0001
`
`PSG2027
`Catalent Pharma Solutions v. Patheon Softgels
`IPR2018-00422
`
`

`

`McGraw-Hill
`A Division ofTheMcGraw-Hill Companies
`
`CHEMISTRY
`
`Copyright © 1998, 1994, 1991, 1988, 1984, 1981 by McGraw-Hill, Inc. All rights reserved.
`Printed in the United States of America. Except as permitted under the United States Copyright
`Act of 1976, no part of this publication may be reproducedor distributed in any form or by any
`means, or stored in a data baseor retrieval system, without the prior written permission of the
`publisher.
`
`This book is printed on acid-free paper.
`
`Photo Credits appear on pages C1—-C3, and on this page by reference.
`
`1234567890 VNH VNH 90987
`
`ISBN 0-07-011644-X
`
`Publisher: James M. Smith
`Sponsoring editor: Kent A. Peterson
`Production supervisor: Paula Keller
`Design manager and cover designer: Charles Carson
`Cover and chapter openers illustrator: Joe Gillians
`Designer: Edward Butler
`Compositor: York Graphic Services, Inc.
`Typeface: 10/12 Times Roman
`Printer and binder: Von Hoffmann Press, Inc.
`
`Library of Congress Cataloging-in-Publication Data
`Chang, Raymond.
`Chemistry / Raymond Chang.—6th ed.
`.
`cm.
`
`Includes index.
`ISBN 0-07-011844-X
`1. Chemistry.
`I. Title.
`QD31.2.C37
`1998
`540—de21
`
`INTERNATIONAL EDITION
`
`97-16758
`
`Copyright © 1998. Exclusive rights by The McGraw-Hill
`Companies, Inc. for manufacture and export. This book cannot be
`
`re-exported from the country to which it is consigned by McGraw-Hill.
`
`
`The International Edition is not available in North America.
`
`Whenorderingthistitle, use ISBN 0-07-115221-0.
`
`hittp://www.mhhe.com
`
`0002
`
`0002
`
`

`

`Contents in Brief
`
`Chapter 1
`
`CHEMISTRY: THE STUDY OF CHANGE
`
`Chapter 2
`
`Atoms, MOLECULES, AND IONS
`
`Chapter 3
`
`Chapter 4
`
`Chapter 5
`
`Chapter 6
`
`Chapter 7
`
`Chapter &
`
`Chapter 9
`
`Mass RELATIONSHIPS IN CHEMICAL REACTIONS
`
`REACTIONS IN AQUEOUS SOLUTION
`
`GASES
`
`‘THERMOCHEMISTRY
`
`QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS
`
`PERIODIC RELATIONSHIPS AMONG THE ELEMENTS
`
`CHEMICAL BonpinG I: Basic CONCEPTS
`
`Chapter 10
`
`CHEMICAL BONDING II: MOLECULAR GEOMETRY AND
`
`Chapter 11
`
`Chapter 12
`
`Chapter 13
`
`HYBRIDIZATION OF ATOMIC ORBITALS
`
`INTERMOLECULAR FORCES AND LIQUIDS AND SOLIDS
`
`PuysicaAL PROPERTIES OF SOLUTIONS
`
`CHEMICAL KINETICS
`
`Chapter 14
`
`CHEMICAL EQUILIBRIUM
`
`Chapter 15
`
`Actps AND BasEs
`
`Chapter 16
`
`Chapter 17
`
`Chapter 18
`
`Chapter 19
`
`Chapter 20
`
`Chapter 21
`
`Chapter 22
`
`Chapter 23
`
`Chapter 24
`
`Chapter 25
`
`Actb-Base EQUILIBRIA AND SOLUBILITY EQUILIBRIA
`
`CHEMISTRY IN THE ATMOSPHERE
`
`ENTROPY, FREE ENERGY, AND EQUILIBRIUM
`
`ELECTROCHEMISTRY
`
`METALLURGY AND THE CHEMISTRY OF METALS
`
`NONMETALLIC ELEMENTS AND ‘THEIR COMPOUNDS
`
`‘TRANSITION METALS AND COORDINATION COMPOUNDS
`
`NUCLEAR CHEMISTRY
`
`ORGANIC CHEMISTRY
`
`SYNTHETIC AND NATURAL ORGANIC POLYMERS
`
`0003
`
`36
`
`68
`
`108
`
`154
`
`202
`
`242
`
`286
`
`328
`
`366
`
`416
`
`466
`
`506
`
`558
`
`596
`
`644
`
`692
`
`724
`
`756
`
`802
`
`830
`
`870
`
`902
`
`938
`
`970
`
`0003
`
`

