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

`

`Mchw~Hill
`
`CHEMISTRY
`
`A Division of The McGraw-Hill Companies
`
`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 reproduced or distributed in any form or by any
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`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 Catalogingwin-Publication Data
`
`Chang, Raymond.
`Chemistry / Raymond Chang—6th ed.
`p.
`cm.
`Includes index.
`ISBN 0-07—011844—X
`
`1. Chemistry.
`QD31.2.C37
`540~d021
`
`I. Title.
`1998
`
`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.
`
`When ordering this title, use ISBN 0—07—115221-0.
`
`http://www.mhhe.com
`
`0002
`
`0002
`
`

`

`Contents in Brief
`
`Chapter 'I
`
`CHEMISTRY: THE STUDY OF CHANGE
`
`Chapter 2
`
`ATOMS, MOLECULES, AND IONS
`
`Chapter 3
`
`MASS RELATIONSHIPS IN CHEMICAL REACTIONS
`
`Chapter 4
`
`REACTIONS IN AQUEOUS SOLUTION
`
`Chapter 5
`
`GASES
`
`Chapter 6
`
`THERMOCHEMISTRY
`
`Chapter 7
`
`QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS
`
`Chapter 8
`
`PERIODIC RELATIONSHIPS AMONG THE ELEMENTS
`
`Chapter 9
`
`CHEMICAL BONDING I: BASIC CONCEPTS
`
`Chapter 10
`
`CHEMICAL BONDING II: MOLECULAR GEOMETRY AND
`HYBRIDIZATION OF ATOMIC ORBITALS
`
`-
`
`Chapter II
`
`INTERMOLECULAR FORCES AND LIQUIDS AND SOLIDS
`
`Chapter 12
`
`PHYSICAL PROPERTIES OF SOLUTIONS
`
`Chapter 13
`
`CHEMICAL KINETICS
`
`Chapter 14
`
`CHEMICAL EQUILIBRIUM
`
`Chapter 15
`
`ACIDS AND BASES
`
`Chapter 16
`
`ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
`
`Chapter 17
`
`CHEMISTRY IN THE ATMOSPHERE
`
`Chapter 18
`
`ENTROPY, FREE ENERGY, AND EQUILIBRIUM
`
`Chapter 19
`
`ELECTROCHEMISTRY
`
`Chapter 20
`
`METALLURGY AND THE CHEMISTRY OF METALS
`
`Chapter 21
`
`NONMETALLIC ELEMENTS AND THEIR COMPOUNDS
`
`Chapter 22
`
`TRANSITION METALS AND COORDINATION COMPOUNDS
`
`Chapter 23
`
`NUCLEAR CHEMISTRY
`
`Chapter 24
`
`ORGANIC CHEMISTRY
`
`Chapter 25
`
`SYNTHETIC AND NATURAL ORGANIC POLYMERS
`
`0003
`
`2
`
`56
`
`68
`
`108
`
`154
`
`202
`
`242
`
`286
`
`528
`
`366
`
`416
`
`466
`
`506
`
`558
`
`596
`
`644
`
`692
`
`724
`
`756
`
`802
`
`830
`
`870
`
`902
`
`938
`
`970
`
`0003
`
`

`

`CHAPTER1 5
`
`Acids and Bases
`
`INTRODUCTION
`
`SOME OF THE MOST IMPORTANT PROCESSES IN CHEMICAL AND BIOLOG—
`
`15.1
`
`BRQNSTED ACIDS AND BASES
`
`ICAL SYSTEMS ARE ACID-BASE REACTIONS IN AQUEOUS SOLUTIONS.
`
`IN
`
`15.2
`
`THE ACID-BASE PROPERTIES OF WATER
`
`THIS FIRST OF TWO CHAPTERS ON THE- PROPERTIES OF ACIDS AND BASES,
`
`15.3
`
`pH—A MEASURE OF ACIDITY
`
`WE \WILL STUDY THE DEFINITIONS OF ACIDS AND BASES, THE pH SCALE,
`
`15.4
`
`STRENGTH OF ACIDS AND BASES
`
`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.
`
`15.5
`
`WEAK ACIDS AND ACID IONIZATION
`CONSTANTS
`
`15.6
`
`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
`
`

