitting enter. Verify that cell B2 has the name KAA by moving to some other cell and coming back to cell B2 and noticing that the name box says KAA. (Practice on another spreadsheet naming cells.) For convenience, we have named cells B3:B5 in a similar manner as KAB, KAC, CP, and KW. Now two rows below this, we have set up column headings as pH, H+ , OH− , Q3, Alpha1, Alpha2, Alpha3, and Equation. In cell A8, under pH is the value we will be trying to calculate. Presently you can enter any guess value for pH that you might think is reasonable, any value between 0 and 14; it is not important. For now let us enter 0. Also, go ahead and name the cell pH (we do not really need to do this, we can keep on referring to it as cell A8, but it is fun to give names to the cells to designate what they are). In cell B8, under H+ we want to calculate the corresponding value of [H+ ]. Excel does not a priori know the relationship between pH and [H+ ]. Since we know that [H+ ] can be expressed as 10−pH , we enter in cell B8, = 10−pH . (If we did not name cell A8 as pH, we would have had to write in cell B8 = 10−A8 .) Again, while this is not essential, for convenience, we name cell B8 as H. In cell C8, to calculate the value of [OH− ], we enter = KW/H. For good measure, we name cell C8 as OH. Next under heading Q3 , we have to put in the expression of Equation 7.81, using current names and so we enter in cell D8: = H∧ 3 + KAA∗ H∧ 2 + KAA∗ KAB∗ H + KAA∗ KAB∗ KAC and name the cell Q (we can name it Q or Q three, but not Q3, remember?). Similarly, we enter the alpha formulas in Equations 7.75 through 7.77 in cells E8:G8 and also name them, respectively, ALFA1, ALFA2, and ALFA3. For example, we have entered in cell F8 = KAA∗ KAB∗ H/Q Finally,we are ready to write our Equation 7.86 in cell H8 as: = (1E10) ∗ (H-OH-CP ∗ (ALFA1 + 2∗ ALFA2 + 3∗ ALFA3)) You will note that the expression following the 1E10 multiplier (remember this is to stop Excel from believing prematurely that it has found a solution) is the expression in Equation 7.86. All we have to do now is to invoke Goal Seek (Data/What-If Analysis/Goal Seek), type in H8 in the Set Cell box, in “To Value” box type 0 (Equation 7.86), and “By changing cell” box enter pH (or A8). You instantly get your solution, the pH is 1.73. In the “Equation” cell (H8), there is merit to squaring the entire parenthetical expression because this makes it have only positive values. This is not important when we are solving a single problem, but it is important when we solve multiple problems at a time; we will discuss this in greater depth in a later section. Now let us solve the second part of the problem in which we put into this solution in addition 0.11 M NaOAc, 0.02 M K2 HPO4 . Introducing another acid-base system would normally seem to be a formidable problem. Actually it is not. We have to do only a modest amount of additional work on the existing spreadsheet to solve this problem. The worked-out solution appears in the spreadsheet Example 7.19b.xlsx on
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the text website, but let us presently modify the Example 7.19.xlsx spreadsheet we have been working on. First, we need to understand the changes. We have added sodium (0.11 M, we will call this CNA), potassium (2 × 0.02 = 0.04 M, we will call this CK) and acetate species (a total of 0.11 M, we will call this COAC). The concentration of our total phosphate species has increased from 0.050 M to 0.070 M and we need to define the dissociation constant for acetic acid (1.75E-5; we shall call this KOAC). The addition of the new species requires that we modify Equation 7.85 to be: [H+ ] + [Na+ ] + [K + ] = [OH− ] + [OAc− ] + [H2 PO4 − ] + 2[HPO4 2− ] + 3[PO4 3− ] (7.87) If we define Q1 to be the relevant denominator for the acetate system (see Equation 7.79), then α1 for the acetate system (ALFA1OAC, the name ALFA1 is already taken by the phosphate system—the two α1 values for these acetate and phosphate systems are not the same—although they are in the same solution and are subject to the same pH, the Ka -values for the two systems are different) will be given by Equation 7.83 and we can write Equation 7.87 as: [H+ ] + [Na+ ] + [K + ] − Kw /[H+ ] − COAc α1OAC − Cp (α1 + 2α2 + 3α3 ) = 0 (7.88) Now going back to the spreadsheet, in cells D1:D4 we put in the values for KOAC, CNA, CK, and COAC, and give these cells the corresponding names.We put the cursor on column H and insert two more columns (Alt-I and C, twice in succession) so we can create headings Q1 and Alpha1OAc. In these two cells we respectively enter, for Q1, = H + KOAC in cell H8 and name it Qone. For Alpha1OAc (cell I8), we enter = KOAC/Q one and also name the cell ALFA1OAC. Now it is a matter of modifying the equation in cell J8 as: = (10000000000) ∗ (H + CK + CNA − OH − COAC ∗ ALFA1OAC − CP ∗ ALFA1 + 2∗ ALFA2 + 3∗ ALFA3 Once again, invoke Goal Seek and ask it to set cell J8 to value 0 by changing the cell pH and you immediately have your answer, the pH is 5.17.
7.12 SALTS OF POLYPROTIC ACIDS—ACID, BASE, OR BOTH? Salts of acids such as H3 PO4 may be acidic or basic. The protonated salts possess both acidic and basic properties (H2 PO4 − , HPO4 2− ), while the unprotonated salt is simply a Brønsted base that hydrolyzes (PO4 3− ). 1. Amphoteric Salts. H2 PO4 − possesses both acidic and basic properties. That is, it is amphoteric. It ionizes as a weak acid and it also is a Brønsted base that hydrolyzes: H2 PO4 − H+ + HPO4 2−
Ka2 =
H2 PO4 − + H2 O H3 PO4 + OH− Kb = =
[H+ ] [HPO4 2− ] = 7.5 × 10−8 [H2 PO4 − ]
(7.89)
Kw [H PO ][OH− ] = 3 4 − Ka1 [H2 PO4 ] 1.00 × 10−14 = 9.1 × 10−13 −2 1.1 × 10
(7.90)
H2 PO4 − acts as both an acid and a base. See the end of Section 7.16 for how to estimate the extent of each reaction using log–log diagrams.
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The solution could, hence, be either alkaline or acidic, depending on which ionization is more extensive. Since Ka2 for the first ionization is nearly 105 greater than Kb for the second ionization, the solution in this case will obviously be acidic. An expression for the hydrogen ion concentration in a solution of an ion such as H2 PO4 − can be obtained as follows. The total hydrogen ion concentration is equal to the amounts produced from the ionization equilibrium in Equation 7.89 and the ionization of water, less the amount of OH− produced from the hydrolysis in Equation 7.90. We can write, then,
or
CH+ = [H+ ]total = [H+ ]H2 O + [H+ ]H2 PO4 − − [OH− ]H2 PO4 −
(7.91)
[H+ ] = [OH− ] + [HPO4 2− ] − [H3 PO4 ]
(7.92)
We have included the contribution from water since it will not be negligible if the pH of the salt solution happens to be near 7—although in this particular case, the solution will be acid, making the water ionization negligible. We can solve for [H+ ] by substituting expressions in the right-hand side of Equation 7.92 from the equilibrium constant expressions 7.61 and 7.62 and Kw to eliminate all but [H2 PO4 − ], the experimental variable, and [H+ ]: [H+ ] =
Ka2 [H2 PO4 − ] [H2 PO4 − ] [H+ ] Kw + − [H+ ] [H+ ] Ka1
(7.93)
from which (by multiplying each side of the equation by [H+ ], collecting the terms containing [H+ ]2 on the left side, and solving for [H+ ]2 ) [H+ ]2 = [H+ ] =
Kw + Ka2 [H2 PO4 − ] 1+
[H2 PO4 − ] Ka1
Ka1 Kw + Ka1 Ka2 [H2 PO4 − ] Ka1 + [H2 PO4 − ]
(7.94)
(7.95)
That is, for the general case HA− , For HA2− , substitute [HA2− ] for [HA− ], Ka2 for Ka1 , and Ka3 for Ka2 .
+
[H ] =
Ka1 Kw + Ka1 Ka2 [HA− ] Ka1 + [HA− ]
(7.96)
This equation is valid for any salt HA− derived from an acid H2 A (or for HA2− derived from H2 A− , etc.) where [H2 PO4 − ] is replaced by [HA− ]. If we assume that the equilibrium concentration [HA− ] is equal to the concentration of salt added, that is, that the extent of ionization and hydrolysis is fairly small, then this value along with the constants can be used for the calculation of [H+ ]. This assumption is reasonable if the two equilibrium constants (Ka1 and Kb ) involving the salt HA− are small and the solution is not too dilute. In many cases, Ka1 Kw Ka1 Ka2 [HA− ] in the numerator and can be neglected. This is the equation we would have obtained if we had neglected the dissociation of water. Furthermore, if Ka1 [HA− ] in the denominator, the equation simplifies to For HA2− , [H+ ] =
Ka2 Ka3 .
[H+ ] =
Ka1 Ka2
(7.97)
Therefore, if the assumptions hold, the pH of a solution of H2 PO4 − is independent of its concentration! This approximation is adequate for our purposes. The equation
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generally applies if there is a large difference between Ka1 and Ka2 . For the case of H2 PO4 − , then,
(7.98) [H+ ] ≈ Ka1 Ka2 = 1.1 × 10−2 × 7.5 × 10−8 = 2.9 × 10−5 M and the pH is approximately independent of the salt concentration (pH ≈ 4.54). This would be the approximate pH of an NaH2 PO4 solution. Similarly, HPO4 2− is both an acid and a base. The K values involved here are Ka2 and Ka3 of H3 PO4 (H2 PO4 − ≡ H2 A and HPO4 2− ≡ HA− ). Since Ka2 >> Ka3 , the pH of a Na2 HPO4 solution can be calculated from
(7.99) [H+ ] ≈ Ka2 Ka3 = 7.5 × 10−8 × 4.8 × 10−13 = 1.9 × 10−10 and the calculated pH is 9.72. Because the pH of amphoteric salts of this type is essentially independent of concentration, the salts are useful for preparing solutions of known pH for standardizing pH meters. For example, potassium acid phthalate, KHC8 H4 O2 , gives a solution of pH 4.0 at 25◦ C. However, these salts are poor buffers against acids or bases; their pH does not fall in the buffer region but occurs at the end point of a titration curve, where the pH can change markedly if either acid or base is added, although dilution does not affect pH as much. 2. Unprotonated Salt. Unprotonated phosphate is a fairly strong Brønsted base in solution and ionizes as follows: K PO4 3− + H2 O HPO4 2− + OH− Kb = w (7.100) Ka3 The constant Ka3 is very small, and so the equilibrium lies significantly to the right. Because Ka3 Ka2 , hydrolysis of HPO4 2− is suppressed by the OH− from the first step, and the pH of PO4 3− can be calculated just as for a salt of a monoprotic weak acid. However, because Ka3 is so small, Kb is relatively large, and the amount of OH− is not negligible compared with the initial concentration of PO4 3− , and the quadratic equation must be solved, that is, PO4 3− is quite a strong base.
Example 7.20 Calculate the pH of 0.100 M Na3 PO4 . Solution
PO4 3− + H2 O HPO4 2− + OH− 0.100 − x
x
x
[HPO4 2− ][OH− ] Kw 1.0 × 10−14 = K = = = 0.020 b [PO4 3− ] Ka3 4.8 × 10−13 1.0 × 10−14 (x)(x) = = 0.020 0.100 − x 4.8 × 10−13 The concentration is only five times Kb , so the quadratic equation is used: x2 + 0.020x − 2.0 × 10−3 = 0
(0.020)2 − 4(−2.0 × 10−3 ) 2 − x = [OH ] = 0.036 M
x=
−0.020 ±
pH = 12.56
KHP is a NIST “standard buffer” (Chapter 13). The pH of its solution is fixed, but it is not buffered.
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The dissociation (hydrolysis) is 36% complete, and phosphate is quite a strong base. See the text website for a program that performs the quadratic equation calculation.
Example 7.21 Calculate the pH of 0.001, 0.002, 0.005, 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, 1.0 M solutions each of H3 PO4 , NaH2 PO4 , Na2 HPO4 , and Na3 PO4 . Neglect activity corrections. Solution
We must calculate ten separate pH values for each of the four compounds, which we will do in batches of 10 at a time using the powerful program, Solver. This problem allows us to explore the powers of Microsoft Excel SolverTM , which can solve for more than one parameter (or more than one equation) at a time. At the time of this writing, Excel 2010 is the current version of Excel and Solver has many more capabilities in this version compared to previous versions, including the ability to save a scenario with up to 32 adjustable parameters and solve for up to 200 parameters at a time (although we do not recommend the latter); frequently it actually requires more time to solve say 50 one-parameter (e.g., in terms of H+ ) equations four times than to solve 200 one-parameter equations at a time. In comparison, Goal Seek can only solve one parameter in a single equation, and does not allow for incorporating constraints in the parameter we want to solve (e.g., if we are solving for pH, it may be helpful for us to tell the computing algorithm that our solution lies within 0 and 14). Solver is not automatically installed when you first install Office 2010 in your computer. After opening Excel, go to File/Options/Add-Ins, click on Go and select the Solver Add-In and click OK to install it. The next time you open Excel, if you go to the Data tab, you should see the icon for Solver in the right-hand corner of the menu bar. Refer to the file Example 7.21.xlsx downloadable from the text website. But presently, just start with an empty spreadsheet.Very similar to what we had done in Example 7.19, in cells B1:B4 we have put down the numerical values of Ka1 , Ka2 , Ka3 , and Kw and named them KAA, KAB, KAC, and KW, respectively. First, we are going to do phosphoric acid (H3 PO4 ) and we write this down as a heading in row 5. Beginning in row 6 and starting at cell A6, we now create 10 columns and title them CP, pH, H+ , CNA, OH, Q3, Alpha1, Alpha2, Alpha3, and Equation. In A7:A16, we serially type in the concentrations our problem states, i.e., 0.001 to 1.0 as enumerated in the problem. In pH column (cell B7) let us enter a placeholder number, e.g., 1 for now. In the H+ column (cell C7) we define is relationship with pH (= 10−B7 ). Then in the CNA column cell D7 we type in zero. (The sodium concentration is zero in all of the pure H3 PO4 solutions.) In the OH− column (cell E7) we enter the relationship between OH− and H+ (given in cell C7) by entering =KW/C7. In the Q3 column (cell F7) we put in (much the same as in Example 7.19): = C7∧ 3 + KAA∗ C7∧ 2 + KAA∗ KAB∗ C7 + KAA∗ KAB∗ KAC Note that rather than defining and using H we are using the cell reference C7 because we will be solving a number of equations, each for the hydrogen ion concentration, in a separate row at one time. The hydrogen ion concentration in each row will be different (in C7:C16), rather than be a single value, and we cannot designate this by naming a single value called H as we did in Example 7.19. Alpha1: Alpha3 in cells G7:I16 are also put in exactly as in Example 7.19 but using C7 rather than H. Our charge balance equation is the same as Equation 7.85, except [Na+ ] is added on the left side, for the sodium containing solutions and when expressed in terms of the equilibrium constants, is the same
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as Equation 7.86, again with [Na+ ] added on the left. Finally, there is the equation in J7 to be put in; we square the charge balance expression (see why below). Although CNA is zero for H3 PO4 , we have retained this term in the charge balance expression so that we can use the same expression for all four of our cases, and multiply the whole by 1010 : = (1E10)∗ (C7 + D7-E7-A7∗ (G7 + 2∗ H7 + 3∗ I7))∧ 2 We now highlight cells B7:J7 and copy and paste (Ctrl-C, Ctrl-V) beginning in cell B8; thus B8:J8 are filled. We could copy the same serially all the way down to B17:J16, but it is more expedient to highlight the two rows (B7:J8) and drag the cursor at the bottom corner of J8 down to J16 and thus fill up the remaining rows. (Note that if we had performed the drag-and-fill operation with only row 7 filled, in all the nonformula cells, by default Excel would have incremented the numerical value by 1 in each step in each of the succeeding rows, e.g., the initial assigned pH cell value would have been 2 in B8, 3 in B9, and so on, until 10 in B16; we do not necessarily want that. By having the same value in cell B8 as in B7, during drag-and-fill, Excel is being told that the increment is zero between the rows in that column and it follows that instruction.) Next we are going to solve the set of 10 equations in J7:J16 by solving for the correct values of pH in cells B7:B16 simultaneously. First, in cell J18 we sum up the value of all of the equation expression values by typing there: = SUM(J7:J16) Now we are ready to invoke Solver. Go to the Data tab and then click on the Solver on the top right-hand corner of the menu bar. Solver will open as a icon drop-down menu box. In the top “Set Objective” box you should already have J18 appearing and highlighted, because you opened Solver with your cursor on this cell. If not, type in J18 here. In order to solve our equations, we want J18 (the charge balance expression) to be zero. If each equation expression is zero, the sum of them must also be zero. In the digital domain, within the numerical resolution we are able to achieve, the best solution will rarely, if ever, be exactly zero but will be able to reach a very small number (especially when you remember that we have already multiplied the expression by 1010 ). Because we solve a number of equations at one time by taking the sum of the expressions, you will understand why we use the square of the expressions rather than the expressions. From the point of view of solving an equation, one side of which is zero, obviously it does not matter if we try to solve x = 0 or x2 = 0. However, if we have two equations x = 0 and y = 0 and try to solve them individually by trying to get x + y = 0, we may never reach the correct solution because for any nonzero value of x and y where x = −y that condition will be made. However, for any real value of x and y, a solution that achieves x2 + y2 = 0 necessarily achieves the solutions x = 0 and y = 0. Next in Solver, there is a choice to set our objective “To” maximize (max), minimize (min), or to a value that we have the option to specify. We can pick either the “min” button (preferred) or “to value 0,” both will have the same impact. Because we have squared our expressions, the value in J18 can never be less than zero, so attempting to minimize it will have the same effect as trying to get it to be zero. Next we click the box “By Changing Variable Cells.” We need to either type in what we want to solve (B7:B16) or to use our cursor to select these cells. This box should now read $B$7:$B$16. The box “Make Unconstrained Variables Non-negative” should already be ticked. Our variable is pH and it should indeed be non-negative for most practical problems. The default method in the “Select a Solving Method” box is “GRG nonlinear” (others are “Simplex LP” and “Evolutionary”) and we will use this in all our work; GRG stands for Generalized Reduced Gradient). You can read about the differences between these solving approaches in Excel.
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In “Subject to the Constraints” box, we want to add that our pH values will be below 14. We click the “Add” button, a new box opens up. In the box “Cell Reference” we type in or select B7:B16. The condition box gives us a choice of =, an integer (int) among others. We are already in the default 100Ka .
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SALTS OF WEAK ACIDS AND BASES 27. If 25 mL of 0.20 M NaOH is added to 20 mL of 0.25 M boric acid, what is the pH of the resulting solution? 28. Calculate the pH of a 0.010 M solution of NaCN. 29. Calculate the pH of a 0.050 M solution of sodium benzoate. 30. Calculate the pH of a 0.25 M solution of pyridinium hydrochloride (pyridine · HCl, C6 H5 NH+ Cl). 31. Calculate the pH of the solution obtained by adding 12.0 mL of 0.25 M H2 SO4 to 6.0 mL of 1.0 M NH3 . 32. Calculate the pH of the solution obtained by adding 20 mL of 0.10 M HOAc to 20 mL of 0.10 M NaOH. 33. Calculate the pH of the solution prepared by adding 0.10 mol each of hydroxylamine and hydrochloric acid to 500 mL water. 34. Calculate the pH of a 0.0010 M solution of sodium salicylate, C6 H4 (OH)COONa. 35. Calculate the pH of a 1.0 × 10−4 M solution of NaCN.
POLYPROTIC ACIDS AND THEIR SALTS 36. What is the pH of 0.0100 M solution of phthalic acid? 37. What is the pH of a 0.0100 M solution of potassium phthalate? 38. What is the pH of a 0.0100 M solution of potassium acid phthalate (KHP)? 39. Calculate the pH of a 0.600 M solution of Na2 S. 40. Calculate the pH of a 0.500 M solution of Na3 PO4 . 41. Calculate the pH of a 0.250 M solution of NaHCO3 . 42. Calculate the pH of a 0.600 M solution of NaHS. 43. Calculate the pH of a 0.050 M solution of the trisodium salt of EDTA (ethylenediaminetetraacetic acid), Na3 HY.
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Bin Wang, Marshall University 44. What is the dominant species in solution: (a) in a diprotic acid (H2 X) system if (i) pH > pKa2 ; (ii) pKa1 < pH < pKa2 ; and (c) pH < pKa1 ? (b) in a triprotic acid system (H3 A) (i) if pH = 1/2 (pK2 + pK3 ), (ii) pH > pKa ?
BUFFERS 45. Calculate the pH of a solution that is 0.050 M in formic acid and 0.10 M in sodium formate. 46. Calculate the pH of a solution prepared by mixing 5.0 mL of 0.10 M NH3 with 10.0 mL of 0.020 M HCl. 47. An acetic acid–sodium acetate buffer of pH 5.00 is 0.100 M in NaOAc. Calculate the pH after the addition of 10 mL of 0.1 M NaOH to 100 mL of the buffer. 48. A buffer solution is prepared by adding 20 mL of 0.10 M sodium hydroxide solution to 50 mL of 0.10 M acetic acid solution. What is the pH of the buffer? 49. A buffer solution is prepared by adding 25 mL of 0.050 M sulfuric acid solution to 50 mL of 0.10 M ammonia solution. What is the pH of the buffer? 50. Aspirin (acetylsalicylic acid) is absorbed from the stomach in the free (nonionized) acid form. If a patient takes an antacid that adjusts the pH of the stomach contents to 2.95 and then takes two 5-grain aspirin tablets (total 0.65 g), how many grams of aspirin are available for immediate absorption from the stomach, assuming immediate dissolution? Also assume that aspirin does not change the pH of the stomach contents. The pKa of aspirin is 3.50, and its formula weight is 180.2.
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51. Tris(hydroxymethyl)aminomethane [(HOCH2 )3 CNH2 —Tris, or THAM] is a weak base frequently used to prepare buffers in biochemistry. Its Kb is 1.2 × 10−6 and pKb is 5.92. The corresponding pKa is 8.08, which is near the pH of the physiological buffers, and so it exhibits good buffering capacity at physiological pH. What weight of THAM must be taken with 100 mL of 0.50 M HCl to prepare 1 L of a pH 7.40 buffer? 52. Calculate the hydrogen ion concentration for Problem 23 if the solution contains also 0.100 M sodium trichloroacetate.
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Bin Wang, Marshall University 53. Use the Henderson–Hasselbalch equation to find the value of [C6 H5 COOH]/[C6 H5 COO− ] in a solution at (a) pH 3.00, and (b) pH 5.00. For C6 H5 COOH, pKa is 4.20.
BUFFERS FROM POLYPROTIC ACIDS 54. What is the pH of a solution that is 0.20 M in phthalic acid (H2 P) and 0.10 M in potassium acid phthalate (KHP)? 55. What is the pH of a solution that is 0.25 M each in potassium acid phthalate (KHP) and potassium phthalate (K2 P)? 56. The total phosphate concentration in a blood sample is determined by spectrophotometry to be 3.0 × 10−3 M. If the pH of the blood sample is 7.45, what are the concentrations of H2 PO4 − and HPO4 2 in the blood?
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Bin Wang, Marshall University 57. A student weighed out 0.6529 g of anhydrous monohydrogen sodium phosphate (Na2 HPO4 ) and 0.2477 g of dihydrogen sodium phosphate (NaH2 PO4 H2 O), then dissolved them into 100 mL of distilled water. What is the pH?
BUFFER INTENSITY 58. A buffer solution contains 0.10 M NaH2 PO4 and 0.070 M Na2 HPO4 . What is its buffer intensity in moles/liter per pH? By how much would the pH change if 10 μL(0.010 mL) of 1.0 M HCl or 1.0 M NaOH were added to 10 mL of the buffer? 59. You wish to prepare a pH 4.76 acetic acid–sodium acetate buffer with a buffer intensity of 1.0 M per pH. What concentrations of acetic acid and sodium acetate are needed?
CONSTANT-IONIC-STRENGTH BUFFERS 60. What weight of Na2 HPO4 and KH2 PO4 would be required to prepare 200 mL of a buffer solution of pH 7.40 that has an ionic strength of 0.20? (See Chapter 6 for a definition of ionic strength.) 61. What volume of 85% (wt/wt) H3 PO4 (sp. gr. 1.69) and what weight of KH2 PO4 are required to prepare 200 mL of a buffer of pH 3.00 that has an ionic strength of 0.20?
α CALCULATIONS 62. Calculate the equilibrium concentrations of the different species in a 0.0100 M solution of sulfurous acid, H2 SO3 , at pH 4.00 ([H+ ] = 1.0 × 10−4 M). 63. Derive Equations 7.75, 7.76, and 7.77 for α1 , α2 , and α3 of phosphoric acid.
DIVERSE SALT EFFECT 64. Calculate the hydrogen ion concentration for a 0.0200 M solution of HCN in 0.100 M NaCl (diverse ion effect). 65. Derive the equivalent of Equation 7.104 for the diverse salt effect on an uncharged weak base B.