`

` ] 5
`
`Acids and Bases
`
`INTRODUCTION
`
`SOME OF THE MOST IMPORTANT PROCESSES IN CHEMICAL AND BIOLOG-
`
`15.1
`
`BRONSTED ACIDS AND BASES
`
`ICAL SYSTEMS ARE ACID-BASE REACTIONS IN AQUEOUS SOLUTIONS.
`
`IN
`
`THIS FIRST OF TWO CHAPTERS ON Tie PROPERTIES OF ACIDS AND BASES,
`WE WILL STUDY THE DEFINITIONS OF ACIDS AND BASES, THE PH SCALE,
`THE IONIZATION OF WEAK ACIDS AND WEAK BASES, AND THE RELA-
`
`:
`|
`:
`
`TIONSHIP BETWEEN ACID STRENGTH AND MOLECULAR STRUCTURE. WE
`
`WILL ALSO LOOK AT OXIDES THAT CAN ACT AS ACIDS OR BASES.
`
`15.2
`
`15.3
`15.4
`15.5
`
`15.6
`
`THE ACID-BASE PROPERTIES OF WATER
`
`pH—A MEASURE OF ACIDITY
`
`STRENGTH OF ACIDS AND BASES
`
`WEAK ACIDS AND ACID IONIZATION
`CONSTANTS
`
`WEAK BASES AND BASE IONIZATION
`CONSTANTS
`
`15.7
`
`THE RELATIONSHIP BETWEEN THE
`
`IONIZATION CONSTANTS OF ACIDS
`
`AND THEIR CONJUGATE BASES
`
`15.8
`
`DIPROTIC AND POLYPROTIC ACIDS
`
`15.9
`
`MOLECULAR STRUCTURE AND THE
`
`STRENGTH OF ACIDS
`
`;
`
`15.10
`
`ACID-BASE PROPERTIES OF SALTS
`
`15.11
`
`ACID-BASE PROPERTIES OF OXIDES
`AND HYDROXIDES
`
`15.12
`
`LEWIS ACIDS AND BASES
`
`0004
`
`0004
`
`

`

`598
`
`ACIDS AND BASES
`
`BRONSTED ACIDS AND BASES
`
`In Chapter 4 we defined a Brgnsted acid as a substance capable of donating a Proton,
`and a Brgnsted base as a substance that can accept a proton. These definitionsare gen.
`erally suitable for a discussion of the properties and reactions of acids and bases.
`An extension of the Brgnsted definition of acids and bases is the conceptof ih,
`conjugate acid-base pair, which can be defined as an acid andits conjugate base or
`a base andits conjugate acid. The conjugate base of a Brgnstedacidis the species that
`remains when one proton has been removed from the acid. Conversely, a conjugate
`acid results from the addition of a proton to a Brgnsted base.
`Every Brgnsted acid has a conjugate base, and every Bronsted base has a conju.
`gate acid. For example, the chloride ion (C1~) is the conjugate base formed from the
`acid HCl, and H,O is the conjugate base of the acid H;0* (hydroniumion). Similarly,
`the ionization of acetic acid can be represented as
`tk
`tL
`H-¢-C—O-H + HO: = H—--C—0D4 ray
`
`
`H
`
`H
`
`H
`
`Conjugate means
`“Joined together.”
`
`H
`
`The formula of a conjugate base
`always has one fewer hydrogen
`atom and one more negative
`charge (or one fewer positive
`charge) than the formula of the
`corresponding acid.
`
`CH;COOH(aq) + H,0() == CH;COO™(aq) + H30*(aq)
`acid,
`base,
`base,
`acids
`The subscripts 1 and 2 designate the two conjugate acid-base pairs. Thus the acetate
`ion (CH;COO_)is the conjugate base of CH,;COOH.Both the ionization of HCI(see
`Section 4.3) and the ionization of CH;COOH are examples of Brg@nsted acid-base re-
`actions.
`The Brgnsted definition also allows us to classify ammonia as a base because of
`its ability to accept a proton:
`
`H
`.
`..
`|
`H-N-H + HO: —=|H-N-H]
`
`H
`
`H
`
`H
`
`+
`
`..
`+H—0 =
`
`NH,(aq) + H,O() === NHitag)
`base;
`acid,
`acid,
`
`+ OH(aq)
`base,
`
`In this case, NHf is the conjugate acid of the base NH3, and the hydroxide ion OH
`is the conjugate base of the acid H,O. Note that the atom in the Brgnsted base that ac-
`cepts a H* ion must havea lonepair.
`In Example 15.1 we identify the conjugate pairs in an acid-base reaction.
`
`
`
`Similar problem: 15.5.
`
`0005
`
`
`
`0005
`
`

`

`13.2
`
`THE ACID-BASE PROPERTIES OF WATER
`
`599
`
`PRACTICE EXERCISE
`Identify the conjugate acid-base pairs for the reaction
`
`
`CN™ + H,O
`
`HCN + OH™
`
`It is acceptable to represent the proton in aqueoussolution either as H* or as
`H30*. The formula H* is less cumbersomein calculations involving hydrogen ion
`concentrations and in calculations involving equilibrium constants, whereas H,0* is
`more useful in a discussion of Brgnsted acid-base properties.
`
` :
`|THE ACID-BASE PROPERTIES OF WATER
`
`Water, as we know,is a unique solvent. Oneof its special propertiesis its ability to act
`either as an acid and as a base. Water functions as a base in reactions with acids such
`as HCl and CH3COOH,andit functions as an acid in reactions with bases such as NH3.
`Tap water and water from under- Water is a very weak electrolyte and therefore a poor conductorofelectricity, butit
`ground sources do conduct
`does undergo ionization to a small extent:
`electricity because they contain
`
`many dissolved ions.
`
`_
`
`(15.1)
`H*(aq) + OH”(aq)
`H,0(!)
`This reaction is sometimescalled the autoionization of water. To describe the acid-base
`properties of water in the Brgnsted framework, we express its autoionization as fol-
`lows (also shown in Figure 15.1):
`
`or
`
`H-O:+ HO: aa + HO:
`i
`i
`
`H,O + H,O ——H,0* + OH
`acid,
`base,
`acid,
`base,
`
`The acid-base conjugate pairs are (1) H,O (acid) and OH™ (base) and (2) H,O* (acid)
`and H,O (base).
`
`Foe eePReeeQL NATER rarest mmwitem dere
`Tn the study of acid-base reactions in aqueous solutions, the hydrogen ion concentra-
`tion is key, because it indicates the acidity or basicity of the solution. Expressing the
`proton as H* rather than H;0*, we can write the equilibrium constant for the au-
`toionization of water, Equation (15.1), as
`_ [H*(0H7)]
`[H,0]
`
`Ko
`
`FIGURE 15.1 Reaction be-
`£€n two water molecules to
`i. hydronium and hydroxide
`
`s,
`
`lo,
`
`+
`
`_ 0006
`
`0006
`
`