`

`§@3
`
`ACIDS AND BASES
`
`m Efiflfigfifi antes ANQ BAEES
`
`______
`
`Conjugate means
`“joined together.”
`
`In Chapter 4 we defined a Bransted acid as a substance capable of donating a protOn
`and a Brpnsted base as a substance that can accept a proton. These definitions are gen:
`erally suitable for a discussion of the properties and reactions of acids and bases.
`An extension of the Brpnsted definition of acids and bases is the concept of the
`conjugate acid-base pair, which can be defined as an acid and its conjugate baSe Or
`a base and its conjugate acid. The conjugate base of a Brpnsted acid is 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 Brpnsted base.
`Every Bransted acid has a conjugate base, and every Brpnsted base has a conju-
`gate acid. For example, the chloride ion (Cl‘) is the conjugate base formed from the
`acid HCl, and H20 is the conjugate base of the acid H3O+ (hydronium ion). Similarly,
`the ionization of acetic acid can be represented as
`
`I'll
`:fi):
`IFII
`:fl):
`H—(‘i~C—o—H + H—ili: : H—c~c-—o :“ + H—?~H]+
`
`
`H
`
`H
`
`H
`
`H
`
`CH3COOH(aq) + H200) : CH3C00‘(aq) + H3O+(aq)
`acrdl
`basez
`base]
`and;
`
`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.
`
`The subscripts l and 2 designate the two conjugate acid—base pairs. Thus the acetate
`ion (CH3COO_) is the conjugate base of CH3COOH. Both the ionization of HCl (see
`Section 4.3) and the ionization of CH3COOH are examples of Brnnsted acid-base re-
`actions.
`‘
`
`The Bronsted definition also allows us to classify ammonia as a base because of
`
`its ability to accept a proton:
`
`..
`..
`1
`H—ITI—H + Iii—(1):: H—l‘lJ—H
`H
`H
`H
`
`..
`+ H—o :—
`
`NH3(aq) + H200) :2 Nruraq)
`basel
`acidz
`acid}
`
`+ OH_(aq)
`base;
`
`In this case, NH: is the conjugate acid of the base NH3, and the hydroxide ion OH“
`is the conjugate base of the acid H20. Note that the atom in the Brpnsted base that ac—
`cepts a H+ ion must have a lone pair.
`ln Example 15.1 we identify the conjugate pairs in an acid~base reaction.
`
`
`
`Similar problem: 15.5.
`
`0005
`
`
`
`0005
`
`

`

`15.2
`
`THE ACID-BASE PROPERTIES OF WATER
`
`599
`
`PRACTICE EXERCISE
`
`Identify the conjugate acid—base pairs for the reaction
`
`
`CN“ + H20
`
`HCN + OH“
`
`It is acceptable to represent the proton in aqueous solution either as HJr or as
`H3O+. The formula H+ is less cumbersome in calculations involving hydrogen ion
`concentrations and in calculations involving equilibrium constants, whereas H30+ is
`more useful in a discussion of Bronsted acid-base properties.
`
` -1 51.72 THE AGED-BASE PROPERTIES OF WATER
`
`Water, as we know, is a unique solvent. One of its special properties is 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, and it 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 conductor of electricity, but it
`ngUI'id sources d0 COBdUCt
`does undergo ionization to a small extent:
`
`electricity because they contain
`many dissolved ions.
`
`.
`
`H200)
`
`H+(‘1Q) + OHTWQ)
`
`(15-1)
`
`This reaction is sometimes called the autoionization of water. To describe the acid—base
`
`properties of water in the Bronsted framework, we express its autoionization as fol—
`lows (also shown in Figure 15.1):
`
`H—o : + H—o: : [H—O—H]
`
`I
`H
`
`|
`H
`
`I
`H
`
`+ H—o :—
`
`01‘
`
`H20 + H20 2 H3O+ + 0H—
`acidl
`basez
`acidz
`basel
`
`The acid—base conjugate pairs are (1) H20 (acid) and OH‘ (base) and (2) H3O+ (acid)
`and H20 (base).
`
`THEIONPRODUCTOFWATER
`
`In 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 H3O+, we can write the equilibrium constant for the au—
`toionization of water, Equation (15.1), as
`
`Kc
`
`: [H+][OH"]
`[H20]
`
`:LGURE 15.1 Reaction be-
`to Sen two water molecules to
`- rm hydronium and hydroxide
`
`IOnS.
`
`+
`
`_ 0006
`
`0006
`
`

