QUICK REFERENCE GUIDE Ions and Ionic Compounds Acids and Bases - Acid (HA) and conjugate base (A ) concentrations, as a function of pH, Table 7.6 Interpreting equilibrium constants (Keq), Table 7.2 Ka and pKa values for selected acids, Table 7.4 Common acids and bases, Table 7.1 pH values of common solutions, Table 7.3 Relative strengths of some acids and their conjugate bases, Table 7.5 Common polyatomic ions, Table 3.2 Some transition metal ions, Table 3.1 The uses of some ionic compounds, Table 3.4 Amino Acids, Proteins, and Enzymes Math a-Amino acids present in proteins, Table 12.1 Amino acids that are essential for humans, Table 14.1 Selected enzyme cofactors, Table 12.2 Atoms Subatomic particles, Table 2.1 The ground state electron distribution for the first 20 elements, Table 2.6 Bonding Lipids Common fatty acids, Table 11.1 Key esters found in some waxes, Table 11.2 Conversion factors and the factor label method, Section 1.6 Logs and antilogs, Chapter Math Support Measurements and significant figures, Section 1.5 Scientific notation, SI and metric prefixes, Section 1.4 Significant figures, Table 1.5 Nucleic Acids Codons in the 5 to 3 sequence of mRNA, Table 13.1 Short tandem repeats (STRs) and the probability of their occurrence, Table 13.2 Bond types, Table 4.1 Organic Compounds Carbohydrates Common molecular shapes, Table 4.2 Formulas and names of alkyl groups, Table 8.3 Physical properties of selected alcohols, ethers, thiols, sulfides, and alkanes, Table 9.1 Physical properties of selected aldehydes and ketones, Table 9.2 Physical properties of selected amines, Table 8.7 Physical properties of selected phenols, Table 8.5 Physical properties of some small carboxylic acids, Table 8.4 Structure, name, and properties of selected hydrocarbons, Table 8.1 The first ten numbering prefixes for IUPAC naming, Table 8.2 Monosaccharides, Table 10.1 Relative sweetness, Table 10.2 Energy Specific heat, Table 1.8 Gases, Liquids, and Solids Density of common substances, Table 1.7 Solutions, colloids, and suspensions, Table 6.5 The solubility of ionic compounds in water, Table 6.3 The vapor pressure of water at various temperatures, Table 6.1 Health Adult body mass index, Table 1.6 Blood pressure guidelines, Table 6.2 Concentration ranges for some blood serum solutes, Table 6.4 Dietary reference intakes (DRIs) for some essential elements, Table 2.4 The biochemical significance of selected elements, Table 2.3 Radioactivity Common forms of nuclear radiation, Table 2.7 Half life and decay type for selected radioisotopes, Table 2.10 Health effects of short term exposure to radiation, Table 2.9 Some uses of radioisotopes in medicine, Table 2.11 Conversion Factors Mass kilogram pound milligram grain ounce = 1000 grams = 2.205 pounds = 453.59 grams = 16 ounces = 1000 micrograms = 65 milligrams = 28.3 grams Length meter kilometer mile inch = 1000 millimeters = 3.281 feet = 39.37 inches = 0.621 mile = 1.609 kilometers = 5280 feet = 2.54 centimeters Volume liter quart gallon = 1000 milliliters = 1.057 quarts = 0.946 liter = 3.785 liters milliliter teaspoon tablespoon = 1 centimeter3 = 1 cubic centimeter = 15 drops = 5 milliliters = 15 milliliters = 0.5 fluid ounces Energy calorie joule = 4.184 joule = 0.2390 calorie Temperature K K °C 0°C °F = °C + 273.15 = -273.15 °C = -459.67°F °F@32 = 1.8 = 273.15K = 32°F = (1.8 * °C) + 32 Pressure atmosphere = 14.7 pounds per square inch = 760 torr = 760 millimeters Hg SI and Metric Prefixes Prefix Symbol Multiplier giga G 1,000,000,000 mega M 1,000,000 kilo k 1,000 hecto h 100 deka da 10 deci d 0.1 centi c 0.01 milli m 0.001 m 0.000001 micro nano n 0.000000001 pico p 0.000000000001 = 109 = 106 = 103 = 102 = 101 = 100 = 10-1 = 10-2 = 10-3 = 10-6 = 10-9 = 10-12 GENERAL, ORGANIC, AND BIOLOGICAL CHEMISTRY An Integrated Approach F o u rth Kenneth W Raymond Eastern Washington University E dition VICE PRESIDENT AND PUBLISHER ASSOCIATE PUBLISHER ACQUISITIONS EDITOR SENIOR PROJECT EDITOR EDITORIAL ASSISTANT EXECUTIVE MARKETING MANAGER MARKETING MANAGER DESIGN DIRECTOR SENIOR PHOTO EDITOR DESIGNERS PRODUCT DESIGNER MEDIA SPECIALIST OUTSIDE PRODUCTION SENIOR PRODUCTION EDITOR Kaye Pace Petra Recter Nicholas Ferrari Jennifer Yee Ashley Gayle Christine Kushner Kristine Ruff Harry Nolan Lisa Gee Thomas Nery and Jim O'Shea Geraldine Osnato Evelyn Brigandi CMPreparé, Rebecca Dunn Elizabeth Swain COVER IMAGE © NREY/iStockphoto This book was set in 10.5/12 Adobe Garamond by CMPreparé and printed and bound by Courier/Kendallville The cover was printed by Courier/Kendallville This book is printed on acid-free paper.∞ Copyright © 2014, 2010, 2008, 2006 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 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 shipping label are available at www.wiley.com/go/returnlabel Outside of the United States, please contact your local representative Library of Congress Cataloging-in-Publication Data Raymond, Kenneth William General, organic, and biological chemistry : an integrated approach / Kenneth W Raymond.–4th ed p cm Includes bibliographical references and index ISBN 978-1-118-835258-8 (cloth) Chemistry—Textbooks Chemistry, Organic–Textbooks Biochemistry–Textbooks I Title QD31.3.R39 2010 540—dc22 2009034009 ISBN: 978-1-118-35258-8 (Main Book) ISBN: 978-1-118-17219-3 (Binder-Ready Version) Printed in the United States of America 14.2 P iii PR E FA C E T his fourth edition of General, Organic, and Biological Chemistry: An Integrated Approach has, like the earlier editions, been written for students preparing for careers in health-related fields such as nursing, dental hygiene, nutrition, occupational therapy, athletic training, and medical technology The text is also suitable for students majoring in other fields where it is important to have an understanding of chemistry and its relationship to living things Students who use this text not need to have a previous background in chemistry but should possess basic math skills For those whose math is a bit rusty, the text provides reviews of the important material While designed for use in one-semester or two quarter General, Organic, and Biochemistry (GOB) courses, instructors have found that it also works well for one-year courses, especially when combined with the supplement Chemistry Case Studies for Allied Health Students by Colleen Kelley and Wendy Weeks In a GOB course it is essential to show how the subject matter relates to the students’ future careers For that reason, this text makes extensive use of real-life examples from the health sciences O R G A N I Z A T I O N Most GOB texts are divided into three distinct parts: general chemistry, organic chemistry, and biochemistry The integrated approach used in this text integrates these subject areas by juxtaposing chapters of related information For example, a study of bonding and compounds (Chapter 3) is followed by a first look at organic compounds (Chapter 4) and then an introduction to inorganic and organic reactions (Chapter 5) Other examples of this integration at the chapter level include the study of acid–base chemistry (Chapter 7) followed by a chapter that includes organic acids and bases (Chapter 8), and the chemistry of alcohols, aldehydes, and ketones (Chapter 9) followed by that of carbohydrates (Chapter 10) Studies have shown that effective learning can take place when material is presented in a context that shows its relationship to the “big picture.” The arrangement of chapters in this text helps students to see how inorganic chemistry and organic chemistry are linked to the biochemistry and health sciences that are so important to their future careers TAKING AN INTEGRATED APPROACH Whether taught in one semester or two, the GOB curriculum is very full Using an integrated approach can shorten the cycle time for returning to similar themes from the different branches of chemistry Having a shorter time interval between when a topic is first presented and when it is reintroduced can help students assimilate the material more readily An added benefit of integrating GOB course material is that students get a better sense of how the chemistry being presented relates to their future careers, and as a result, their interest and motivation are enhanced BENEFITS OF AN INTEGRATED APPROACH iii iv preface TRANSITIONING TO AN