`

`600
`
`ACIDS AND BASES
`
`Recall that in pure water,
`[H,0] = 35.5 M (see p. 564).
`
`Tf you could randomly remove
`and examine ten particles (H,O,
`H*, or OH™) per second from a
`liter of water, it would take you
`two years, working nonstop, to
`find one H* ion!
`
`
`
`Since a very small fraction of water molecules are ionized, the concentration of Water
`[H,O] remainsvirtually unchanged. Therefore
`(15.2)
`K.[H,0] = K, = [H*][0H7]
`The equilibrium constant K,, is called the ion-product constant, which is the produc,
`of the molar concentrations of H* and OH™ionsat a particular temperature.
`In pure water at 25°C, the concentrations of H* and OH” ionsare equal and found
`to be [H*] = 1.0 X 1077 M and [OH] = 1.0 X 1077 M. Thus, from Equation (15.2),
`at 25°C
`
`w = (1.0 X 1077)1.0 X 1077) = 1.0 x 107
`
`Whether we have pure water or a solution of dissolved species, the followingrelation
`always holds at 25°C:
`
`K,, = {H*][OH™] = 1.0 x 107"
`
`Whenever [H*] = [OH™], the aqueous solution is said to be neutral. In an acidic go.
`lution there is an excess of Ht ions and [H*] > [OH]. In a basic solutionthereis an
`excess of hydroxide ions, so [H*] < [OH™]. In practice we can change the concen.
`tration of either H* or OH™ ions in solution, but we cannot vary both of them inde-
`pendently. If we adjust the solution so that [H*] = 1.0 x 107° M, the OH™ concen-
`tration must change to
`
`
`
`[OH™] =
`
`Ky _ 10x 1074
`Aw ~ VAI
`-8
`(H#*]~
`10x19 10x 10°M
`
`Be aware that the calculations shown above, and indeedall the calculations in-
`volving solution concentrations discussed in Chapters 14-16, are subject to error be-
`cause we have implicitly assumed ideal behavior. In reality, ion-pair formation and
`other types of intermolecular interactions may affect the actual concentrations of species
`in solution and hence also the equilibrium constant values. The situation is analogous
`to the relationships between ideal gas behavior and the behavior of real gases discussed
`in Chapter 5. Depending on temperature, volume, and amountand type of gas present,
`the measured gas pressure may differ from that calculated using the ideal gas equa-
`tion. Similarly, the actual, or “effective,” concentration of a solute may not be what we
`think it should be from the amountof substanceoriginally dissolved in solution. Just
`as we have the van der Waals and other equations to reconcile discrepancies between
`the ideal gas equation and nonideal gas behavior, we can account for nonideal behav-
`ior in solution. But for our purposes, it is acceptable to ignore deviations from ideal-
`ity. In most cases this approach will give us a good approximation of the chemical
`processes that actually take place in the solution phase.
`An application of Equation (15.2) is given in Example 15.2.
`
`EXAMPLE 15.2
`The concentration of OH™ ions in a certain household ammonia cleaning solution
`is 0.0025 M. Calculatethe concentration of H™ ions.
`
`Answer Rearranging Equation (15.2), we write
` Ht} =. ake Oe Oe
`
`=poreeeeeeee (TE A
`{OH]
`60025
`4.0 X 10°"
`Many household cleaning fluids
`confain ammonia.
`
`0007
`
`0007
`
`