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LOGARITHMIC CONCENTRATION DIAGRAMS You can use the HOAc spreadsheet exercise on the text website as a guide for Problem 66 (Spreadsheet Problems). Prepare a spreadsheet for Problem 69 using α-values—see the text website for HOAc log plots using α-values. 66. Construct the log–log diagram for a 10−3 M solution of acetic acid. 67. From the diagram in Problem 66, estimate the pH of a 10−3 M solution of acetic acid. What is the concentration of acetate ion in this solution? 68. For Problem 66, derive the expression for log[OAc− ] in acid solution and calculate the acetate concentration at pH 2.00 for a 10−3 M acetic acid solution. Compare with the value estimated from the log–log diagram. 69. Construct the log–log diagram for a 10−3 M solution of malic acid by preparing a spreadsheet using α values. 70. From the diagram in Problem 69, estimate the pH and concentrations of each species present in (a) 10−3 M malic acid and (b) 10−3 M sodium malate solution. 71. For Problem 69, derive the expressions for the HA− curves in the acid and alkaline regions. 72. Derive expressions for (a) log[H3 PO4 ] between pH = pKa1 and pKa2 , (b) log[H2 PO4 − ] between pH = pKa2 and pKa3 , (c) log[HPO4 2− ] at between pH = pKa2 and pKa1 , and (d) log [PO4 3− ] at between pH = pKa3 and pKa2 . Check with representative points on the curves. 73. Construct a log–log diagram for 0.001 M H3 PO4 using α values. Start with the spreadsheet for Figure 7.2 (given in the text website). Compare the chart with Figure 7.16.3 on the text website (Addendum to Section 7.16). Vary the H3 PO4 concentration and see how the curves change. 74. The Stig Johansson pH calculator has been shown to give pH calculations of NIST standard buffers that are within a few thousandths of a pH unit of the NIST values. The NIST buffers are given in Table 13.2 in Chapter 13. Use the calculator in Reference 15 to calculate the ActpH of the NIST phosphate buffer consisting of 0.025 M KH2 PO4 and 0.025 M Na2 HPO4 (footnote e) at 50◦ C, and compare with the NIST value of 6.833. Use pKw = 13.26, pK1 = 2.25, pK2 = 7.18, and pK3 = 12.36 for 50◦ C. Don’t forget to enter the temperature. 75. Use the Stig Johansson pH calculator to calculate the pH in Problem 41. 76. Use the Stig Johansson pH calculator to calculate the pH in Problem 43.
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor George S. Wilson, University of Kansas 77. Many geochemical processes are governed by simple chemical equilibria. One example is the formation of stalactites and stalagmites in a limestone cave, and is a good illustration of Henry’s law. This is illustrated in the diagram below:
Rain
Soil
Stalactite Cave atmosphere
Stalagmite Pooled water/ River
Rainwater percolates through soil. Due to microbial activity in the soil, the gaseous CO2 concentration in the soil interstitial space (expressed as the partial pressure of CO2 , pCO2 ,
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in atmospheres), is 3.2 × 10−2 atm, significantly higher than that in the ambient atmosphere (3.9 × 10−4 atm; CO2 concentration in the ambient atmosphere is presently increasing by 2 × 10−6 atm each year, see http://CO2now.org for the current atmospheric CO2 concentration). Water percolating through the soil reaches an equilibrium (called Henry’s law equilibrium) with the soil interstitial pCO2 as given by Henry’s law: [H2 CO3 ] = KH pCO2 where H2 CO3 is the aqueous carbonic acid concentration and KH is the Henry’s law constant for CO2 , 4.6 × 10−2 M/atm at the soil temperature of 15◦ C. The CO2 -saturated water effluent from the soil layer then percolates through fractures and cracks in a limestone layer, whereupon it is saturated with CaCO3 . This CaCO3 saturated water drips from the ceiling of the cave. Because of the diurnal temperature variation outside the cave, the cave “breathes”: the CO2 concentration in the cave atmosphere is essentially the same as in ambient air (3.9 × 10−4 atm). Show that when the water dripping from the ceiling re-equilibrates with the pCO2 concentration in the cave atmosphere, some of the calcium in the drip water will re-precipitate as CaCO3 , thus forming stalactites and stalagmites. Assume cave temperature to be 15◦ C as well. At this temperature the successive dissociation constants of H2 CO3 are: Ka1 = 3.8 × 10−7 and Ka2 = 3.7 × 10−11 , Kw is 4.6 × 10−15 and Ksp of CaCO3 is 4.7 × 10−9 . See the text website (as well as the Solutions Manual) for a detailed solution of this complex problem. Corresponding Goal Seek calculations are also given on the website.
PROFESSOR’S FAVORITE CHALLENGE: Contributed by Professor Noel Motta, University of Puerto Rico, Rio Piedras + 78. The following 5 mathematical expressions can be used to approximately calculate H concen
−14 Ca K2 K3 Ca K1 K2 K K 10 w a (c) (d) K +Ca (e) K +Ca trations in different contexts: (a) Ka Ca (b) √ K Kb Cb
b
1
2
For each of the following salts in 0.10 M aqueous solutions, match the most appropriate expression to calculate [H+ ]: (i) K2 HPO4 , (ii) NH4 CN, (iii) CH3 NH3 Cl, (iv) Na2 CO3 , (v) NaHSO3 , (vi) CH3 COONH(CH3 )3 , (vii) Na2 H2 Y (where: H4 Y = EDTA, H4 C10 H12 N2 O8 ).
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Noel Motta, University of Puerto Rico 79. A given polyproticacid Hn X has the following fractional composition (alpha-values) vs. pH: 1.2
1
0.8
𝛂 0.6
0.4
0.2
0 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
pH
What is n? If 15.0 mL of 0.10 M Hn X is titrated with 0.10 M NaOH, the titration curve should clearly show: (a) only an equivalence point at V = 15.0 mL; (b) an equivalence point at V = 15.0 mL, and another at V = 30.0 mL; (c) only an equivalence point at V = 30.0 mL; (d) an equivalence point at V = 15.0 mL, and another at V = 45.0 mL; (e) an equivalence point at V = 15.0 mL, another at V = 30.0 mL, and another at V = 45.0 mL.
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280
Recommended References ACID–BASE THEORIES, BUFFERS 1. R. P. Buck, S. Rondini, A. K. Covington, F. G. K Baucke, C. M. A. Brett, M. F. Camoes, M. J. T. Milton, T. Mussini, R. Naumann, K. W. Pratt, and others, “Measurement of pH. Definition, Standards, and Procedures,” Pure and Applied Chemistry, 74 (2002) 2169. From oceanography paper as the authoritative reference on the modern definition and measurement of pH. 2. H. Kristensen, A. Salomen, and G. Kokholm, “International pH Scales and Certification of pH,” Anal. Chem., 63 (1991) 885A. 3. R. G. Bates, “Concept and Determination of pH,” in I. M. Kolthoff and P. J. Elving, eds., Treatise on Analytical Chemistry, Part I, Vol. 1. New York: Wiley-Interscience, 1959, pp. 361–401. 4. N. W. Good, G. D. Winget, W. Winter, T. N. Connally, S. Izawa, and R. M. M. Singh, “Hydrogen Ion Buffers for Biological Research,” Biochemistry, 5 (1966) 467. 5. D. E. Gueffroy, ed., A Guide for the Preparation and Use of Buffers in Biological Systems. La Jolla, CA: Calbiochem, 1975. 6. D. D. Perrin and B. Dempsey, Buffers for pH and Metal Ion Control. New York: Chapman and Hall, 1974.
EQUILIBRIUM CALCULATIONS 7. S. Brewer, Solving Problems in Analytical Chemistry. New York: Wiley, 1980. Describes iterative approach for solving equilibrium calculations. 8. J. N. Butler, Ionic Equilibria. A Mathematical Approach. Reading, MA: Addison-Wesley, 1964. 9. W. B. Guenther, Unified Equilibrium Calculations. New York: Wiley, 1991. 10. D. D. DeFord, “The Reliability of Calculations Based on the Law of Chemical Equilibrium,” J. Chem. Ed., 31 (1954) 460. 11. E. R. Nightingale, “The Use of Exact Expressions in Calculating H+ Concentrations,” J. Chem. Ed., 34 (1957) 277. 12. R. J. Vong and R. J. Charlson, “The Equilibrium pH of a Cloud or Raindrop: A Computer-Based Solution for a Six-Component System,” J. Chem. Ed., 62 (1985) 141. 13. R. deLevie, A Spreadsheet Workbook for Quantitative Chemical Analysis. New York: McGrawHill, 1992. 14. H. Freiser, Concepts and Calculations in Analytical Chemistry. A Spreadsheet Approach. Boca Raton, FL: CRC Press, 1992.
WEB pH CALCULATORS
(Source: Stig Johansson/ www.phcalculation.se/Courtesy of Late Stig Johannson.)
15. www.phcalculation.se. A program that calculates the pH of complex mixtures of strong and weak acids and bases using concentrations (ConcpH) or activities (ActpH) is described. Equilibrium concentrations of all species are calculated. The latter also calculates activity coefficients of the species and the ionic strength. The website for this programs is reproduced on the text’s website and the program can be downloaded from there. 16. www2.iq.usp.br/docente/gutz/Curtipot.html. A free program that calculates the pH and paH of mixtures of acids and bases, and also titration curves and distribution curves, alpha plots, and more. The Curtipot program is available on the text’s website.
SEAWATER pH SCALES 17. I. Hansson, “A New Set of pH-Scales and Standard Buffers for Seawater”. Deep-Sea Research 20 (1973) 479. 18. A. G. Dickon. ”pH Scales and Proton-Transfer Reactions in Saline Media Such as Sea Water”. Geochim. Cosmochim. Acta 48 (1984) 2299. 19. R. E. Zeebe and D. Wolf-Gladrow, CO2 in Seawater: Equilibrium, Kinetics, Isotopes, Elsevier Science B.V., Amsterdam, Netherlands, 2001. 20. T. D. Clayton and R. H. Byrne, “Spectrophotometric Seawater pH Measurements: Total Hydrogen Ion Concentration Scale Calibration of m-Cresol Purple and At-Sea Results,” Deep-Sea Research, 40 (1993) 2115–2129. 21. M. P. Seidel, M. D. DeGrandpre, and A. G. Dickson, “A Sensor for in situ Indicator-Based Measurements of Seawater pH,” Mar. Chem., 109 (2008) 18–28.
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Chapter Eight ACID–BASE TITRATIONS
Chapter 8 URLs
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Calculating acid–base titration curves
Derivative titrations, Goal Seek, Solver, p. 304 , website (8.11) Buffer intensity, buffer capacity, p. 307
●
Strong acids, strong bases (Table 8.1), p. 282
●
●
Charge balance approach strong acid–strong base, Goal Seek, Solver, p. 285
●
Titration of Na2 CO3 , p. 296
●
Titration of polyprotic acids (Table 8.3), p. 300
●
Spreadsheet calculations, pp. 283, 285
●
Weak acids, weak bases (Table 8.2), p. 290
●
Charge balance approach weak acid–strong base, Goal Seek, Solver, p. 293
●
Titration of amino acids, p. 309
Spreadsheet calculations, weak acid–strong base, p. 293
●
Kjeldahl analysis of nitrogen-containing compounds, proteins, p. 310
● ●
●
Indicators (key equations: 8.12, 8.13), pp. 289
●
Spreadsheets for deriving titration curves, p. 302, 8.11 (website, Master Polyprotic Acid Titration Using Solver), Problem 52 (website Universal Acid Titrator)
In Chapter 7, we introduced the principles of acid–base equilibria. These are important for the construction and interpretation of titration curves in acid–base titrations. In this chapter, we discuss the various types of acid–base titrations, including the titration of strong acids or bases and of weak acids or bases. The shapes of titration curves obtained are illustrated. Through a description of the theory of indicators, we discuss the selection of a suitable indicator for detecting the completion of a particular titration reaction. The titrations of weak acids or bases with two or more titratable groups and of mixtures of acids or bases are presented. The important Kjeldahl analysis method is described for determining nitrogen in organic and biological samples. What are some uses of acid–base titrations? They are important in a number of industries. They provide precise measurements. While manual titrations are tedious, automated titrators are often used that can perform titrations effortlessly and accurately. Acid–base titrations are used in the food industry to determine fatty acid content, acidity of fruit drinks, total acidity of wines, acid value of edible oils, and acetic acid content of vinegar. They are used in biodiesel production to determine the acidity of waste vegetable oil, one of the primary ingredients in biodiesel production; waste vegetable oil must be neutralized before a batch may be processed to remove free fatty acids that would normally react to make soap instead of biodiesel. Ammonia is very toxic to aquatic life; aquaria test for ammonia content. In the plating industry, boric acid concentration in nickel plating solutions is determined. In the metal industries,
(Courtesy of Metrohm AG.)
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acid in etching solutions is determined. In the environmental arena, alkalinity or acidity of city and sewage water is determined. In the petroleum industry, the acid number of engine oil is determined, and acetic acid in vinyl acetate. In the pharmaceutical industry, sodium hydrogen carbonate in stomach antacids is determined. Acid–base titrations can be used to determine the percent purity of chemicals. The saponification value of carboxylic acids is used to determine the average chain length of fatty acids in fat, by measuring the mass in milligrams of KOH required to saponify the carboxylic acid in one gram of fat.
8.1 Strong Acid versus Strong Base—The Easy Titrations
Only a strong acid or base is used as the titrant.
The equivalence point is where the reaction is theoretically complete.
An acid–base titration involves a neutralization reaction in which an acid is reacted with an equivalent amount of base. By constructing a titration curve, we can easily explain how the end points of these titrations can be detected. The end point signals the completion of the reaction. A titration curve is constructed by plotting the pH of the solution as a function of the volume of titrant added. The titrant is always a strong acid or a strong base. The analyte may be either a strong base or acid or a weak base or acid. In the case of a strong acid versus a strong base, both the titrant and the analyte are completely ionized. An example is the titration of hydrochloric acid with sodium hydroxide: (8.1) H+ + Cl− + Na+ + OH− → H2 O + Na+ + Cl− The H+ and OH− combine to form H2 O, and the other ions (Na+ and Cl− ) remain unchanged, so the net result of neutralization is conversion of the HCl to a neutral solution of NaCl. The titration curve for 100 mL of 0.1 M HCl titrated with 0.1 M NaOH is shown in Figure 8.1, plotted from the spreadsheet exercise setup below. The calculations of titration curves simply involve computation of the pH from the concentration of the particular species present at the various stages of the titration, using the procedures given in Chapter 7. The volume changes during the titration must be taken into account when determining the concentration of the species. Table 8.1 summarizes the equations governing the different portions of the titration curve. We use f to denote the fraction of analyte, which has been titrated by titrant. In Figure 8.1, at the beginning of the titration (f = 0), we have 0.1 M HCl, so the initial pH is 1.0. As the titration proceeds (0 < f < 1), part of the H+ is removed from solution as H2 O. So the concentration of H+ gradually decreases. At 90% neutralization (f = 0.9) (90 mL NaOH), only 10% of the H+ remains. Neglecting the volume change, the H+ concentration at this point would be 10−2 M, and the pH would have risen by only one pH unit. (If we correct for volume change, it will be slightly higher—see the spreadsheet below.) However, as the equivalence point is approached (the point at which a stoichiometric amount of base is added), the H+ concentration is rapidly reduced until at the equivalence point (f = 1), when the neutralization is complete, a neutral solution of NaCl remains and the pH is 7.0. As we continue to add NaOH (f > 1), the OH− concentration rapidly increases from 10−7 M at the equivalence point and levels off between 10−2 and 10−1 M; we then have a solution of NaOH plus NaCl. Thus, the pH remains fairly constant on either side of the equivalence point, but it changes markedly very near the equivalence point. This large change allows the determination of the completion of the reaction by measurement of either the pH or some property that changes with pH (e.g., the color of an indicator or potential of an electrode).
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8.1 STRONG ACID VERSUS STRONG BASE— —THE EASY TITRATIONS
100 mL 0.1M HCl vs. 0.1M NaOH 14 12
10 Phenolphthalein transition range
pH
8
6 4
2 0 0
20
40
60 80 mL NaOH
100
120
140
283
A B C D E F G 1 100.00 mL of 0.1000 M HCl vs. 0.1000 M NaOH 2 mLHCl= 100.00 MHCl= 0.1000 0.1000 Kw= 1.00E–14 3 MNaOH= [H+] [OH−] pOH pH 4 mLNaOH 0.1 1.00 5 0.00 6 10.00 0.0818182 1.09 7 20.00 0.0666667 1.18 8 30.00 0.0538462 1.27 9 40.00 0.0428571 1.37 10 50.00 0.0333333 1.48 11 60.00 0.025 1.60 12 70.00 0.0176471 1.75 13 80.00 0.0111111 1.95 14 90.00 0.0052632 2.28 15 95.00 0.0025641 2.59 16 98.00 0.0010101 3.00 17 99.00 0.0005025 3.30 18 99.20 0.0004016 3.40 19 99.40 0.0003009 3.52 20 99.60 0.0002004 3.70 21 99.80 0.0001001 4.00 22 99.90 5.003E–05 4.30 23 99.95 2.501E–05 4.60 24 100.00 0.0000001 7.00 2.5E−05 4.60 9.40 25 100.05 26 100.10 5E–05 4.30 9.70 27 100.20 1E–04 4.00 10.00 28 100.40 0.0002 3.70 10.30 29 100.80 0.0004 3.40 10.60 30 101.00 0.0005 3.30 10.70 31 102.00 0.00099 3.00 11.00 32 105.00 0.00244 2.61 11.39 33 110.00 0.00476 2.32 11.68 34 120.00 0.00909 2.04 11.96 35 140.00 0.01667 1.78 12.22 36 Formulas for cells in boldface: 37 Cell B5: [H+] = (mLHCl x MHCl − mLNaOH x MNaOH)/(mLHCl + mLNaOH) = ($B$2*$D$2−A5*$B$3)/($B$2+A5) Copy through Cell B23 38 39 Cell E5 = pH = Copy through Cell E24 (−LOG10(B5)) 40 Cell B24 = [H+] = Kw1/2 = SQRT(D3) 41 Cell C25 = [OH−] = (mLNaOH x MNaOH − mLHCl x MHCl)/(mLHCl + mLNaOH) 42 = (A25*$B$3−$B$2*$D$2)/($B$2+A25) Copy to end 43 Cell D25 = pOH = −log[OH−] = (−LOG10(C25)) Copy to end 44 Cell E25 = pH = 14 − pH = Copy to end 14−D25
Fig. 8.1.
Titration curve for 100 mL of 0.1 M HCl versus 0.1 M NaOH.
Table 8.1
Equations Governing a Strong-Acid (HX) or Strong-Base (BOH) Titration Fraction f Titrated
Strong Acid Present
Strong Base Equation
Present
Equation
f =0
HX
[H+ ] = [HX]
BOH
[OH− ] = [BOH]
0 1,where f is the fraction titrated) to solve for pH; rather, the same charge balance equations are (iteratively) solved. Let us rework the problem that we have just considered: the titration of 100 mL of 0.1 M HCl with 0.1 M NaOH. The generic approach to any problem that we wish to solve by the charge balance method involves the following steps: 1. Write down all the species present in the solution at any time. In the present case, except the initial point where no NaOH has been added, at all other times we have present in solution: H2 O, H+ , OH− , Na+ and Cl− . 2. Write down any equilibrium expression that may exist. Since strong acids and strong bases are fully dissociated, the only equilibrium of concern here is the ionization of water: [H+ ][OH− ] = Kw Realize also that the after accounting for the total volume, [Cl− ] at any point represents the initial amount of HCl present and similarly, after accounting for the total volume, [Na+ ] at any point represents the total amount of NaOH added until that point. 3. Write a charge balance expression, putting the concentrations of all charged species on the left-hand side of an equation, multiplying each concentration by the magnitude of the charge on that ion (in the present case these are all 1) and setting the difference of the sum of positive charges and the sum of negative charges to be zero. In the present case, this amounts to: ([Na+ ] + [H+ ]) − ([OH− ] + [Cl− ]) = 0
(8.2)
4. Convert all parameters into either known values that may involve concentrations, volumes (or fraction titrated), equilibrium constants, etc., and/or a single variable, most commonly H+ . Let us call the initial volume of the acid taken VA and the concentration of the acid taken CA eq/L (For NaOH and HCl, there is no difference between molar units M and equivalents/liter). If VB volume of base of concentration CB eq/L has been added, by definition the fraction titrated will be: (8.3) VB CB = f VA CA Alternatively, VB can be expressed as VB = f VA CA /CB
(8.4)
In the present case, since CA = CB = 0.1 eq/L, VB can be written as f VA . The total volume at any point (VA + VB ) can therefore be written as: VA + VB = VA + f VA = VA (1 + f )
(8.5)
Now we can make the dilution correction for the concentration of chloride, which starts out with CA but is diluted during the titration because of the added volume of the titrant. Thus, at any point: [Cl− ] = (VA CA )/(VA + VB ) = (VA CA )/VA (1 + f ) = CA /(1 + f )
(8.6)
We also must realize because of the reaction stoichiometry, [Na+ ], which directly reflects the amount of NaOH added, must be [Na+ ] = f [Cl− ] = f CA /(1 + f )
(8.7)
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Recognizing that [OH− ] = Kw /[H+ ], we can now write our desired equation by putting Equations 8.6 and 8.7 into Equation 8.2: f CA /(1 + f ) + [H+ ] − Kw /[H+ ] − CA /(1 + f ) = 0
(8.8)
We can put the first and the last term together: [H+ ] − Kw /[H+ ] + (f CA − CA )/(1 + f ) This simplifies to a more pleasing form: 1−f Kw + C =0 [H ] − + + 1+f A H
(8.9)
Given that Kw and CA are known, [H+ ] or pH can be computed for each specified value of f using Goal Seek. Using Solver, multiple solutions for each point in a multipoint solution can be computed. Construct an Excel sheet (or start with the spreadsheet 8.2.xlsx available on the book’s website—Section 8.2 Charge balance and Solver for HCl vs. NaOH). In cell A1 type in CA and in B1 type in 0.1 and name this CA. Similarly type in KW in cell A2 and 1E-14 in cell B2 and define that as KW. Beginning in row 5, type in the column headings f, pH, H+ , Kw / H+ , and Equation. Enter 0 in cell A6 under f and 0.05 in cell A7. Then highlighting these two cells, drag down from the right bottom corner to fill through to cell A30. This should fill in these cells in column A with f values from 0 to 1.2 with a step of 0.05. Enter any value, say zero, in cell B6. Type in C6 the value of [H+ ] as = 10∧ −B6 (this should now read 1 if we entered zero in cell B6). In cell D6, type in = KW/B6. Now we are ready to write Equation 8.9. We enter: = C6 − (D6 + CA ∗ (1 − A6)/(1 + A6)). Remember previous experience with numerical solutions in Excel? We square the whole thing to solve multiple problems in multiple rows at the same time (see Example 7.21). Then we multiply that with 1E10 to prevent premature solutions. Now the ultimate expression in cell E6 reads: = 10000000000∗ (C6 − (D6 + CA∗ (1 − A6)/(1 + A6)))∧ 2. Now highlight cell E6 and invoke Data/What-if Analysis/Goal Seek. Execute Goal Seek by seeking to set cell E6 to value 0 by changing cell B6. If you typed it correctly, Goal Seek would find the best value for cell B6 to be 1.00 and cell E6 will have a low residual, of the order of 0.0005. We can solve each row from cell B7 to cell B30 in this fashion each time solving for the pH value for the specific value of f. We can basically also try to solve all the rows at a time by using Solver. To do this, let us select cells B6:E6 and copy (Ctrl+C). Then paste the same into cells B7:E7 (Ctrl+V, beginning in B7). We then select both rows (avoid column A) B6:E7 and drag down from the right-hand corner to E30 and fill up the rest of the rows. In cell E32 the entered formula = SUM(E6:E30) sums up the values of cells E6:E30. Now we invoke Solver (Data/Analysis/Solver). Set the objective cell as E32. The variable cells to be changed are B6:B30. We need to add constraints to permissible values of B6:B30, these will be within 0–14. Using the Add button we bring up the “Add constraint” box, type in B6:B30 in the cell reference box, leave the operator box at the default “=” and enter 0 in the constraint box. This time click on OK and you will return to the main Solver menu. We now click on Solver Options and enter 1E-10 on the precision box, 500 seconds on the maximum time box, and 100,000 on the maximum iterations. Click OK, return to the main window, and click on “Solve.” Solver will open the Results pane and provide some preliminary (and incorrect) solutions. You will have to execute Solver several times until the value in cell E32 no longer decreases. To facilitate this, check the box on the Results pane that says “Return to Solver Parameters Dialog.” Click OK. You may need to shuttle between the
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Solver pane and the Results pane 4–5 times before you will see the value in cell E32 no longer decreases (about 0.00578). Solver has done the best it can in this summation mode and actually all of the values except for f = 1.00 are correctly solved. The f = 1.00 value is unique because at this point [Na+ ] and [Cl− ] are exactly equal and the charge balance is determined by [H+ ] and [OH− ]. Obviously, for Equation 8.2 to be satisfied, at this point [H+ ] = [OH− ] and pH = 7. However, even if Solver tries a pH value of 6, ([H+ ] = 10−6 ) the difference between [H+ ] and [OH− ] will still be very small, comparable to the residuals left when a correct solution is obtained with the other rows. As a result, Solver returns a solution of pH = 6.38 for f = 1. This row can be examined by itself. For example, if we multiply 1E20 instead of 1E10 so that the residual remains high enough to prevent a premature solution and then invoke Goal Seek seeking to set cell E26 to zero, by changing cell B26, it will immediately reach the correct solution of pH being 7.00. In general it is advisable to solve the exact neutralization point pH by itself at the end. You can then construct the titration curve by highlighting cells A6:B30, click on Insert/Charts/Scatter, and select the plot with “Smooth Lines and Markers” (top right). The titration curve is plotted in terms of pH as a function of the fraction titrated. You can move the chart to a separate sheet if you like by right-clicking on the frame of the chart, then Move Chart/New Sheet/OK.