`

`600
`
`ACIDS AND BASES
`
`’.
`
`Recall that in pure wazer,
`[H20] 2 55-5 M (365 P4 564)
`
`Since a very small fraction of water molecules are ionized, the concentration of Water
`[H20] remains virtually unchanged. Therefore
`
`If you could randomly remove
`and “amine ten Pailides (H20,
`14+, or OH“) per second from a
`liter of water. it would take you
`two years. working nonstop, to
`find one H+ ion!
`
`KJHZO] = Kw = [H+][OH‘]
`
`(15.2)
`
`The equilibrium constant KW is called the ion-product constant, which is the Product
`of the molar concentrations of H+ and 0H_ ions at a particular temperature.
`In pure water at 25°C, the concentrations of H+ and OH‘ ions are equal and fOUnd
`to be [H+] = 1.0 X 10—7 M and [OH_] = 1.0 X 10—7 M. Thus, from Equation (152)
`at 25°C
`
`’
`
`w = (1.0 X 10—7)(l.0 X 10—7) = 1.0 X 10—14
`.
`.
`'
`.
`Whether we have pure water or a solution of dissolved spooles, the followmg relation
`always holds at 25°C:
`
`Kw = [H+][OH’] = 1.0 x 10—14
`
`Whenever [H+] = [OH’], the aqueous solution is said to be neutral. In an acidic so—
`lution there is an excess of H+ ions and [H+] > [OH‘]. In a basic solution there is 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 10—6 M, the OH“ concen-
`tration must change to
`
`
`
`[OH—l =
`
`
`Kw _ 1.0><10‘14
`_—._.
`[Hr]
`1.0><10—6
`
`—8
`LOX“) M
`
`Be aware that the calculations shown above, and indeed all 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 amount and 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 amount of substance originally dissolved in solution. lust
`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. Calculate“ the concentration of I—l+ ions.
`
` Answer Rearranging' Equation (15.2), we write
`
`Many household cleaning fluids
`contain ammonia.
`
`.
`
`
`.
`_
`t
`1.0 x 10“”
`. KW
`..:
`IH+l I ‘[OH"] 3 W— : 4.0 X 10 12 M
`
`0007
`
`0007
`
`

`

`15.3
`
`pH—A MEASURE or ACIDlTY
`
`601
`
`Similar problems: 15.1.5, l5.l6.
`
`Comment Since [11+] </ [OHT], 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 HCl Solution whose hydrogen ion con—
`centration is 1.3 M.
`
`“pH—A MEASURE OF ACIDITY
`
`The pH of concentrated acid solu—
`tions can be negative. For exam,—
`ple, the pH of a 2.0 M HCl
`solution is —O.30.
`
`FIGURE 15.2 A pH meter is
`f°meIonly used in the laboratory
`f? determine the pH of a solu-
`‘0'7. Although many pH meters
`frave Scales marked with values
`0m l
`to 14, pH values can, in
`l
`QC,» be less than 7 and greater
`”‘On I4.
`
`Because the concentrations of H+ and OH‘ ions in aqueous solutions are frequently
`very small numbers and therefore inconvenient to work with, Soren Sorensenl in 1909
`proposed a more practical measure called pH. The pH of a solution is defined as the
`negative logarithm of the hydrogen ion concentration (in mol/L):
`
`pH = —log [H+]
`
`(15.3)
`
`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 number for 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 numerical part of 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:
`Basic solutions:
`
`Neutral solutions:
`
`[H+] > 1.0 x 10*‘7 M, pH < 7.00
`[H"] < 1.0 x 10—7 M, pH > 7.00
`
`[11“] = 1.0 x 10‘7 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 common fluids. As you can see, the pH of body
`fluids varies greatly, depending on location and function. The low pH (high acidity) of
`
`lSoren Peer Lauritz Sorensen (1868—1939). Danish biochemist. Sorensen originally wrote the symbol as pH and called p
`the “hydrogen ion exponent” (Wasserstofj‘ionexponent); 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
`
`