`

`15.3
`
`pH—A MEASURE OF ACIDITY
`
`601
`
`Similar problems: 15.15, 15.16.
`
`Comment Since [H*] <[OH™], the solution is basic, as we would expect from the
`earlier discussion of the reaction of ammonia with water.
`
`PRACTICE EXERCISE
`Calculate the concentration of OH™ ions in a HCI solution whose hydrogen ion con-
`centration is 1.3 M.
`
`REY pHa MEASURE OF ACIDITY
`Because the concentrations of H* and OH™ ions in aqueoussolutions are frequently
`very small numbers and therefore inconvenient to work with, Soren Sorensen’ in 1909
`proposed a more practical measure called pH. The pH ofa solution is defined as the
`negative logarithm of the hydrogen ion concentration (in mol/L):
`
`pH = —log [H*]
`
`(15.3)
`
`The pH of concentrated acid solu-
`tions can be negative. For exam-
`ple, the pH of a 2.0 M HCl
`solution is —0.30.
`
`Keep in mind that Equation (15.3) is simply a definition designed to give us conve-
`nient numbers to work with. The negative logarithm gives us a positive numberfor pH,
`which otherwise would be negative due to the small value of [H*]. Furthermore, the
`term [H*] in Equation (15.3) pertains only to the numericalpartof the expression for
`hydrogen ion concentration, for we cannot take the logarithm of units. Thus, like the
`equilibrium constant, the pH of a solution is a dimensionless quantity.
`Since pH is simply a way to express hydrogen ion concentration, acidic and ba-
`sic solutions at 25°C can be distinguished by their pH values, as follows:
`Acidic solutions:|[Ht] > 1.0 x 10°? M, pH < 7.00
`
`Basic solutions:
`[fH*]<1.0 x 10°7 M, pH > 7.00
`Neutral solutions:
`[H*] = 1.0 X 1077 M, pH = 7.00
`Notice that pH increases as [H*] decreases.
`In the laboratory, the pH of a solution is measured with a pH meter (Figure 15.2).
`Table 15.1 lists the pHs of a number of commonfluids. As you cansee, the pH of body
`fluids varies greatly, depending on location and function. The low pH (highacidity) of
`
`FIGURE 15.2 A pH meter is
`ommonly used in the laboratory
`i, ¢fermine the pH of a solu-
`on. Although many pH meters
`Qe scales marked with values
`"°M 1 fo 14, pH values can,in
`"Ct, be less than 1 and greater
`than 14,
`
`Soren Peer Lauritz Sorensen (1868-1939). Danish biochemist. Sorensen originally wrote the symbol as Dy and called p
`the “hydrogen ion exponent” (Wasserstoffionexponent), it is the initial letter of Potenz (German), puissance (French), and
`
`power(English). It is now customary to write the symbol as pH.
`
`0008
`
`0008
`
`

`

`602
`
`ACIDS AND BASES
`
`TABLE 15.1
`The pHs of
`
`Some CommonFluids
`
`SAMPLE
`
`pH VALUE
`
`Gastric juice in
`the stomach
`
`Lemon juice
`Vinegar
`Grapefruit juice
`Orange juice
`Urine
`
`Water exposed
`to aur*
`Saliva
`Milk
`
`Pure water
`Blood
`Tears
`Milk of ©
`magnesia
`Household
`ammonia
`
`1.0-2.0
`
`2.4
`3.0
`3.2
`3.5
`4.8-7.5
`
`5.5
`
`6.4-6.9
`6.5
`
`7.0
`7.35-7.45
`74
`10.6
`
`11.5
`
`*Water exposedto air for a long period of
`time absorbs atmospheric CO, to form car-
`bonic acid, H.CO3.
`
`In each case the pH has only two
`significant figures. The two fig-
`ures to the right of the decimal in
`3.49 tell us that there are two sig-
`nificant figures in the original
`number (see Appendix 4).
`
`Similar problems: 15.17, 15.18.
`
`gastric juices facilitates digestion whereas a higher pH of blood is necessary for the
`transport of oxygen. These pH-dependentactions will be illustrated in Chemistry |in
`Action essays in this and the next chapter.
`A pOHscale analogous to the pH scale can be devised using the negative loga.
`rithm of the hydroxide ion concentration of a solution. Thus we define pOH as
`
`pOH = —log [OH™]
`
`(15.4)
`
`Now consider again the ion-product constant for water:
`[H*][OH7] = K,, = 1.0 x 107-#
`
`Taking the negative logarithm of both sides, we obtain
`—(log [H*] + log [OH~]) = —log (1.0 x 107)
`—log [H*] — log [OH] = 14.00
`
`From the definitions of pH and pOH weobtain
`
`pH + pOH = 14,00
`
`(15.5)
`
`Equation (15.5) provides us with another way to express the relationship between the
`H* ion concentration and the OH™ ion concentration.
`The following examplesillustrate calculations involving pH.
`
`EXAMPLE 15.3
`The concentration of H* ionsin a bottle of table wine was 3.2 X 107 Mrightaf.
`ter the cork was removed. Only half of the wine was consumed. The otherhalf, af-
`ter it had been standing open to the air for a month, was found to have a hydrogen
`ion coricentration equal to 1.0 X 107? M. Calculate the.pH of the wine on these two
`occasions.
`Answer Whenthe bottle wasfirst opened, [H*] = 3.2 X 10~4.M, which we sub-
`stitute in Equation (15.3).
`
`pH =—log {H*]
`= —log (3.2 X 1074) = 3.49
`On the second occasion, fH*] = 1,0 1073 M,so that
`pH = =log (1.0 x 107%) = 3.00
`Comment The increase in hydrogen ion concentration (or decreasein pH).is largely
`the result of the conversion of some of the alcohol (ethanol) to acetic acid, a reac-
`tion that takes place in the presence of molecular oxygen.
`PRACTICE EXERCISE
`Nitric acid (HNO3)is used in the production of fertilizer, dyes, drugs, and explo-
`sives. Calculate the pH of a HNO, solution having a hydrogen ion concentration of
`0.76 M.
`
`EXAMPLE 15.4
`The pH of rainwater collected |in a certain region of the northeastern United States
`ona particular day was 4.82. Calculate the H* ion concentration of the rainwater
`
`i
`:
`
`|
`
`i
`:
`
`0009
`
`0009
`
`