8.3 Detection of the End Point: Indicators The goal is for the end point to coincide with the equivalence point.
Carrying out the titration process is of little value unless we can tell exactly when the acid has completely neutralized the base, i.e., when the equivalence point has been reached. Therefore, we wish to determine accurately when the equivalence point is reached. The point at which the reaction is observed to be complete is called the end point. A measurement is chosen such that the end point coincides with or is very close to the equivalence point. The difference between the equivalence point and the end point is referred to as the titration error; as with any measurement, we want to minimize error. The most obvious way of determining the end point is to measure the pH at different points of the titration and make a plot of this versus milliliters of titrant. This is done with a pH meter, which is discussed in Chapter 13. It is often more convenient to add an indicator to the solution and visually detect a color change. An indicator for an acid–base titration is a weak acid or weak base that is highly colored. The color of the ionized form is markedly different from that of the nonionized form. One form may be colorless, but at least one form must be colored. These substances are usually composed of highly conjugated organic constituents that give rise to the color (see Chapter 16). Assume the indicator is a weak acid, designated HIn, and assume that the nonionized form is red while the ionized form is blue: Hln H+ + In− (red)
(blue)
(8.10)
We can write a Henderson–Hasselbalch equation for this, just as for other weak acids: pH = pKIn + log Your eyes can generally discern only one color if it is 10 times as intense as the other.
[In− ] [HIn]
(8.11)
The indicator changes color over a pH range. The transition range depends on the ability of the observer to detect small color changes. With indicators in which both forms are colored, generally only one color is perceived if the ratio of the concentration
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8.3 DETECTION OF THE END POINT: INDICATORS
289
of the two forms is 10:1, if the molar absorptivities (Chapter 16), i.e., color intensities, of each are not too different; only the color of the more concentrated form is visually sensed. From this information, we can calculate the pH transition range required to go from one color to the other. When only the color of the nonionized form is seen, 1 . Therefore, [In− ]/[HIn] = 10 pH = pKa + log
= pKa − 1
1 10
(8.12)
When only the color of the ionized form is observed, [In− ]/[HIn] = pH = pKa + log
10 1 ,
and
= pKa + 1
10 1
(8.13)
So the pH in going from one color to the other has changed from pKa − 1 to pKa + 1. This is a pH change of 2, and most indicators require a transition range of about two pH units. During this transition, the observed color is a mixture of the two colors. Midway in the transition, the concentrations of the two forms are equal, and pH = pKa . Obviously, then, the pKa of the indicator should be close to the pH of the equivalence point. Calculations similar to these can be made for weak base indicators, and they reveal the same transition range; the pOH midway in the transition is equal to pKb , and the pH equals 14 − pKb . Hence, a weak base indicator should be selected such that pH = 14 − pKb . Many find it convenient to treat weak base indicators in terms of their conjugate acids and use pKa values exclusively. Figure 8.4 illustrates the colors and transition ranges of some commonly used indicators. The range may be somewhat less in some cases, depending on the colors; some colors are easier to see than others. The transition is easier to see if one form of the indicator is colorless. For this reason, phenolphthalein is usually used as an indicator for strong acid–base titrations when applicable (see Figure 8.1, titration of 0.1 M HCl). In dilute solutions, however, phenolphthalein falls outside the steep portion of the titration curve (Figure 8.2), and an indicator such as bromothymol blue must be used. A similar situation applies to the titration of NaOH with HCl (Figure 8.3). A more complete list of indicators is given on the inside back cover. See also http://en.wikipedia.org/wiki/PH_indicator, which gives pH ranges for color changes of indicators, and shows the colors for acid and base forms. Indicator
Crystal violet Cresol red
2
yellow
4
12
blue
red
yellow
yellow
blue
red
Methyl red Methyl purple
yellow
purple
green yellow
Bromothymol blue
blue
red
Litmus
blue yellow
Cresol red
Alizarin yellow R
10
yellow yellow
Bromcresol green
Thymolphthalein
8
blue red
Methyl orange
Phenolphthalein
6
violet
yellow
Bromphenol blue
Thymol blue
See the inside back cover for a comprehensive list of indicators.
pH 0
Methyl violet
Choose an indicator with a pKa near the equivalence point pH.
red
yellow
red
yellow
blue
colorless
red violet
colorless yellow
Fig. 8.4.
blue red
pH transition ranges and colors of some common indicators.
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290 Two drops (0.1 mL) of 0.004 M indicator (0.1% solution with fw = 250) is equal to 0.01 mL of 0.04 M titrant.
Since an indicator is a weak acid or base, the amount added should be kept minimal so that it does not contribute appreciably to the pH and so that only a small amount of titrant will be required to cause the color change. That is, the color change will be sharper when the concentration is lower because less acid or base is required to convert it from one form to the other. Of course, sufficient indicator must be added to impart an easily discernible color to the solution. Generally, a few tenths percent solution (wt/vol) of the indicator is prepared and two or three drops are added to the solution to be titrated.
8.4 Standard Acid and Base Solutions See Chapter 2 for special procedures required to prepare and standardize acid and base solutions.
Hydrochloric acid is usually used as the strong acid titrant for the titration of bases, while sodium hydroxide is the usual titrant for acids. Most chlorides are soluble, and few side reactions are possible with HCl. It is convenient to handle. Neither of these is a primary standard and so solutions of approximate concentrations are prepared, and then they are standardized by titrating a primary base or acid. Special precautions are required in preparing the solutions, particularly the sodium hydroxide solution. The preparation and standardization of hydrochloric acid and sodium hydroxide titrants is presented in Chapter 2.
8.5 Weak Acid versus Strong Base—A Bit Less Straightforward The titration curve for 100 mL of 0.1 M acetic acid titrated with 0.1 M sodium hydroxide is shown in Figure 8.5. The neutralization reaction is HOAc + Na+ + OH− → H2 O + Na+ + OAc− The curve is flattest and the buffering capacity the greatest at the midpoint of a weak-acid—strong base or weak base—strong acid titration, when the weak acid or base is half neutralized.
(8.14)
The acetic acid, which is only a few percent ionized, depending on the concentration, is neutralized to water and an equivalent amount of the salt, sodium acetate. Before the titration is started, we have 0.1 M HOAc, and the pH is calculated as described for weak acids in Chapter 7. Table 8.2 summarizes the equations governing the different portions of the titration curve, as developed in Chapter 7. As soon as the titration is started, some of the HOAc is converted to NaOAc, and a buffer system is set up. As the titration proceeds, the pH slowly increases as the ratio [OAc− ]/[HOAc] changes. At the midpoint of the titration, [OAc− ] = [HOAc], and the pH is equal to pKa . At the equivalence point, we have a solution of NaOAc. Since this is a Brønsted base
Table 8.2
Equations Governing a Weak-Acid (HA) or Weak-Base (B) Titration Fraction f Titrated
Weak Acid Present
Equation
f =0
HA
[H+ ] =
0= 0 and B6:B126 0 to < 100%)
H2 A/HA−
f = 1 (100%) (1st eq. pt.)
HA−
CHA− (Eq. 7.45) CH2 A + (or CHA− + [H ] and CH2 A− [H+ ] if a strong acid)
[H+ ] ≈ Ka1 Ka2 (Eq. 7.84) pH = pKa1 + log
(or Eq. 7.83 if H2 A ∼ strong acid) 1 < f < 2 (> 100 to < 200%)
HA− /A2−
pH = pKa2 + log
f = 2 (200%) (2nd eq. pt.)
A
f > 2 (> 200%)
OH− /A2−
and
2−
CA2− (Eq. 7.45, Ex. 7.16, 7.24) CHA−
Kw · CA2− (Eq. 7.32) KA2 (or Eq. 7.29, Ex. 7.20, quadratic if A2− ∼ strong base) −
[OH ] ≈
[OH− ] = [excess titrant]
HA− + H2 O H2 A + OH−
The equilibrium concentration of H2 A is decreased from the calculated analytical concentration by an amount equal to [H+ ] and increased by an amount equal to [OH− ]: [H2 A] = CH2 A − [H+ ] + [OH− ] That of HA− is decreased by an amount equal to [OH− ] and increased by an amount equal to [H+ ]: [HA− ] = CHA− + [H+ ] − [OH− ] Since the solution is acid, we can neglect [OH− ]: [H2 A] = CH2 A − [H+ ] [HA− ] = CHA− + [H+ ] (For simplification, we could have written the same equations just from the H2 A dissociation above.) So, for a fairly strong acid, we must substitute in the Ka1 = [H+ ] [HA− ]/[H2 A] expression: [H2 A] = CH2 A − [H+ ], and [HA− ] = CHA− + [H+ ], and solve a quadratic equation for the buffer region, where CH2 A and CHA− are the calculated concentrations resulting from the acid–base reaction at a given point in the titration: [H+ ]2 + (Ka1 + CHA− )[H+ ] − Ka1 CH2 A = 0 Also, if Kb1 for A2− is fairly large, then just beyond the second equivalence point, OH− from hydrolysis of A2− cannot be ignored compared to the concentration of the OH− from excess titrant. So we continue to use the Kb1 expression, Kb1 = Kw /Ka2 = [HA− ] [OH− ]/[A2− ] where [A2 ] = CA2− − [OH− ], and [HA− ] = [OH− ], and solve the quadratic equation. After addition of, say, 0.5 mL, of NaOH, there is enough excess OH− to suppress the hydrolysis of A2− , and we can calculate the pH just from the excess OH− concentration. Finally, for HA− at the first equivalence point, we may have to use the more exact Equation 7.96 instead of 7.97 for calculation of [H+ ] to
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get the correct pH, since Ka1 may not be negligible compared to [HA− ] (Ka1 Kw in the numerator will probably still be negligible). So, Ka1 Ka2 [HA− ] [H+ ] = Ka1 + [HA− ] And very near the equivalence points, even more complicated expressions may be required. You can appreciate that these types of considerations, each different for each region of a titration, can quickly get cumbersome. In contrast, if the systematic approach using the charge balance equation is used for calculations, then no simplifying assumptions or approximations are necessary and a single charge balance equation applies at any point in the titration. We will not construct a diprotic acid titration curve here, but you probably have realized that constructing a titration curve for a polyprotic acid by the charge balance method is quite starightforward. The website for the book in Chapter 8 has a master spreadsheet that allows you to construct a titration curve for up to a tetraprotic acid, titled Master polyprotic acid titration spreadsheet.xlsx. Note that the approach in this spreadsheet calculates pH as a constant increment of titrant is added (or f is varied with constant increments). Try with some of the polyprotic acids for which dissociation constants are given in Appendix C, Table C.1. While the Master titration spreadsheet approach mimics what one does experimentally, i.e., add a little titrant and measure the pH, computationally an alternative approach, where we ask how much titrant we need to add to get to a certain pH is much easier. This approach is discussed in Section 8.11 and is also embodied in the professor’s favorite spreadsheet Exercise 8.52. For the titration of weak polyprotic bases, see Section 8.11b Supplement on the text’s website. A master spreadsheet for easy calculations is given. See also the website video, Excel H3PO4 titration, for construction of a phosphoric acid titration curve. And the applet constructed by Professor Constantinos Efstathiou of the University of Athens illustrates titrations of 26 different acids and 22 different bases: http://www.chem.uoa.gr/applets/AppletTitration/Appl_Titration2.html. Video: Excel H3 PO4 titration curve
8.10 Mixtures of Acids or Bases One acid should be at least 104 times weaker than the other to titrate separately.
Mixtures of acids (or bases) can be titrated stepwise if there is an appreciable difference in their strengths. In general, one Ka value must be at least 104 times greater than the other to clearly see the individual end points separately. If one of the acids is a strong acid, a separate end point will be observed for the weak acid only if Ka is about 10−5 or smaller. See, for example, Figure 8.12, where a break is seen for the HCl. The stronger acid will titrate first and will give a pH break at its equivalence point. This will be followed by titration of the weaker acid and a pH break at its equivalence point. The titration curve for a mixture of hydrochloric acid and acetic acid versus sodium hydroxide is shown in Figure 8.12. At the equivalence point for HCl, a solution of HOAc and NaCl remains, and so the equivalence point is acidic. Beyond the equivalence point, the OAc− /HOAc buffer region is established, and this markedly suppresses the pH break for HCl, leading to only a small pH change in the equivalence point of HCl, compared to when HCl is titrated in the absence of HOAc. The remainder of the titration curve is identical to Figure 8.5 for the titration of HOAc. If two strong acids are titrated together, there will be no differentiation between them, and only one equivalence point break will occur, corresponding to the titration of both acids. The same is true for two weak acids if their Ka values are not
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0.04
Excess OH–
12 pH, First Derivative (pH unit/mL)
10
pH
8 HOAc e.p.
HCI e.p.
6
4
2
20
Fig. 8.12.
40
60
80 100 120 mL 0.1 M NaOH
140
160
Titration Plot (pH)
8 0 Second Derivative
4
–0.02
180
Titration curve for 50 mL of mixture of 0.1 M HCl and 0.2 M HOAc with 0.1 M NaOH.
0.02
Second Derivative, (pH unit)2/mL2
12
First Derivative –0.04
0 0
40 80 Volume Base Added, mL
Fig. 8.13.
120
The titration plot and the two derivative plots for the H3 PO4 -NaOH titration generated in the spreadsheet 8.11.xlsx (text’s website Sec. 8.11 Derivative titrations—Easy method).
too different. For example, a mixture of acetic acid, Ka = 1.75 × 10−5 , and propionic acid, Ka = 1.3 × 10−5 , would titrate together to give a single equivalence point. Put the dissociation constants for citric acid, a triprotic acid, in the Master titration spreadsheet on the website and construct a titration plot to observe how many equivalence points are visible. With H2 SO4 , the first proton is completely dissociated and the second proton has a Ka of about 10−2 . Therefore, the second proton is ionized sufficiently to titrate as a strong acid, and only one equivalence point break is found. The same is true for a mixture of a strong acid and a weak acid with a Ka in the neighborhood of 10−2 . The first ionization constant of sulfurous acid, H2 SO3 , is 1.3 × 10−2 , and the second ionization constant is 5 × 10−6 . Therefore, in a mixture with HCl, the first proton of H2 SO3 would titrate along with the HCl, and the pH at
the equivalence point would be determined by the HSO3 − remaining; that is, [H+ ] = Ka1 Ka2 , since HSO3 − is both an acid and a base. This would be followed by titration of the second proton to give a second equivalence point. The volume of titrant required to reach the first end point would always be greater than that in going from the first to the second since the first includes the titration of both acids. The amount of H2 SO3 could be determined from the amount of base required for the titration of the second proton. The amount of HCl could be found by subtracting from the first end point the volume of base required to titrate the second proton of H2 SO3 , which is equal to the volume required to titrate the first proton. In reality, this titration would find little practical use because H2 SO3 is volatilized and lost from solution as SO2 gas in a strongly acid solution. Phosphoric acid in mixture with a strong acid acts in a manner similar to the above example. The first proton titrates with the strong acid, followed by titration of the second proton to give a second equivalence point; the third proton is too weakly ionized to be titrated. The titration of a polyprotic acid is essentially the same as that of the titration of a mixture of monoprotic acids of the corresponding Ka values, where the separate monoprotic acids all have the same concentration. Figure 8.13 shows the titration of phosphoric acid. The derivatives of titration curves are discussed in Section 8.11.
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Example 8.3 A mixture of HCl and H3 PO4 is titrated with 0.1000 M NaOH. The first end point (methyl red) occurs at 35.00 mL, and the second end point (bromthymol blue) occurs at a total of 50.00 mL (15.00 mL after the first end point). Calculate the millimoles HCl and H3 PO4 present in the solution. Solution
The second end point corresponds to that in the titration of one proton of H3 PO4 (H2 PO4 − → HPO4 − ). Therefore, the millimoles H3 PO4 is the same as the millimoles NaOH used in the 15.00 mL for titrating that proton: mmolH3 PO4 = MNaOH × mLNaOH = 0.1000 mmol/mL × 15.00 mL = 1.500 mmol The HCl and the first proton of H3 PO4 titrate together. A 15.00-mL portion of base was used to titrate the first proton of H3 PO4 (same as for the second proton), leaving 20.00 mL used to titrate the HCl. Therefore, mmolHCl = 0.1000 mmol/mL × (35.00 − 15.00) mL = 2.000 mmol
Similarly, mixtures of bases can be titrated if their strengths are sufficiently different. Again, the difference in Kb values must be at least a factor of 104 . Also, if one of them is a strong base, the weak base must have a Kb no greater than about 10−5 to obtain separate end points. For example, sodium hydroxide does not give a separate end point from that for the titration of CO3 2− to HCO3 − when titrated in the presence of Na2 CO3 (Kb1 = 2.1 × 10−4 ).
8.11 Equivalence Points from Derivatives of a Titration Curve In instrumental monitoring of a titration as with a pH electrode (or some other potentiometric electrodes, see Chapter 13), the end point is frequently determined from a derivative of the titration curve. You will note that in all of the foregoing titration plots, the rate at which pH changes with added titrant is the greatest at the equivalence point. For an acid sample, titrated with a base, this rate of change is the first derivative of the pH vs. VB plot, or dpH/dVB . The pH increases during such a titration and dpH/dVB is always positive—it reaches its maximum value at the end point. Conversely, when a base is titrated with an acid, the pH decreases monotonically during the titration as acid (volume VA ) is added, and dpH/dVA is always negative and reaches the lowest (maximum negative) value at the end point. If you have had calculus already, then you know the condition of a maximum or minimum for any function is that the derivative of that function will be zero when the maximum or minimum is reached. Although the following discussion is solely centered on titrating an acid with a base, essentially the same considerations apply when a base is titrated with an acid. To reiterate, if we have pH data for the titration of an acid as a function of the volume of base added (pH vs. VB ), a plot of the first derivative dpH/dVB against VB would show a maximum at the equivalence point and a plot of the second derivative, denoted d2 pH/dVB 2 (meaning the rate of change of dpH/dVB with VB ), versus VB
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will go through zero at the end point (or equivalence point for a calculated curve). In general, derivatives reveal subtle features that are not always observable on the original plots, although the generation of such derivatives (differentiation) from real data does make for a more noisy result. By strict definitions of calculus, differentiation pertains to increments (of volume base added for example) that are infinitesimally small. However, as long as the volume increments or the resulting pH increments are not too large, the use of pH/VB (this means the difference in two adjacent pH values, pH, brought about by the addition of a finite increment of base, VB ) instead of the theoretically desired dpH/dVB , produces equally good results. We could use any of the titration data that we have generated thus far, except that we have used a constant increment of base in this titration and near the equivalence point the pH change is too large for a good derivative plot. We have two choices: we can either generate data that will provide for higher resolution (lower volume increments) in the neighborhood of the equivalence point or we can generate titration data with high resolution throughout the titration (say, a thousand points per titration). The latter approach does not require that we select a region near the equivalence point (which requires that we already know where the equivalence point is!) to produce high-resolution data. Producing a high-resolution titration plot with 1000 points is a rather tall order if this has to be done by Goal Seek or Solver in Excel using the approach we have thus far used. On the other hand, if our primary desire is to generate a titration plot, it does not have to be done by constant volume increments of the titrant. If we know the concentration of an acid, it is rather trivial by charge balance principles to calculate how much base we need to add to get to a certain pH. Consider the titration of VA mL of a triprotic acid of molar concentration CA being titrated by CB molar NaOH, VB mL having been added at any point. Charge balance requires that [Na+ ] = [OH− ] + CA (α1 + 2α2 + 3α3 ) − [H+ ] If VB mL of NaOH has been added with the total volume being VA + VB , the above equation becomes VB CB /(VA + VB ) = [OH− ] − [H+ ] + VA CA (α1 + 2α2 + 3α3 )/(VA + VB ) Multiply both sides by (VA + VB ): VB CB = (VA + VB ) ([OH− ] − [H+ ]) + VA CA (α1 + 2α2 + 3α3 ) Separating the VA and VB multipliers on the first term on the right and transposing: VB (CB − ([OH− ] − [H+ ])) = VA ([OH− ] − [H+ ]) + VA CA (α1 + 2α2 + 3α3 ) or VB = VA {([OH− ] − [H+ ]) + CA (α1 + 2α2 + 3α3 )}/(CB + [H+ ] − [OH− ]) (8.26) If we know the dissociation constants and we specify pH (and hence [H+ ]), we can easily calculate the alpha values and all terms on the right side of Equation 8.26 are readily calculated. Let us illustrate this with 0.10 molar solution of H3 PO4 (Ka1 = 1.1 × 10−2 , Ka2 = 7.5 × 10−8 , Ka3 = 4.8 × 10−13 ) with 0.17 M NaOH. Refer to the spreadsheet “Sec. 8.11 derivative titrations easy method.xlsx” in the text website for the following discussion. In cells A1:B7, we put in the values of CA , CB , VA , Ka1 , Ka2 , Ka3 (these named as KAA, KAB, and KAC so Excel will accept these names) and Kw and define these names. Titles are put in cells D8:K8, for VB , pH, [H+ ], Q, α1 , α2 , α3 and [OH− ]. In row 9, we enter any trial value for pH in cell E9, express cell F9 as 10∧ −E9 and Q, α1 , α2 , α3 as their customary expressions, respectively, in cells G9:J9 in terms of the Ka ’s and [H+ ] (F9). Finally in cell K9, we express [OH− ] as Kw /[H+ ]. Only this first row of
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calculations, row 9, with no base added, is solved differently from the subsequent rows to know what the starting pH is. For this row and this row only, we enter the charge balance expression in cell A9 as [OH− ] − [H+ ] + CA (α1 + 2α2 + 3α3 ), multiply by 1E10 and use Goal Seek to set cell A9 to zero by changing cell E9 (pH). Once this is solved, we enter the next 0.01 pH unit higher pH in (rounded to the nearest 0.01 pH unit) in cell E10 and another 0.01 higher pH unit in cell E11. We then select cells E10 and E11 and drag from the bottom right-hand corner of cell E11 down to populate the column up to the desired pH value (in the illustrated case we stopped at a pH of 12.55). We now select cells F9:K9 and double-click at the bottom right corner of cell K9 to fill the rest of the data space in columns F through K. In cell D9, the first entry for VB , enter zero. But beginning in cell D10, enter the expression in Equation 8.26; this will now directly solve for VB . Now double-click on the bottom right corner of cell D10 to fill up the D column. Between columns D and E (VB and pH), you generated the titration plot! Now we are ready to compute the first derivative. Consider that in cell M10 we have calculated VB as the difference between the VB values (D10-D9), and in cell N10 we have similarly calculated the pH value as (E10-E9). Thence, in cell O10 we have similarly calculated the pH/VB value as N10/M10. We want to plot these pH/VB values against VB —but we have two VB values—one in cell D9, one in cell D10. Since our pH/VB data are derived from both VB values, we actually use an average of the two. We have so designated column L as “Average VB ” and calculated it as the average of cells D9 and D10 (VB,av ). We are now ready to fill up column L, so we double-click on bottom right corner of cell L10 to fill up the column. Plot now the standard titration plot (pH vs. VB ) and right-click on the plot, choose Select Data and Add, and add new X and Y data corresponding to VB and pH/VB . If you wish, you can put the second plot on the right hand Y-axis by right-clicking on the trace to select it, choosing “Format data series” and choosing to plot it using a secondary axis. The chart labeled “Tit plot & 1st Deriv” shows such an overlaid plot. You can readily see that how sharp inflection points indicate the two equivalence points in this titration. In the following Professor’s Favorite Example with the real example for red wine titration given in Figure 8.14, the exact location of the second equivalence point will be far more difficult to discern without a derivative plot. Various titrations are extensively used in industrial laboratories, but few titrations are done manually. Extensive use is made of robotic sample handling and automatic titrators (often referred to as autotitrators) in which the titrant is dispensed by a motorized buret (see Chapter 14). Potentiometric sensors (see Chapter 13), of which a pH electrode is a hallmark example, do not exhibit particularly fast response times. If the titrant was delivered continuously at a fast rate, the system may go by the equivalence point too fast and would not get an accurate reading. On the other hand, if you delivered the titrant continuously at a very slow rate, it will take too long to complete a titration. Such instruments continuously calculate the first derivative and as its value begins to increase, the system lowers the rate of titrant addition such that the titrant is added slower as the equivalence point is approached. This way both accuracy and analytical throughput are maintained.