INTEGRATED APPROACH For instructors, making the transition from the traditional approach to an integrated one should not pose a problem The integration of material takes place at the chapter level, and required introductory material is always presented before a new organic chemistry or biochemistry topic is begun For example, instead of introducing carboxylic acids, phenols, and amines in their traditional position—late in the group of chapters devoted to organic chemistry—this text places these organic acids and bases (Chapter 8) directly after the introduction to acids and bases (Chapter 7) Supplements to the text can also assist with making the transition to an integrated text These include Chemistry Case Studies for Allied Health Students, an instructor’s manual, an instructor’s solutions manual, PowerPoint lecture slides, and a test bank KEY FEATURES OF THE FOURTH EDITION In terms of organization, some major changes have been made to this edition of the text A number of these modifications were suggested by reviewers and by instructors who have used previous editions Many reviewers recommended moving the chemistry of hydrocarbons from Chapter to a chapter later in the text In this fourth edition, hydrocarbons appear in Chapter The latter part of Chapter now introduces the key organic families One new feature of the text is the “Did you Know?” paragraphs that briefly highlight topics that relate to the chemistry being presented in each chapter Numerous end of chapter problems, sample problems, and practice problems have been added or revised in the Fourth Edition Other changes that will be noted by those familiar with the text include: Chapter • A new chapter section titled “Measurement in General, Organic, and Biochemistry” shows how the topics presented in Chapter relate to these three fields of chemistry • This chapter now includes a discussion of the kinetic molecular theory, phase changes, heat of fusion, and heat of vaporization In previous editions of the text this material appeared in Chapter Chapter • A new chapter section related to trace elements has been added • In earlier editions, Chapter included a section on fission and fusion This chapter section has been dropped in the new edition • Three new Health Links were added: Stable Isotopes and Drug Testing, Lead, and Radioisotopes for Sale Chapter • The Health Link Pass the Salt, Please was added Chapter • The chapter-opening vignette was changed • Hydrocarbon chemistry (Sections 4.4–4.8 in earlier editions) has been moved to Chapter In its place, a new chapter section related to organic families was added Chapter • The Biochemistry Link The Henderson-Hasselbalch Equation was added Chapter • A new section (Section 8.8 Reactions of Hydrocarbons) gathers topics that, in previous editions, were presented in earlier chapters Section 8.8 also introduces the alkane halogenation and aromatic substitution • The topics of decarboxylation (Section 8.9) and phenol oxidation (Section 8.8) have been removed • There is now a greater emphasis on skeletal structures than in previous editions Chapter • The treatment of nucleophilic substitution (Section 9.2) was trimmed and is now tied to the alkane halogenation reactions introduced in Chapter Chapter 10 • Examples of simple glycosides have been added as part of the introduction to glycosidic bonds Chapter 11 • The structure of esters and their hydrolysis are reviewed just before the discussion of triglycerides and saponification Chapter 12 • The discussion of DG was expanded by introducing the concept of DGo′ Chapter • Two new Health Links were added: Tamiflu and Relenza; and Immunotherapy Chapter 13 • A discussion of inhaled anesthetics and their solubility in blood was added Chapter • The Health Link Lupus was added • The Biochemistry Link Glowing Cats was added Chapter 14 • The effect of pressure and temperature on equilibrium is now described • Figures were modified and sample and practice problems were added preface v PR O B L E M S O L V I N G Learning to anything requires practice, and in chemistry this practice involves solving problems This text offers students ample opportunities to so SAMPLE PROBLEM 1.13 Unit conversions a An over-the-counter (nonprescription) cough syrup contains 7.5 mg of dextromethorphan in every mL The recommended dose of dextromethorphan for a 44 lb child is 10.0 mg How many milliliters of cough syrup should be given? b For a 55 lb child, the recommended dose of dextromethorphan is 12.5 mg How many milliliters of cough syrup should be given? Sample Problems and Practice Problems Each major topic is followed by a sample problem and a related practice problem The solution to each sample problem is accompanied by a strategy to use when solving the problem The answers to practice problems are given at the end of the chapter STRATEGY In part a, you are being asked to convert from a 10 mg dose of dextromethorphan to milliliters of cough syrup For the cough syrup, the relationship between these units (7.5 mg dextromethorphan mL ) can be used to make a conversion factor SOLUTION a 10.0 mg dextromethorphan b 12.5 mg dextromethorphan PRACTICE PROBLEM mL cough syrup 7.5 mg dextromethorphan mL cough syrup 7.5 mg dextromethorphan mL cough syrup mL cough syrup 1.13 The 44 lb child is given a cold tablet that contains mg of dextromethorphan and is then given mL of the cough syrup mentioned in Sample Problem 1.13a Has the child received greater than the recommended dose? END OF CHAPTER PROBLEMS Answers to problems whose numbers are printed in color are given in Appendix C More challenging questions are marked with an asterisk 4.1 One of the alkenes is nonpolar and the other is polar Which is which? F C End of Chapter Problems H H F C H 4.2 F F C H One of the aromatic compounds is nonpolar and the other is polar Which is which? H F Problems are paired and Appendix C provides answers for the odd-numbered problems Each chapter includes multistep Learning Group problems designed to be worked with other students and Thinking It Through problems that ask students to go a bit further with one or more of the concepts presented in the chapter-opening vignette H C C C C C C F F F H C C C C C C F F molecules? a CH3CH2NH2 O CH3CH2NH2 O c 4.86 Which of the molecules in Problem 4.85 can form a hydrogen bond with water? 4.87 Of the pairs of molecules in Problem 4.83, which interact primarily through London forces? 4.88 Of the pairs of molecules in Problem 4.85, which interact primarily through London forces? 4.89 Of the pairs of molecules in Problem 4.81, which can interact through dipole–dipole forces, but not hydrogen bonds? 4.90 Of the pairs of molecules in Problem 4.82, which can interact through dipole–dipole forces, but not hydrogen bonds? HealthLink | PRION DISEASES 4.91 Are covalent bonds broken when PrPc is converted into PrPsc? Explain 4.92 Suggest a way to reduce the spread of mad cow disease between cattle BiochemistryLink | ETHYLENE, A PLANT HORMONE 4.93 During ripening, bananas produce small amounts of ethylene When bananas are shipped, why should they not be shipped in closed containers? 4.94 Ethylene gas can be produced from petroleum and then stored in metal cylinders When food processors want to ripen bananas, they expose the fruit to this manufactured ethylene Would you expect plants to react differently to ethylene made from petroleum than to ethylene that they have produced themselves? HealthLink | SUNSCREENS 4.95 What properties are important for molecules used as sunscreens? 