`

`
`
`
`
`Similar problem: 15.19.
`
`
`
`15.4
`
`STRENGTHS OF ACIDS AND BASES
`
`603
`
`Similar problem: 15.18.
`
`REZG strencru OF ACIDS AND BASES
`Strong acids are strong electrolytes which, for practical purposes, are assumedto ion-
`ize completely in water (Figure 15.3). Most of the strong acids are inorganicacids: hy-
`drochloric acid (HCI), nitric acid (HNO3), perchloric acid (HC1O,), and sulfuric acid
`(H)SO,):
`
`HCl(aq) + H,O() —> H,0*(aq) + Cl (aq)
`HNO;3(aq) + H.O@) —— H,0°*(aq) + NO3(aq)
`HCIO,(ag) + H,OU) ——» H,0*(ag) + ClOz (aq)
`H,SO,(ag) + H,0(1) ——> H30°(ag) + HSOq(aq)
`
`
`
`
`
`Note that H,SO, is a diprotic acid; we show only thefirst stage of ionization here. At
`equilibrium, solutions of strong acids will not contain any nonionized acid molecules.
`Most acids are weak acids, which ionize only to a limited extent in water. At equi-
`librium, aqueoussolutions of weak acids contain a mixture of nonionized acid mole-
`cules, H,O7 ions, and the conjugate base. Examples of weak acids are hydrofluoric
`acid (HF), acetic acid (CH;COOH), and the ammonium ion (NH7). The limited ion-
`
`0010
`
`0010
`
`

`

`604
`
`ACIDS AND BASES
`
`Before
`ionization
`
`HA
`
`At
`equilibrium
`
`Ht A
`
`
`
`
`
`
`
`(c) i
`
`FIGURE 15.3 The extent of
`ionization of (a) a strong acid
`that undergoes 100 percent ion-
`ization, {b) a weak acid, and
`{c] a very weak acid.
`
`
`
`ization of weak acidsis related to the equilibrium constant for ionization, which We
`will study in the next section.
`Like strong acids, strong bases are all strong electrolytes that ionize completely
`in water. Hydroxidesof alkali metals and certain alkaline earth metals are strong baggy
`{All alkali metal hydroxides are soluble. Of the aikaline earth hydroxides, Be(OH),
`and Mg(OH), are insoluble; Ca(OH), and Sr(OH), are slightly soluble; and Ba(Ok),
`is soluble]. Some examples of strong basesare:
`
`_
`H,0
`NaOH(s) ——> Na* (ag) + OH (aq)
`H,O
`KOH(s) —> K* (aq) + OH (aq)
`HO
`Ba(OH).(s) ——> Ba*t (aq) + 20H”(aq)
`
`
`
`Strictly speaking, these metal hydroxides are not Brgénsted bases because they cannot
`accept a proton. However, the hydroxide ion (OH) formed when they ionize is g
`Brgnsted base because it can accept a proton:
`
`H,0* (aq) + OH~ (ag) —> 2H,0(1)
`
`Thus, when we call NaOH or any other metal hydroxide a base, we are actually re-
`ferring to the OH species derived from the hydroxide.
`Weak bases, like weak acids, are weak electrolytes. Ammonia is a weak base.It
`ionizes to a very limited extent in water:
`
`NH3(aq) + H,O() === NHj(aq) + OH(aq)
`
`Note that NH3 does not ionize like an acid because it does not split up to form ions
`the way, say, HCI does.
`Table 15.2 lists some important conjugate acid-base pairs, in order oftheir rela-
`tive strengths. Conjugate acid-base pairs have the following properties:
`
`»
`
`If an acid is strong, its conjugate base has no measurable strength. Thus the Cl7
`ion, which is the conjugate base of the strong acid HCl, is an extremely weak base.
`
`° HO”is the strongest acid that can exist in aqueous solution. Acids stronger than
`H,0*react with water to produce H;O* andtheir conjugate bases. Thus HCI, which
`is a stronger acid than H,0*, reacts with water completely to form H,O07 and Cl”:
`
`HCl(aq) + H,O() —> H,0* (ag) + Cl~(aq)
`
`Acids weaker than H,0* react with water to a much smaller extent, producing
`H,0°*and their conjugate bases. For example, the following equilibrium lies pri-
`marily to the left:
`
`HF(aq) + H,0@)
`
`H,07 (aq) + F” (aq)
`
`
`
`
`
`* The OH™ ionis the strongest base that can exist in aqueous solution. Bases stronger
`than OH™ react with water to produce OH” and their conjugate acids. For exam-
`ple, the oxide ion (O*~) is a stronger base than OH”, so it reacts with water com-
`pletely as follows:
`
`0?" (aq) + H,O() —> 20H“ (aq)
`
`For this reason the oxide ion does not exist in aqueous solutions.
`
`The following example showscalculations of pH for a solution containing a strong
`acid and a solution of a strong base.
`
`0011
`
`0011
`
`