BUFFER INTENSITY Note that the first derivative dpH/dVB has a direct relationship to the buffer intensity β, first discussed in Chapter 7, Section 7.8. It was defined in Equation 7.94 as dCB /dpH where dCB is the infinitesimal increase in base concentration added to affect the change in pH of dpH. Note that dVB is readily converted to the corresponding amount of base in moles by multiplying with CB . This is then readily converted to the
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corresponding increment in concentration by dividing with the total volume: dCB = (CB dVB )/(VA + VB ) When using real data with finite increments, we will have to use VB,Av instead of VB and approximate (it will be exactly equal only if it was a true differential) β as: β ≈ CB /pH = {(CB VB )/(VA + VB,av )}pH
(8.27)
β is calculated according to Equation 8.27 in column P. For a monoprotic acid base system, the strict expression for β is (see the book’s website Sec. 8.11 Supplement–Buffer intensity derivation for a detailed derivation and discussion, some familiarity with elementary differential calculus will be needed): β = 2.303 (Cα0 α1 + [H+ ] + [OH− ])
(8.28)
where C is the total concentration of the buffering species in the solution. Under conditions where [H+ ] and [OH− ] are negligible relative to Cα0 α1 , Equation 8.28 is simplified and noting that due to dilution, C = CA VA /(VA + VB ), we have β = 2.303 CA α0 α1 /(VA + VB )
(8.29)
Note that this is the limiting maximum value, attainable only when the increments are infinitesimally small. Refer again to the spreadsheet “Sec. 8.11 derivative titn master spreadsheet–Goal Seek.xlsx” in the text website. In the chart “Tit plot & Beta vs VB” (if you don’t see the tab for this, click on the left pointing arrow to the left of the tabs to scroll to it) we have the titration plot (pH vs. VB ) and β computed both according to Equations 8.27 (column P) and 8.29 (column Q). In using Equation 8.29, we recognize rather than a monoprotic acid, here the second proton of a diprotic acid is being neutralized, so in calculating the data in column Q, we use α1 α2 rather than α0 α1 . Also, we limit the computation to the range where only these species are important, where both α1 and α2 is at least 0.1 (10% of the total concentration). Observe that there is an excellent agreement between the results obtained with Equations 8.27 and 8.29. If you compare the actual numerical values in columns P and Q, you will find that maximum β is reached at a pH of 7.06 for both computations, and the finite difference computed value in column P is 99.9% of the maximum calculated in column Q. The exact pH at which β reaches a maximum value is slightly before pKa2 because of two reasons: (a) this is a titration system where the concentration does not remain constant but decreases continuously as the acid is titrated and the pH increases; (b) although the other dissociations are not close, they can still have a minute effect, and in this case, pKa1 is slightly closer to pKa2 than is pKa3 . BUFFER INTENSITY COMPUTATIONS DEPEND ON THE pH RESOLUTION: BUFFER CAPACITY If we examine the “Beta vs pH” plot, in the same spreadsheet it is clear that even at its maximum where β is relatively unchanging, it is constant only over a rather small range. Although at the maximum we observe a value of ∼ 31 meq/L per pH unit, this depends on the pH resolution of the present calculations being 0.01 pH unit. For practical purposes one is typically interested in how much base or acid is needed to cause a larger change in pH, e.g., 1 pH unit. Because β does not remain constant across a significant range, specifying β values thus calculated may not directly address the question as to how much the pH will change if a significant amount of acid or base is added. For example, if we had a buffer solution corresponding to that at the maximum β in our plot and we added ∼ 31 meq/L base (without volume change), the pH will increase by more than 1 pH unit because the plot also shows that across 1 pH unit range β will not remain at this maximum value.
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We can examine how β changes as a function of the pH resolution we use. In columns T:AF, AH:AT, and AV:BH we essentially copy and paste our previous calculations and then repeat them with pH resolutions of 0.1, 0.5, and 1 unit. The results are plotted in the “Buffer Intensity and Capacity” chart. You can readily see that while there is no significant change in β in going from a pH resolution of 0.01 to 0.1, it rapidly decreases when the pH resolution is decreased to 0.5 and 1. The apparent shift in the location of the maximum is an artifact of Excel’s attempt to draw a smooth curve through too few data points (the points are shown for the two lowest resolution traces). The decrease in β over a wider range without this artifact shift in pH location can be observed by using the Moving Average function within Data/Data Analysis in Excel. (Like Solver, the “Data Analysis” Toolpak is not automatically installed. You must install it from Excel options.) The chart also shows a 200-point running average of pH and β from the 0.01 pH resolution data set; note that the position of the maximum does not actually move but β decreases. When referring to changes over a significant pH interval, the term buffer capacity, rather than buffer intensity, should be used. In going from pH 7.1 to pH 8.1, the buffer capacity of our solution for example will be ∼ 21 meq/L (see cell BH17). THE SECOND DERIVATIVE The second derivative 2 pH/VB 2 passes through zero at the end point and makes it electronically simpler to locate the end point in this fashion. In principle it also permits locating the end point with a precision greater than the resolution of the data points. We calculate the second derivative (column R) simply by obtaining the difference between two successive rows of the first derivative (column O). Note that to compute the second derivative, the original pH and VB data that we ultimately use are in rows 9, 10, and 11 and for this reason the value is located centrally in row 10 and when plotted against VB , it is plotted against the VB data also in row 10 (and not the VB,av data in column L). The original titration plot and both the first and second derivative plots are shown together in the worksheet “1st and 2nd deriv.” The titration plot and both the derivative plots are shown in Figure 8.13.
Professor’s Favorite Example ´ Contributed by Professor Alberto Rojas-Hernandez, Universidad ´ Autonoma Metropolitana-Iztapalapa, Mexico Acid–base titration of red wine: Titrable acidity (TA), tartaric acid content (g/L) and maximum buffer capacity. Red wines are composed of several substances with acid–base properties that may be considered among three main groups: carboxylic acids, carboxylic-polyphenolic, and polyphenolic acids. The common carboxylic acids are tartaric (pKa1 = 3.2, pKa2 = 4.3), malic (pKa1 = 3.4, pKa2 = 5.1), and citric (pKa1 = 3.1, pKa2 = 4.7, pKa3 = 6.4). The common carboxylic-polyphenolic acids are gallic (pKa1 = 4.4, pKa2 = 8.6, pKa3 = 11.2, pKa4 = 12.0) and caffeic (pKa1 = 4.4, pKa2 = 8.6, pKa3 = 11.5). Finally the common polyphenolic acids are tannic (with several pKa values between 6 and 10), as well as flavonols and flavonoids like anthocyanines (that give color to these wines). Figure 8.14 shows the titration of 25 mL of red wine with 0.1926 M NaOH, obtained with an automatic titrator. The titration shows two equivalence points. The first equivalence point corresponds to the titration of the carboxylic groups, often
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40 30 20 10
8.95 12.05 10 15 v /mL
5 pH
Buffer capacity /mmo L–1 per pH
50
13 12 11 10 9 8 7 6 5 4 3
d(pH)/dv /mL–1
pH
8.12 TITRATION OF AMINO ACIDS— —THEY ARE ACIDS AND BASES
309
50.0 40.0 30.0 20.0 10.0 0.0 0
2
4
20
10
12
14
pH
Fig. 8.15.
dpH/dv
8
6
Buffer capacity curve as a function of pH for the sample of red wine, for which titration is shown in Figure 8.14.
Fig. 8.14.
Titration curve for 25 mL of red wine titrated with 0.1926 M NaOH (markers) and first derivative of the curve (solid line).
called titrable acidity (TA). The second equivalence point corresponds with the true total acidity [TAc, see for example, Journal of Food Composition and Analysis, Composition study of methods for determination of titrable acidity in wine, 16 (2003) 555–562] and the second step is due to titration of the phenolic groups. If we assume that the only carboxylic acid present in the wine is tartaric acid (HOOC-CHOH-CHOH-COOH = H2 tar), the mass of tartaric acid (mH2 tar ) in the 25 mL red wine sample is: mH2 tar ≈ (8.95 mL)(0.1926 M)(1mmol H2 tar/2mmol NaOH)(150.1 mg/mmol) = 129.4 mg, equivalent to 5.18 g/L of tartaric acid. The buffer capacity of the same red wine sample is shown in Figure 8.15. The buffer capacity has a maximum value for pH ≈ 3.85, which is near the initial sample pH. The buffer capacity is 40 mmol/L per pH. This means that the amount of NaOH or HCl that should be added to change the initial pH of the wine by 1 unit is approximately (40 mmol/1000 mL)(25 mL) = 1 mmol (see Equation 7.50).
8.12 Titration of Amino Acids—They Are Acids and Bases Amino acids are important in pharmaceutical chemistry and biochemistry. These are amphoteric substances that contain both acidic and basic groups (i.e., they can act as acids or bases). The acid group is a carboxylic acid group (—CO2 H), and the basic group is an amine group (—NH2 ). In aqueous solutions, these substances tend to undergo internal proton transfer from the carboxylic acid group to the amino group because the RNH2 is a stronger base than RCO2 − . The result is a zwitterion: − R—CH—CO 2
NH3 + Since they are amphoteric, these substances can be titrated with either a strong acid or a strong base. Many amino acids are too weak to be titrated in aqueous solutions, but some will give adequate end points, especially if a pH meter is used to construct a titration curve.
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We can consider the conjugate acid of the zwitterion as a diprotic acid, which ionizes stepwise: + + − − R—CH—CO 2 H H + R—CH—CO 2 H + R—CH—CO 2
NH3 +
NH3 + zwitterion
conjugate acid of zwitterion
For the zwitterion,
[H+ ] ≈ Ka1 Ka2 .
NH2 conjugate base of zwitterion
(8.30) Ka1 and Ka2 values are frequently tabulated for amino acids (see Table C.1 in Appendix C). The values listed represent the successive ionization of the protonated form (i.e., the conjugate acid of the zwitterion); it ionizes to give first the amphoteric zwitterion and second to give the conjugate base, which is the same as a salt of a weak acid that hydrolyzes. Acid–base equilibria of amino acids are therefore treated just as for any other diprotic acid. The hydrogen ion concentration of the zwitterion is calculated in the same way as for any amphoteric salt, such as HCO3 − , as we described in Chapter 7; that is,
(8.31) [H+ ] = Ka1 Ka2 The titration of amino acids is not unlike the titration of other amphoteric substances, such as HCO3 − . In the latter case, titration with a base gives CO3 2− , with an intermediate CO3 2− /HCO3 − buffer region, and titration with an acid gives H2 CO3 , with an intermediate HCO3 − /H2 CO3 buffer region. When the zwitterion of an amino acid is titrated with strong acid, a buffer region is first established, consisting of the zwitterion (the “salt”) and the conjugate acid. Halfway to the equivalence point, pH = pKa1 (just as with HCO3 − /H2 CO3 ); and at the equivalence point, the pH is determined by the conjugate acid (and Ka1 , as with H2 CO3 ). When the zwitterion is titrated with a strong base, a buffer region of conjugate base (the “salt”) and zwitterion (now the “acid”) is established. Halfway to the equivalence point, pH = pKa2 (as with CO3 2− /HCO3 − ); and at the equivalence point, the pH is determined by the conjugate base (whose Kb = Kw /Ka2 , as with CO3 2− ). Amino acids may contain more than one carboxyl or amine group; in these cases, they may yield stepwise end points like other polyprotic acids (or bases), provided the different groups differ in K’s by at least a factor of 104 and are still strong enough to be titrated.
8.13 Kjeldahl Analysis: Protein Determination Danish chemist Johan Kjeldahl (1849–1900) developed the standard method for protein analysis, as the head of the chemistry department of Carlsberg Laboratory (of Carlsberg Brewery). His laboratory technique for nitrogen and protein analysis is still the universally accepted method for this analysis.
An important method for accurately determining nitrogen in proteins and other nitrogen-containing compounds is the Kjeldahl analysis. The quantity of protein can be calculated from a knowledge of the percent nitrogen contained in it. Although other more rapid methods for determining proteins exist, the Kjeldahl method is the standard on which all other methods are based. The material is digested with sulfuric acid to decompose it and convert the nitrogen to ammonium hydrogen sulfate: H2 SO4
Ca Hb Nc −−−→ aCO2 ↑ + 12 bH2 O + cNH4 HSO4 catalyst
The solution is cooled, concentrated alkali is added to make the solution alkaline, and the volatile ammonia is distilled into a solution of standard acid, which is in excess. Following distillation, the excess acid is back-titrated with standard base.
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OH−
cNH4 HSO4 −−−→ cNH3 ↑ +cSO4 2− cNH3 + (c + d)HCl → cNH4 Cl + dHCl dHCl + dNaOH → 12 dH2 O + dNaCl mmol N(c) = mmol reacted HCl = mmol HCl taken ×(c + d) − mmol NaOH(d) mmol Ca Hb Nc = mmol N × 1/c The digestion is speeded up by adding potassium sulfate to increase the boiling point and by a catalyst such as a selenium or copper salt. The amount of the nitrogen-containing compound is calculated from the weight of nitrogen analyzed by multiplying it by the gravimetric factor.
Example 8.4 A 0.2000-g sample containing urea, O ||
NH2 —C—NH2 is analyzed by the Kjeldahl method. The ammonia is collected in 50.00 mL of 0.05000 M H2 SO4 , and the excess acid is back-titrated with 0.05000 M NaOH, a procedure requiring 3.40 mL. Calculate the percent urea in the sample. Solution
The titration reaction is H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O mmol NaOH consumed = 3.40 mL × 0.05 mmol/mL = 0.17 mmol mmol H2 SO4 neutralized = 0.17/2 mmol = 0.085 mmol mmol H2 SO4 originally taken = 0.05000 mmol/mL × 50.00 mL = 2.500 mmol mmol H2 SO4 neutralized by NH3 = 2.500 − 0.085 mmol = 2.415 mmol The reaction with the NH3 is 2NH3 + H2 SO4 → (NH4 )2 SO4 Two millimoles NH3 react with one millimole H2 SO4 and two millimoles NH3 come from one millimole urea. Therefore, millimoles of H2 SO4 neutralized by ammonia, 2.415 mmol, is the same as the amount of urea. Multiplying by 60.05 mg/mmol urea, we have 2.415 mmol × 60.05 mg/mmol = 145.02 mg urea % urea by weight: 145.02/200.0 × 100% = 72.51%
A large number of different proteins contain very nearly the same percentage of nitrogen. The gravimetric factor for conversion of weight of N to weight of protein for normal mixtures of serum proteins (globulins and albumin) and protein in feeds is 6.25, (i.e., the proteins contain 16% nitrogen). When the sample is made up almost entirely of gamma globulin, the factor 6.24 is more accurate. If it contains mostly albumin, 6.27 is preferred.
Many proteins contain nearly the same amounts of nitrogen.
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In the conventional Kjedahl method, two standard solutions are required, the acid for collecting the ammonia and the base for back-titration. A modification can be employed that requires only standard acid for direct titration. The ammonia is collected in a solution of boric acid. In the distillation, an equivalent amount of ammonium borate is formed: (8.32) NH3 + H3 BO3 NH4 + + H2 BO3 − Boric acid is too weak to be titrated, but the borate, which is equivalent to the amount of ammonia, is a fairly strong Brønsted base that can be titrated with a standard acid to a methyl red end point. The boric acid is so weak it does not interfere, and its concentration need not be known accurately. Also, boric acid is so weak that it is not significantly ionized, it is nonconductive. In contrast, ammonium borate is ionized and is conductive. In a present-day adaptation, instead of titration, the conductivity of the ammonium borate formed is measured as a measure of ammonia absorbed.
Example 8.5 A 0.300-g feed sample is analyzed for its protein content by the modified Kjeldahl method. If 25.0 mL of 0.100 M HCl is required for titration, what is the percent protein content of the sample? Solution
Since this is a direct titration with HCl which reacts 1 : 1 with NH3 , the millimoles of NH3 (and therefore of N) equals the millimoles of HCl. Multiplication by 6.25 gives the milligrams of protein. % protein =
0.10 mmol/mL HCl×25.0 mL HCl×14.01 mg N/mmol HCl×6.25 mg protein/mg N 300 mg ×100% = 73.0%
The boric acid method (a direct method) is simpler and is usually more accurate since it requires the standardization and accurate measurement of only one solution. However, the end point break is not so sharp, and the indirect method requiring backtitration is usually preferred for micro-Kjeldahl analysis. A macro-Kjeldahl analysis of blood requires about 5 mL blood, while a micro-Kjeldahl analysis requires only about 0.1 mL. We have confined our discussion to those substances in which the nitrogen exists in the −3 valence state, as in ammonia. Such compounds include amines and amides. Compounds containing oxidized forms of nitrogen, such as organic nitro and azo compounds, must be treated with a reducing agent prior to digestion in order to obtain complete conversion to ammonium ion. Reducing agents such as iron(II) or thiosulfate are used. Inorganic nitrate and nitrite are not converted by such treatment to ammonia.
8.14 Titrations Without Measuring Volumes Can titrations be done without measuring volumes? Yes. Professor Michael DeGrandpre at the University of Montana has developed “tracer monitored” titrations (TMT) in which the spectrophotometric monitoring of titrant dilution, rather than volumetric increment, places the burden of analytical performance solely on the spectrophotometer. Most modern titration systems are fully automated with precision
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pumps. The TMT method is insensitive to pump precision and reproducibility of automatic titrators, and can allow less precise and less expensive pumps. An inert indicator tracer (e.g., dye) is added to the titrant. Dilution of a pulse of titrant in a titration vessel is tracked using the total tracer concentration, measured spectrophotometrically (an instrumental technique that measures the concentration of a colored substance from the amount of light absorbed based on linearity defined by Beer’s law, see Chapter 16). The concentration of the tracer at the detected end point (e.g., from pH measurement), derived from Beer’s law, is used to calculate the relative proportions of titrant and sample. In conventional titrations, the analyte concentration in the sample, [analyte], is given by Q[titrant]Vtit.ep (8.33) [analyte] = Vsample where Q is the reaction stoichiometry (moles analyte : moles titrant), [titrant] is the titrant concentration, Vtit,ep is the volume of the titrant at the end point, and Vsample is the original volume of the sample taken. The TMT method is based on determining the dilution factor, Dep , of the titrant at the end point, determined using the inert tracer: Dep =
Vtit.ep (Vtit.ep + Vsample )
=
[tracer]ep [tracer]tit
(8.34)
where [tracer]ep is the tracer concentration in the mixture at the end point, and [tracer]tit is the tracer concentration in the titrant. The sample analyte concentration is then given by Q[titrant] [analyte] = (8.35) 1 Dep − 1 So we only need to determine Dep . Any inert tracer can be used, for example, a fluorescent dye can be used in a low concentration in the titrant. The technique can be applied to other types of titrations, e.g., complexometric or redox titrations. It will also be realized, that the same principle can be applied if the sample is spiked with an indicator dye at the beginning of the titration and the titrant simply dilutes the dye level as the titration progresses. Either way, the method eliminates the need for volumetric measurements. However, Beer’s law (see Chapter 16) must be obeyed through the dye concentration range of interest. Details of the method are given in References 10 and 11.
Questions 1. What is the minimum pH change required for a sharp indicator color change at the end point? Why? 2. What criterion is used in selecting an indicator for a particular acid–base titration? 3. At what pH is the buffering capacity of a buffer the greatest? 4. Is the pH at the end point for the titration of a weak acid neutral, alkaline, or acidic? Why? 5. What would be a suitable indicator for the titration of ammonia with hydrochloric acid? Of acetic acid with sodium hydroxide? 6. Explain why boiling the solution near the end point in the titration of sodium carbonate increases the sharpness of the end point. 7. What is the approximate pK of the weakest acid or base that can be titrated in aqueous solution? 8. What must be the difference in the strengths of two acids in order to differentiate between them in titration?
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9. Distinguish between a primary standard and a secondary standard. 10. What is a zwitterion? 11. What percent nitrogen is contained in a typical protein? 12. What is the preferred acid for titrating bases? Why?
Problems STANDARDIZATION CALCULATIONS 13. A hydrochloric acid solution is standardized by titrating 0.4541 g of primary standard tris(hydroxymethyl)aminomethane. If 35.37 mL is required for the titration, what is the molarity of the acid? 14. A hydrochloric acid solution is standardized by titrating 0.2329 g of primary standard sodium carbonate to a methyl red end point by boiling the carbonate solution near the end point to remove carbon dioxide. If 42.87 mL acid is required for the titration, what is its molarity? 15. A sodium hydroxide solution is standardized by titrating 0.8592 g of primary standard potassium acid phthalate to a phenolphthalein end point, requiring 32.67 mL. What is the molarity of the base solution? 16. A 10.00-mL aliquot of a hydrochloric acid solution is treated with excess silver nitrate, and the silver chloride precipitate formed is determined by gravimetry. If 0.1682 g precipitate is obtained, what is the molarity of the acid?
INDICATORS 17. Write a Henderson–Hasselbalch equation for a weak-base indicator, B, and calculate the required pH change to go from one color of the indicator to the other. Around what pH is the transition?
TITRATION CURVES You may wish to use spreadsheets for some of these calculations. 18. Calculate the pH at 0, 10.0, 25.0, and 30.0 mL of titrant in the titration of 50.0 mL of 0.100 M NaOH with 0.200 M HCl. 19. Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 25.0 mL of 0.200 M HA with 0.100 M NaOH. Ka = 2.0 × 10−5 . 20. Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL of titrant in the titration of 50.0 mL of 0.100 M NH3 with 0.100 M HCl. 21. Calculate the pH at 0, 25.0, 50.0, 75.0, 100, and 125% titration in the titration of both protons of the diprotic acid H2 A with 0.100 M NaOH, starting with 100 mL of 0.100 M H2 A. Ka1 = 1.0 × 10−3 , Ka2 = 1.0 × 10−7 . 22. Calculate the pH at 0, 25.0, 50.0, 75.0, 100, and 150% titration in the titration of 100 mL of 0.100 M Na2 HPO4 with 0.100 M HCl to H2 PO4 − .
QUANTITATIVE DETERMINATIONS 23. A 0.492-g sample of KH2 PO4 is titrated with 0.112 M NaOH, requiring 25.6 mL: H2 PO4 − + OH− → HPO4 2− + H2 O What is the percent purity of the KH2 PO4 ? 24. What volume of 0.155 M H2 SO4 is required to titrate 0.293 g of 90.0% pure LiOH? 25. An indication of the average formula weight of a fat is its saponification number, expressed as the milligrams KOH required to hydrolyze (saponify) 1 g of the fat:
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where R can be variable. A 1.10-g sample of butter is treated with 25.0 mL of 0.250 M KOH solution. After the saponification is complete, the unreacted KOH is back-titrated with 0.250 M HCl, requiring 9.26 mL. What is the saponification number of the fat and what is its average formula weight (assuming the butter is all fat)? 26. A sample containing the amino acid alanine, CH3 CH(NH2 )COOH, plus inert matter is analyzed by the Kjeldahl method. A 2.00-g sample is digested, the NH3 is distilled and collected in 50.0 mL of 0.150 M H2 SO4 , and a volume of 9.0 mL of 0.100 M NaOH is required for back-titration. Calculate the percent alanine in the sample. 27. A 2.00-mL serum sample is analyzed for protein by the modified Kjeldahl method. The sample is digested, the ammonia is distilled into boric acid solution, and 15.0 mL of standard HCl is required for the titration of the ammonium borate. The HCl is standardized by treating 0.330 g pure (NH4 )2 SO4 in the same manner. If 33.3 mL acid is required in the standardization titration, what is the concentration of protein in the serum in g% (wt/vol)?
QUANTITATIVE DETERMINATIONS OF MIXTURES 28. A 100-mL aliquot of a solution containing HCl and H3 PO4 is titrated with 0.200 M NaOH. The methyl red end point occurs at 25.0 mL, and the bromthymol blue end point occurs at 10.0 mL later (total 35.0 mL). What are the concentrations of HCl and H3 PO4 in the solution? 29. A 0.527-g sample of a mixture containing NaCO3 , NaHCO3 , and inert impurities is titrated with 0.109 M HCl, requiring 15.7 mL to reach the phenolphthalein end point and a total of 43.8 mL to reach the modified methyl orange end point. What is the percent each of Na2 CO3 and NaHCO3 in the mixture? 30. Sodium hydroxide and Na2 CO3 will titrate together to a phenolphthalein end point (OH− → H2 O; CO3 2− → HCO3 − ). A mixture of NaOH and Na2 CO3 is titrated with 0.250 M HCl, requiring 26.2 mL for the phenolphthalein end point and an additional 15.2 mL to reach the modified methyl orange end point. How many milligrams NaOH and Na2 CO3 are in the mixture? 31. Sodium carbonate can coexist with either NaOH or NaHCO3 but not with both simultaneously, since they would react to form Na2 CO3 . Sodium hydroxide and Na2 CO3 will titrate together to a phenolphthalein end point (OH− → H2 O; CO3 2− → HCO3 − ). A mixture of either NaOH and Na2 CO3 or of Na2 CO3 and NaHCO3 is titrated with HCl. The phenolphthalein end point occurs at 15.0 mL and the modified methyl orange end point occurs at 50.0 mL (35.0 mL beyond the first end point). The HCl was standardized by titrating 0.477 g Na2 CO3 , requiring 30.0 mL to reach the modified methyl orange end point. What mixture is present and how many millimoles of each constituent are present? 32. What would be the answers to Problem 31 if the second end point had occurred at 25.0 mL (10.0 mL beyond the first end point)? 33. A mixture containing only BaCO3 and Li2 CO3 weighs 0.150 g. If 25.0 mL of 0.120 M HCl is required for complete neutralization (CO3 2− → H2 CO3 ), what is the percent BaCO3 in the sample? 34. A sample of P2 O5 contains some H3 PO4 impurity. A 0.405-g sample is reacted with water (P2 O5 + 3H2 O → 2H3 PO4 ), and the resulting solution is titrated with 0.250 M NaOH (H3 PO4 → Na2 HPO4 ). If 42.5 mL is required for the titration, what is the percent of H3 PO4 impurity?