`

`692
`
`ACIDS AND BASES
`
`The pHs of
`TABLE 15.1
`
`Some Common Fluids
`
`SAMPLE
`
`Gastric juice in
`the stomach
`
`pH VALUE
`
`1.0—2.0
`
`gastric juices facilitates digestion whereas a higher pH of blood1s necessary fot the
`transport of oxygen. These pH—dependent actions will be illustrated1n Chemistry111
`Action essays in this and the next chapter.
`A pOH scale analogous to the pH scale can be devised using the negative 10ga~
`rithm of the hydroxide ion concentration of a solution. Thus we define pOH as
`
`Lemon juice
`
`Vinegar
`
`2.4
`
`3.0
`
`Grapefruit juice
`Orange juice
`Urine
`
`3.2
`3.5
`4.8—7.5
`
`Water exposed
`to air*
`Saliva
`Milk
`
`Pure water
`Blood
`Tears
`
`Milk of
`magnesia
`Household
`ammonia
`
`5.5
`
`6.4—6.9
`6.5
`
`7.0
`7.35—7.45
`7.4
`
`10.6
`
`1 1.5
`
`*Water exposed to air for a long period of
`time absorbs atmospheric C02 to form car-
`bonic acid, H2C03.
`
`pOH = —log [OH’]
`
`(15.4)
`
`Now consider again the ion-product constant for water:
`
`[H+][OH"] = Kw = 1.0 x 10‘14
`
`Taking the negative logarithm of both sides, we obtain
`
`—(log [11+] + log [OH—1) = -log (1.0 x 10‘”)
`
`—log [H+] — 10g [OH‘] = 14.00
`
`From the definitions of pH and pOH we obtain
`
`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 examples illustrate calculations involving pH.
`
`EXAMPLE 15-.3
`
`The concentration of 11+ions in a bottle of table wine was 3.2 X 10'4 Mright af~
`ter the cork was removed. Only half of the wine was consttmed._The otherhalf, af-
`ter it had been standing open to the air for a month. was found to haveahy-drogen
`ion concentration equal to 1.0 X 10'3 M. Calculate thepH of the wine on these two
`occasions.
`
`Answer When the bottle was first opened; [HT] '2: 3.2 .X 10"4 M. which we sub—L '
`stitute in Equation (15.3).
`
`111 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).
`
`pH =~ “log [Ht]
`= ~10g (3.2 x 10—4) '= 3.49
`
`On the secondocca’s-ion, [H‘L]; _=,l.0 X 110‘"3 M. so that
`
`.ij = —'log (1.0x 10—3) = 3.00
`
`Similar problems: 15.17, 15.18.
`
`tion’th’at takes place in the presence of molecular oxygen.
`
`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 read—
`
`PRACTICE EXERCISE
`Nitric acid (HN03) is usedin the production of fertilizer, dyes, drugs and explo—
`sives. Calculate the pH of a HNO3 solution having a hydrogenion concentration of
`0.76 M.
`
`EXAMPLE 15.4
`
`The pH of rainwater collected in a certain region of the northeastern United States
`on a particular day was 4. 82 Calculate the H+ion conCentration of the rainwater
`
`0009
`
`0009
`
`

`

`
`
`
`
`15.4
`
`STRENGTHS OF ACIDS AND BASES
`
`603
`
`Similar problem: 15.19.
`
`
`
`Similar problem: 15.18.
`
`“grammes or acres AND BASES
`
`Strong acids are strong electrolytes which, for practical purposes, are assumed to ion—
`ize completely in water (Figure 15.3). Most of the strong acids are inorganic acids: hy-
`drochloric acid (HCl), nitric acid (HNO3), perchloric acid (HClO4), and sulfuric acid
`(H2304):
`
`HCl(aq) —— H200) —> H3O"(aq) + C1‘(aq)
`
`HNO3(aq) + H200) ———9 H3O"(aq) + NO3—(aq)
`
`HClO4(aq) -- H200) ~——> H3O“(aq) + C10; (aq)
`
`
`
`
`
`H2504(aq) “ H200) “—9 P13016161) + H30; (6161)
`
`Note that H2804 is a diprotic acid; we show only the first 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, aqueous solutions of weak acids contain a mixture of nonionized acid mole—
`cules, H3O+ ions, and the conjugate base. Examples of weak acids are hydrofluoric
`acid (HF), acetic acid (CH3COOH), and the ammonium ion (NHI). The limited ion—
`
`0010
`
`0010
`
`