`

`15.4
`
`STRENGTHS OF ACIDS AND BASES
`
`605
`
`TABLE 15.2 Relative Strengths of Conjugate Acid-Base Pairs
`CONJUGATE BASE
`ACID
`HCI10, (perchloric acid)
`ClO4 (perchlorate ion)
`3 HI (hydroiodic acid)
`I” Godide ion)
`& |HBr (hydrobromic acid)
`Br” (bromide ion)
`e HC! (hydrochloric acid)
`Cl” (chloride ion)
`8
`= H,SO, (sulfuric acid)
`HSO; (hydrogen sulfate ion)
`6
`a
`HNO,(nitric acid)
`NO; (nitrate ion)
`Ge
`5
`H,0* (hydronium ion)
`HO (water)
`g
`5
`HSO; (hydrogen sulfate ion)
`SOZ- (sulfate ion)
`5
`a
`HF (hydrofluoric acid)
`F™ (fluoride ion)
`a
`
`$|.,|HNO, (nitrousacid) NOs (nitrite ion) s
`
`3 3 HCOOH(formic acid)
`HCOO™ (formate ion)
`2
`<x |
`“4
`CH;3COOH(acetic acid)
`CH3COO™ (acetate ion)
`ea
`iS NH? (ammonium ion)
`NH; (ammonia)
`HCN(hydrocyanic acid)
`CN” (cyanide ion)
`H,O (water)
`OH™ (hydroxide ion)
`NH; (ammonia)
`NH, (amide ion)
`
`
`
`
`
`A
`
`EXAMPLE 15.6
`Calculate the pH of (a) a 1.0 X 10-7 M HCIsolution and (6) a 0.020 M@ Ba(OH),
`solution.
`Answer
`
`(a) Since HClis astrong acid, it is completely ionizedin solution:
`HCl(aq) —> H* (ag) + C¥(aq)
`
`The concentrations of all the species (HCI, H*, and Cl) before and after ioniza-
`tion can be represented as follows:
`
`Initial (AZ):
`Change (M):
`Final (M):
`
`HClag)
`1.0 x 1073
`-—1.0 x 1077
`0.0
`
`—>»
`
`H*(aqg)
`0.0
`+1.0x1073
`1.0 x 1073
`
`+ Cl (aq)
`0.0
`+1.0« 1073
`1.0 x 1073
`
`A positive (+) changerepresents an increase and a negative (—) change indicates
`a decrease in concentration. Thus
`
`[H*]=1.0x 107M
`pH = —log (1.0 x 107)
`= 3.00
`
`(b) Ba(OH), is a strong base; each Ba(OH), unit produces two OH™ ions:
`Ba(OH),(aq) —> Ba2* (ag) + 20H™(aq)
`
`The changes in the concentrationsof all the species can be represented as follows:
`Ba(OH),(aq) ——> Ba** (ag) + 20H ~(ag)
`0.020
`0.00
`0.00
`—0.020
`+0.020
`+2(0.020)
`
`0.00
`0.020
`0.040
`
`Initial (44):
`Change(M):
`Final (4):
`
`mere
`
`0012
`
`0012
`
`

`

`
`
`|
`
`|
`
`|
`
`|
`
`)i
`
`} }
`
`
`
`606
`
`ACIDS AND BASES
`
`Thus
`
`Therefore
`
`[OH™] = 0.040 M@
`pOH = —log 0.040 = 1.40
`
`pH = 14.00 — pOH
`14.00 — 1.40
`
`= 12.60
`
`Similar problem: 15.18.
`
`Similar problem: 15.35.
`
`Comment Note that in both (a) and (b) we have néglected the contribution of the
`autoionization of water to [H*] and [OH™] because 1.0 X 1077 M is so small com-
`pared with 1.0 x 1077 M and 0.040 M.
`PRACTICE EXERCISE
`Calculate the pH of a 1.8 X 10-7 M Ba(OH), solution.
`If we know therelative strengths of two acids, we can predict the position of equi-
`librium between one of the acids and the conjugate base of the other, as illustrated in
`Example 15.7.
`
`EXAMPLE 15.7
`
`Predict the direction of the following reaction in aqueous solution:
`
`HNO,(ag) + CN (aq) === HCN(ag) + NO; (aq)
`
`In Table 15.2 we see that HNO,is a stronger acid than HCN. Thus CN7
`Answer
`is a stronger base than NOZ. Thenet reaction will proceedfrom left to right as writ-
`ten because HNO,is a better proton donor than HCN (and CN7is a better proton
`acceptor than NOZ).
`
`PRACTICE EXERCISE
`Predict whether the equilibrium constantfor the following reaction is greater than
`or smaller than 1:
`
`CH;COOH(aqg) + HCOO~(ag) == CH,COO (aq) + HCOOH(aq)
`
`GE weak ACIDS AND ACID IONIZATION CONSTANTS
`
`As we have seen, there are relatively few strong acids. The vast majority of acids are
`weak acids. Consider a weak monoprotic acid, HA. Its ionization in water is repre-
`sented by
`
`or simply
`
`HA(aq) + H,0(U) —— H,0*(aq) + A7(aq)
`
`
`
`
`HA(aq)
`
`H* (aq) + A“ (aq)
`
`All concentrations in this equation
`are equilibrium concentrations.
`
`The equilibrium expression for this ionization is
`_ [H*][A7]
`[HA]
`
`a
`
`0013
`
`0013
`
`