PROFESSOR’S FAVORITE PROBLEMS Contributed by Professor Tarek Farhat, The University of Memphis 35. A bubbler containing sodium hydroxide solution is used as a filter to remove CO2 gas from an air stream that purges a glove box. The following reaction takes place: 2NaOH + CO2 (g) → Na2 CO3 + H2 O
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If the bubbler contained 100 mL of the NaOH solution and the NaOH solution was initially at a pH of 12, calculate the drop in pH after passing an air stream (1 atm pressure, 25 ◦ C) at a flow rate of 5.0 mL/min that contains 0.5% CO2 gas by volume at 25 ◦ C and 1 atm for (a) one hour, and (b) five hours? Contributed by Professor Peter R. Griffiths, University of Idaho 36. Which curve most closely approximates that expected for the complete titration of a solution of sodium hydrogen sulfate, NaHSO4 with aqueous NaOH? (Dissociation constant for HSO− 4 = 1.2 × 10−2 ) (A)
14
(B)
pH
14 pH
7
0 (C)
7
volume NaOH
14
(D)
pH
volume NaOH
14 pH
7
7
volume NaOH
volume NaOH
The following problems were contributed by Professor Noel Motta, University of Puerto Rico-R´ıo Piedras 37. The following curve may correspond to the titration of 14 12 10
pH
8 6 4 2 0
5
10 15 0.080 M NaOH volume, mL
20
25
(a) 5.00 mL of a 0.080 M diprotic acid, H2 A (b) 5.00 mL of a mixture consisting of 0.080 M H2 A and 0.080 M HA− (c) 5.00 mL of a mixture consisting of two acids with the same Ka , 0.080 M HA, and 0.16 M HB (d) 5.00 mL of a mixture consisting of two acids with the same concentration (0.080 M), but different Ka 38. Changing titration conditions can have a significant influence on its feasibility as the following example illustrates. The titration of 0.010 M NH4 + with NaOH 0.010 M is shown at left. Select the set of conditions under which the curve at right will be observed for the titration of NH4 + with NaOH.
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12
12
10
10
8
8 pH
14
pH
14
6
6
4
4
2
2
5
10
15 20 25 30 35 0.010 M NaOH volume, mL
40
45
50
5
10
15
20 25 30 35 x M NaOH volume, mL
(a) Titration of 50.0 mL 0.0050 M NH4 + vs 0.010 M NaOH (b) Titration of 25.0 mL 0.50 M NH4 + vs 0.50 M NaOH (c) Titration of 25.0 mL 0.010 M NH4 + vs 1.0 M NaOH (d) Titration of 12.5 mL 0.020 M NH4 + vs 0.01 M NaOH (e) Titration of 12.5 mL 0.0050 M NH4 + vs 0.010 M NaOH 39. In the following example, by changing the titration conditions on a weak acid as compared to a strong acid, a difference at the pre-equivalence regions for the two is evident, while no difference at the post-equivalence regions is observed:
13
14
HWeak 0.50 M
HStrong 0.50 M
HWeak 0.025 M Case A
pH 7
10 NaOH volume, mL
Case B
pH 7
1 0
HStrong 0.025 M
20
1 0
10 NaOH volume, mL
20
In both Case A and Case B, 0.50 M and 0.025 M acids are titrated with NaOH 0.50 M and NaOH 0.025 M, respectively. Case A corresponds to the weak acid titration (represented as HWeak) and Case B corresponds to the strong acid titration (represented as HStrong). Explain why the pH behavior in Case A is different from Case B in the pre-equivalence region when changing conditions, while they are the same in the post-equivalence region. 40. An analyst titrated two solutions with HCl 0.0100 M. The analyst knew that one of them contained a mixture of CO3 2− and OH− and the other contained a mixture of CO3 2− and HCO3 − . However, the analyst did not know which solution corresponded to each of those two compositions. To find out, she performed a potentiometric titration on 10.00 mL of each of the two solutions and obtained the following titration curves. Determine compositions of A and B.
40
45
50
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14 12
12.00
10
10.00
A
8
B
pH
8.00
4 4.00
2
2.00
0 0
0.00 0.00
10
20
30
40
50
HCI 0.01 M volume (mL) 10.00
20.00
30.00
40.00
50.00
60.00
70.00
HCI 0.0100 M volume (mL)
41. A newly synthesized triprotic acid showed the following general titration curve: 13 12 11 10 9 8 pH
pH
6 6.00
7 6 5 4 3 2 1 0 NaOH Volume, mL
To determine the MW of the acid, a student dissolved a pure sample (400.0 mg) in a small volume of water, then transferred it to a 100.0 mL volumetric flask and made up to the mark with distilled water. The student then took a 25.00-mL aliquot of this solution and titrated it with 15.90 mL 0.0750 M NaOH to reach a phenol red end point. The MW of the acid is: (a) 111.8, (b) 167.7 (c) 218.5 (d) 286.2 (e) 335.4 42. For which of the following diprotic acids a titration curve with 0.100 M NaOH will show just one equivalence point with 2:1 stoichiometry? (a) Resorcinol: pK1 : 9.30; pK2 : 11.06, (b) Leucine: pK1 : 2.33; pK2 : 9.74, (c) Maleic acid: pK1 : 1.82; pK2 : 6.59, (d) Salicylic acid: pK1 : 2.97; pK2 : 13.74, (e) Tartaric acid: pK1 : 3.036; pK2 : 4.366 43. For which of the following amino acids a titration with 0.10 M NaOH will show three well-defined end points?
60
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I. Aspartic Acid II. Arginine III. Histidine
pK1
pK2
pK3
1.99 1.82 1.70
3.90 8.99 6.02
10.00 12.48 9.08
(a) I only, (b) II only, (c) III only, (d) II and III, (e) I, II, and III The following problems were contributed by Professor Bin Wang, Marshall University 44. (a) A 0.9872 g sample of unknown potassium hydrogen phthalate (KHP) required 28.23 mL of 0.1037 M NaOH for neutralization. What is the percentage of KHP in the sample? 45. A 0.2027 g sample of finely powered limestone (mainly CaCO3 ) was dissolved in 50.00 mL of 0.1035 M HCl. The solution was heated to expel CO2 produced by the reaction. The remaining HCl was then titrated with 0.1018 M of NaOH and it required 16.62 mL. Calculate the percentage of CaCO3 in the limestone sample. The following problem was contributed by Professor Wen-Yee Lee, The University of Texas at El Paso 46. The Kjeldahl procedure was used to analyze 500 μL of a solution containing 50.0 mg protein/mL solution. The liberated NH3 was collected in 5.00 mL of 0.0300M HCl. The remaining acid required 10.00 mL 0.010 M NaOH for complete titration. What is the weight percent of nitrogen in the protein? The following problems were contributed by Professor Christopher Harrison, San Diego State University 47. (a) What is the pH of a solution that contains 12.00 mg of H3 PO4 in one liter of pure water? (b) How many milligrams of citric acid will you need to prepare one liter of a solution that has exactly twice the molar concentration of the H3 PO4 solution? (c) How many milligrams of sodium hydroxide will you need to add to the citric acid solution to get to the same pH as the phosphoric acid solution? How easy will it be to do this? (See the text website for Excel calculations.) 48. A buffer is typically made from 30 millimoles of sodium hydrogensulfite (NaHSO3 ) and 90 millimoles of disodium sulfite (Na2 SO3 ) per liter. You have run out of these reagents. At your disposal you have solid sodium carbonate, solid disodium hydrogen phosphate, 1 M dimethylamine, solid sodium hydroxide, and 1 M nitric acid. Using some or all of these chemicals, prepare a buffer with the same pH and formal concentration as the sulfite-hydrogensulfite buffer. What is the simplest combination of chemicals and how many grams or milliliters of each will be needed (ignoring the water needed)? Relevant dissociation constant data are in Appendix C. 49. What volume of a stock HNO3 solution containing 25% (w/w) nitric acid (density of 1.18 g/mL) is needed to make 500 mL HNO3 solution of pH 2.45? The following Professor’s favorite final exam problem was contributed by Professor Michael D. Morris, University of Michigan 50. The pKa of finalic acid is 6.50. (a) What is the pH of an equimolar solution of finalic acid and sodium finalate? Assume that both solutions are at 0. 1 M. (b) Would you expect the buffer capacity to be higher or lower if you decrease the concentration of each constituent to 0.01 M? Why? (c) What is the effect of water dissociation on the pH of these solutions? (d) Only considering the pH of the solution, would it be safe to drink a 0.1 M solution of finalic acid? How about a 0.01 M solution?
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Professor’s Favorite Puzzle The following puzzle was contributed by Professor Fred Hawkridge, Virginia Commonwealth University 51. You want to prepare a pH 7.0 phosphate buffer that will be exactly 0.100 M in phosphate concentration. You look up the Ka values for H3 PO4 and decide that you will need mostly Na2 HPO4 that may need a little bit of acid added until the pH is right. However, if you took exactly 0.100 M Na2 HPO4 , the phosphate concentration will be altered when you add the acid. How are you going to make the buffer?
PROFESSOR’S FAVORITE SPREADSHEET EXERCISES 52. Professor Alexander Scheeline, University of Illinois, has developed a spreadsheet Universal Acid Titrator (see the book’s website, 8.52 Universal Acid Titrator.xlsx). You can type in the pKa value(s) for any acid (up to a decaprotic acid!) in row 11, or pKa value(s) in row 9, and generate a titration plot (pH vs.Volume base) for titration with strong, monoprotic base like NaOH. Delight Professor Scheeline by playing with the spreadsheet and generating different titration curves. (We recommend that when the higher K’s are irrelevant, directly enter 0 for these K values in row 11.) Rather than computing the pH for a given amount of base added, this approach, also based on charge balance, computes how much base you have to add to get to a given pH. Here are the following challenge questions: (a) Why is it so much easier to directly compute the needed base amounts without involving solver or Goal Seek rather than computing the pH for a given amount of base added? (b) Especially at low pH values why are some of the calculated base volumes negative? (They are never plotted.) (c) For any acid of your choice (or your creation), calculate the pH when no base has been added (V = 0) by using detailed calculation or Goal Seek or Solver. Then use the current spreadsheet to calculate how much base you will need to get to this pH. See if you indeed get zero. (d) For the more adventurous: After computing the titration curve with this spreadsheet (that ignores activity corrections), do a successive approximations calculation at each pH that takes ionic strength and activity corrections into account, and compute a revised titration curve. Compare the activity corrected and uncorrected curves: where does activity correction matter most?
The following problems were contributed by Professor Steven Goates, Brigham Young University Excel answers are given on the text’s website. 53. Develop a spreadsheet program to calculate the change of pH with volume of base added for the theoretical titration of an acid with a strong base. Calculate the titration curve for the titration of KHP with NaOH described below, and graph your results. (Even though you will plot pH on the y-axis and volume of titrant on the x-axis, as in the case of Problem 52 above you may find it much more convenient in your calculation to use pH as the independent variable and solve for volume of base added as the dependent variable.) You should calculate the volume of base required to reach increasingly higher pH values at intervals of 0.100 pH units up to a final pH of 12.500. For this problem, assume all activity coefficients to be 1, pH = − log aH+ , the KHP is 100.00 % pure, the NaOH solution is carbonate-free, and volumes of solution are additive. The dissociation constant data are available in Appendix C. Specifically calculate for the following titration (but have the spreadsheet in a manner where all these values can be readily changed): A 0.994 mL aliquot of 0.4103 M KHP solution was placed in a titration vial and 5.00 mL distilled water was added. The solution was then titrated with 0.2102 M NaOH solution. What effect is there on the titration curve if 6 mL instead of 5 mL water is added? 54. Modify your calculation of Problem 53 to include activity coefficients. Go through the calculations at least two rounds. First time, as in Problem 53, calculate ionic strength μ (Equation 6.18). From this ionic strength, calculate the activity coefficients by the extended Debye-H¨uckel equation given in your textbook (Equation 6.19). Assume ion size parameters of 9, 3.5, 6, and 6 A for H+ , OH− , HP− , and P2− , respectively. Convert the known activity
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constants Kw , Ka1 , and Ka2 to the corresponding values given in terms of concentrations because mass balance equations must ultimately be in terms of concentrations, and not activities. Using these constants in terms of concentration, calculate the new values of [H+ ], [OH− ], α1 , and α2 and ionic strength again. Then repeat the calculations and finally, plot your calculated Vb vs. pH for Problems 53 and 54 on the same graph, and discuss any differences.
Recommended References INDICATORS 1. R. W. Sabnis, Handbook of Acid–Base Indicators. Boca Raton, FL: CRC Press, 2008. 2. G. Gorin, “Indicators and the Basis for Their Use,” J. Chem. Ed., 33 (1956) 318. 3. E. Bishop, Indicators. Oxford: Pergamon, 1972.
TITRATION CURVES 4. R. K. McAlpine, “Changes in pH at the Equivalence Point,” J. Chem. Ed., 25 (1948) 694. 5. A. K. Covington, R. N. Goldberg, and M. Sarbar, “Computer Simulation of Titration Curves with Application to Aqueous Carbonate Solutions,” Anal. Chim. Acta, 130 (1981) 103.
TITRATIONS 6. Y.-S. Chen, S. V. Brayton, and C. C. Hach, “Accuracy in Kjeldahl Protein Analysis,” Am. Lab., June (1988) 62. 7. R. M. Archibald, “Nitrogen by the Kjeldahl Method,” in D. Seligson, ed., Standard Methods of Clinical Chemistry, Vol. 2. New York: Academic, 1958, pp. 91–99.
WEB pH TITRATION CALCULATOR 8. www2.iq.usp.br/docente/gutz/Curtipot.html. A free program that calculates the pH and paH of mixtures of acids and bases, and also titration curves and distribution curves, alpha plots, and more. 9. www.phcalculation.se. A free program that calculates pH of complex mixtures of strong and weak acids and bases using concentrations (ConcpH) or activities (ActpH). Equilibrium concentrations of all species are calculated. The latter also calculates activity coefficients of the species and the ionic strength. A variation of the program allows calculation of points in a titration curve by adding different volumes of titrant to the analyte solution.
TITRATIONS WITHOUT MEASURING VOLUMES 10. T. R. Martz, A. G. Dickson, and M. D. DeGrandpre, “Tracer Monitoring Titrations: Measurement of Total Alkalinity,” Anal. Chem., 78 (2006) 1817. 11. M. D. DeGrandpre, T. R. Martz, R. D. Hart, D. M. Elison, A. Zhang, and A. G. Bahnson, “Universal Tracer Monitored Titrations,” Anal. Chem., 83 (2011) 9217.
(Source: Stig Johansson/www. phcalculation.se/Courtesy of late Stig Johannson.)
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Chapter Nine COMPLEXOMETRIC REACTIONS AND TITRATIONS “Simple things should be simple. Complex things should be possible.” —Alan Kay
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Formation constants, p. 322
●
Indicators for EDTA titrations, p. 334
●
EDTA equilibria (key equations: 9.5–9.8), p. 326
●
αM and β values (key equations: 9.17–9.21), p. 336
Many metal ions form slightly dissociated complexes with various ligands (complexing agents). The analytical chemist makes judicious use of complexes to mask undesired reactions. The formation of complexes can also serve as the basis of accurate and convenient titrations for metal ions in which the titrant is a complexing agent. Complexometric titrations are useful for determining a large number of metals. Selectivity can be achieved by appropriate use of masking agents (addition of other complexing agents that react with interfering metal ions, but not with the metal of interest) and by pH control, since most complexing agents are weak acids or weak bases whose equilibria are influenced by the pH. In this chapter, we discuss metal ions, their equilibria, and the influence of pH on these equilibria. We describe titrations of metal ions with the very useful complexing agent EDTA, the factors that affect them, and indicators for the titrations. The EDTA titration of calcium plus magnesium is commonly used to determine water hardness. In the food industry, calcium is determined in cornflakes. In the plating industry, nickel is determined in plating solutions by complexometric (also called chelometric) titration, and in the metals industry in etching solutions. In the pharmaceutical industry, aluminum hydroxide in liquid antacids is determined by similar titrations. Nearly all metals can be accurately determined by complexometric titrations. Complexing reactions are useful for gravimetry, spectrophotometry, and fluorometry, and for masking interfering ions.
9.1 Complexes and Formation Constants—How Stable Are Complexes? Complexes play important roles in many chemical and biochemical processes. For example, the heme molecule in blood holds iron tightly because the nitrogen atoms of the heme form strong ligating or complexing bonds. In general, the nitrogen atom derived from an amino group is a good donor atom or complexer. The iron [as iron(II)] in turn bonds readily with oxygen to transport oxygen gas from the lungs to elsewhere in the body and then easily releases it because oxygen is a poor donor atom or complexer. Carbon monoxide kills because it is a strong complexer and 322
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displaces oxygen; it binds to heme 200 times more strongly than does oxygen, forming carboxyhemoglobin. Many cations will form complexes in solution with a variety of substances that have a pair of unshared electrons (e.g., on N, O, S atoms in the molecule) capable of satisfying the coordination number of the metal cation. [The metal cation is a Lewis acid (electron pair acceptor), and the complexer is a Lewis base (electron pair donor).] The number of molecules of the complexing agent, called the ligand, will depend on the coordination number of the metal cation and on the number of complexing groups on the ligand molecule. Ammonia is a simple complexing agent with one pair of unshared electrons that will complex copper ion:
2
Cu
NH3 H3N: Cu:NH3 NH3
4:NH3
323
Most ligands contain O, S, or N as the complexing atoms.
2
Here, the copper ion acts as a Lewis acid, and the ammonia is a Lewis base. The Cu2+ (hydrated) ion is pale blue in solution, while the ammonia (the ammine) complex is deep blue. A similar reaction occurs with the green hydrated nickel ion to form a deep blue ammine complex. Ammonia will also complex with silver ion to form a colorless complex. Two ammonia molecules complex with each silver ion in a stepwise fashion, and we can write an equilibrium constant for each step, called the formation constant Kf : Ag+ + NH3 Ag(NH3 )+
Kf 1 =
[Ag(NH3 )+ ] [Ag+ ] [NH3 ]
(9.1)
= 2.5 × 103 Ag(NH3 )+ + NH3 Ag(NH3 )2 +
Kf 2 =
[Ag(NH3 )2 + ] [Ag(NH3 )+ ] [NH3 ]
(9.2)
= 1.0 × 104 The overall reaction is the sum of the two steps, and the overall formation constant is the product of the stepwise formation constants: Ag+ + 2NH3 Ag(NH3 )2 +
Kf = Kf 1 · Kf 2 =
[Ag(NH3 )2 + ] [Ag+ ] [NH3 ]2
(9.3)
= 2.5 × 107 For the formation of a simple 1:1 complex, for example, M + L = ML, the formation constant is simply Kf = [ML]/[M][L]. The formation constant is also called the stability constant Ks , or Kstab . We could write the equilibria in the opposite direction, as dissociations. If we do this, the concentration terms are inverted in the equilibrium constant expressions. The equilibrium constants then are simply the reciprocals of the formation constants, and they are called instability constants Ki , or dissociation constants Kd : Ag(NH3 )2 + Ag+ + 2NH3
Kd =
+
2
1 [Ag ] [NH3 ] = 4.0 × 10−8 = Kf [Ag(NH3 )2 + ]
(9.4)
You can use either constant in calculations, as long as you use it with the proper reaction and the correct expression. You will note that the dissociation of an acid HA H+ + A− is in fact very similar to the dissociation of a metal ligand complex
Kf = Ks = 1/Ki or 1/Kd .
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ML M + L However, the convention is to write the first as a dissociation reaction and the second as an association reaction.
Example 9.1 A divalent metal M2+ reacts with a ligand L to form a 1:1 complex: [ML2+ ] [M2+ ] [L] Calculate the concentration of M2+ in a solution prepared by mixing equal volumes of 0.20 M M2+ and 0.20 M L. Kf = 1.0 × 108 . M2+ + L ML2+
Kf =
Solution
We have added stoichiometrically equal amounts of M2+ and L. The complex is sufficiently strong that their reaction is virtually complete. Since we added equal volumes, the initial concentrations were halved by dilution. Let x represent [M2+ ]. At equilibrium, we have ML2+ M2+ + L x x 0.10 − x ≈ 0.10 Essentially, all the M2+ (original concentration 0.20 M) was converted to an equal amount of ML2+ , with only a small amount of uncomplexed metal remaining. Substituting into the Kf expression, 0.10 = 1.0 × 108 (x)(x) x = [M2+ ] = 3.2 × 10−5 M You can also solve the equation (0.10 − x)/x2 = Kf by Goal Seek in Excel. This will apply even when Kf is not so high and you cannot really assume that x is so small that 0.10−x can be approximated as 0.10.
Example 9.2 Silver ion forms a stable 1:1 complex with triethylenetetraamine, called “trien” [NH2 (CH2 )2 NH(CH2 )2 NH(CH2 )2 NH2 ]. Calculate the silver ion concentration at equilibrium when 25 mL of 0.010 M silver nitrate is added to 50 mL of 0.015 M trien. Kf = 5.0 × 107 . Solution
Ag+ + trien Ag(trien)+
Kf =
[Ag(trien)+ ] [Ag+ ] [trien]
Calculate the millimoles Ag+ and trien added: mmol Ag+ = 25 mL × 0.010 mmol/mL = 0.25 mmol mmol trien = 50 mL × 0.015 mmol/mL = 0.75 mmol The equilibrium lies far to the right, so you can assume that virtually all the Ag+ reacts with 0.25 mmol of the trien (leaving 0.50 mmol trien in excess) to form 0.25 mmol complex. Calculate the molar concentrations: [Ag+ ] = x = mol/L unreacted [trien] = (0.50 mmol/75 mL) + x = 6.7 × 10−3 + x ≈ 6.7 × 10−3
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[Ag(trien)+ ] = 0.25 mmol/75 mL − x = 3.3 × 10−3 − x ≈ 3.3 × 10−3 Try neglecting x compared to the other concentrations: 3.3 × 10−3 = 5.0 × 107 (x) (6.7 × 10−3 ) x = [Ag+ ] = 9.8 × 10−9 M We were justified in neglecting x.
Try with Excel Goal Seek: Let us assume that in the same example instead of trien there is some other ligand L, which reacts the same way but has an association constant of only 5. What will be the concentrations of AgL+ , Ag+ , and L?
9.2 Chelates: EDTA—The Ultimate Titrating Agent for Metals Simple complexing agents such as ammonia are rarely used as titrating agents because a sharp end point corresponding to a stoichiometric complex is generally difficult to achieve. This is because the stepwise formation constants are frequently close together and are not very large, and a single stoichiometric complex cannot be observed. Certain complexing agents that have two or more complexing groups on the molecule, however, do form well-defined complexes and can be used as titrating agents. Schwarzenbach demonstrated that a remarkable increase in stability is achieved if a bidentate ligand (one with two complexing groups) is used (see Reference 4 for his many contributions). For example, he showed replacing ammonia with the bidentate ethylenediamine, NH2 CH2 CH2 NH2 (en), results in a highly stable Cu(en)2 2+ complex. The most generally useful titrating agents are aminocarboxylic acids, in which the amino nitrogen and carboxylate groups serve as ligands. The amino nitrogens are more basic and are protonated (−NH3 + ) more strongly than the carboxylate groups. When these groups bind to metal atoms, they lose their protons. The metal complexes formed with these multidentate complexing agents are often 1:1, regardless of the charge on the metal ion, because there are sufficient complexing groups on one molecule to satisfy all the coordination sites of the metal ion. An organic agent that has two or more groups capable of complexing with a metal ion is called a chelating agent. The complex formed is called a chelate. The chelating agent is called the ligand. Titration with a chelating agent is called a chelometric titration, perhaps the most important and practical type of complexometric titrations. The most widely used chelating agent in titrations is ethylenediaminetetraacetic acid (EDTA). The formula for EDTA is
HO2CCH2
O2CCH2
H H NCH2CH2 N
CH2CO2H
CH2CO2
Each of the two nitrogens and each of the four carboxyl groups contains a pair of unshared electrons capable of complexing with a metal ion. Thus, EDTA contains six
The term chelate is derived from the Greek term meaning “clawlike.” Chelating agents literally wrap themselves around a metal ion.
The protons in EDTA are displaced upon complexing with a metal ion. A negatively charged chelate results. Note that in the structure we have depicted the uncharged EDTA molecule as the double zwitterion (see Section 8.12); this is the form in which it really exists. This is why EDTA is most commonly available as the disodium salt, where the two ionized carboxylic acid groups form the salt.
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complexing groups. We will represent EDTA by the symbol H4 Y. It is a tetraprotic acid, and the hydrogens in H4 Y refer to the four ionizable hydrogens belonging to the four carboxylic acid groups. At sufficiently low pH, the nitrogens can also be protonated and this diprotonated EDTA can be considered a hexaprotic acid. However, this occurs at a very low pH and EDTA is almost never used under such conditions. It is the unprotonated ligand Y4− that forms complexes with metal ions, that is, the protons are displaced by the metal ion upon complexation. THE CHELON EFFECT——THE MORE COMPLEXING GROUPS, THE BETTER For more discussion of the design of chelating agents, see C. N. Reilley, R. W. Schmid, and F. S. Sadek, “Chelon Approach to Analysis (I). Survey of Theory and Application,” J. Chem. Ed., 36 (1959) 555. Illustrated experiments are given in a second article in J. Chem. Ed., 36 (1959) 619.
The chelon effect is an entropy effect.