`

`604
`
`ACIDS AND BASES
`
`Before
`ionization
`
`HA
`FIG-‘1"
`
`At
`equilibrium
`
`H+
`
`A—
`
`
`
`
`
`
`
`(c) _._
`
`._F—
`
`FHGURE 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 acids is 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. Hydroxides of alkali metals and certain alkaline earth metals are strong bages~
`[All alkali metal hydroxides are soluble. Of the alkaline earth hydroxides, Be(0H)2
`and Mg(OH)2 are insoluble; Ca(OH)2 and Sr(OH)2 are slightly soluble; and l3a(QH)2
`is soluble]. Some examples of strong bases are:
`
`
`
`_
`H20
`NaOH(s) —-+ Na+(aq) + OH (aq)
`H20
`KOH(s) —_> K+(aq) + OH—(aq)
`H20
`Ba(OI—I)2(s) a Ba2+(aq) -- 20H_(aq)
`
`
`
`Strictly speaking, these metal hydroxides are not Bronsted bases because they Cannot
`accept a proton. However, the hydroxide ion (0-?) formed when they ionize is a
`Brensted base because it can accept a proton:
`
`H3O+(aq) + OH_(aq) H 2H20(l)
`
`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) + H200) a NHQKaq) + OH_(aq)
`
`Note that NH3 does not ionize like an acid because it does not split up to form ions
`the way, say, HCl does.
`Table 15.2 lists some important conjugate acid-base pairs, in order of their rela—
`tive strengths. Conjugate acid—base pairs have the following properties:
`
`a
`
`If an acid is strong, its conjugate base has no measurable strength. Thus the Cl"
`ion, which is the conjugate base of the strong acid HCl, is an extremely weak base.
`
`- H3O" is the strongest acid that can exist in aqueous solution. Acids stronger than
`H3O" react with water to produce H3O+ and their conjugate bases. Thus HCl, which
`is a stronger acid than H3O+, reacts with water completely to form H3O+ and Cl":
`
`HCl(aq) + nzoa) a H3O+(aq) + Cl—(aq)
`
`Acids weaker than H3O+ react with water to a much smaller extent, producing
`H3O+ and their conjugate bases. For example, the following equilibrium lies pri—
`marily to the left:
`
`HFW‘I) + H200)
`
`H30+(aq) + F 1619)
`
`
`
`
`
`0 The OH‘ ion is 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 (02') is a stronger base than OH‘, so it reacts with water com-
`pletely as follows:
`
`oz—(aq) + H200) e 20H_(aq)
`
`For this reason the oxide ion does not exist in aqueous solutions.
`
`The following example shows calculations of pH for a solution containing a strong
`acid and a solution of a strong base.
`
`001 1
`
`0011
`
`

`

`15.4
`
`STRENGTHS OF ACIDS AND BASES
`
`605
`
`TABLE 15.2 Relative Strengths of Coniugate Acid-Base Pairs
`ACH)
`CONJUGATE EASE
`
`
`
`
`
`Acidstrengthincreases
`
`
`
`Strongacids
`
`
`
`Weakacids
`
`"'
`
`
`
`HC104 (perchloric acid)
`HI (hydroiodic acid)
`
`HBr (hydrobromic acid)
`HCl (hydrochloric acid)
`
`H2804 (sulfuric acid)
`HN03 (nitric acid)
`
`H3O+ (hydronium ion)
`HSO; (hydrogen sulfate ion)
`HF (hydrofluoric acid)
`HN02 (nitrous acid)
`HCOOH (formic acid)
`
`CH3COOH (acetic acid)
`
`NH;F (ammonium ion)
`HCN (hydrocyanic acid)
`H20 (water)
`
`NH3 (ammonia)
`
`C104— (perchlorate ion)
`1‘ (iodide ion)
`Br" (bromide ion)
`
`Cl‘ (chloride ion)
`
`HSOZ (hydrogen sulfate ion)
`N03— (nitrate ion)
`H20 (water)
`
`802— (sulfate ion)
`F _ (fluoride ion)
`
`N02— (nitrite ion)
`HCOO‘ (forrnate ion)
`
`CH3COO_ (acetate ion)
`NH3 (ammonia)
`CN— (cyanide ion)
`
`OH‘ (hydroxide ion)
`NH; (amide ion)
`
`
`
`
`
`
`
`Basestrengthincreases
`
`.w
`
`EXAMPLE 1 5.6
`Calculate the pH of (a) a 1.0 X 10—3 _M HCl. solution and (b) a 0.020 M Ba(OH)2
`solution.
`
`
`
`Answer
`
`(a) Since HCl is a strong acid, it is completely ionized in solution:
`
`HCl(aq) —.+ H+(aq)’+ C1" (MI)
`
`The concentrations 'Of all the species (HCl, H+, and Cl‘) before and after ioniza—
`tion canbe represented as follows:
`
`Initial (M):
`Change (M):
`
`;Final (M):
`
`HCl(aq) fl H+(aq)
`1.0 X 10“3
`(0.0
`+1.0 ><10‘3
`—1.0 >< 10—3
`
`+ Cl_(aq).
`0.0
`+1.0 x 10"3
`
`0.0
`
`1.0 x 10*3
`
`1.0 x 10—3
`
`A positive (+) change represents an increase and a negative (—) change indicates
`a decrease in concentration. Thus
`
`111+] = 1.0 x10“3 M
`pH = ~1og (1.0 x 10—3)
`= 3.00
`
`(b) Ba(OH)2 is a strong base; each Ba(OH)2 unit produces two OH‘ ions:
`
`B.a(OH’)2(aq) ——» Baz+(aq) + 20H‘(aq)
`
`The changes in the concentrations of all the species can be represented as folloWs:
`
`Initial (M):
`Change (M):
`
`Final (M)1
`
`Ba(OH)2(aq) -~—> Ba2+(aq) + 20H“'(aq’)’r
`0.020
`0.00
`‘ 0.00
`
`-’-0.~020
`+0020
`+2(0;02_0)
`
`0.00
`
`0.020
`
`0.040
`
`0012
`
`0012
`
`