4.96 When applied to the skin of mice, forskolin, a compound present in an Asian plant, was shown to hydrogen, should have an octet of valence electrons a OH b NH4 c CN 4.14 Draw each polyatomic ion Each atom, except for hydrogen, should have an octet of valence electrons b HPO42 c H2PO4 4.15 Draw each of the following Each atom should have an octet of valence electrons b SO32 4.16 Draw each of the following Each atom should have an octet of valence electrons b SH a PO33 4.17 Draw two different molecules that have the formula 4.1 STRUCTURAL FORMULAS 4.85 Will hydrogen bonds form between each pair of c C3H4 4.13 Draw each polyatomic ion Each atom, except for a SO3 H C2H6O 4.3 hydrogen bond with water? c C2H2 4.12 Draw each molecule a C3H8 b C3H6 a PO43 H H 4.84 Which of the molecules in Problem 4.83 can form a b C 4.11 Draw each molecule a C2H6 b C2H4 Indicate the number of covalent bonds that each nonmetal atom is expected to form a C b O c P d Br 4.4 Indicate the number of covalent bonds that each nonmetal atom is expected to form a Se b H c I d N 4.5 Draw the structural formula of the molecule that contains atoms increasethe thefollowing production of melanin Which, you a.suppose, one oxygen atom twoofhydrogen atomsstudy? were the and results this scientific b.a.one atommore and one iodine atom Thehydrogen mice tanned quickly c.b.one nitrogen atom three hydrogen The mice did notand sunburn as easily atoms 4.6 Draw themice structural formula of the molecule that c The were less susceptible to skin cancer contains the following atoms seleniumGatom two fluorine atoms 4.5a one LEARNING ROUPand PROBLEMS b one phosphorus atom and three hydrogen atoms 4.97 c.a.one Tohydrogen which organic family the molecule atom and onedoes bromine atom belong? CH3CH CH 2CH2of 2CH 2OH 4.7 Draw the Lewis structure each molecule formula of the molecule in a.b.F2Give the molecular b O2 part a 4.8 Draw the Lewis structure of each molecule c Can two of the molecules in part a interact through a I2 b N2 London forces? 4.9 Draw thetwo Lewis structure of each molecule d Can of the molecules in part a interact through a CH b NFforces? 2S dipole–dipole 4.10 Draw thetwo Lewis structure of each molecule e Can of the molecules in part a interact through a OCl b CS2 hydrogen bonds? f Draw a molecule that has the same molecular formula as the molecule in part a but belongs to a different family of organic compounds g Can two of the molecules in part f interact through London forces? h Can two of the molecules in part f interact through dipole–dipole forces? i Can two of the molecules in part f interact through hydrogen bonds? 4.98 a To which organic family does the molecule belong? O CH3CH2CH2CH2C OH b Give the molecular formula of the molecule in part a c Can two of the molecules in part a interact through London forces? d Can two of the molecules in part a interact through dipole–dipole forces? e Can two of the molecules in part a interact through hydrogen bonds? f Draw a molecule that has the same molecular formula as the molecule in part a but is an ester g Can two of the molecules in part f interact through London forces? h Can two of the molecules in part f interact through dipole–dipole forces? i Can two of the molecules in part f interact through hydrogen bonds? j Draw a molecule that has the same molecular formula as the molecule in part a but is both an aldehyde and an ether 4.18 Draw three different molecules that have the formula C3H9N 4.19 Write a condensed structural formula for each molecule H H H H ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ a H¬ C ¬ C ¬ C ¬ C ¬ H H H H H H H ƒ ƒ ƒ ƒ b H¬ C ¬ C ¬ N¬ H ƒ H H H 4.20 Write a condensed structural formula for each molecule H H H ƒ ƒ ƒ ƒ ƒ ƒ H ƒ a H¬ C ¬ C ¬ C ¬ O¬ C ¬ H H H H H H H ƒ ƒ ƒ ƒ ƒ ƒ b H¬ C ¬ C ¬ C ¬ F H H H ƒ H 262 Chapter 7 Acids, Bases, and Equilibrium Sample Problem 7.12 Understanding buffers A buffer can be prepared using H2PO4- and its conjugate base a Write the acid–base reaction equation for H2PO4- and H2O reacting to form the conjugate base of H2PO4- and H3O + b Write a reaction equation that shows what takes place when H3O + is added to this buffer Strategy Part a gives the formula of the two reactants and one of the two products The conjugate base of H2PO4- has one less H + than H2PO4- In part b, think about what happens to the rate of the forward or reverse reaction if the concentration of a reactant or product is changed Solution a H2PO4 - + H2O N HPO4 2- + H3O+ b An increase in the concentration of H3O + causes the rate of the reverse reaction to increase H2PO4 - + H2O d HPO4 2- + H3O+ Practice Problem 7.12 a For the buffer described in Sample Problem 7.12, write a reaction equation that shows what takes place when OH- is added to this buffer b Over what pH range is this buffer effective? (See Table 7.4.) T h e H e n d e r s o n – H a s s e l b a l c h E q u a t i o n BiochemistryLink We have seen that a buffer can be prepared from a weak acid and its conjugate base, and that the concentration of these two substances is connected to the pH of the solution and the pKa of the acid (Table 7.6) Taking a more mathematical approach to these concepts allows us to more accurately describe the makeup of a buffer Beginning with the equation for the Ka of the acid HA, HA + H2O N A - + H3O + [A - ][H3O + ] Ka = [HA] taking the logarthim of both sides of the equation, and rearranging gives what is known as the Henderson–Hasselbalch equation [A - ] pH = pKa + log [HA] The Henderson–Hasselbalch equation To see how this equation can be used, let us consider the acid HF (pKa = 3.18) [F - ][H3O + ] HF + H2O N F - + H3O + Ka = [HF] [F - ] and pH = pKa + log [HF] • What is the pH when [F - ] = [HF]? If [F - ] = [HF], then [F - ] = Inserting this and the value for the pKaof HF [HF] into the Henderson–Hasslebalch equation gives a pH of 3.18 This agrees with the prediction in Table 7.6 that if the values of the pH and the pKa for an acid are the same, then [HA] = [A-] pH = pKa + log [F - ] = 3.18 + log = 3.18 + = 3.18 [HF] • What is the pH when [HF] is two times larger than [F-]? In mathematical terms, this relationship can be expressed [F - ] as [HF] = 2[F - ] or = 0.5 Solving the Henderson– [HF] Hasselbalch equation gives a pH of 2.88 As predicted by Table 7.6, when [HA] [A-], pH pKa pH = pKa + log [F - ] = 3.18 + log 0.5 = 3.18 + -0.30 = 2.88 [HF] • What is the pH when [HF] is half that of [F-]? In mathematical terms, this relationship can be expressed as [F - ] 2[HF] = [F -] or = Solving the Henderson–Hasselbalch [HF] equation gives a pH of 3.48 When [HA] [A-], pH pKa pH = pKa + log [F - ] = 3.18 + log 0.5 = 3.18 + 0.30 = 3.48 [HF] 7.11 Maintaining the pH of Blood Serum 263 7.11 Ma i n t a i n i n g t h e p H o f b l o o d s e r u m The pH of blood serum normally falls between 7.35 and 7.45 To keep enzymes functioning properly, it is important that the pH not move out of this range Enzymes, which catalyze chemical reactions that take place in the body, are proteins consisting of chains of amino acids, some of which contain acidic or basic groups Altering the pH can change the structure of these groups (Figure 7.12) in a way that affects the noncovalent interactions that are so important for maintaining enzyme shape and function The adverse health effects associated with blood serum pHs that fall outside of the 7.35–7.45 range largely result from changes to enzyme structure Maintaining the appropriate blood serum pH involves dealing with acids that are produced by metabolism (the reactions that take place within living things) These acids include fatty acids, lactic acid, phosphoric acid, and members of a group of compounds called ketone bodies Some acids are also present in the foods that we eat Another significant source of acid is the CO2 produced by cells In moving from the cells where it is produced to the lungs where it is exhaled (Figure 6.13), most of the CO2 is carried in the blood as H2CO3 or HCO3- H2CO3 is produced when CO2 reacts with water, and HCO3- is formed in the acid–base reaction that takes place between H2CO3 and water H2O + CO2 N H2CO3 H2CO3 + H2O N HCO3- + H3O+ To deal with these acids and to maintain a constant serum pH, the body relies on two general approaches One involves using buffers to minimize pH changes The other is to control the H3O + concentration in the blood through breathing and by action of the kidneys Let us begin by considering the buffers present in serum The most important of these consists of H2CO3 and its conjugate base HCO3- H2CO3 + H2O N HCO3- + H3O+ Acid Conjugate base Salt bridge CH2CO2 CH2CO2 – – ± NH (CH2)4 (CH2)4 (a) ■■ Figure NH2 (b) 7.12 The effect of changing pH on enzyme structure In enzymes and other proteins, a change in pH can switch groups between their acidic and conjugate base forms This can disrupt ionic bonds (known also as salt bridges) and other noncovalent interactions that are essential for maintaining function (a) Near pH an ionic bond holds a portion of the protein chain in the shape required for the enzyme to act as a catalyst (b) Above pH 9, i