Multidentate chelating agents form stronger complexes with metal ions than do similar bidentate or monodentate ligands. This is the result of thermodynamic effects in complex formation. Chemical reactions are driven by decreasing enthalpy (liberation of heat, negative H) and by increasing entropy (increased disorder, positive S). Recall from Chapter 6, Equation 6.7, that a chemical process is spontaneous when the free energy change, G, is negative, and G = H − T S. The enthalpy change for ligands with similar groups is often similar. For example, four ammonia molecules complexed to Cu2+ and four amino groups from two ethylenediamine molecules complexed to Cu2+ will result in about the same release of heat. However, more disorder or entropy is created by the dissociation of the Cu(NH3 )4 2+ complex in which five species are formed than in the dissociation of the Cu(H2 NCH2 CH2 NH2 )2 2+ complex, in which three species are formed. Hence, S is greater for the former dissociation, creating a more negative G and a greater tendency for dissociation. Thus, multidentate complexes are more stable (have larger Kf values), largely because of the entropy effect. This is known as the chelon effect or chelate effect. It is pronounced for chelating agents such as EDTA, which have sufficient ligand atoms to occupy up to six coordination sites on metal ions. EDTA EQUILIBRIA We can represent EDTA as having four Ka values corresponding to the stepwise dissociation of the four protons1 : H4 Y H+ + H3 Y−
Ka1 = 1.0 × 10−2 =
[H+ ] [H3 Y− ] [H4 Y]
(9.5)
H3 Y− H+ + H2 Y2−
Ka2 = 2.2 × 10−3 =
[H+ ] [H2 Y2− ] [H3 Y− ]
(9.6)
H2 Y2− H+ + HY3−
Ka3 = 6.9 × 10−7 =
[H+ ] [HY3− ] [H2 Y2− ]
(9.7)
[H+ ] [Y4− ] (9.8) [HY3− ] Polyprotic acid equilibria are treated in Section 7.9, and you should review this section before proceeding with the following discussion. Figure 9.1 illustrates the fraction of each form of EDTA as a function of pH. Since the anion Y4− is the ligand species in complex formation, the complexation equilibria are affected markedly by the pH. H4 Y has a very low solubility in water, and so the disodium salt Na2 H2 Y · 2H2 O is generally used, in which two of the acid groups HY3− H+ + Y4−
Ka4 = 5.5 × 10−11 =
1 As stated before, the nitrogens on the EDTA molecule can protonate, and so there are in reality six dissociation
steps and six Ka values, the first two being ∼1.5 and 0.032. The two nitrogens are more basic than the carboxyl oxygens, and so protonate more readily. The nitrogen protonation does affect the solubility of EDTA in acid.
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1.0 H 4Y
Fraction α
0.8
H 2Y 2−
Y 4−
HY 3−
H 3Y −
0.6
0.4
Fig. 9.1.
0.2
2
4
6
8
10
12
Fraction of EDTA species as function of pH. In this treatment we ignore further protonation of the nitrogens beyond H4 Y.
14
pH
are neutralized. This salt dissociates in solution to give predominantly
H2 Y2− ; the pH + of the solution is approximately 4 to 5 (theoretically 4.4 from [H ] = Ka2 Ka3 ). The following ladder diagram shows the distribution of EDTA species at varied pH. This ladder diagram was contributed by Professor Galina Talanova, Howard University. As noted above, the nitrogens on EDTA can protonate in very acidic solution, and when considering this, we can write six Ka values, in which case the pK3 − pK6 values in the ladder diagram correspond to Ka1 − Ka4 in Equations 9.5 through 9.8 (note the protonated nitrogens occur around pH 0 and 1.5): Dominating species Y4− pK6
pH = pKa6 = 10.24 HY3−
Higher (monobasic)
pK5
pH
pK4
pH = pKa5 = 6.16 H2Y
2−
pH = pKa4 = 2.66 H3Y−
Lower (monoacidic)
pK3
pH = pKa3 = 2.0 H4Y
pK2
pH = pKa2 = 1.5 H5Y+
pK1
H6Y2+
pH = pKa1 = 0.0
FORMATION CONSTANT Consider the formation of the EDTA chelate of Ca2+ . This can be represented by: Ca2+ + Y4− CaY2−
(9.9)
The formation constant for this is Kf =
[CaY2− ] [Ca2+ ] [Y4− ]
(9.10)
The values of some representative EDTA formation constants are given in Appendix C.
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Two videos on the website entitled H4 Y alpha plot Excel 1 and H4 Y alpha plot Excel 2 illustrate two approaches for generation of this plot using Excel.
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EFFECT OF pH ON EDTA EQUILIBRIA——HOW MUCH IS PRESENT AS Y4− ? The equilibrium in Equation 9.9 is shifted to the left as the hydrogen ion concentration is increased, due to competition for the chelating anion by the hydrogen ion. The dissociation may be represented by: H+
H+
H+
H+
3− −− 2− −− − −− − CaY2− Ca2+ + Y4− − −− − − HY − −− − H2 Y − −− − H3 Y − −− − H4 Y ! CH4 Y
Note that CH4 Y = [Ca2+ ]. Or, from the overall equilibrium: Ca2+ + H4 Y CaY2− + 4H+ Video: H4 Y alpha plot Excel 1
Video: H4 Y alpha plot Excel 2
Protons compete with the metal ion for the EDTA ion. To apply Equation 9.10, we must replace [Y4− ] with α4 CH4 Y as the equilibrium concentration of Y4− .
According to Le Chˆatelier’s principle, increasing the acidity will favor the competing equilibrium, that is, the protonation of Y4− (all forms of EDTA are present in equilibrium, but some are very small; see Figure 9.1). Decreasing the acidity will favor formation of the CaY2− chelate. From a knowledge of the pH and the equilibria involved, Equation 9.10 can be used to calculate the concentration of free Ca2+ under various solution conditions (e.g., to interpret a titration curve). The Y4− concentration is calculated at different pH values as follows (see Chapter 7, polyprotic acids). If we let CH4 Y represent the total concentration of all forms of uncomplexed EDTA, then CH4 Y = [Y4− ] + [HY3− ] + [H2 Y2− ] + [H3 Y− ] + [H4 Y] and we can also write that
[Y4− ] = α4 CH4 Y
(9.11) (9.12a)
where α4 is the fraction of CH4 Y that exists as Y4− and α4 is given by (revisit Chapter 7; the discussion of how to arrive at this is in Equations 7.78 through 7.84): α4 = Ka1 Ka2 Ka3 Ka4 /Q4 = Ka1 Ka2 Ka3 Ka4 /([H+ ]4 + Ka1 [H+ ]3 +Ka1 Ka2 [H+ ]2 + Ka1 Ka2 Ka3 [H+ ] + Ka1 Ka2 Ka3 Ka4 )
(9.12b)
Similar equations can be derived for the fraction of each of the other EDTA species α0 , α1 , α2 , and α3 , as in Chapter 7. (This is the way Figure 9.1 was constructed.) We can use Equation 9.12b, then, to calculate the fraction of the EDTA that exists as Y4− at a given pH; and from a knowledge of the concentration of uncomplexed EDTA (CH4 Y ), we can calculate the free Ca2+ using Equation 9.10.
Example 9.3 Calculate the fraction of EDTA that exists as Y4− at pH 10, and from this calculate pCa in 100 mL of solution of 0.100 M Ca2+ at pH 10 after adding 100 mL of 0.100 M EDTA. Solution
Equation 7.82 states Q4 = [H+ ]4 + Ka1 [H+ ]3 + Ka1 Ka2 [H+ ]2 + Ka1 Ka2 Ka3 [H+ ] + Ka1 Ka2 Ka3 Ka4 Putting in [H+ ] = 10−10 and the Ka values as listed in Equations 9.5 through 9.8, we get Q4 = 2.35 × 10−21 The numerator in Equation 9.12b is Ka1 Ka2 Ka3 Ka4 = 8.35 × 10−22
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Thus, from Equation 9.12b, α4 = 8.35 × 10−22 /2.35 × 10−21 = 0.36 Stoichiometric amounts of Ca2+ and EDTA are added to produce an equivalent amount of CaY2− , less the amount dissociated: mmol Ca2+ = 0.100 M × 100 mL = 10.0 mmol mmol EDTA = 0.100 M × 100 mL = 10.0 mmol We have formed 10.0 mmol CaY2− in 200 mL, or 0.0500 M: Ca2+ + EDTA CaY2− x
x
0.0500 M − x ≈ 0.0500 M (since Kf is large)
where x represents the total equilibrium EDTA concentration in all forms, CH4 Y . [Y4− ], needed to apply Equation 9.10, is equal to α4 CH4 Y . Hence, we can write Equation 9.10 as: [CaY2− ] Kf = 2+ [Ca ]α4 [CH4 Y ] From Appendix C, Kf = 5.0 × 1010 . Hence, 5.0 × 1010 =
0.0500 (x)(0.36)(x)
x = 1.7 × 10−6 M pCa = 5.77
CONDITIONAL FORMATION CONSTANT——USE FOR A FIXED pH The term conditional formation constant is used under specified conditions (e.g., at a certain pH) and can be convenient for calculations. It takes into account that only a fraction of the total EDTA species are present as Y4− . We can substitute α4 CH4 Y for [Y4− ] in Equation 9.10: [CaY2− ] (9.13) Kf = [Ca2+ ]α4 CH4 Y We can then rearrange the equation to yield Kf α4 = Kf =
[CaY2− ] [Ca2+ ]CH4 Y
(9.14)
where Kf is the conditional formation constant and is dependent on α4 and, hence, the pH. We can use this equation to calculate the equilibrium concentrations of the different species at a given pH in place of Equation 9.10.
Example 9.4 The formation constant for CaY2− is 5.0 × 1010 . At pH 10, α4 is calculated (Example 9.3) to be 0.36 to give a conditional constant (from Equation 9.14) of 1.8 × 1010 . Calculate pCa in 100 mL of a solution of 0.100 M Ca2+ at pH 10 after addition of (a) 0 mL, (b) 50 mL, (c) 100 mL, and (d) 150 mL of 0.100 M EDTA.
The conditional formation constant applies for a specified pH.
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330 Solution
(a) pCa = −log[Ca2+ ] = − log ×1.00 × 10−1 = 1.00 (b) We started with 0.100 M × 100 mL = 10.0 mmol Ca2+ . The millimoles of EDTA added are 0.100 M × 50 mL = 5.0 mmol. In general, a rule of thumb for a binary association reaction (A + B C) is that the limiting reagent can be assumed to have reacted quantitatively (>99.9%) if the effective equilibrium constant is greater than ∼106 . The conditional formation constant in this case is larger, so the reaction 9.9 will be far to the right. Hence, we can neglect the amount of Ca2+ from the dissociation of CaY2− and the number of millimoles of free Ca2+ is essentially equal to the number of unreacted millimoles: mmol Ca2+ = 10.0 − 5.0 = 5.0 mmol [Ca2+ ] = 5.0 mmol/150 mL = 0.033 M pCa = − log 3.3 × 10−2 = 1.48 (c) At the equivalence point, we have converted all the Ca2+ to CaY2− . We must, therefore, use Equation 9.14 to calculate the equilibrium concentration of Ca2+ . The number of millimoles of CaY2− formed is equal to the number of millimoles of Ca2+ started with, and [CaY2− ] = 10.0 mmol/200 mL = 0.0500 M. From the dissociation of CaY2− , [Ca2+ ] = CH4 Y = x, and the equilibrium [CaY2− ] = 0.050 M − x. But since the dissociation is slight, we can neglect x compared to 0.050 M. Therefore, from Equation 9.14, 0.050 = 1.8 × 1010 (x)(x) x = 1.7 × 10−6 M = [Ca2+ ] pCa = − log 1.7 × 10−6 = 5.77 Compare this value with that calculated using Kf in Example 9.3, instead of Kf . (d) The concentration CH4 Y is equal to the concentration of excess EDTA added (neglecting the dissociation of CaY2− , which will be even smaller in the presence of excess EDTA). The millimoles of CaY2− will be as in (c). Hence, [CaY2− ] = 10.0 mmol/250 mL = 0.0400 M mmol excess CH4 Y = 0.100 M × 150 mL − 0.100 M × 100 mL = 5.0 mmol CH4 Y = 5.0 mmol/250 mL = 0.020 M 0.040 = 1.8 × 1010 [Ca ](0.020) 2+
[Ca2+ ] = 1.1 × 10−10 M pCa = − log 1.1 × 10−10 = 9.95
The pH can affect stability of the complex (i.e., Kf ) by affecting not only the form of the EDTA but also that of the metal ion. For example, hydroxo species may form (M2+ + OH− → MOH+ ). That is, OH− competes for the metal ion just as H+ competes for the Y4− . Figure 9.2 (prepared from a spreadsheet—see Problem 23) shows how Kf changes with pH for three metal–EDTA chelates with moderate (Ca) to strong (Hg) formation constants. The calcium chelate is obviously too weak to be titrated in acid solution (Kf < 1), while the mercury chelate is strong enough to be
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log K′f vs. pH for EDTA chelates 25.0
Kf = 6.3 × 1021
20.0
Kf = 1.1 × 1018
HgY2−
15.0 Kf = 5 × 1010
PbY2−
log Kf~
10.0
5.0 CaY2− 0.0 0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
−5.0
−10.0
Fig. 9.2.
Effect of pH on Kf values for EDTA chelates.
−15.0 pH
titrated in acid. At pH 13, all Kf values are virtually equal to the Kf values because α4 is essentially unity; that is, the EDTA is completely dissociated to Y4− . The curves are essentially parallel to one another because at each pH, each Kf is multiplied by the same α4 value to obtain Kf .
9.3 Metal–EDTA Titration Curves A titration is performed by adding the chelating agent to the sample; the reaction occurs as in Equation 9.9. Figure 9.3 shows the titration curve for Ca2+ titrated with EDTA at pH 10. Before the equivalence point, the Ca2+ concentration is nearly equal to the amount of unchelated (unreacted) calcium since the dissociation of the chelate is slight (analogous to the amount of an unprecipitated ion). At the equivalence point and beyond, pCa is determined from the dissociation of the chelate at the given pH as described in Example 9.3 or 9.4, using Kf or Kf . The effect of pH on the titration is apparent from the curve in Figure 9.3 for titration at pH 7. 12 pH 10
10
8 pCa
pH 7 6
4
2
Fig. 9.3. 0
20
40
60 80 mL EDTA
100
120
140
Titration curves for 100 mL 0.1 M Ca2+ versus 0.1 M Na2 EDTA at pH 7 and pH 10.
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You will recall that in pH titrations, it is simpler to calculate how much titrant is needed to get to a specified pH than to calculate the reverse, i.e., what is the pH if a certain amount of titrant is added. Revisit Section 8.11 and Problem 8.52 how to handle these. The same is true for metal ion (M)—EDTA (L) titrations. As is commonly the case, we can assume that the titration is being conducted in a buffered medium of constant pH in which the conditional formation constant is Kf . We have taken VM mL of a metal of analytical concentration CM to titrate. At any given point we have added VL mL of titrant L of concentration CL and we wish to calculate for a given amount of free metal ion concentration (that we express as pM, as in the case of hydrogen ion concentrations being expressed as pH). In terms of pM = − log[M], what is VL ? Based on mass balance approaches, it is readily shown (see Section 9.3’s website supplement) that VL =
VM (C)M − [M])(1 + Kf [M]) CL (Kf + [M]) + [M] (1 + Kf [M])
(9.15)
We calculate [M] as 10−pM and then calculate VL . To generate the data for Figure 9.3,we took Kf to be 5.01 × 10−10 (Appendix C, Table C4) and calculate α4 to be 0.355 and 4.80 × 10−4 at pH 10 and 7, respectively, resulting in Kf values of 1.8 × 1010 and 2.4 × 107 . The calculation is illustrated in the spreadsheet Figure 9.3a.xlsx in the text website for pH 7 and 10. It is interesting to note, however, that unlike acid–base titrations where ionization of water can complicate matters, in the present case there are mass balance equations for both M-containing species and L-containing species. This results in a quadratic equation as shown below that can be explicitly solved. Consider Equation 9.14. For a generic metal ion M and a generic ligand L we can write Kf =
[ML] [M] [LT ]
Where LT indicates all L-containing species other than ML. Let us denote the total volume at any point, VM + VL (both in units of mL), as VT ; CM VM is the original mmol metal taken and CL VL is the total mmol of ligand taken at any point. Mass balance for M requires that the original amount of metal taken, less the amount present as ML, divided by VT must equal [M]: [M] = (CM VM − [ML]VT )/VT A similar mass balance for L yields: [LT ] = (CL VL − [ML]VT )/VT A combination of the three equations above gives Kf (CM VM − [ML]VT )(CL VL − [ML]VT ) = [ML]VT 2 This is a quadratic equation in [ML], where, in the usual notations, a = Kf VT 2 , b = −VT (Kf (CM VM + CL VL ) + VT ) and c = Kf CM VM CL VL
Only some metal chelates are stable enough to allow titrations in acid solution; others require alkaline solution.
Generation of Figure 9.3 using the explicit equation above is illustrated in Figure 9.3b.xlsx in the website for the book. Note that of the two solutions to a quadratic equation, the negative root is meaningful in this problem; else one ends up with negative concentrations. The more stable the chelate (the larger Kf ), the farther to the right will be the equilibrium of the reaction (Equation 9.9), and the larger will be the end-point break. Also, the more stable the chelate, the lower the pH at which the titration can be performed (Figure 9.2). This is important because it allows the titration of some
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metals in the presence of others whose EDTA chelates are too weak to titrate at the lower pH. Figure 9.4 shows the minimum pH at which different metals can be titrated with EDTA. The points on the curve represent the pH at which the conditional formation constant Kf for each metal is 106 (log Kf = 6), which was arbitrarily chosen as the minimum needed for a sharp end point. Note that the smaller the Kf , the more alkaline the solution must be to obtain a Kf of 106 (i.e., the larger α4 must be). Thus, Ca2+ with Kf only about 1010 requires a pH of ∼≥ 8. The dashed lines in the figure divide the metals into separate groups according to their formation constants. One group is titrated in a highly acidic (pH < ∼3) solution, a second group at pH ∼3 to 7, and a third group at pH > 7. At the highest pH range, all the metals will react, but not all can be titrated directly due to precipitation of hydroxides. For example, titration of Fe3+ or Th4+ is not possible without the use of back-titration or auxiliary complexing agents to prevent hydrolysis. At the intermediate pH range, the third group will not titrate, and the second group of metals can be titrated in the presence of the third group. And finally, in the most acidic pH range, only the first group will titrate and can be determined in the presence of the others. Masking can be achieved by precipitation, by complex formation, by oxidation– reduction, and kinetically. A combination of these techniques is often used. For example, Cu2+ can be masked by reduction to Cu(I) with ascorbic acid and by complexation with I− . Lead can be precipitated with sulfate when bismuth is to be titrated. Most masking is accomplished by selectively forming a stable, soluble complex. Hydroxide ion complexes aluminum ion [Al(OH)4 − or AlO2 − ] so calcium can be titrated. Fluoride masks Sn(IV) in the titration of Sn(II). Ammonia complexes 28
Fe 3+
26
In 3+ Th 4+
24
Hg 2+
Sc 3+
1
22
log K MY n −4
Ga 3+ Lu 3+
20
VO 2+ Cu 2+ Zn 2+ Al 3+
Ni 2+ 18 Y 3+ Pb 2+ Sm 3+
16
La 3+
Cd 2+
Fe 2+
Co 2+
2
Mn 2+
14
12 Ca 2+ 10
Sr 2+ Mg 2+
Fig. 9.4.
3
8 0
2
4
6
8 pH
10
12
14
Minimum pH for effective titration of various metal ions with EDTA. (Reprinted with permission from C. N. Reilley and R. W. Schmid, Anal. Chem., 30 (1958) 947. Copyright by the American Chemical Society.)
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copper so it is difficult to titrate Cu(II) with EDTA in an ammoniacal buffer. Other metals can be titrated in the presence of Cr(III) because its EDTA chelate, although very stable, forms only slowly.
9.4 Detection of the End Point: Indicators—They Are Also Chelating Agents
O
+
N O−
O S O−
N N HO
O
OH
Na+
We can measure the pM potentiometrically if a suitable electrode is available, for example, an ion-selective electrode (see Chapter 13), but it is simpler if an indicator can be used. Indicators used for complexometric titrations are themselves chelating agents. They are usually dyes of the o, o -dihydroxy azo type. Eriochrome Black T is a typical indicator. It contains three ionizable protons, so we will represent it by H3 In. This indicator can be used for the titration of Mg2+ with EDTA. A small amount of indicator is added to the sample solution, and it forms a red complex with part of the Mg2+ ; the color of the uncomplexed indicator is blue. As soon as all the free Mg2+ is titrated, the EDTA displaces the indicator from the magnesium, causing a change in the color from red to blue: MgIn− + H2 Y2− → MgY2− + HIn2− + H+ (red)
(colorless)
(colorless)
(blue)
(9.16)
Eriochrome Black T
Water hardness is expressed as ppm CaCO3 and represents the sum of calcium and magnesium.
This will occur over a range of pMg values, and the change will be sharper if the indicator is kept as dilute as possible but is still sufficient to give a good color change. Of course, the metal–indicator complex must be less stable than the metal–EDTA complex, or else the EDTA will not displace it from the metal. On the other hand, it must not be too weak, or the EDTA will start replacing it at the beginning of the titration, and a diffuse end point will result. In general, the Kf for the metal–indicator complex should be 10 to 100 times less than that for the metal–titrant complex. The formation constants of the EDTA complexes of calcium and magnesium are too close to differentiate between them in an EDTA titration, even by adjusting pH (see Figure 9.4). So they will titrate together, and the Eriochrome Black T end point can be used as above. This titration is used to determine total hardness of water, (Ca2+ plus Mg2+ —see Experiment 11 on the text website). Eriochrome Black T cannot be used to indicate the endpoint of a direct EDTA titration of calcium in the absence of magnesium, however, because the indicator forms too weak a complex with calcium to give a sharp end point. Therefore, a small measured amount of Mg2+ is added to the Ca2+ solution; as soon as the Ca2+ and the small amount of free Mg2+ are titrated, the end-point color change occurs as above. (The Ca2+ titrates first since its EDTA chelate is more stable.) A correction is made for the amount of EDTA used for titration of the Mg2+ by performing a “blank” titration on the same amount of Mg2+ added to the buffer. It is more convenient to add, instead, about 2 mL of 0.005 M Mg–EDTA rather than MgCl2 . This is prepared by adding together equal volumes of 0.01 M MgCl2 and 0.01 M EDTA solutions and adjusting the ratio with dropwise additions until a portion of the reagent turns a dull violet when treated with pH 10 buffer and Eriochrome Black T indicator. When this occurs, one drop of 0.01 M EDTA will turn the solution blue, and one drop of 0.01 M MgCl2 will turn it red. If we add Mg–EDTA to the sample, the Ca2+ in the sample displaces the EDTA from the Mg2+ (since the Ca–EDTA is more stable) so that the Mg2+ is free to react with the indicator. At the end point, an equivalent amount of EDTA displaces the indicator from the Mg2+ , causing the color change, and no correction is required for the added Mg–EDTA. This procedure is used in Experiment 11 described in the text website.
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An alternative method is to add a small amount of Mg2+ to the EDTA solution. This immediately reacts with EDTA to form MgY2− with very little free Mg2+ in equilibrium. This, in effect, reduces the molarity of the EDTA. So the EDTA is standardized after adding the Mg2+ by titrating primary standard calcium carbonate (dissolved in HCl and pH adjusted). When the indicator is added to the calcium solution, it is pale red. But as soon as the titration is started, the indicator is complexed by the magnesium and turns wine red. At the end point, it changes to blue, as the indicator is displaced from the magnesium. No correction is required for the Mg2+ added because it is accounted for in the standardization. This solution should not be used to titrate metals other than calcium. High-purity EDTA can be prepared from Na2 H2 Y · 2H2 O by drying at 80◦ C for 2 h. The waters of hydration remain intact; this can be used as a primary standard for preparing a standard EDTA solution. The titration of calcium and magnesium with EDTA is done at pH 10, using an ammonia–ammonium chloride buffer. The pH must not be too high or the metal hydroxide may precipitate, causing the reaction with EDTA to be very slow. Calcium can actually be titrated in the presence of magnesium by raising the pH to 12 with strong alkali; Mg(OH)2 precipitates and does not titrate. Since Eriochrome Black T and other indicators are weak acids, their colors will depend on the pH because their ionized species have different colors. For example, with Eriochrome Black T, H2 In− is red (pH < 6), HIn2− is blue (pH 6 to 12), and In3− is yellow orange (pH > 12). Thus, indicators can be used over definite pH ranges. It should be emphasized, though, that although complexometric indidators respond to pH, their mechanism of action does not involve changes in pH, as the solution is buffered. But the pH affects the stability of the complex formed between the indicator and the metal ion, as well as that formed between EDTA and the metal ion. An indicator is useful for indication of titrations of only those metals that form a more stable complex with the titrant than with the indicator at the given pH. This may sound complex but suitable indicators are known for many titrations with several different chelating agents. Calmagite gives a somewhat improved end point over Eriochrome Black T for the titration of calcium and magnesium with EDTA. It also has a longer shelf life. Xylenol orange is useful for titration of metal ions that form very strong EDTA complexes and are titrated at pH 1.5 to 3.0. Examples are the direct titration of thorium(IV) and bismuth(III), and the indirect determination of zirconium(IV) and iron(III) by back-titration with one of the former two metals. There are many other indicators for EDTA titrations. The work of Flaschka and Barnard (Reference 6) gives many examples of EDTA titrations, and you are referred to this excellent source for detailed descriptions of procedures for different metals. There are a number of other useful reagents for complexometric titrations. A notable example is ethyleneglycol bis (β-aminoethyl ether)-N, N, N , N -tetraacetic acid (EGTA). This is an ether analog of EDTA that will selectively titrate calcium in the presence of magnesium:
HOOCCH2
OOCCH2
H H NCH2CH2OCH2CH2OCH2CH2N
CH2COOH
CH2COO
Complexing agents having ether linkages have a strong tendency to complex the alkaline earths heavier than magnesium. Log Kf for calcium–EGTA is 11.0, while that for magnesium–EGTA is only 5.2. For other chelating agents and their applications, see References 4 and 6.
HO
335
O OH
S
O N
N
HO
Calmagite
O OH
HO O O S O
N
O OH
OH O
N HO
OH O
Xylenol orange
EGTA allows the titration of calcium in the presence of magnesium.