`

`606
`
`ACIDS AND BASES
`
`
`
`Thus
`
`Therefore
`
`[OH‘] = 0.040 M
`
`pOI-I = ~-log 0.040 = 1.40
`
`pH = 14.00 ~ pOH
`= 14.00 - 1.40
`
`= 12.60
`
`Similar problem: 15 .18.
`
`Comment Note that in both (a) and (b) we have neglected the contribution of the
`autoionization of water to [111+] and [Off] because 1.0 X 10—7 M is so small com-
`pared with 1.0 x 10*3 M and 0.040 M.
`
`PRACTICE EXERCISE
`
`Calculate the pH of a 1.8 X '10—2 M Ba(OH)2 solution.
`
`If we know the relative strengths of two acids, we can predict the position of equj_
`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:
`
`HN02(aq) + CN”(aq) ‘4____ HCN(aq) + NO§(aq)
`
`Similar problem: 15.35.
`
`In Table 15.2 we see that HNO; is a stronger acid than HCN. Thus CN“
`Answer
`is a stronger base than NO; . The net reaction will proceed from left to right as writ-
`ten becauseHNOg is a better proton donor than HCN. (and CN" is a better pro-ton
`acceptor than N02“).
`
`PRACTECE EXERCISE
`
`Predict whether the equilibrium constant for the following reaction is greaterthan
`or smaller than 1:
`
`CH3COOH(aq) + Hcoo-(aq) :2 CH3COO‘(aq) + HCOOI—Kaq)
`
`EWEAK ACIDS AND ACID IONIZATEON cowsmmrs
`
`As we have seen, there are relatively few strong acids. The vast majority of acids are
`weak acids. Consider a weak monoprotic acid, HA. lts ionization in water is repre-
`sented by
`
`HAWD + H200) : H30+(at1) + ATM)
`
`or simply
`
`
`
`
`HA(aq)
`
`H+(aq) + A7614)
`
`The equilibrium expression for this ionization is
`
`All concentrations in this equation
`are equilibrium concentrations.
`
`: [H+][A“]
`[HA]
`
`Ka
`
`0013
`
`
`
`0013
`
`