`

`
`
`15.8 WEAK ACIDS AND ACID IONIZATION CONSTANTS
`
`607
`
`where K,, the acid ionization constant, is the equilibrium constant for the ionization
`of an acid. At a given temperature, the strength of the acid HA is measured quantita-
`tively by the magnitude of K,. The larger K,, the stronger the acid——that is, the greater
`the concentration of H* ions at equilibrium due to its ionization. Keep in mind, how-
`ever, that only weak acids have K, values associated with them.
`Table 15.3 lists a number of weak acids and their K, values at 25°C in order of
`decreasing acid strength. Although all these acids are weak, within the group there is
`great variation in their strengths. For example, K, for HF (7.1 x 107%) is about 1.5
`million times that for HCN (4.9 x 107).
`Generally, we can calculate the hydrogen ion concentration or pH of an acid so-
`lution at equilibrium, given the initial concentration of the acid and its. K, value.
`Alternatively, if we know the pH of a weak acid solution and its initial concentration,
`we can determine its K,. The basic approach for solving these problems, which deal
`with equilibrium concentrations, is the same one outlined in Chapter 14. However, be-
`cause acid ionization represents a major category of chemical equilibrium in aqueous
`solution, we will develop a systematic procedure for solving this type of problem that
`will also help us to understand the chemistry involved.
`
`lenization Constants ef Some Weak Acids at 25°C
`
` TABLE 15.3
`
`
`NAME OF ACID
`FORMULA
`STRUCTURE
`K,
`CONJUGATE BASE
`Ky
`Hydrofluoric acid
`HF
`H—F
`7.1 x 1074
`F-
`14x 107!
`Nitrous acid
`HNO,
`O=N—O—H
`4.5 x 1074
`NOZ
`2.2 x 107!
`Acetylsalicylic acid
`CoHgO,
`O
`3.0 x 1074
`CoH,O7
`3.3 x 107!
`(aspirin)
`
`loOre.
`
`
`
`I
`
`O
`
`Formic acid
`Ascorbic acid*
`
`HCOOH
`CoHg0¢
`
`1.7 x 1074
`8.0 x 1075
`
`HCOO~
`CgH,05
`
`5.9 x 107!
`1.3 x 107!
`
`
`
`H-U—0-H
`_OH
`H—O_
`foe
`qo
`\cxo
`‘c
`con~o
`buon
`
`(C)-e-0-n
`5.6 x 1071°
`CH,;COO-
`1.8 x 10>
`cu,—_o-H
`CH;COOH
`Acetic acid
`
`
`
`
`Hydrocyanic acid 49x 107!|CNHCN H—C=N 2.0 x 1075
`
`
`
`Phenol 1.3.x 107!|CgH,O7CsH;OH ©) 1.7 X 1075
`
`
`
`Benzoic acid 6.51075|C6HsCOO-C,H<;COOH 0 1.5 x 107
`
`
`
`
`
`O
`
`O—H
`
`T a
`
`. Pee
`.
`:
`.
`Scorbic acid it is the upper left hydroxyl group that is associated with this ionization constant.
`
`0014
`
`
`
`0014
`
`

`

`
`
`
`
`608
`
`ACIDS AND BASES
`
`The sign ~ means “approximately
`equal to.” An analogy of the ap-
`proximation is a track loaded with
`coal. Losing a few lumpsof coal
`on a delivery trip will not
`significantly change the overall
`mass of the load.
`
`Suppose we are asked to calculate the pH of a 0.50 M HFsolution at 25°C. The
`ionization of HF is given by
`
`
`HF(aq) ——= H*(aq) + F(aq)
`
`From Table 15.3 we write
`
`K= {H*][F7] =7.1 x 1074
`3
`[HF]
`
`Thefirst step is to identify all the species present in solution that may affectits
`pH. Because weak acids ionize to a small extent, at equilibrium the major species pres-
`ent are nonionized HF and some H* and F™ ions. Another major species is H,O,but
`its very small K,, (1.0 X 107'4) meansthat wateris not a significant contributorto the
`H™ ion concentration. Therefore, unless otherwise stated, we will always ignore the
`H* ions produced by the autoionization of water. Note that we need not be concerned
`with the OH™ ionsthat are also present in solution. The OHconcentration can be de-
`termined from Equation (15.2) after we have calculated [H*].
`We can summarize the changes in the concentrations of HF, H‘, and F” accord-
`ing to the steps shown on p. 576 as follows:
`
`HF(aq) == H* (aq) + F(aq)
`0.50
`0.00
`0.00
`
`Xx
`+x
`+x
`
`Initial (M):
`Change (M):
`
`Equilibrium (/):
`0.50 — x
`x
`x
`The equilibrium concentrations of HF, H*, and F, expressed in termsof the un-
`known x, are substituted into the ionization constant expression to give
`
`~ @)@)_
`K.= 959 yp 7 TX 10
`
`-4
`
`Rearranging this expression, we write
`x +71 1074x — 3.6 X 10°*=0
`
`This is a quadratic equation which can be solved using the quadratic formula (see
`Appendix 4). Or wecan try using a shortcut to solve for x. Because HF is a weak acid
`and weak acidsionize only to a slight extent, we reason that x must be small compared
`to 0.50. Therefore we can make the approximation
`
`Now the ionization constant expression becomes
`
`0.50 — x ~ 0.50
`
`Rearranging, we get
`
`x? = (0.50)(7.1 X 1074) = 3.55 x 1074
`x = V3.55 X 107-4 = 0.019 M
`
`Thus we have solved for x without having to use the quadratic equation. At equilib-
`rium, we have:
`
`[HF] = (0.50 — 0.019) M = 0.48 M
`[H*] = 0.019 M@
`[F7] = 0.019 M
`
`0015
`
`0015
`
`