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N
N
NH NH
With the exception of the alkali metals, nearly every metal can be determined with high precision and accuracy by complexometric titration. These methods are much more rapid and convenient than gravimetric procedures and are therefore preferred, except in those few instances when greater accuracy is required. In more recent years, however, all of these metal determination methods are giving way to atomic and mass spectrometric measurements (see Chapter 17). Complexometric titrations in the clinical laboratory are limited to those substances that occur in fairly high concentrations since volumetric methods are generally not too sensitive. The most important complexometric titration is the determination of calcium in blood. Chelating agents such as EDTA are used in the treatment of heavy-metal poisoning, for example, when children ingest chipped paint that contains lead. The calcium chelate (as Na2 CaY) is administered to prevent complexation and use of Ca-EDTA, rather than Na2 EDTA, prevents leaching of calcium in the bones. Heavy metals such as lead form more stable EDTA chelates than calcium does and will displace the calcium from the EDTA. The chelated lead is then excreted via the kidneys. A table of formation constants of some EDTA chelates appears in Appendix C, Table C4.
S Dithizone
Chelates are used in gravimetry, spectrophotometry, fluorometry, solvent extraction, and chromatography. HO N N OH Dimethylglyoxime
9.5 Other Uses of Complexes Analytical chemists can use the formation of complexes to advantage in ways other than in titrations. For example, metal ion chelates may be formed and extracted into a water-immiscible solvent for separation by solvent extraction. Complexes of metal ions with the chelating agent dithizone, for example, are useful for extractions. The chelates are often highly colored. Their formation can then serve as the basis for spectrophotometric or atomic spectroscopic determination of metal ions. Complexes that fluoresce may also be formed. Metal chelates may sometimes precipitate. The nickel–dimethylglyoxime precipitate is an example used in gravimetric analysis. Table 10.2 lists several other metal chelate precipitates. Complexation equilibria may influence chromatographic separations, and we have mentioned the use of complexing agents as masking agents to prevent interfering reactions. For example, in the solvent extraction of vanadium with the chelating agent oxine (8-quinolinol—see Chapter 10), the extraction of copper is avoided by chelating it with EDTA, thus preventing the formation of its oxine chelate. Many metal chelates are intensely colored. It is common practice today to separate metal ions by chromatography and then introduce a chromogenic (meaning color-forming) chelating ligand that reacts with the different metals nonselectively; the product is then spectrophotometrically detected. The lack of selectivity in this case is a virtue, as the same chelating agent can be used to detect a variety of metals that have already been separated. All of these complexing reactions are pH dependent, and pH adjustment and control (with buffers) are always necessary to optimize the desired reaction or to minimize undesired reactions.
9.6 Cumulative Formation Constants β and Concentrations of Specific Species in Stepwise Formed Complexes While EDTA reacts with metals on a 1:1 basis, many ligands, especially those that have a more limited number of binding sites, will react with a metal ion in a stepwise fashion, one ligand being added at a time. Ammonia complexes with the nickel ion
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337
Ni2+ , for example, in 6 steps, forming ultimately Ni(NH3 )6 2+ . The stepwise formation constants for a metal reacting with a ligand L can be written as: M + L ML, ML + L ML2 ,
[ML]/[M][L] = Kf 1 [ML2 ]/[ML][L] = Kf 2
and so on to MLn−1 + L MLn ,
[MLn ]/[MLn−1 ][L] = Kfn
If you compare with a polyprotic acid Hn A, you will notice that H is analogous to L and likewise M to A. But we are not only writing these equilibria as associations, rather than dissociations, we also have the order of the stepwise equilibria reversed. For the dissociation of the acid, we would write as the first step the dissociation of Hn A to Hn−1 A− and H+ and designate that dissociation constant as Ka1 . So, 1/Kfn corresponds to Ka1 , 1/Kfn−1 to Ka2 , and so on until 1/Kf 1 , which corresponds to Kan . In dealing with association equilibria, the cumulative constants, designated as β, are also often used. The cumulative constant for formation of MLn is designated βn and it is simply the product of the Kf values until that point, thus: βn = Kf 1 Kf 2 Kf 3 . . . Kfn
(9.17)
For example, for the cumulative formation of ML3 , the equilibrium and the corresponding constant will be written as M + 3L ML3
[ML3 ]/[M][L]3 = Kf 1 Kf 2 Kf 3 = β3
Note that β1 is the same as Kf 1 . Although β0 has no physical meaning, for mathematical convenience (the reason why will be apparent in the next section), β0 is taken to be 1. In a metal-ligand system containing M, L, and various MLn species, the sum of the concentrations of all the metal species, including the free metal, often called the analytical concentration of the metal, is typically designated by CM and is given by CM = [M] + [ML] + [ML2 ] + [ML3 ] + · · · · + [MLn ] This is readily expanded to CM = [M] + β1 [M] [L] + β2 [M][L]2 + β3 [M][L]3 + · · · · + βn [M] [L]n CM = [M] (1 + β1 [L] + β2 [L]2 + β3 [L]3 + · · · · + βn [L]n )
(9.18)
Should we consider β0 to be 1, Equation 9.18 can be written in a more compact form i=n βi L i CM = [M]
(9.19)
i=0
One fraction, that of the free metal ion, denoted αM , is of frequent interest, and it will be readily realized from Equation 9.18 that −1 i=n i βi L (9.20) αM = [M]/CM = i=0
Since [M] can thus also be expressed as [M] = CM /
i=n
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βi L
i
(9.21)
i=0
the concentration of any other species such as MLi can also be readily calculated as βi [M]Li . Illustrative examples follow.
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Example 9.5 Given that Cu2+ forms a tetrammine complex with log βI values of 3.99, 7.33, 10.06, and 12.03 for i = 1 − 4, respectively, and given that from Table C4 the formation constant for the Cu2+ −EDTA complex is 6.30 × 1018 , calculate the conditional formation constant for the Cu2+ −EDTA complex in an NH3 − NH4 Cl buffer, which has a pH of 10 and [NH3 ] = 1.0 M. Comment on the feasibility of an EDTA titration of Cu2+ under these conditions. Solution
The conditional formation constant depends not only on what fraction of the total EDTA exists as Y4− (we have already previously calculated that αY4− = 0.35 at pH 10), it also depends on what fraction of the metal, αCu , will exist as the free Cu2+ not complexed by ammonia (EDTA complexation is not taken into this account). In other words, the conditional formation constant can be written as: Kf = Hence
Kf [CuY2− ] [CuY2− ] 1 = = αCu αY4− CCu CH4 Y αCu αY4 [Cu2+ ][Y4− ] Kf = αCu αY4− Kf
We calculate αCu from Equation 9.20: αCu = (1 + 103.99 × 1.0 + 107.33 × 1.02 + 1010.06 × 1.03 + 1012.03 × 1.04 )−1 = (1 + 9770 + 2.14 × 107 + 1.15 × 1010 + 1.07 × 1012 )−1 = 9.26 × 10−13 Therefore Kf = 9.26 × 10−13 × 0.35 × 6.3 × 1018 = 1.2 × 106 The minimum conditional formation constant at which a titration is feasible is 106 . So it will be just marginally possible to do a titration under these conditions.
Example 9.6 In 100 mL of a 0.010 M NH3 solution, 11.9 mg of NiCl2 · 6H2 O was dissolved. The log βi values for NH3 complexation of Ni2+ are 2.67, 4.79, 6.40, 7.47, 8.10, 8.01 for i = 1 − 6, respectively. Compute the concentrations of the various Ni(NH3 )i 2+ species. Solution
The FW of NiCl2 · 6H2 O is 237.7 g/mol, 11.9 mg is 11.9/238 = 0.0500 mmol, dissolved in 0.1 L, CNi = 5.00 × 10−4 M. A simplistic solution assumes that the equilibrium free NH3 concentration (L) is 1.0 × 10−2 M; hence, we can calculate from Equation 9.19. (Note the shortened algebraic notation, e.g., for β1 L1 : β1 L1 = 102.67 × 10−2 = 10(2.67−2) ): αNi = (1 + 10(2.67−2) + 10(4.79−4) + 10(6.40−6) + 10(7.47−8) + 10(8.10−10) + 10(8.01−12) )−1 = (1 + 4.68 + 6.17 + 2.51 + 0.30 + 0.01 + 1.00 × 10−4 )−1 = 6.8 × 10−2
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PROBLEMS
339
[Ni2+ ] = CNi αNi = 5.00 × 10−4 × 6.8 × 10−2 = 3.4 × 10−5 M [Ni(NH3 )2+ ] = [Ni2+ ]β1 L = 3.4 × 10−5 × 4.68 = 1.6 × 10−4 M [Ni(NH3 )2 2+ ] = [Ni2+ ]β2 L2 = 3.4 × 10−5 × 6.17 = 2.1 × 10−4 M [Ni(NH3 )3 2+ ] = [Ni2+ ]β3 L3 = 3.4 × 10−5 × 2.51 = 8.5 × 10−5 M [Ni(NH3 )4 2+ ] = [Ni2+ ]β4 L4 = 3.4 × 10−5 × 0.30 = 1.0 × 10−5 M [Ni(NH3 )5 2+ ] = [Ni2+ ]β5 L5 = 3.4 × 10−5 × 0.01 = 3.4 × 10−7 M [Ni(NH3 )6 2+ ] = [Ni2+ ]β6 L6 = 3.4 × 10−5 × 1.0 × 10−4 = 3.4 × 10−9 M This solution is unfortunately only approximate because it ignores the consumption of NH3 in forming the Ni(NH3 )n complexes. 0.16 mM Ni(NH3 )2+ , 0.21 mM Ni(NH3 )2 2+ 0.085 mM Ni(NH3 )3 2+ and 0.01 mM Ni(NH3 )4 2+ respectively ties up 0.16, 0.42, 0.26, and 0.04 mM NH3 , totaling 0.88 mM NH3 , when the original total NH3 present was 10 mM, this consumption is not insignificant. We can solve this problem without such errors by Goal Seek. We set up the problem by first expressing αM and [M] in the usual manner and then the total Ligand concentration (LT = 0.010) as the sum of all ligand containing terms: LT − ([L] + [ML] + 2∗ [ML2 ] + 3∗ [ML3 ] + 4∗ [ML4 ] + 5∗ [ML5 ] + 6∗ [ML6 ]) = 0 The result is multiplied by 1010 and Goal Seek is invoked to make this result a target of zero by changing the value of L. The Excel spreadsheet example is given in the website supplement as Example 9.6.xlsx. The result for [Ni2+ ] is 3.97 × 10−5 M. And additional check for the metal, CM − ([L] + [ML] + [ML2 ] + [ML3 ] + [ML4 ] + [ML5 ] + [ML6 ]) = 0 can be put in to verify that this result also is satisfied. See the text’s website for a video of solving this problem using Goal Seek, which gives the same answer 3.97 × 10−5 M for [Ni2+ ].
Video: Example 9.6
Questions 1. Distinguish between a complexing agent and a chelating agent. 2. Explain the principles of indicators used in chelometric titrations. 3. Why is a small amount of magnesium salt added to the EDTA solution used for the titration of calcium with an Eriochrome Black T indicator? 4. Why would it be difficult to titrate Cu2+ with EDTA in a strongly ammoniacal medium?
Problems COMPLEX EQUILIBRIUM CALCULATIONS (Kf ) 5. Calcium ion forms a weak 1:1 complex with nitrate ion with a formation constant of 2.0. What would be the equilibrium concentrations of Ca2+ and Ca(NO3 )+ in a solution prepared by adding 10 mL each of 0.010 M CaCl2 and 2.0 M NaNO3 ? Neglect diverse ion effects. 6. The formation constant of the silver–ethylenediamine complex, Ag(NH2 CH2 CH2 NH2 )+ , is 5.0 × 104 . Calculate the concentration of Ag+ in equilibrium with a 0.10 M solution of the complex. (Assume no higher order complexes.)
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7. What would be the concentration of Ag+ in Problem 6 if the solution contained also 0.10 M ethylenediamine, NH2 CH2 CH2 NH2 ? 8. Silver ion forms stepwise complexes with thiosulfate ion, S2 O3 2− , with Kf 1 = 6.6 × 108 and Kf 2 = 4.4 × 104 . Calculate the equilibrium concentrations of all silver species for 0.0100 M AgNO3 in 1.00 M Na2 S2 O3 . Neglect diverse ion effects.
CONDITIONAL FORMATION CONSTANTS 9. The formation constant for the lead–EDTA chelate (PbY2− ) is 1.10 × 1018 . Calculate the conditional formation constant (a) at pH 3 and (b) at pH 10. 10. Using the conditional constants calculated in Problem 9 calculate the pPb (−log[Pb2+ ]) for 50.0 mL of a solution of 0.0250 M Pb2+ (a) at pH 3 and (b) at pH 10 after the addition of (1) 0 mL, (2) 50 mL, (3) 125 mL, and (4) 200 mL of 0.0100 M EDTA. 11. The conditional formation constant for the calcium–EDTA chelate was calculated for pH 10 in Example 9.4 to be 1.8 × 1010 . Calculate the conditional formation constant at pH 3. Compare this with that calculated for lead at pH 3 in Problem 9. Could lead be titrated with EDTA at pH 3 in the presence of calcium? 12. Calculate the conditional formation constant for the nickel–EDTA chelate in an ammoniacal pH 10 buffer containing [NH3 ] = 0.10 M. See Example 9.6 for the relevant complexation constants with ammonia.
STANDARD SOLUTIONS 13. Calculate the weight of Na2 H2 Y · 2H2 O required to prepare 500.0 mL of 0.05000 M EDTA. 14. An EDTA solution is standardized against high-purity CaCO3 by dissolving 0.3982 g CaCO3 in hydrochloric acid, adjusting the pH to 10 with ammoniacal buffer, and titrating. If 38.26 mL was required for the titration, what is the molarity of the EDTA? 15. Calculate the titer of 0.1000 M EDTA in mg CaCO3 /mL. 16. If 100.0 mL of a water sample is titrated with 0.01000 M EDTA, what is the titer of the EDTA in terms of water hardness/mL?
QUANTITATIVE COMPLEXOMETRIC DETERMINATIONS 17. Calcium in powdered milk is determined by dry ashing (see Chapter 1) a 1.50-g sample and then titrating the calcium with EDTA solution, 12.1 mL being required. The EDTA was standardized by titrating 10.0 mL of a zinc solution prepared by dissolving 0.632 g zinc metal in acid and diluting to 1 L (10.8 mL EDTA required for titration). What is the concentration of calcium in the powdered milk in parts per million? 18. Calcium is determined in serum by microtitration with EDTA. A 100-μL sample is treated with two drops of 2 M KOH, Cal-Red indicator is added, and the titration is performed with 0.00122 M EDTA, using a microburet. If 0.203 mL EDTA is required for titration, what is the level of calcium in the serum in mg/dL and in meq/L? 19. In the Liebig titration of cyanide ion, a soluble complex is formed; at the equivalence point, solid silver cyanide is formed, signaling the end point: 2CN− + Ag+ → Ag(CN)2 − −
(titration)
+
Ag(CN)2 + Ag → Ag[Ag(CN)2 ] (end point) A 0.4723-g sample of KCN was titrated with 0.1025 M AgNO3 , requiring 34.95 mL. What is the percent purity of the KCN? 20. Copper in saltwater near the discharge of a sewage treatment plant is determined by first separating and concentrating it by solvent extraction of its dithizone chelate at pH 3 into methylene chloride and then evaporating the solvent, ashing the chelate to destroy the organic portion, and titrating the copper with EDTA. Three 1-L portions of the sample are each extracted with 25-mL portions of methylene chloride, and the extracts are combined in a 100-mL volumetric flask and diluted to volume. A 50-mL aliquot is evaporated, ashed, and titrated. If the EDTA solution has a CaCO3 titer of 2.69 mg/mL and 2.67 mL is required for titration of the copper, what is the concentration of copper in the seawater in parts per million?
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21. Chloride in serum is determined by titration with Hg(NO3 )2 ; 2Cl− + Hg2+ HgCl2 . The Hg(NO3 )2 is standardized by titrating 2.00 mL of a 0.0108 M NaCl solution, requiring 1.12 mL to reach the diphenylcarbazone end point. A 0.500-mL serum sample is treated with 3.50 mL water, 0.50 mL 10% sodium tungstate solution, and 0.50 mL of 0.33 M H2 SO4 solution to precipitate proteins. After the proteins are precipitated, the sample is filtered through a dry filter into a dry flask. A 2.00-mL aliquot of the filtrate is titrated with the Hg(NO3 )2 solution, requiring 1.23 mL. Calculate the mg/L chloride in the serum. (Note: mercury is rarely used today due to its toxicity. The problem is illustrative.)
PROFESSOR’S FAVORITE PROBLEM Contributed by Professor Bin Wang, Marshall University 22. A 0.1021 g sample containing ZnO was titrated using a standard EDTA solution with Erichrome Black T as indicator. It took 25.52 mL of 0.0100 M EDTA to reach the end point. What is the percentage of ZnO in the sample?
SPREADSHEET PROBLEMS See the textbook website, Chapter 9, for suggested setups. 23. Prepare a spreadsheet for Figure 9.2, log Kf vs. pH for the EDTA chelates of calcium, lead, and mercury. This will require calculating α4 for EDTA and the Kf values for the chelates of calcium, lead, and mercury. Calculate at 0.5 pH intervals. Compare your plot with Figure 9.2. 24. Prepare a spreadsheet for the titration of 100.00 mL of 0.1000 M Hg2+ with 0.1000 M Na2 EDTA at pH 10 (similar to pCa vs. mL EDTA—Figure 9.3). Start out at pHg = 1 and go up to pHg = 17 with a resolution of 0.1. See Figure 9.3.xlsx. (See the text website for the solution.) 25. Prepare a spreadsheet to plot the seven fractional values (from αM to αML6 ) present in a Ni2+ − NH3 system as a function of [NH3 ]. Plot the results from 0.001 to 1.0 M NH3 , with a resolution of 0.001 M, use logarithmic scaling for the X-axis (ammonia concentration).
Recommended References 1. Stability Constants of Metal-Ion Complexes. Part A: Inorganic Ligands, E. Hogfeldt, ed., Part B: Organic Ligands, D. D. Perrin, ed. Oxford: Pergamon, 1979, 1981. 2. A. Martell and R. J. Motekaitis, The Determination and Use of Stability Constants. New York: VCH, 1989. 3. J. Kragten, Atlas of Metal-Ligand Equilibria in Aqueous Solution. London: Ellis Horwood, 1978. 4. G. Schwarzenbach, Complexometric Titrations. New York: Interscience, 1957. 5. H. Flaschka, EDTA Titrations. New York: Pergamon, 1959. 6. H. A. Flaschka and A. J. Barnard, Jr., “Titrations with EDTA and Related Compounds,” in C. L. Wilson and D. W. Wilson, eds. Comprehensive Analytical Chemistry, Vol. 1B. New York: Elsevier, 1960. 7. F. J. Welcher, The Analytical Uses of Ethylenediaminetetraacetic Acid. Princeton: Van Nostrand, 1958. 8. J. Stary, ed., Critical Evaluation of Equilibrium Constants Involving 8-Hydroxyquinoline and Its Metal Chelates. Oxford: Pergamon, 1979. 9. H. A. Flaschka, Chelates in Analytical Chemistry, Vols. 1–5. New York: Dekker, 1967–1976.
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Chapter Ten GRAVIMETRIC ANALYSIS AND PRECIPITATION EQUILIBRIA “Some loads are light, some heavy. Some people prefer the light to the heavy . . . ” —Mao Tse-tung
Learning Objectives WHAT ARE SOME OF THE KEY THINGS WE WILL LEARN FROM THIS CHAPTER? ●
Steps of a gravimetric analysis: precipitation, digestion, filtration, washing, drying, weighing, calculation, p. 343
●
Gravimetric calculations (key equations: 10.1, 10.3, 10.5), pp. 349, 350
Gravimetry is among the most accurate analytical techniques (but it is tedious!). T. W. Richards used it to determine atomic weights. He received the Nobel Prize in 1914 for his work. See Z. Anorg. Chem., 8 (1895), 413, 419, and 421 for some of his careful studies on contamination. See also http://nobelprizes.com.
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●
The solubility product, the common ion effect, p. 355
●
The diverse ion effect (key equation: 10.10), p. 361
Gravimetric analysis is one of the most accurate and precise methods of macroquantitative analysis. In this process the analyte is selectively converted to an insoluble form. The separated precipitate is dried or ignited, possibly to another form, and is accurately weighed. From the weight of the precipitate and a knowledge of its chemical composition, we can calculate the weight of analyte in the desired form. Gravimetric analysis is capable of exceedingly precise analysis. In fact, gravimetric analysis was used to determine the atomic masses of many elements to six figure accuracy. Theodore W. Richards at Harvard University developed highly precise and accurate gravimetric analysis of silver and chlorine. He used these methods to determine the atomic weights of 25 elements by preparing pure samples of the chlorides of the elements, decomposing known weights of the compounds, and determining the chloride content by gravimetric methods. For this work, he was the first American to receive the Nobel Prize. He was the ultimate analytical chemist! Gravimetry does not require a series of standards for calculation of an unknown since calculations are based only on atomic or molecular weights. Only a precise analytical balance is needed for measurements. Gravimetric analysis, due to its high degree of accuracy, can also be used to calibrate other instruments in lieu of reference standards. While it is tedious and time consuming, it may find use where very precise results are needed, for example, in determining the iron content of an ore, whose price is determined by the iron content. It is used to determine the chloride content of cement. In environmental chemistry, sulfate is precipitated with barium ion, and in the petroleum field, hydrogen sulfide in desulfurization waste water is precipitated with silver ion. This chapter describes the specific steps of gravimetric analysis, including preparing the solution in proper form for precipitation, the precipitation process and how to obtain the precipitate in pure and filterable form, the filtration and washing of the precipitate to prevent losses and impurities, and heating the precipitate to convert it to a weighable form. It gives calculation procedures for computing the quantity of analyte from the weight of precipitate, following the principles introduced in Chapter 5. It also provides some common examples of gravimetric analysis. Finally, it discusses the solubility product and associated precipitation equilibria.
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10.1 How to Perform a Successful Gravimetric Analysis A successful gravimetric analysis consists of a number of important operations designed to obtain a pure and filterable precipitate suitable for weighing. You may wish to precipitate silver chloride from a solution of chloride by adding silver nitrate. There is more to the procedure than simply pouring in silver nitrate solution and then filtering.
Accurate gravimetric analysis requires careful manipulation in forming and treating the precipitate.
WHAT STEPS ARE NEEDED? The steps required in a gravimetric analysis, after the sample has been dissolved, can be summarized as follows: 1. 2. 3. 4.
5. 6. 7. 8.
Preparation of the solution Precipitation Digestion Filtration
Washing Drying or igniting Weighing Calculation
These operations and the reasons for them are described below. FIRST PREPARE THE SOLUTION The first step in performing gravimetric analysis is to prepare the solution. Some form of preliminary separation may be necessary to eliminate interfering materials. Also, we must adjust the solution conditions to maintain low solubility of the precipitate and to obtain it in a form suitable for filtration. Proper adjustment of the solution conditions prior to precipitation may also mask potential interferences. Factors that must be considered include the volume of the solution during precipitation, the concentration range of the test substance, the presence and concentrations of other constituents, the temperature, and the pH. Although preliminary separations may be required, in other instances the precipitation step in gravimetric analysis is sufficiently selective that other separations are not required. The pH is important because it often influences both the solubility of the analytical precipitate and the possibility of interferences from other substances. For example, calcium oxalate is insoluble in basic medium, but at low pH the oxalate ion combines with the hydrogen ions to form a weak acid and begins to dissolve. 8-Hydroxyquinoline (oxine—also known as 8-quinolinol) can be used to precipitate a large number of elements, but by controlling pH, we can precipitate elements selectively. Aluminum ion can be precipitated at pH 4, but the concentration of the anion form of oxine is too low at this pH to precipitate magnesium ion; magnesium oxinate has a much greater solubility product, the solubility product concept is discussed later in the chapter. .. N
N + H+ Al/3
1 3
.. N Al3+ +
−
O
O N
Usually, the precipitation reaction is selective for the analyte.
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8-Hydroxyquinoline can be used in combination with pH adjustments for selective precipitation of different metals. Al3+ can be selectively precipitated over Mg2+ at pH 4.
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A higher pH is required to shift the ionization step to the right in order to precipitate magnesium. If the pH is too high, however, magnesium hydroxide will precipitate, causing interference. The effects of the other factors mentioned above will become apparent as we discuss the precipitation step. THEN DO THE PRECIPITATION— —BUT UNDER THE RIGHT CONDITIONS
During the precipitation process, supersaturation occurs (this should be minimized!), followed by nucleation and precipitation.