`

`
`
` TABtE ”€53
`
`15.5 WEAK ACIDS AND ACID IONlZATION CONSTANTS
`
`@Q?
`
`where Ka, 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 Ka. The larger Ka, 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 Ka values associated with them.
`Table 15.3 lists a number of weak acids and their K, values at 250C 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 10—4) is about 1.5
`million times that for HCN (4.9 X 10—10).
`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 itsK,1 value.
`Alternatively, if we know the pH of a weak acid solution and its initial concentration,
`we can determine its Ka. 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.
`
`Ionization Constants of Some Weak Acids at 25°C
`
`NAME OF ACID
`FORMULA
`STRUCTURE
`Ka
`CONJUGATE BASE
`Kb
`
`
`
`
`
`
`
`
`Hydrofluoric acid
`
`HF
`
`H—F
`
`Nitrous acid
`
`Acetylsalicylic acid
`(aspirin)
`
`HNo2
`
`C9H804
`
`Formic acid
`
`Ascorbic acid*
`
`HCOOH
`
`061—1806
`
`O=N—O—H
`
`O
`—(|‘|‘,———O—H
`©—O—C——CH3
`
`(ll)
`
`0
`H—d—o—H
`
`/0H
`H——O\
`C C—o
`H\C C
`CfiOH\0/
`C|1H2OH
`
`7.1 x 10—4
`
`4.5 x 10—4
`
`3.0 X 10—4
`
`1.7 ><'10—4
`
`8.0 x 10-5
`
`BenZoic acid
`
`CGHSCOOH
`
`o
`
`6.5 x 10—5
`
`Acetic acid
`ychroCyanic acid
`
`CH3COOH
`HCN
`A
`
`@l':_o_a
`
`o
`CH3—d—O—-H
`H——CEN
`
`1.8 x 10—5
`4.9 x 10—10
`
`
`
`F‘
`
`No;
`
`C9H7OZ
`
`1.4 x 10—11
`
`2.2 x 10—11
`
`3.3 X 10‘11
`
`Hcoo—
`
`C6H7Og
`
`5.9 x 10“11
`
`1.3 x 10“10
`
`Cénscoo—
`
`1.5 x 10—10
`
`CH3COO“
`CN”
`
`5.6 x 10—10
`2.0 x 10—5
`
`C6H50—
`
`7.7 >< 1075
`
`O—H
`
`M w
`0r
`a
`.
`_
`.
`.
`.
`.
`.
`.
`.
`Scorblc acrd it is the upper left hydroxyl group that IS assocrated With this lonization constant.
`
`Phenol
`
`
`
`
`cfinson C
`
`1.3 x 10—10
`
`0014
`
`0014
`
`

`

`
`
`608
`
`ACIDS AND BASES
`
`Suppose we are asked to calculate the pH of a 0.50 M HF solution at 25°C. The
`ionization of HF is given by
`
`
`HFWI) _ H+(aq) + 1376161)
`
`From Table 15.3 we write
`
`Ka
`
`: [H+][F”] = 7.1 x 10—4
`[HF]
`
`The first step is to identify all the species present in solution that may affect its
`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 H20, but
`its very small KW (1.0 X 10—14) means that water is not a significant contributor to 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‘ ions that are also present in solution. The OH" concentration can be de_
`tennined 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:
`
`Initial (M):
`
`Change (M):
`
`
`HF(aq) <—— H+(aq) + F “(61(1)
`0.50
`0.00
`0.00
`
`—x
`+x
`+x
`
`Equilibrium (M):
`
`0.50 — x
`
`x
`
`x
`
`The equilibrium concentrations of HF, H+, and F" , expressed in terms of the un-
`known it, are substituted into the ionization constant expression to give
`
`_M__
`Ka—0_50_x-7.1><10
`
`—4
`
`Rearranging this expression, we write
`
`x2 + 7.1x10—4x — 3.6 x10‘4 = 0
`
`This is a quadratic equation which can be solved using the quadratic formula (see
`Appendix 4). Or we can try using a shortcut to solve for x. Because HF is a weak acid
`and weak acids ionize only to a slight extent, we reason that x must be small compared
`to 0.50. Therefore we can make the approximation
`
`The sign “~' means “approximately
`equal to.” An analogy of the ap~
`proximation is a truck loaded with
`coal. Losing a few lumps of coal
`on a delivery trip will not
`significantly change the overall
`mass of the load.
`
`Now the ionization constant expression becomes
`
`0.50 - x 130.50
`
`Rearranging, we get
`
`x2 1 =
`0.50—x
`0.5
`
`“X10
`
`—4
`
`x2 = (0.50)(7.1 x 10—4) = 3.55 x 10—4
`x = V3.55 x 10— = 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
`
`[F‘] = 0.019 M
`
`0015
`
`0015
`
`