NH3 + (acid) is converted into i NH2 (conjugate base) The ionic bond no longer exists and the shape and catalytic ability of the enzyme change 264 Chapter 7 Acids, Bases, and Equilibrium As shown by the equations below, small amounts of H3O + or OH- can be consumed by this buffer: Addition of H3O+: H2CO3 + H2O — HCO3- + H3O+ Addition of OH-: H2CO3 + OH- ¡ HCO3- + H2O For this particular buffer system to work, both H2CO3 and HCO3- must be available As we saw above, H2CO3 is produced when CO2 formed in cells reacts with H2O Some HCO3- is formed by loss of H + from H2CO3 The kidneys, by releasing or taking up HCO3-, also control blood concentrations of this buffer component Dihydrogen phosphate ion (H2PO4-) and hydrogen phosphate ion (HPO42-) are another buffer system that is present in blood (see Sample Problem 7.12) Proteins, which are weak acids, provide yet another buffer system that helps to maintain a relatively constant blood pH The respiratory system also plays an important role in maintaining the pH of blood serum If, in spite of the action of buffers, the pH of the blood becomes too acidic, the respiratory center in the brain signals for faster and deeper breathing This leads to more CO2 being exhaled and a drop in its partial pressure (PCO2) As shown in the first reaction equation below, loss of CO2 leads to a net reduction in the concentration of H2CO3 The second reaction equation shows that a decrease in the concentration of H2CO3 leads to a loss of H3O + Blood serum becomes less acidic H2O + CO2 — H2CO3 H2CO3 + H2O — HCO3- + H3O+ A good example of this is what happens when you exercise Exercise increases metabolism, which leads to an increased cellular production of CO2 Blood buffer systems are unable to deal with the excess H3O + that forms as a result However, the increased breathing rate that accompanies exercise helps to move the serum pH back into the normal range The respiratory system can also deal with serum that is too basic In this case the respiratory center in the brain signals for slower and shallower breathing, which causes PCO2 in the blood to rise This leads to the production of more H2CO3 and a drop in the serum pH The kidneys also play a role in maintaining an appropriate pH If serum is too acidic, the kidneys release HCO3- into the blood This component of the H2CO3/HCO3buffer system helps to reduce the H3O + concentration Additionally, the kidneys remove H3O + from the body by releasing it into urine If the pH of blood serum becomes too high, the kidneys the opposite: HCO3- is removed from the blood and H3O + is added The effects of the H2CO3/HCO3- buffer system, respiration, and the kidneys on blood pH are summarized in Figure 7.13 Even with all of these controls, the pH of blood can still sometimes fall outside of the normal range These acid/base disorders are classified as being either respiratory or metabolic in nature (Table 7.7) Respiratory disorders occur when exhaling does not remove CO2 from the blood at the same rate that it is produced in cells Metabolic disorders arise from an inability to deal with acids produced by metabolism, problems with controlling the blood concentration of HCO3-, ingestion of compounds that are acids or bases, or ingestion of compounds that can be converted into acids or bases Mild cases of acidosis, a low blood serum pH, can result in light-headedness Severe cases produce a depression of the central nervous system that can lead to coma and death Respiratory acidosis, characterized by a low pH, an elevated blood PCO2, and a normal or higher than normal HCO3- concentration, is caused by any condition that interferes with the ability of the lungs to exchange gases—specifically, to remove CO2 from the blood Pneumonia, emphysema, cystic fibrosis, shallow breathing, or holding your breath can cause this form of acidosis As described above, a rise in PCO2 produces higher concentrations of H2CO3, which can upset the H2CO3 >HCO3- buffer system enough to lower the pH of blood The serum concentration of HCO3- is sometimes higher than normal in respiratory acidosis because the kidneys are trying to compensate for the inability of the lungs to control pH Releasing more HCO3- into the blood provides a buffer component to help remove unwanted H3O + 7.11 Maintaining the pH of Blood Serum 265 ■■ Figure (a) Acidosis: Low blood serum pH Problem: Controlling plasma pH The H2CO3 >HCO3- buffer system, respiration, the kidneys, and other buffer systems all play a role in controlling the pH of blood plasma (a) The response to a drop in pH (b) The response to a rise in pH H3O+ concentration is too high Carbonic acid– hydrogen carbonate buffer system 7.13 H2O+CO2 ∆ H2CO3+H2O ∆ HCO3-+H3O+ _ Release HCO3 in blood Solutions: Other buffers absorb H3O+ Faster, deeper breathing reduces CO2 Excrete H3O+ into urine Lungs Kidneys (b) Alkalosis: High blood serum pH Problem: H3O+ concentration is too low Carbonic acid– hydrogen carbonate buffer system H2O+CO2 ∆ H2CO3+H2O ∆ HCO3-+H3O+ _ Remove HCO3 from blood Other buffers release H3O+ Solutions: Release H3O+ into blood Slower, shallower breathing increases CO2 Lungs Kidneys Table | 7.7 Acid/base disorders pH P CO2 HCO - concentration 7.35–7.45 35–45 torr 22–26 mEq/L Respiratory acidosis lower higher higher, if compensating Metabolic acidosis lower lower, if compensating lower Respiratory alkalosis higher lower lower, if compensating Metabolic alkalosis higher higher, if compensating higher Condition Normal range in blood plasma th Source: Page 1055, Human Anatomy and Physiology, ed By Elaine N Marieb © 2004 by Pearson Education, Inc Reprinted by permission 266 Chapter 7 Acids, Bases, and Equilibrium 14.11 Did You Know ? High in the mountains the partial pressure of O2 is low, so climbers must breathe faster to get enough oxygen This can lead to altitude alkalosis, a higher than normal blood serum pH that can lead to dizziness, numbness, muscle spasms, and loss of consciousness Rapid breathing removes too much CO2 from the blood, which produces an imbalance in the H2CO3/HCO3- blood buffer Climbing slowly to allow the body to adjust can slow the onset of altitude alkalosis Metabolic acidosis, distinguished by a low pH, a low HCO3 concentration, and a PCO2 that is normal, or lower if the lungs are compensating, can be caused by ketone bodies (produced from starvation or poorly controlled diabetes), lactic acid (produced by strenuous exercise), excessive alcohol consumption (alcohol is metabolized into acetic acid), diarrhea (excessive loss of HCO3 -), and certain kidney problems The symptoms of alkalosis, a high blood serum pH, include headaches, nervousness, cramps, and, in severe cases, convulsions and death Respiratory alkalosis (a high pH, a low PCO2 , and an HCO3 -concentration that is normal, or lower if the kidneys are compensating) occurs when CO2 is exhaled from the body more quickly than it is produced by cells A net loss of CO2 causes a reduction in the concentration of H2CO3 present in the blood, and too large of a decrease leads to a rise in pH Hyperventilation, which can be brought on by anxiety, central nervous system damage, aspirin poisoning, fever, and other factors are the usual causes Metabolic alkalosis (a high pH, a high HCO3 concentration, and a PCO2 that is normal, or higher if the lungs are compensating) can result from excessive use of antacids (weak bases) and from constipation (greater than normal amounts of HCO3 end up in the blood) Sample Problem 7.13 The blood buffer system Constipation causes blood serum to carry higher than normal levels of HCO3- a What effect does excess HCO3- have on the H2CO3/HCO3- buffer system and on the pH of blood serum? b How would the respiratory system respond to a higher than normal concentration of HCO3-? Strategy The H2CO3/HCO3- buffer system involves equilibrium between H2CO3 and its conjugate base Apply Le Châtelier’s principle to solve part a of this problem Figure 7.13 will help you answer part b Solution a An increase in the HCO3- concentration will upset the equilibrium between H2CO3/HCO3-, causing a net reverse reaction to take place This will reduce the concentration of HCO-3 and H3O+ (the pH will rise) H2CO3 + H2O — HCO3- + H3O + b To offset a rise in pH, breathing is slowed This causes CO2 levels in the blood to rise, leading to