`

`
`
`13.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS
`
`609
`
`and the pH of the solution is
`
`pH = —log (0.019) = 1.72
`
`How goodis this approximation? Because K, values for weak acids are generally
`knownto an accuracy of only +5%, it is reasonable to require x to be less than 5% of
`0.50, the number from which it is subtracted. In other words, the approximationis valid
`if the following expression is equal to or less than 5%:
`
`0.019 M
`
`Thus the approximation we madeis acceptable.
`Now consider a different situation. If the initial concentration of HF is 0.050 M,
`and we use the above procedure to solve for x, we would get 6.0 X 107? M. However,
`the following test shows that this answer is not a valid approximation becauseit is
`greater than 5% of 0.050 M:
`
`6.0X 103M
`0.050 M x 100% = 12%
`
`In this case there are two waysto get an accurate value for x: by solving the quadratic
`equation and by applying the method of successive approximation.
`
`THE QUADRATIC EQUATION
`
`We start by writing the ionization expression in terms of the unknown x:
`
`+71 X 1074x — 3.6 x 1075 =0
`
`This expressionfits the quadratic equation ax* + bx + c = 0. Using the quadratic for-
`mula, we write
`
`x=
`
`—b+ Vb* — 4ac
`2a
`
`_ -7.1.x 1074 + V7.1 x 1074)? — 40(—3.6 X 1075)
`7
`2(1)
`
`—7.1 X 107~* + 0.012
`2
`
`=56X 1073 Mor -64xX10°M
`
`The second solution (x = —6.4 X 107? M) is physically impossible because the con-
`centration of ions produced as a result of ionization cannot be negative. Choosing x =
`5.6 X 107? M, we can solve for [HF], [H*], [F-] as follows:
`
`fHF] = (0.050 — 5.6 x 1077) M = 0.044 M
`fH*]=5.6xX 107M
`[F°] =56x 103M
`
`The pH ofthe solution, then, is
`
`pH = —log 5.6 X 1073 = 2.25
`
`0016
`
`x2—_a —4
`0.050 — x
`7.1 X10
`
`
`
`
`
`
`
`
`0016
`
`

`

`610
`
`ACIDS AND BASES
`
`
`
`THE METHODOF SUCCESSIVE APPROXIMATION_
`
`With this method, wefirst solve forx by assuming 0.050 —~x <0.0.050aassshown above
`Next, we use the approximate value of x (6.0 X 107? M) to find a more exact Value og
`
`the concentration for HF:
`
`[HF] = (0.050 — 6.0 x 1077) M = 0.044 M
`
`Substituting this value of [HF] in the expression for K,, we write
`
`=
`
`—4
`
`71X10
`0.044
`x=56X10°M
`
`Using 5.6 X 1073 M for x, we can recalculate [HF] and solve for x again. This time
`wefind that the answeris still 5.6 X 10-3 M so there is no need to proceed further. In
`general, we apply the method of successive approximation until the value of x obtaineg
`for the last step does not differ from the value formed in the previous step. In most
`cases, you will need to apply this method only twice to get the correct answer.
`In summary, the main steps for solving weak acid ionization problemsare:
`
`> 1.
`
`Identify the major species that can affectthepH ofthe solution. In most cases we
`
`can ignore the ionization of water. We omit the hydroxide ion becauseits concen-
`tration is determined by that of the H* ion
`2. Express the equilibrium concentrations of these species in terms of the initial con-
`centration of the acid and a single unknown x, which represents the changein con-
`centration.
`3. Write the acid ionization constant (K,) in terms of the equilibrium concentrations,
`First solve for x by the approximate method.If the aprroximation is not valid, use
`the quadratic equation or the method of successive approximation to solve for x.
`4. Having solved for x, calculate the equilibrium concentrations of all species and/or
`the pH ofthe solution.
`
`
`
`Nitrous acid
`
`Example 15.8 provides another illustration of the above procedure.
`
`EXAMPLE 15.8
`
`Calculate the pH of a 0.036 M nitrous acid (HNO) solution:
`
`
`HNO,(aq)
`
`H* (aq) + NO3(aq)
`
`Answer
`Step I: The species that can affect the pH of the solution are HNO,, H™, and the
`conjugate base NO; . We ignore water’s contribution to [H*].
`Step 2: Letting x be the equilibrium concentration of Ht and NOions in mol/L,
`we summarize:
`
`Initial (M1):
`Change (M4):
`Equilibrium (14);
`
`HNO,(aqg) ===> H* (aq) + NOFZ(aq)
`0.036
`0.00
`0.00:
`Sarl:
`+x
`+X
`
`0.036 — x
`x
`x
`
`Step 3: From Table 15.3 we. write
`
`0017
`
`0017
`
`

`

`15.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS
`
`611
`
`_ [H*}—NOz] _
`K,= Tino,
`= 45% 10
`Py
`x
`is
`Dea 745 10
`
`ei
`
`i
`
`Applying the approximation 0.036 — x ~ 0.036, we obtain
`
`= 4.5 x 1074
`
`x
`x
`0.036-—x 0.036
`pees ieoge lee
`x= 40x 10° M
`
`To test the approximation,
`
`4.0 x 1073 M
`BOI AX 100% = 11%
`
`Since this is greater than 5%, our approximation isnot valid and we must
`solve the quadratic equation, as: follows:
`
`x +45 xX 1074 - 1.62 x 10°77 =0
`
`
`—4.5 x 1074 + V(4.5 x 1074)? = 4(1)(1.62 X 1075)
`
`JOSeeEeee eel arc een
`21)
`= 3.8 x 10°? Mor —4.3 x 103M
`
`The secondsolution is physically impossible, since the concentrationof ions
`producedas a result of ionizati