`

`
`
`
`
`
`
`
`
`
`15.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS
`
`ago
`
`and the pH of the solution is
`
`pH = —log (0.019) = 1.72
`
`How good is this approximation? Because Ka values for weak acids are generally
`known to an accuracy of only i5%, 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 approximation is valid
`if the following expression is equal to or less than 5%:
`
`0.019 M
`0.50M X 100% = 3.8%
`
`Thus the approximation we made is 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 10“3 M. However,
`the following test shows that this answer is not a valid approximation because it is
`greater than 5% of 0.050 M:
`
`6.0 x 10‘3 M
`0.050M X 100% =12%
`
`In this case there are two ways to 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:
`
`x2—=
`0.050 —x
`
`—4
`
`7.1 X 10
`
`x2 + 7.1 ><10_4x — 3.6 >< 10“5 = 0
`
`This expression fits the quadratic equation not2 + bx + c = 0. Using the quadratic for-
`mula, we write
`
`x:
`
`—b i Vb2 — 4616
`2a
`
`
`
`
`—7.1 x 10—4 : \/(7.1 x 10—4)2 — 4(1)(—3.6 x 10*)
`2(1)
`
`H
`
`—7.1 x 10‘4 : 0.012
`2
`
`= 5.6 X 10‘3 M or —6.4 x 10‘3 M
`
`The second solution (x = —6.4 X 10"3 M) is physically impossible because the con—
`centration of ions produced as a result of ionization cannot be negative. Choosing x =
`5.6 X 10‘3 M, we can solve for [HF], [H+], [F‘] as follows:
`
`[HF] = (0.050 — 5.6 x 10*) M = 0.044 M
`
`[H+] = 5.6 x 10‘3 M
`
`[13“] = 5.6 X 10*3 M
`
`The pH of the solution, then, is
`
`pH = —log 5.6 >< 10‘3 = 2.25
`
`0016
`
`0016
`
`

`

`610
`
`ACIDS AND BASES
`
`
`
`THE” METHOD I__O_F SUCCESSNE APPROXIMATION
`With this method, we first solve for x by assuming 0.050 — x ~ 0500. as shown above
`Next, we use the approximate value of x (6.0 X 103M) to find a more exact value of
`the concentration for HF:
`
`[HF] = (0.050 — 6.0 X 10—3) M = 0.044 M
`
`Substituting this value of [HP] in the expression for Ka, we write
`
`x2
`m— 7.1 X 10
`
`—4
`
`x = 5.6 ><10"3 M
`
`!
`
`1'
`
`.
`
`!
`
`3
`ii‘
`
`l
`
`’
`
`E
`
`J
`
`Using 5.6 X 103 M for x, we can recalculate [HF] and solve for x again. This time
`we find that the answer is still 5. 6 X 103 M so thereis no need to proceed further In
`general, we apply the method of successive approximation until the value Of )6 Obtained
`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 problems are:
`
`. 1.
`
`Identify the major species that can affect the pH ofthe solution. In most cases We
`
`can ignore the ionization of water. We omit the hydroxide ion because its 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 change in con—
`centration.
`
`First solve for x by the approximate method. If the apiroximation 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 of the solution.
`
`Example 15.8 provides another illustration of the above procedure.
`
`3. Writetheacidionizationconstant(Ka)intermsoftheequilibriumconcentrations.
`
`
`EXAMPLE 15.8
`
`
`Calculate the pH of a 0.036 M nitrous acid (HN02) solution:
`
`HN02(aq)
`
`H+(aq) + NOEWI)’
`
`Nitrous acid
`
`Answer
`
`Step I: The species that can affect the pH of the solution are-HNOZ, Hi and the
`conjugate base NO; . We ignore water’s contribution to [FF].
`Step 2: Letting x be the equilibrium concentration of H+ and N02— ions in mol/L,
`'we summarize:
`
`Initial (M):
`Change (M);
`
`HNOZ(aq) :z: H+(aq) '+ NO§(aq)
`0.036
`0.00
`0.00
`
`-x
`+x
`+x
`
`Equilibrium (M);
`
`0.036 - x
`
`x
`
`x
`
`Step 3: From Table 15.3 we. write
`
`0017
`
`0017
`
`

`

`15.5 WEAK ACIDS AND ACID IONIZATION CONSTANTS
`
`6] 'I
`
`_ [Humor] _
`K3 —
`[HNOZ]
`— 4.5 x 10
`2
`
`-4
`
`x
`@0035fl x
`
`=
`
`4.5 >