the formation of H2CO3 H2O + CO2 ¡ H2CO3 An increase in the concentration of H2CO3 causes a net forward reaction for the H2CO3/HCO3- equilibrium, and the pH drops H2CO3 + H2O ¡ HCO3- + H3O + Practice Problem 7.13 In some forms of kidney disease, the kidneys are unable to release HCO3- into the blood a What effect does lower than normal concentrations of HCO3- have on the H2CO3/ HCO3- buffer system and on the pH of blood serum? b How would the respiratory system respond to a lower than normal concentration of HCO3-? Chapter Objectives 267 At the library Revisited T Media Bakery o help with the digestion of food and to activate digestive enzymes, your stomach produces gastric juice, a pH 1.5 solution that contains HCl Overeating or stress can cause too much gastric juice to be released, which can lead to stomach upset For temporary relief, many people take antacids The active ingredients in antacids are weak bases, most commonly CaCO3, MgCO3, NaHCO3, Mg(OH)2, and Al(OH)3 Some examples include Antacid Weak Base Alka Seltzer NaHCO3 Di-Gel, Mylanta, Maalox Al(OH)3 and Mg(OH)2 Milk of Magnesia Mg(OH)2 Tums, Alka-2 CaCO3 Gaviscon Al(OH)3 and MgCO3 Rolaids AlNa(OH)2CO3 Riopan AlMg(OH)5 The carbonate, hydrogen carbonate, or hydroxide ions provided by these bases react with and neutralize some of the H3O+ that is present in gastric juice CO32-(aq) + H3O+(aq) ¡ HCO3-(aq) + H2O(I) HCO3-(aq) + H3O+(aq) ¡ H2CO3(aq) + H2O(I) OH-(aq) + H3O+(aq) ¡ 2H2O(I) Excessive use of antacids can lead to alkalosis, a higher than normal blood pH (Section 7.11) Besides alkalosis, some of the bases used as antacids can have other side effects—Al3+ and Ca2+ in large doses cause constipation, while Mg2+ has a laxative effect THINKING IT THROUGH Suggest an explanation for the fact that the antacids Di-Gel, Mylanta, and Maalox contain both Al(OH)3 and Mg(OH)2 Chapter objectives Chapter Objectives Sample and End of Practice Chapter Objective Summary Section Problems Problems List the common characteristics of acids and bases Acids dissolve some metals, turn litmus pink, and have a sour taste Bases feel slippery or soapy, turn litmus blue, and have a bitter taste Describe Brønsted–Lowry acids and bases and explain how they differ from their conjugates According to the Brønsted–Lowry definition, acids 7.2 release H+ and bases accept H+ In the equation for a Brønsted–Lowry acid–base reaction, an acid and a base appear on each side of the reaction arrow Each acid has a corresponding conjugate base (they differ by an H+) on the other side of the reaction equation 7.1 7.1 7.3–7.6 7.7–7.20 268 Chapter 7 Acids, Bases, and Equilibrium Sample and End of Practice Chapter Objective Summary Section Problems Problems Write the equilibrium constant expression for a reversible reaction and use Le Châtelier’s principle to explain how an equilibrium responds to being disturbed When the rates of the forward and reverse reactions are the same, a reversible reaction has reached equilibrium For the reaction aA + bB N cC + d D the equilibrium constant expression is written Le Châtelier’s principle states that when a reversible reaction is pushed out of equilibrium, the reaction responds to restore equilibrium Increasing the concentration of a reactant will, for example, cause a net forward reaction to occur until equilibrium is reestablished Use H3O+ concentration and pH to identify a solution as being acidic, basic, or neutral 7.3, 7.4 7.2–7.5 7.21–7.42 In a neutral solution, [H3O+] = * 10-7 M and pH = 7, in an acidic solution [H3O+] * 10-7 M and pH 7, and in a basic solution [H3O+] * 10-7 M and pH 7.5, 7.6 7.6–7.8 7.43–7.70 Use Ka and pKa values to assess the relative strength of acids and state the relationship between the strength of an acid and the strength of its conjugate base The larger the value of Ka or the smaller the value of pKa, the stronger an acid A strong acid has a weak conjugate base and a weak acid has a strong conjugate base 7.7 7.9 7.71–7.80 Describe the processes of neutralization and titration Neutralization involves reacting an acid with a base to produce water, a salt, and a neutral solution The concentration of an unknown acidic solution can be determined through titration, in which just enough of a basic solution of known concentration is added to neutralize the acid 7.8 7.10 7.81–7.90 Keq = [C]c[D]d [A]a[B]b Explain how the pH of a In a solution containing an acid (HA) and its conjugate 7.9, solution can affect the relative base (A-), when pH = pKa, [HA] = [A-] When 7.10 concentrations of an acid and pH pKa, [HA] [A-] and when pH pKa, its conjugate base and [HA] [A-] Buffers, solutions that resist changes in describe buffers pH when small amounts of acid or base are added, can be made by combining a weak acid and its conjugate base Buffers are most effective when the pH is within one unit of the pKa for the weak acid 7.11, 7.12 7.91–7.104 Describe the role of buffers, respiration, and the kidneys in maintaining a stable blood serum pH 7.13 Carbonic acid and hydrogen carbonate ion make up the most important of the blood serum buffers H2CO3 + H2O N HCO3 - + H3O + The hydrogen carbonate ion present in blood is formed when CO2 produced by metabolism reacts with water Respiration helps to determine serum concentration of CO2 and, indirectly, the pH of blood The kidneys help control pH by regulating the amount of HCO3- present in blood and by adding or removing H3O+ 7.11 7.105–7.112 End of Chapter Problems 269 End of Chapter Problems Answers to problems whose numbers are printed in color are given in Appendix C More challenging questions are marked with an asterisk 7.1 a Write the equilibrium constant expression for the reaction A N B b For this reaction, Keq = The drawing below represents an equilibrium mixture of A and B Explain B A 7.2 Brønsted–Lowry Acids and Bases B c If additional Bs are added, the mixture is no longer in equilibrium Explain B A B B B B 7.5 Name each of the following as an acid and as a binary compound a HCl b HBr 7.6 Name each of the following as an acid and as a binary compound a HF b HI d To adjust to equilibrium, which will take place for the mixture in part c: a net forward reaction or a net reverse reaction? e Based on your answer to part d, redraw the picture in part c as it will appear once the mixture reaches equilibrium 7.2 a Write the equilibrium constant expression for the reaction C N D b For this reaction, Keq = 0.25 The mixture of C and D shown below is not in equilibrium Which is too high, [C] or [D]? C C D C D c To adjust to equilibrium, which will take place: a net forward reaction or a net reverse reaction? d Based on your answer to part c, redraw the picture in part b as it will appear once the mixture reaches equilibrium 7.1 Acids and Bases 7.3 Which of the following observations would indicate that a compound might be an acid? a Turns litmus pink d Has a sour taste b Turns litmus blue e Dissolves a metal c Has a bitter taste f Feels slippery 7.4 Which of the observations in Problem 7.3 would indicate that a compound might be a base? 7.7 Give the formula for the conjugate base of each acid a H2CO3 c H2SO4 b HCO3- d H2PO4 7.8 7.9 7.10 Give the formula for the conjugate base of each acid a HSO4- c HPO42b H3PO4 d HNO3 Name the conjugate base of each acid in Problem 7.7 Name the conjugate base of each acid in Problem 7.8 7.11 Give the formula for the conjugate acid of each base a F- c NH3 b CN- d NO2 7.12 Give the formula for the conjugate acid of each base a OH- c CH3NH2 b Cl- d PO32 7.13 Water is amphoteric (can act as an acid or a base) a Write a chemical equation that shows water acting as an acid b In this reaction, what is the conjugate base of water? 7.14 Water is amphoteric (can act as an acid or a base) a Write a chemical equation that shows water acting as a base b In this reaction, what is the conjugate acid of water? 