After preparing the solution, the next step is to do the precipitation. Again, certain conditions are important. The precipitate should first be sufficiently insoluble that the amount lost due to solubility will be negligible. It should consist of large crystals that can be easily filtered. All precipitates tend to carry some of the other constituents of the solution with them. This contamination should be negligible. Keeping the crystals large can minimize this contamination. We can achieve an appreciation of the proper conditions for precipitation by first looking at the precipitation process. When a solution of a precipitating agent is added to a test solution to form a precipitate, such as in the addition of AgNO3 to a chloride solution to precipitate AgCl, the actual precipitation occurs in a series of steps. The precipitation process involves heterogeneous equilibria and, as such, is not instantaneous (see Chapter 6). The equilibrium condition is described by the solubility product, discussed at the end of the chapter. First, supersaturation occurs, that is, the solution phase contains more of the dissolved salt than it can carry at equilibrium. This is a metastable condition, and the driving force will be for the system to approach equilibrium (saturation). This is started by nucleation. For nucleation to occur, a minimum number of particles must come together to produce microscopic nuclei of the solid phase. The higher the degree of supersaturation, the greater the rate of nucleation. The formation of a greater number of nuclei per unit time will ultimately produce more total crystals of smaller size. The total crystal surface area will be larger, and there will be more danger that impurities will be adsorbed (see below). Although nucleation should theoretically occur spontaneously, it is usually induced, for example, on dust particles, scratches on the vessel surface, or added seed crystals of the precipitate (not in quantitative analysis). Following nucleation, the initial nucleus will grow by depositing other precipitate particles to form a crystal of a certain geometric shape. Again, the greater the supersaturation, the more rapid the crystal growth rate. An increased growth rate increases the chances of imperfections in the crystal and trapping of impurities. Von Weimarn discovered that the particle size of precipitates is inversely proportional to the relative supersaturation of the solution during the precipitation process: Relative supersaturation =
Q−S S
where Q is the concentration of the mixed reagents before precipitation occurs, S is the solubility of the precipitate at equilibrium, and Q − S is the degree of supersaturation. This ratio, (Q − S)/S, relative supersaturation, is also called the von Weimarn ratio. As previously mentioned, when a solution is supersaturated, it is in a state of metastable equilibrium, and this favors rapid nucleation to form a large number of small particles. That is, High relative supersaturation → many small crystals (high surface area)
Low relative supersaturation → fewer, larger crystals (low surface area)
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Obviously, then, we want to keep Q low and S high during precipitation. Several steps are commonly taken to maintain favorable conditions for precipitation:
1. Precipitate from dilute solution. This keeps Q low. 2. Add dilute precipitating reagents slowly, with effective stirring. This also keeps Q low. Stirring prevents local excesses of the reagent. 3. Precipitate from hot solution. This increases S. The solubility should not be too great or the precipitation will not be quantitative (with less than 1 part per thousand remaining). The bulk of the precipitation may be performed in the hot solution, and then the solution may be cooled to make the precipitation quantitative. 4. Precipitate at as low a pH as is possible to maintain quantitative precipitation. As we have seen, many precipitates are more soluble in acid medium, and this slows the rate of precipitation. They are more soluble because the anion of the precipitate (which comes from a weak acid) combines with protons in the solution.
Most of these operations can also decrease the degree of contamination. The concentration of impurities is kept lower and their solubility is increased, and the slower rate of precipitation decreases their chance of being trapped. The larger crystals have a smaller specific surface area (i.e., a smaller surface area relative to the mass) and so have less chance of adsorbing impurities. Note that the most insoluble precipitates do not make the best candidates for pure and readily filterable precipitates. An example is hydrous iron oxide (or iron hydroxide), which forms a gelatinous precipitate of large surface area. When the precipitation is performed, a slight excess of precipitating reagent is added to decrease the solubility by mass action (common ion effect) and to assure complete precipitation. A large excess of precipitating agent should be avoided because this increases chances of adsorption on the surface of the precipitate, in addition to being wasteful. If the approximate amount of analyte is known, a 10% excess of reagent is generally added. Completeness of precipitation is checked by waiting until the precipitate has settled and then adding a few drops of precipitating reagent to the clear solution above it. If no new precipitate forms, precipitation is complete.
Here is how to minimize supersaturation and obtain larger crystals.
Very insoluble precipitates are not the best candidates for gravimetric analysis! They supersaturate too easily.
Don’t add too much excess precipitating agent. This will increase adsorption.
Check for completeness of precipitation!
DIGEST THE PRECIPITATE TO MAKE LARGER AND MORE PURE CRYSTALS We know that very small crystals with a large specific surface area have a higher surface energy and a higher apparent solubility than large crystals. This is an initial rate phenomenon and does not represent the equilibrium condition, and it is one consequence of heterogeneous equilibria. When a precipitate is allowed to stand in the presence of the mother liquor (the solution from which it was precipitated), the large crystals grow at the expense of the small ones. This process is called digestion, or Ostwald ripening, and is illustrated in Figure 10.1. Small particles have greater surface energy associated with a greater surface area and display somewhat greater solubility than larger particles. The small particles tend to dissolve and reprecipitate on the surfaces of the larger crystals. In addition, individual particles agglomerate to effectively share a common counterion layer (see below), and the agglomerated particles finally cement together by forming connecting bridges. This noticeably decreases surface area.
Ostwald ripening improves the purity and crystallinity of the precipitate.
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346
Counter layer
H+
Primary adsorptive layer
NO − 5
Na + Ag+ + NO−3
Na+ + NO−3
Cl− NO−3
NO−3 Ag + Na +
Fig. 10.1.
AgCl
AgCl
Cl−
+
AgCl
Ostwald ripening.
Peptization is the reverse of coagulation (the precipitate reverts to a colloidal state and is lost). It is avoided by washing with an electrolyte that can be volatized by heating.
AgCl Ag + − Cl −
AgCl
K+
Na +
Na +
K+
H+
AgCl
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Cl−
Cl−
+
−
+
−
+
−
−
+
−
+
−
+
+
−
+
−
+
−
−
+
−
+
−
+
Cl−
Cl−
NO − 3
Cl−
+ = Ag + − − = Cl
Cl− Colloidal AgCl H+
Na +
Fig. 10.2.
Representation of silver chloride colloidal particle and adsorptive layers when Cl− is in excess.
Also, imperfections of the crystals tend to disappear, and adsorbed or trapped impurities tend to go into solution. Digestion is usually done at elevated temperatures to speed the process, although in some cases it is done at room temperature. It improves both the filterability of the precipitate and its purity. Many precipitates do not give a favorable von Weimarn ratio, especially very insoluble ones. Hence, it is impossible to yield a crystalline precipitate (small number of large particles), and the precipitate is first colloidal (large number of small particles). Colloidal particles are very small (1 to 100 nm) and have a very large surfaceto-mass ratio, which promotes surface adsorption. They are formed by virtue of the precipitation mechanism. As a precipitate forms, the ions are arranged in a fixed pattern. In AgCl, for example, there will be alternating Ag+ and Cl− ions on the surface (see Figure 10.2). While there are localized + and − charges on the surface, the net surface charge is zero. However, the surface does tend to adsorb the ion of the precipitate particle that is in excess in the solution, for example, Ag+ if precipitating Cl− with an excess of Ag+ ; this imparts a charge. (With crystalline precipitates, the degree of such adsorption will generally be small in comparison with particles that tend to form colloids.) The adsorption creates a primary layer that is strongly adsorbed and is an integral part of the crystal. It will attract ions of the opposite charge in a counterlayer or secondary layer so the particle will have an overall neutral charge. There will be solvent molecules interspersed between the layers. Normally, the counterlayer completely neutralizes the primary layer and is close to it, so the particles will collect together to form larger particles; that is, they will coagulate. However, if the secondary layer is loosely bound, the primary surface charge will tend to repel like particles, maintaining a colloidal state. When coagulated particles are filtered, they retain the adsorbed primary and secondary ion layers along with solvent. Washing the particles with water increases the extent of solvent (water) molecules between the layers, causing the secondary layer to be loosely bound, and the particles revert to the colloidal state. This process is called peptization and is discussed in more detail below where we describe washing the precipitate. Adding an electrolyte will result in a closer secondary layer and will promote coagulation. Heating tends to decrease adsorption and the effective charge in the adsorbed layers, thereby aiding coagulation. Stirring will also help.
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While all colloidal systems cause difficulties in analytical determinations, some are worse than others. Depending on the affinity of the dispsersed material for water, colloidal systems can be classified into hydrophilic (water loving) and hydrophobic (do not like water). While the former ones tend to produce stable dispersions in water, the latter ones tend to aggregate. Coagulation of a hydrophobic colloid is fairly easy and results in a curdy precipitate. An example is silver chloride. Coagulation of a hydrophilic colloid, such as hydrous ferric oxide, is more difficult, and it produces a gelatinous precipitate that is difficult to filter because it tends to clog the pores of the filter. In addition, gelatinous precipitates adsorb impurities readily because of their very large surface area. Sometimes a reprecipitation of the filtered precipitate is required. During the reprecipitation, the concentration of impurities in solution (from the original sample matrix) has been reduced to a low level, and adsorption will be very small. Despite the colloidal nature of silver chloride, the gravimetric determination of chloride is one of the most accurate determinations compared to other techniques such as titrimetry. In fact, it was used for atomic weight determination by T. W. Richards, who used nephelometry (light scattering) to correct for colloidal silver chloride.
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AgCl forms a hydrophobic colloid (a sol), which readily coagulates. Fe2 O3 · xH2 O forms a hydrophilic colloid (a gel) with large surface area.
IMPURITIES IN PRECIPITATES Precipitates tend to carry down from the solution other constituents that are normally soluble, causing the precipitate to become contaminated. This process is called coprecipitation. The process may be equilibrium based or kinetically controlled. There are a number of ways in which a foreign material may be coprecipitated. 1. Occlusion and Inclusion. In the process of occlusion, material that is not part of the crystal structure is trapped within a crystal. For example, water may be trapped in pockets when AgNO3 crystals are formed, and this can be removed to a degree by dissolution and recrystallization. If such mechanical trapping occurs during a precipitation process, the water will contain dissolved impurities. Inclusion occurs when ions, generally of similar size and charge, are trapped within the crystal lattice (isomorphous inclusion, as with K+ in NH4 MgPO4 precipitation). These are not equilibrium processes. Occluded or included impurities are difficult to remove. Digestion may help some but is not completely effective. The impurities cannot be removed by washing. Purification by dissolving and reprecipitating is helpful.
Occlusion is the trapping of impurities inside the precipitate.
2. Surface Adsorption. As we have already mentioned, the surface of the precipitate will have a primary adsorbed layer of the lattice ions in excess. This results in surface adsorption, the most common form of contamination. For example, after barium sulfate is completely precipitated, the lattice ion in excess will be barium, and this will form the primary layer. The counterion will be a foreign anion, say, nitrate two for each barium. The net effect then is an adsorbed layer of barium nitrate, an equilibrium-based process. These adsorbed layers can often be removed by washing, or they can be replaced by ions that are readily volatilized. Gelatinous precipitates are especially troublesome, though. Digestion reduces the surface area and, therefore, the adsorbed amount.
Surface adsorption of impurities is the most common source of error in gravimetry. It is reduced by proper precipitation technique, digestion, and washing.
3. Isomorphous Replacement. Two compounds are said to be isomorphous if they have the same type of formula and crystallize in similar geometric forms. When their lattice dimensions are about the same, one ion can replace another in a crystal, resulting in a mixed crystal. This process is called isomorphous replacement or isomorphous substitution. For example, in the precipitation of Mg2+ as magnesium
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ammonium phosphate, K+ has nearly the same ionic size as NH4 + and can replace it to form magnesium potassium phosphate. Isomorphous replacement, when it occurs, causes major interference, and little can be done about it. Precipitates in which it occurs are seldom used analytically. Chloride cannot be selectively determined by precipitation as AgCl, for example, in the presence of other halides and vice versa. Mixed crystal formation is a form of equilibrium precipitate formation, although it may be influenced by the rate of precipitation. Such a mixed precipitate is akin to a solid solution. The mixed crystal may be spatially homogeneous if the crystal is in equilibrium with the final solution composition (homogeneous coprecipitation) or heterogenous if it is in instaneous equilibrium with the solution as it forms (heterogenous coprecipitation), as the solution composition changes during precipitation. 4. Postprecipitation. Sometimes, when the precipitate is allowed to stand in contact with the mother liquor, a second substance will slowly form a precipitate with the precipitating reagent. This is called postprecipitation. For example, when calcium oxalate is precipitated in the presence of magnesium ions, magnesium oxalate does not immediately precipitate because it tends to form supersaturated solutions. But it will precipitate if the solution is allowed to stand too long before being filtered. Similarly, copper sulfide will precipitate in acid solution in the presence of zinc ions without zinc sulfide being precipitated, but eventually zinc sulfide will precipitate. Postprecipitation is a slow equilibrium process. WASHING AND FILTERING THE PRECIPITATES——TAKE CARE OR YOU MAY LOSE SOME
Test for completeness of washing.
Coprecipitated impurities, especially those on the surface, can be removed by washing the precipitate after filtering. The precipitate will be wet with the mother liquor, which is also removed by washing. Many precipitates cannot be washed with pure water, because peptization occurs. This is the reverse of coagulation, as previously mentioned. The process of coagulation discussed above is at least partially reversible. As we have seen, coagulated particles have a neutral layer of adsorbed primary ions and counterions. We also saw that the presence of another electrolyte will cause the counterions to be forced into closer contact with the primary layer, thus promoting coagulation. These foreign ions are carried along in the coagulation. Washing with water will dilute and remove foreign ions, and the counterion will occupy a larger volume, with more solvent molecules between it and the primary layer. The result is that the repulsive forces between particles become strong again, and the particles partially revert to the colloidal state and pass through the filter. This can be prevented by adding an electrolyte to the wash liquid, for example, HNO3 or NH4 NO3 for AgCl precipitate (but not KNO3 since it is nonvolatile—see below). The electrolyte must be one that is volatile at the temperature to be used for drying or ignition, and it must not dissolve the precipitate. For example, dilute nitric acid is used as the wash solution for silver chloride. The nitric acid replaces the adsorbed layer of Ag+ anion− , and it is volatilized when dried at 110◦ C. Ammonium nitrate is used as the wash electrolyte for hydrous ferric oxide. It is decomposed to NH3 , HNO3 , N2 , and oxides of nitrogen when the precipitate is dried by ignition at a high temperature. When you wash a precipitate, you should conduct a test to determine when the washing is complete. This is usually done by testing the filtrate for the presence of an ion of the precipitating reagent. After several washings with small volumes of the wash liquid, a few drops of the filtrate are collected in a test tube for the testing. For example, if chloride ion is determined by precipitating with silver nitrate reagent, the filtrate is tested for silver ion by adding sodium chloride or dilute HCl. We describe the technique of filtering in Chapter 2.
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DRYING OR IGNITING THE PRECIPITATE If the collected precipitate is in a form suitable for weighing, it must be heated to remove water and to remove the adsorbed electrolyte from the wash liquid. This drying can usually be done by heating at 110 to 120◦ C for 1 to 2 h. Ignition at a much higher temperature is usually required if a precipitate must be converted to a more suitable form for weighing. For example, magnesium ammonium phosphate, MgNH4 PO4 , is decomposed to the pyrophosphate, Mg2 P2 O7 , by heating at 900◦ C. Hydrous ferric oxide, Fe2 O3 · xH2 O, is ignited to the anhydrous ferric oxide. Many metals that are precipitated by organic reagents (e.g., 8-hydroxyquinoline) or by sulfide can be ignited to their oxides. The technique of igniting a precipitate is also described in Chapter 2.
Drying removes the solvent and wash electrolytes.
10.2 Gravimetric Calculations—How Much Analyte Is There? The precipitate we weigh is usually in a different form than the analyte whose weight we wish to report. The principles of converting the weight of one substance to that of another are given in Chapter 5 (Section 5.8), using stoichiometric mole relationships. We introduced the gravimetric factor (GF), which represents the weight of analyte per unit weight of precipitate. It is obtained from the ratio of the formula weight of the analyte to that of the precipitate, multiplied by the moles of analyte per mole of precipitate obtained from each mole of analyte, that is, GF =
fw analyte (g/mol) a × (mol analyte/mol precipitate) fw precipitate (g/mol) b
(10.1)
= g analyte/g precipitate So, if Cl2 in a sample is converted to chloride and precipitated as AgCl, the weight of Cl2 that gives 1 g of AgCl is g Cl2 = g AgCl ×
1 fw Cl2 (g Cl2 /mol Cl2 ) × (mol Cl2 /mol AgCl) fw AgCl (g AgCl/mol AgCl) 2
= g AgCl × GF (g Cl2 /g AgCl) = g AgCl × 0.24737 (g Cl2 /g AgCl)
Example 10.1 Calculate the grams of analyte per gram of precipitate for the following conversions: Analyte P K2 HPO4 Bi2 S3
Precipitate Ag3 PO4 Ag3 PO4 BaSO4
Solution g P/g Ag3 PO4 = GF =
1 at wt P (g/mol) = (mol P/mol Ag3 PO4 ) fw Ag3 PO4 (g/mol) 1 1 30.97 (g P/mol) × = 0.07399 g P/g Ag3 PO4 418.58 (g Ag3 PO4 /mol) 1
Grams precipitate × GF gives grams analyte.
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g K2 HPO4 /g Ag3 PO4 = GF = g Bi2 S3 /g BaSO4 = GF =
fw K2 HPO4 (g/mol) 1 × (mol K2 HPO4 /mol Ag3 PO4 ) fw Ag3 PO4 (g/mol) 1 174.18 (g K2 HPO4 /mol) 1 × = 0.41612 g K2 HPO4 /g Ag3 PO4 418.58 (g Ag3 PO4 /mol) 1 1 fw Bi2 S3 (g/mol) × (mol Bi2 S3 /mol BaSO4 ) fw BaSO4 (g/mol) 3 1 514.15 (g Bi2 S3 /mol) × = 0.73429 g Bi2 S3 /g BaSO4 233.40 (g BaSO4 /mol) 3
In gravimetric analysis, we are generally interested in the percent composition by weight of the analyte in the sample, that is, weight of substance sought (g) × 100% (10.2) % substance sought = weight of sample (g) We obtain the weight of substance sought from the weight of the precipitate and the corresponding weight/mole relationship (Equation 10.1): Weight of substance sought (g) = weight of precipitate (g) fw substance sought (g/mol) fw precipitate (g/mol) a × (mol substance sought/mol precipitate) b ×
= weight of precipitate (g) × GF (g sought/g precipitate) (10.3) Calculations are usually made on a percentage basis: %A=
Check units!
gA gsample
× 100%
(10.4)
where gA represents the grams of analyte (the desired test substance) and gsample represents the grams of sample taken for analysis. We can write a general formula for calculating the percentage composition of the substance sought: % sought =
weight of precipitate (g) × GF (g sought/g precipitate) × 100% weight of sample (g) (10.5)
Example 10.2 Orthophosphate (PO4 3− ) is determined by weighing as ammonium phosphomolybdate, (NH4 )PO4 · 12MoO3 . Calculate the percent P in the sample and the percent P2 O5 if 1.1682 g precipitate (ppt) were obtained from a 0.2711-g sample. Perform the % P calculation using the gravimetric factor and just using dimensional analysis.
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Solution
1.1682 g ppt × %P= =
(NH4 )3 PO4 · 12MoO3 0.2711 g sample
(g P/g ppt) × 100%
1.1682 g × (30.97/1876.5) × 100% = 7.111% 0.2711 g 1.1682 g ppt ×
% P2 O5 = =
P
P2 O5 (g P2 O5 /g ppt) 2(NH4 )3 PO4 · 12MoO3 × 100% 0.2711 g sample
1.1682 g × [141.95/(2 × 1876.5)] × 100% 0.2711 g
= 16.30% Let’s do the same calculation using dimensional analysis for the % P setup. 4 1.982 g (NH4 )2 PO · 12MoO × (30.97/1867.5)g P/g (NH ) PO · 12MoO 3 42 4 3 %P= 0.2771 g sample × 100% = (7.111 g P/g sample) × 100% = 7.111% P
Note that the (NH4 )2 PO4 · 12MoO3 species cancel one another (dimensional analysis), leaving only g P in the numerator. When we compare this approach with the gravimetric factor calculation, we see that the setups are really identical. However, this approach better shows which units cancel and which remain.
Example 10.3 An ore is analyzed for the manganese content by converting the manganese to Mn3 O4 and weighing it. If a 1.52-g sample yields Mn3 O4 weighing 0.126 g, what would be the percent Mn2 O3 in the sample? The percent Mn? Solution
3Mn2 O3 (g Mn2 O3 /g Mn3 O4 ) 2Mn3 O4 × 100% 1.52 g sample
0.126 g Mn3 O4 × % Mn2 O3 = =
0.126 g × [3(157.9)/2(228.8)] × 100% = 8.58% 1.52 g 3Mn (g Mn/g Mn3 O4 ) Mn3 O4 × 100% 1.52 g sample
0.126 g Mn3 O4 × % Mn = =
0.126 g × [3(54.94)/228.8] × 100% = 5.97% 1.52 g
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The following two examples illustrate some special additional capabilities of gravimetric computations.
Example 10.4 What weight of pyrite ore (impure FeS2 ) must be taken for analysis so that the BaSO4 precipitate weight obtained will be equal to one-half that of the percent S in the sample? Solution
If we have A% of S, then we will obtain
1 2
A g of BaSO4 . Therefore,
1 S A(g BaSO4 ) × (g S/g BaSO4 ) 2 BaSO4 A% S = × 100% g sample or 1 32.064 × 1% S = 2 233.40 × 100% g sample g sample = 6.869 g
PRECIPITATE MIXTURES— —WE NEED TWO WEIGHTS
Example 10.5 A mixture containing only FeCl3 and AlCl3 weighs 5.95 g. The chlorides are converted to the hydrous oxides and ignited to Fe2 O3 and Al2 O3 . The oxide mixture weighs 2.62 g. Calculate the percent Fe and Al in the original mixture. Solution
There are two unknowns, so two simultaneous equations must be set up and solved. Let x = g Fe and y = g Al. Then, for the first equation, g FeCl3 + g AlCl3 = 5.95 g FeCl3 AlCl3 x +y = 5.95 g Fe Al 133.34 162.21 +y = 5.95 g x 55.85 26.98
(1)
2.90x + 4.94y = 5.95 g
(4)
g Fe2 O3 + g Al2 O3 = 2.62 g Al2 O3 Fe2 O3 +y = 2.62 g x 2Fe 2Al 101.96 159.69 +y = 2.62 g x 2 × 55.85 2 × 26.98
(5)
1.43x + 1.89y = 2.62 g
(8)
See the website supplement to look up Example 10.5 Solving Simultaneous Equations that solves two-variable simultaneous equations.
(2) (3)
(6) (7)
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Solving (4) and (8) simultaneously for x and y: x = 1.07 g y = 0.58 g % Fe =
1.07 g × 100% = 18.0% 5.95 g
% Al =
0.58 g × 100% = 9.8% 5.95 g
10.3 Examples of Gravimetric Analysis Some of the most precise and accurate analyses are based on gravimetry. There are many examples, and you should be familiar with some of the more common ones. These are summarized in Table 10.1, which lists the substance sought, the precipitate formed, the form in which it is weighed, and the common elements that will interfere and must be absent. We do not present more details because gravimetry is not currently used often (unless the much longer time and increased labor requirements can be justified by the necessity of high precision and accuracy). You should consult more advanced texts and comprehensive analytical reference books for details on these and other determinations.
10.4 Organic Precipitates All the precipitating agents we have talked about so far, except for oxine, cupferrate, and dimethylglyoxime (Table 10.1), have been inorganic in nature. There are also a large number of organic compounds that are very useful precipitating agents for
Table 10.1
Some Commonly Employed Gravimetric Analyses Substance Analyzed Fe
Precipitate Formed
Precipitate Weighed
Fe(OH)3 Fe cupferrate Al(OH)3 Al(ox)3 a
Fe2 O3 Fe2 O3 Al2 O3 Al(ox)3
Ca
CaC2 O4
CaCO3 or CaO
Mg Zn Ba SO4 2− Cl−
MgNH4 PO4 ZnNH4 PO4 BaCrO4 BaSO4 AgCl
Mg2 P2 O7 Zn2 P2 O7 BaCrO4 BaSO4 AgCl
Ag PO4 3− Ni
AgCl MgNH4 PO4 Ni(dmg)2 b
AgCl Mg2 P2 O7 Ni(dmg)2
Al
a ox
= Oxine (8-hydroxyquinoline) monoanion. = Dimethylglyoxime monoanion.
b dmg
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Interferences Many. Al, Ti, Cr, etc. Tetravalent metals Many. Fe, Ti, Cr, etc. Many. Mg does not interfere in acidic solution All metals except alkalis and Mg All metals except alkalis All metals except Mg Pb NO3 − , PO4 3− , ClO3 − Br− , I− , SCN− , CN− , S2− , S2 O3 2− Hg(I) MoO4 2− , C2 O4 2− , K+ Pd
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Chelates are described in Chapter 9.
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metals. Some of these are very selective, and others are very broad in the number of elements they will precipitate. Organic precipitating agents have the advantages of giving precipitates with very low solubility in water and a favorable gravimetric factor. Most of them are chelating agents that form slightly soluble, uncharged chelates with the metal ions. A chelating agent is a type of complexing agent that has two or more groups capable of complexing with a metal ion. The complex formed is called a chelate. See Chapter 9 for a more thorough discussion of chelates. Since chelating agents are weak acids, the number of elements precipitated, and thus the selectivity, can usually be regulated by adjustment of the pH. The reactions can be generalized as (the underline indicates what is precipitated): Mn+ + nHX MXn + nH+
Metal chelate precipitates (which give selectivity) are sometimes ignited to metal oxides for improved stoichiometry.
There may be more than one ionizable proton on the organic reagent. The weaker the metal chelate, the higher the pH needed to achieve precipitation. Some of the commonly used organic precipitants are listed in Table 10.2. Some of these precipitates are not stoichiometric, and more accurate results are obtained by igniting them to form the metal oxides. Some, such as sodium diethyldithiocarbamate, can be used to perform group separations, as is done with hydrogen sulfide. You should consult specialized
Table 10.2
Some Organic Precipitating Agents Reagent
Structure
Dimethylglyoxime
α -Benzoinozime (cupron)
Metals Precipitated
CH3
C
NOH
CH3
C
NOH
OH NOH CH C
N O
Ammonium nitrosophenylhydroxylamine (cupferron)
N O NH4
8-Hydroxyquinoline (oxine)
N
Sodiu