7.15 HCO3- is amphoteric a Write the equation for the reaction that takes place between HCO3- and the acid HCl b Write the equation for the reaction that takes place between HCO3- and the base OH- 7.16 HSO4- is amphoteric a Write the equation for the reaction that takes place between HSO4- and the acid HCl b Write the equation for the reaction that takes place between HSO4- and the base OH- 7.17 Identify the Brønsted–Lowry acids and bases for the forward and reverse reactions of each a F-(aq) + HCl(aq) N HF(aq) + Cl-(aq) b CH3CO2H(aq) + NO3 -(aq) N CH3CO2 -(aq) + HNO3(aq) 270 Chapter 7 Acids, Bases, and Equilibrium 7.18 Identify the Brønsted–Lowry acids and bases for the forward and reverse reactions of each a PO4 3-(aq) + NH4 +(aq) N NH3(aq) + HPO42(aq) b HCN(aq) + H2PO4 (aq) N CN-(aq) + H3PO4(aq) * 7.19 Complete each acid–base reaction For the forward and reverse reactions, identify each acid and its conjugate base a NH4+ + SO42- N b CN- + HI N * 7.20 Complete each acid–base reaction For the forward and reverse reactions, identify each acid and its conjugate base a OH- + HNO3 N b HPO42- + CO32- N 7.3 Equilibrium * 7.21 Which of the following statements are correct at * 7.22 7.23 7.24 7.25 equilibrium? a The concentration of reactants is always equal to the concentration of products b No reactants are converted into products c The rate of the forward reaction is equal to the rate of the reverse reaction Which of the following statements are correct at equilibrium? a The forward and reverse reactions stop b The concentration of reactants and products does not change c The rate of the forward reaction is greater than the rate of the reverse reaction Balance the chemical equation and write the equilibrium constant expression CH4 + H2O N H2 + CO Balance the chemical equation and write the equilibrium constant expression H2 + N2 N N2H4 Balance the chemical equation and write the equilibrium constant expression SO2 + O2 N SO3 7.26 Balance the chemical equation and write the equilibrium constant expression CS2 + H2 N CH4 + H2S 7.27 Write the equilibrium constant expression for each reaction In part b, H2O is a solvent a C(s) + CO2( g ) N 2CO( g ) b NH3(aq) + H2O( l ) N NH4 +(aq) + OH-(aq) 7.28 Write the equilibrium constant expression for each reaction In part b, H2O is a solvent a PCl5( g ) N PCl3( g ) + Cl2( g ) b H2SO4(aq) + 2H2O( l ) N 2H3O+(aq) + SO4 2-(aq) 7.29 Explain why the concentrations of solids not appear in equilibrium constant expressions 7.30 Explain why the concentrations of solvents not appear in equilibrium constant expressions 7.31 Write the reaction equation from which each equilibrium constant expression is derived (assume that no solids or solvents are present) [NOCl]2 [HBr]2 a Keq = b Keq = [H2][Br2] [NO] [Cl2] 7.32 Write the reaction equation from which each equilibrium constant expression is derived (assume that no solids or solvents are present) [SO3]2 [CH4][H2O] a Keq = b K = eq [O2][SO2]2 [CO][H2]3 7.33 Keq for the reaction below has a value of 4.2 * 10-4 CH3NH2(aq) + H2O( l ) N CH3NH3 +(aq) + OH-(aq) a Write the equilibirium constant expression for the b Which are there more of at equilibrium, products reaction or reactants? 7.34 When steam is passed over coal and allowed to come to equilibrium at 800°C, the equilibrium constant for the reaction has a value of 8.1 * 10-2 C(s) + H2O( g ) N CO( g ) + H2( g ) a Explain why the concentration of water appears in the equilibrium constant expression for this reaction, when it is often omitted b Write the equilibrium constant expression for the reaction c Which are there more of at equilibrium, reactants or products? 7.35 For each reaction, which is greater at equilibrium, the concentration of reactants or products? a HPO42- + CN- N PO43- + HCN Keq = 8.6 * 10-4 b H3PO4 + CN N H2PO4 + HCN Keq = 1.5 * 107 7.36 For each reaction, which is greater at equilibrium, the concentration of reactants or products? a HF + HCO3- N F- + H2CO3 Keq = 1.5 * 103 + b NH4 + HCO3 N NH3 + H2CO3 Keq = 1.5 * 10-3 7.4 Le Châtelier’s Principle 7.37 The enzyme carbonic anhydrase catalyzes the rapid conversion of CO2 and H2O into H2CO3 CO2( g ) + H2O( l ) N H2CO3(aq) a Write the equilibrium constant expression for this b What effect, if any, does doubling the amount of reaction carbonic anhydrase have on an equilibrium mixture of H2CO3 , CO2 , and H2O? Explain End of Chapter Problems 271 7.38 Carbon disulfide (CS2), used in the manufacture of some synthetic fabrics, is prepared by heating sulfur (S2) and charcoal (C) S2( g ) + C(s) N CS2( g ) a For the equilibrium above, what is the effect of increasing [CS2]? Of decreasing [CS2]? Of decreasing [S2]? b What would be the effect of continually removing CS2 from the reaction? 7.39 When carbon monoxide reacts with hydrogen gas, methanol (CH3OH) is formed CO( g ) + 2H2( g ) N CH3OH( g ) a For the equilibrium above, what is the effect of increasing [CO]? Of decreasing [H2]? Of increasing [CH3OH]? b What would be the effect of continually removing CH3OH from the reaction? 7.40 Nitrogen gas and hydrogen gas react to form ammonia N2( g ) + 3H2( g ) N 2NH3( g ) a For the equilibrium above, what is the effect of increasing [NH3]? b What is the effect of increasing [H2]? c What is the effect of decreasing [N2]? 7.41 The reaction below is exothermic If the reaction is at equilibrium, what would be the effect of H2( g ) + O2( g ) N H2O( g ) 7.42 a decreasing the concentration of O2? b increasing the concentration of H2? c increasing the pressure? d decreasing the volume? e decreasing the temperature? f increasing the temperature? The reaction below is endothermic If the reaction is at equilibrium, what would be the effect of O2( g ) N O3( g ) a decreasing the concentration of O2? b decreasing the concentration of O3? c decreasing the pressure? d increasing the volume? e decreasing the temperature? f increasing the temperature? 7.5 Ionization of Water 7.43 Calculate the H3O+ concentration present in water when a [OH-] = 8.4 * 10-3 M b [OH-] = 2.9 * 10-9 M c [OH-] = 5.8 * 10-1 M 7.44 Calculate the H3O+ concentration present in water when a [OH-] = 4.8 * 10-8 M b [OH-] = 6.6 * 10-2 M c [OH-] = 1.5 * 10-12 M 7.45 In Problem 7.43, indicate whether each solution is acidic, basic, or neutral 7.46 In Problem 7.44, indicate whether each solution is acidic, basic, or neutral 7.47 Calculate the OH- concentration present in water when a [H3O+] = 9.1 * 10-7 M b [H3O+] = 1.3 * 10-3 M c [H3O+] = 8.8 * 10-2 M 7.48 Calculate the OH- concentration present in water when a [H3O+] = 6.2 * 10-4 M b [H3O+] = 8.5 * 10-8 M c [H3O+] = 1.9 * 10-11 M 7.49 In Problem 7.47, indicate whether each solution is acidic, basic, or neutral 7.50 In Problem 7.48, indicate whether each solution is acidic, basic, or neutral 7.6 The pH Scale 7.51 Calculate the pH of a solution in which a [H3O+] = 1 * 10-5 M b [H3O+] = 3.9 * 10-2 M c [H3O+] = 1 * 10-7 M d [H3O+] = 7.0 * 10-5 M 7.52 Calculate the pH of a solution in which a [H3O+] = 4.9 * 10-3 M b [H3O+] = 4.9 * 10-4 M c [H3O+] = 2.2 * 10-13 M d [H3O+] = 8.5 * 10-7 M 7.53 In Problem 7.51, indicate whether each solution is acidic, basic, or neutral 7.54 In Problem 7.52, indicate whether each solution is acidic, basic, or neutral 7.55 Calculate the pH of a solution in which a [OH-] = 6.8 * 10-7 M b [OH-] = 1 * 10-7 M c [OH-] = 7.0 * 10-5 M d [OH-] = 1 * 10-1 M 7.56 Calculate the pH of a solution in which a [OH-] = 4.3 * 10-2 M b [OH-] = 9.7 * 10-13 M c [OH-] = 5.9 * 10-6 M d [OH-] = 2.4 * 10-4 M 7.57 In Problem 7.55, indicate whether each solution is acidic, basic, or neutral 7.58 In Problem 7.56, indicate whether each solution is acidic, basic, or neutral 7.59 What is the concentration of H3O+ in a solution if the pH is a 7.00 c 9.37 b 1.74 d 10.3 272 Chapter 7 Acids, Bases, and Equilibrium 7.60 What is the concentration of H3O+ in a solution if the pH is a 5.54 c 2.94 b 13.8 d 3.5 7.61 What is the concentration of OH- in a solution if the pH is a 4.12 b 7.45 c 2.80 d 11.61 7.62 What is the concentration of OH- in a solution if the pH is a 2.77 b 0.50 c 14.19 d 8.88 * 7.63 Alkali metals react with water to produce alkaline (basic) solutions Na, for example, reacts with water to form NaOH and H2(g) Write a balanced chemical equation for this reaction * 7.64 Alkaline earth metals react with water to produce alkaline (basic) solutions Ca, for example, reacts with water to form Ca(OH)2 and H2(g) Write a balanced chemical equation for this reaction * 7.65 In Problem 7.63, what is reduced and what is oxidized? * 7.66 In Problem 7.64, what is reduced and what is oxidized? 7.67 A 1.00 mL sample of blood serum has a pH of 7.35 a What is the concentration of H3O+? b What is the concentration of OH-? c How many moles of H3O+ are present? d How many moles of OH- are present? e How many H3O+ ions are present? f How many OH- ions are present? 7.68 A 5.00 mL sample of blood serum has a pH of 7.45 a What is the concentration of H3O+? b What is the concentration of OH-? c How many moles of H3O+ are present? d How many moles of OH- are present? e How many H3O+ ions are present? f How many OH- ions are present? 7.69 Sometimes the basicity of a solution is reported using pOH (pOH = -log[OH-]) What is the a [OH-] if pOH = 5.20 b [H3O+] if pOH = 4.33 c pH if pOH = 8.50 7.70 Sometimes the basicity of a solution is reported using pOH (pOH = -log[OH-]) What is the a [OH-] if pOH = 2.50 b [H3O+] if pOH = 13.44 c pH if pOH = 9.84 7.7 Acid and Base Strength 7.71 Write the chemical equation for the reaction of each weak acid with water Write the corresponding acidity constant expression a NH4+ b HPO42 7.72 Write the chemical equation for the reaction of each weak acid with water Write the corresponding acidity constant expression a CH3NH3+ b. HCO3- 7.73 a The Ka of NH4+ equals 5.6 * 10-10 For the reaction in Problem 7.71a, what are there more of at equilibrium, reactants or products? b The Ka of HPO42- equals 4.2 * 10-13 For the reaction in Problem 7.71b, what are there more of at equilibrium, reactants or products? c Which is the stronger acid, NH4+ or HPO42-? 7.74 a The pKa of CH3NH3+ equals 10.62 For the reaction in Problem 7.72a, what are there more of at equilibrium, reactants or products? b The pKa of HCO3- equals 10.25 For the reaction in Problem 7.72b, what are there more of at equilibrium, reactants or products? c Which is the stronger acid, CH3NH3+ or HCO3-? 7.75 Calculate the pKa of each acid and indicate which is the stronger acid a HClO, Ka = 3.0 * 10-8 b C2O4H-, Ka = 6.4 * 10-5 7.76 Calculate the pKa of each acid and indicate which is the stronger acid a C2O4H2 , Ka = 5.9 * 10-2 b C6O2H8 , Ka = 1.7 * 10-5 7.77 Write the formula of the conjugate base of each acid in Problem 7.75 Which of the two is the stronger base? 7.78 Write the formula of the conjugate base of each acid in Problem 7.76 Which of the two is the stronger base? * 7.79 0.10 M solutions of each of the following acids are prepared: acetic acid (Ka = 1.8 * 10-5) and hydrofluoric acid (Ka = 6.6 * 10-4) Which solution will have the lower pH? Explain 7.80 0.50 M solutions of each of the following acids are prepared: HCN (pKa = 9.31) and H2CO3 (pKa = 6.36) a Which solution will have the lower pH? Explain b Which solution will have the lower [OH-]? Explain 7.8 Neutralizing Acids and Bases 7.81 a How many moles of NaOH are present in 31.7 mL of 0.155 M NaOH? b How many moles of HCl are present in a 15.0 mL sample that is neutralized by the 31.7 mL of 0.155 M NaOH? c What is the molar concentration of the HCl solution described in part b of this question? 7.82 a How many moles of NaOH are present in 71.3 mL of 0.551 M NaOH? b How many moles of HCl are present in a 25.0 mL sample that is neutralized by the 71.3 mL of 0.551 M NaOH? c What is the molar concentration of the HCl solution described in part b of this question? 7.83 It requires 17.2 mL of 0.100 M KOH to titrate 75.0 mL of an HCl solution of unknown concentration Calculate the initial HCl concentration 7.84 It requires 35.7 mL of 0.250 M KOH to titrate 50.0 mL of an HCl solution of unknown concentration Calculate the initial HCl concentration End of Chapter Problems 273 7.85 It requires 25.9 mL of 0.100 M HCl to titrate 15.0 mL of an NaOH solution of unknown concentration What is the NaOH concentration? * 7.86 It requires 45.6 mL of 0.200 M HCl to titrate 25.0 mL of an NaOH solution of unknown concentration What is the NaOH concentration? * 7.87 It requires 12.2 mL of 0.100 M NaOH to neutralize 10.0 mL of an unknown H2SO4 solution Calculate the initial H2SO4 concentration H2SO4 + NaOH N Na2SO4 + 2H2O * 7.88 It requires 23.9 mL of 15.0 mM HCl to neutralize 20.0 mL of an unknown Ca(OH)2 solution Calculate the initial Ca(OH)2 concentration 2HCl + Ca(OH)2 N CaCl2 + 2H2O * 7.89 It requires 19.2 mL of 0.25 M NaOH to neutralize 15.0 mL of an unknown H3PO4 solution a Balance the reaction equation b Calculate the initial H3PO4 concentration H3PO4 + NaOH N Na3PO4 + H2O * 7.90 It requires 21.1 mL of 12.0 mM H2SO4 to neutralize 15.0 mL of an unknown Ca(OH)2 solution a Balance the reaction equation b Calculate the initial Ca(OH)2 concentration H2SO4 + Ca(OH)2 N CaSO4 + H2O 7.9 Effect of pH on Acid and Conjugate Base Concentrations 7.91 Ammonium ion (NH4+, pKa = 9.25) reacts with water to produce ammonia (NH3) and hydronium ion (H3O+) NH4+ + H2O N NH3 + H3O + a At pH 9.25, which of the following is true? [NH4+] = [NH3], [NH4+] [NH3], [NH4+] [NH3] b At pH 5.25, which of the following is true? [NH4+] = [NH3], [NH4+] [NH3], [NH4+] [NH3] c At pH 13.25, which of the following is true? [NH4+] = [NH3], [NH4+] [NH3], [NH4+] [NH3] 7.92 Lactic acid (CH3CH(OH)CO2H, pKa = 3.85) reacts with water to produce lactate ion (CH3CH(OH)CO2-) and hydronium ion (H3O+) CH3CH(OH)CO2H + H2O N CH3CH(OH)CO2- + H3O + a At pH 3.85, which of the following is true? [lactic acid] = [lactate ion], [lactic acid] [lactate ion], [lactic acid] [lactate ion] b At pH 11.61, which of the following is true? [lactic acid] = [lactate ion], [lactic acid] [lactate ion], [lactic acid] [lactate ion] c At pH 1.66, which of the following is true? [lactic acid] = [lactate ion], [lactic acid] [lactate ion], [lactic acid] [lactate ion] 7.93 a The pKa of NH4+ equals 9.25 For the reaction in Problem 7.71a, what is there more of at pH 7.0, NH4+ or NH3? b The pKa of HPO42- equals 7.21 For the reaction in Problem 7.71b, what is there more of at pH 10.0, HPO42- or PO43-? 7.94 a The Ka of CH3NH3+ equals 2.4 * 10-11 For the reaction in Problem 7.72a, what is there more of at pH 7.0, CH3NH3+ or CH3NH2? b The Ka of HCO3- equals 5.6 * 10-11 For the reaction in Problem 7.72b, what is there more of at pH 3.0, HCO3- or CO32-? * 7.95 At low pH, the amino acid aspartic acid has the structure shown below + H3NCHCO2H ƒ CH2CO2H Aspartic acid a Redraw aspartic acid, showing the acidic –CO2H and –NH3+ functional groups in their conjugate base form b In amino acids, –CO2H groups have pKa’s in the range 1.8–4.3 and –NH3+ groups have pKa’s in the range 9.1–12.5 Draw aspartic acid as it would appear at pH What is the net charge on aspartic acid at this pH? * 7.96 At low pH the amino acid lysine has the structure shown below + H3NCHCO2H ƒ + CH2(CH2)3NH3 Lysine a Redraw lysine, showing the acidic -CO2H and -NH3+ functional groups in their conjugate base form b In amino acids, –CO2H groups have pKa’s in the range 1.8–4.3 and –NH3+ groups have pKa’s in the range 9.1–12.5 Draw lysine as it would appear at pH What is the net charge on lysine at this pH? 7.97 An equilibrium mixture of H2CO3 and HCO3- has a pH of 6.5 What happens to the H2CO3 and HCO3- concentrations when the pH of the solution is adjusted to 8.5? 7.98 An equilibrium mixture of CH3CO2H and CH3CO2- has a pH of 4.8 What happens to the CH3CO2H and CH3CO2- concentrations when the pH of the solution is adjusted to 2.8? 7.10 Buffers 7.99 A buffer consisting of a weak acid and its conjugate base is prepared Over what pH range will the buffer be most effective? 7.100 Which weak acid or weak acids from Table 7.4 would be the best choice if you wished to prepare buffers with the following pH values? a 7.0 b 10.0 c 4.0 274 Chapter 7 Acids, Bases, and Equilibrium 7.101 If you wish to maintain a pH of 6.0, which is the better buffer, H2CO3 and HCO3- or H2PO4- and HPO42-? 7.102 If you wish to maintain a pH of 5.0, which is the better buffer, NH4+ and NH3 or CH3CO2H and CH3CO2-? 7.103 A buffer can be prepared using NH4+ and NH3 a Write an equation for the reaction that takes place when H3O+ is added to this buffer b Write an equation for the reaction that takes place when OH- is added to this buffer 7.104 A buffer can be prepared using lactic acid [CH3CH(OH)CO2H] and its conjugate base, lactate ion [CH3CH(OH)CO2-] a Write an equation for the reaction that takes place when H3O+ is added to this buffer b Write an equation for the reaction that takes place when OH- is added to this buffer 7.11 Maintaining the pH of Blood Serum 7.105 Explain how the H2CO3 >HCO3- buffer system helps maintain blood serum at a constant pH 7.106 At which pH is the H2CO3 >HCO3- buffer system most effective? 7.107 Explain how respiration helps maintain blood serum at a constant pH 7.108 In metabolic acidosis, the serum PCO2 levels can be lower than normal Explain this form of compensation by the respiratory system 7.109 Explain how the kidneys help maintain blood serum at a constant pH 7.110 In respiratory alkalosis, the serum concentration of HCO3- can be lower than normal Explain this form of compensation by the kidneys 7.111 Explain how hyperventilation can lead to respiratory alkalosis 7.112 One treatment for respiratory alkalosis is to breathe into a paper bag Explain how this helps restore the pH of the blood to its normal value BiochemistryLink | D i v i n g M a m m a l s , Oxygen, and Myoglobin 7.113 In terms of Le Châtelier’s principle, describe how the equilibrium Mb + O2 N MbO2 responds to increases in the concentration of a Mb c MbO2 b O2 7.114 Studies have shown that the value of Keq for the reaction in the previous problem is identical for humans and diving animals Why is the MbO2 in diving animals able to provide more O2 than the MbO2 in humans? Math Support—Logs and Antilogs 7.115 Calculate the logarithm of each number Assume that each is a measured quantity a 1 * 10-9 c 3.4 * 10-2 12 b 1 * 10 d 9.7 * 104 7.116 Calculate the logarithm of each number Assume that each is a measured quantity a 1 * 103 c 2.2 * 10-13 -6 b 1 * 10 d 8.7 * 109 7.117 Calculate the antilogarithm of each number a c 7.9 b -3 d 15.3 7.118 Calculate the antilogarithm of each number a -1 c 3.5 b 12 d 10.9 BiochemistryLink | P l a n t s a s p H Indicators 7.119 Someone tells you that the spice turmeric is a natural pH indicator How might you test this claim, using only substances that you are likely to have available at home? 7.120 When you pick a particular flower and immerse it in a pH 7 solution, there is no change in the color If you place the same flower in a pH 9 solution, the color changes Why does this happen? BiochemistryLink | T h e H e n d e r s o n – Hasselbalch Equation * 7.121 A buffer may be prepared using acetic acid (pKa = 4.74) and its conjugate base What is the pH of this buffer when CH3CO2H + H2O N CH3CO2 - + H3O + a the concentration of CH3CO2H is twice that of CH3CO2- ? b the concentration of CH3CO2H is one-third that of CH3CO2- ? c [CH3CO2H] = 0.15 M and [CH3CO2-] = 0.62 M? d [CH3CO2H] = 0.33 M and [CH3CO2-] = 0.095 M? * 7.122 A buffer may be prepared using dihydrogen phosphate (pKa = 7.21) and its conjugate base What is the pH of this buffer when H2PO4 - + H2O N HPO4 - + H3O + a the concentration of H2PO4- is one-fifth that of HPO42- ? b the concentration of H2PO4- is four times greater than that of HPO42- ? c [H2PO4-] = 0.68 M and [HPO42-] = 0.86 M? d [H2PO4-] = 0.55 M and [HPO42-] = 0.15 M? * 7.123 A buffer may be prepared using ammonium ion (Ka = 5.6 * 10-10) and its conjugate base For this buffer, NH4 + + H2O N NH3 + H3O + a What is the concentration of NH4+ if [NH3] = b What is the concentration of NH4+ if [NH3] = 0.15 M and pH = 8.25? 0.36 M and pH = 9.30? Solutions to Practice Problems 275 c What is the concentration of NH3 if [NH4+] = 0.50 M and pH = 8.55? * 7.124 A buffer may be prepared using hydrogen carbonate ion (Ka = 5.6 * 10-11) and its conjugate base For this buffer, HCO3 - + H2O N CO3 - + H3O + a what is [HCO3-] if [CO32-] = 0.053 M and pH = b what is [HCO3-] if [CO32-] = 0.36 M and pH = c what is [CO32-] and [HCO3-] if the combined 10.90? f For carbonic acid (H2CO3), Ka = 4.4 * 10-7 Which g Which is the stronger base, the conjugate base of hy- h Assuming that the reaction in your answer to part a is the stronger acid, hypochlorous acid or carbonic acid? pochlorous acid or the conjugate base of carbonic acid? is at equilibrium, what will happen to restore equilibrium if the pH is lowered by addition of H3O+, a net forward reaction or a net reverse reaction? i Over what pH range would a solution containing hypochlorous acid and its conjugate base be effective as a buffer? 9.44? concentration of CO32- and HCO3- = 0.50 M and pH = 10.25? * 7.125 Buffers are effective when the pH is within one unit of the pKa of the weak acid (pH = pKa ; 1) An HCO3-/ CO32- buffer has a pH of 11.25 At this pH, how many times more CO32- is present than HCO32-? For HCO3- pKa = 10.25 HCO3 - + H2O N CO3 - + H3O + * 7.126 For the buffer in Problem 7.125, at pH 9.25 how many times less CO32- is present than HCO32-? Learning Group Problems 7.127 a Write the balanced chemical equation for the reversible acid–base reaction that takes place between nitrous acid (HNO2) and water b Write the equilibrium constant expression for this reaction c Name and draw the Lewis structure of the conjugate base of nitrous acid d For nitrous acid, Ka = 4.0 * 10-4 What is the value of pKa? e Which are there more of once the reaction in part a has reached equilibrium, reactants or products? f For carbonic acid (H2CO3), Ka = 4.4 * 10-7 Which is the stronger acid, nitrous acid or carbonic acid? g Which is the stronger base, the conjugate base of nitrous acid or the conjugate base of carbonic acid? h Assuming that the reaction in your answer to part a is at equilibrium, what will happen to restore equilibrium if the pH is lowered by addition of H3O+, a net forward reaction or a net reverse reaction? i Over what pH range would a solution containing nitrous acid and its conjugate base be effective as a buffer? 7.128 a Write the balanced chemical equation for the reversible acid–base reaction that takes place between hypochlorous acid (HOCl) and water b Write the equilibrium constant expression for this reaction c Draw the Lewis structure of the conjugate base of hypochlorous acid d For hypochlorous acid, Ka = 3.5 * 10-8 What is the value of pKa? e Which are there more of once the reaction in part a has reached equilibrium, reactants or products? SOLUTIONS TO PRACTICE PROBLEMS 7.1 a HSO4 - + HPO4 2- N H2SO4 + PO4 Base Acid b HPO4 2- Base + OH N H2O + PO4 Acid 7.2 Acid - Base Acid 3- Base a H2PO4 - + HSO4 - N H3PO4 + SO4 2- Base Acid Acid Base - H2PO4 and H3PO4 are conjugates HSO4- and SO42- are conjugates b H2PO4 - + CN- N HPO4 2- + HCN Acid Base - Base Acid H2PO4 and HPO4 are conjugates CN- and HCN are conjugates [NO2]4[O2] [CO][H2O] 7.3 a. Keq = ; b. Keq = [H2][CO2] [N2O5]2 7.4 Reactant concentration 7.5 a. forward; b. forward; c. reverse; d. reverse; e. forward 7.6 a. 1.0 * 10-9 M; b. 1.2 * 10-4 M; c. 2.6 * 10-14 M; d. 2.3 * 10-6 M 7.7 a. 7.30; b. 13.0; c. 1.0; d. 11.67 7.8 a [OH-] = 1.6 * 10-7 M (basic) b [OH-] = 1.3 * 10-5 M (basic) c [OH-] = 2.0 * 10-11 M (acidic) 7.9 a Ka for HNO2 = * 10-4 and Ka for H2SO3 = * 10-2 H2SO3 is the stronger acid b The conjugate base of HNO2 is NO2- and the conjugate base of H2SO3 is HSO3- c NO2- is the stronger of the two bases 7.10 0.52 M 2- - - 7.11 a CH3CHCO2 ƒ NH2 b CH3CHCO2 + ƒ NH3 7.12 a H2PO4 - + OH - ¡ HPO4 2- + H2O b. 6.21–8.21 7.13 a A lower than normal HCO3- concentration causes a net forward reaction in the H2CO3/HCO3- buffer system This leads to a drop in the pH of blood serum b. To offset a drop in pH, the rate of breathing would increase Derek Berwin/The Image Bank/Getty Images in the spring It is early spring and the air is filled with pollen You are acutely aware of this because your eyes are watery and itchy, your nose is running, and you feel just awful On a trip to the drugstore, you stop and read the labels on antihistamines, hoping to find one that will alleviate your allergy symptoms Some antihistamines claim to be non-drowsy, others warn you to not operate heavy machinery while taking them, and all contain drugs with very long names You suspect that antihistamines work against histamine, but not know what histamine is ... 0.0000000000 01 = 10 9 = 10 6 = 10 3 = 10 2 = 10 1 = 10 0 = 10 -1 = 10 -2 = 10 -3 = 10 -6 = 10 -9 = 10 -12 GENERAL, ORGANIC, AND BIOLOGICAL CHEMISTRY An Integrated Approach F o u rth Kenneth W Raymond Eastern... Acids 426 13 .4 DNA Structure 509 13 .5 Denaturation 512 13 .6 Nucleic Acids and Information Flow 514 13 .7 DNA Replication 515 11 .2 Waxes 427 13 .8 Transcription and RNA 517 11 .3 Triglycerides ... energy of motion 1. 2 1. 3 1. 7, 1. 8, 1. 13 1. 26 Describe and give examples of physical properties and physical change Physical properties, including odor and melting point, and physical changes, including