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18 Acid-Base Equilibria
792
Source: © McGraw-Hill Education/Mark A. 
Dierker, photographer
 18.1 Acids and Bases in Water
Arrhenius Acid-Base Definition
Acid-Dissociation Constant (Ka)
Relative Strengths of Acids 
and Bases
 18.2 Autoionization of Water and 
the pH Scale
Autoionization and Kw
The pH Scale
 18.3 Proton Transfer and the 
Brønsted-Lowry Acid-Base 
Definition
Conjugate Acid-Base Pairs
Net Direction of Acid-Base 
Reactions
 18.4 Solving Problems Involving 
Weak-Acid Equilibria
Finding Ka Given Concentrations
Finding Concentrations Given Ka
Extent of Acid Dissociation
Polyprotic Acids
 18.5 Molecular Properties and Acid 
Strength
Nonmetal Hydrides
Oxoacids
Hydrated Metal Ions
 18.6 Weak Bases and Their Relation 
to Weak Acids
Ammonia and the Amines
Anions of Weak Acids
Relation Between Ka and Kb
 18.7 Acid-Base Properties of Salt 
Solutions
Salts That Yield Neutral Solutions
Salts That Yield Acidic Solutions
Salts That Yield Basic Solutions
Salts of Weakly Acidic Cations and 
Weakly Basic Anions
Salts of Amphiprotic Anions
 18.8 Generalizing the Brønsted-Lowry 
Concept: The Leveling Effect
 18.9 Electron-Pair Donation and the 
Lewis Acid-Base Definition
Molecules as Lewis Acids
Metal Cations as Lewis Acids
Overview of Acid-Base Definitions
› role of water as solvent (Section 4.1)
› writing ionic equations (Section 4.2)
› acids, bases, and acid-base reactions (Section 4.4)
› proton transfer in acid-base reactions (Section 4.4)
› properties of an equilibrium constant (Section 17.2)
› solving equilibrium problems (Section 17.5)
Concepts and Skills to Review Before You Study This Chapter
It’s lunch time: how about a meal of citric and ascorbic acids (lemons and oranges), oxalic acid (spinach), folic acid (broccoli), 
and lactic acid (yogurt), all washed down with some phosphoric acid (soft drink)—if 
you eat too much, you may need some magnesium or aluminum hydroxide base 
(antacid)! Acids and bases are in many common consumer products (Table 18.1) and 
are indispensable in academic and industrial research.
Acids gives substances such as lemon juice and vinegar a sour taste. In fact, sourness 
was a defining property of an acid since the 17th century: an acid was any substance that 
had a sour taste; reacted with active metals, such as aluminum and zinc, to produce 
hydrogen gas; and turned certain organic compounds specific colors. (We discuss indica-
tors in this chapter and Chapter 19.) Similarly, a base was any substance that had a bitter 
taste and slippery feel and turned the same organic compounds different colors. Moreover, 
it was known that when an acid and a base react, each cancels the properties of the other 
in a process called neutralization. Although these early definitions described distinctive 
properties, they gave way to others based on molecular behavior. As science progresses, 
limited definitions are replaced by broader ones that explain more phenomena.
793
IN THIS CHAPTER . . . We develop three definitions of acids and bases that explain an expanded 
range of reactions, and we apply equilibrium principles to understand acid-base behavior.
› We begin with the Arrhenius acid-base definition, which relies on formulas and behavior in water.
› We examine acid dissociation to see how variation in acid strength is expressed by a new 
equilibrium constant.
› We introduce the pH scale to measure the acidity of aqueous solutions.
› We discuss proton transfer in the Brønsted-Lowry acid-base definition, which expands the 
meaning of “base”, along with the scope of acid-base reactions.
› We apply a systematic approach to solving acid-base equilibrium problems.
› We examine the molecular structures of acids to rationalize their relative strengths.
Substance Use
Acids
Acetic acid, CH3COOH Flavoring, preservative (vinegar)
Citric acid, H3C6H5O7 Flavoring (lemon juice)
Ascorbic acid, H2C6H6O6 Vitamin C; nutritional supplement
Aluminum salts, NaAl(SO4)2⋅12H2O In baking powder, with sodium hydrogen 
 carbonate
Bases
Sodium hydroxide (lye), NaOH Oven and drain cleaners
Ammonia, NH3 Household cleaner
Sodium carbonate, Na2CO3 Water softener, grease remover
Sodium hydrogen carbonate, NaHCO3 Fire extinguisher, rising agent in cake mixes 
 (baking soda), mild antacid
Sodium phosphate, Na3PO4 Cleaner for surfaces before painting or 
 wallpapering
Table 18.1 Some Common Acids and Bases and Their Household Uses
Source: © McGraw-Hill Education/Stephen Frisch, photographer
794 Chapter 18 • Acid-Base Equilibria
› We examine weak bases and their interdependence with weak acids.
› We determine the relative acidity of salt solutions from the reactions of the salts’ cations 
and anions with water.
› We see that the designations “acid” and “base” depend on the substances’ relative strengths 
and on the solvent.
› We discuss the Lewis acid-base definition, which greatly expands the meanings of “acid” 
and “acid-base reaction”.
 18.1 ACIDS AND BASES IN WATER
Most laboratory work with acids and bases involves water, as do most environmental, 
biological, and industrial applications. Recall from our discussion in Chapter 4 that 
water is the product in all reactions between strong acids and strong bases, which 
the net ionic equation for any such reaction shows:
 HX(aq) + MOH(aq) ⟶ MX(aq) + H2O(l) [molecular]
 H+(aq) + X−(aq) + M+ (aq) + OH−(aq) ⟶ M+(aq) + X−(aq) + H2O(l ) [total ionic]
 H+ (aq) + OH− (aq) ⟶ H2O(l) [net ionic]
where M+ is a metal ion and X− is a nonmetal ion.
Furthermore, whenever an acid dissociates in water, solvent molecules participate 
in the reaction:
HA(g or l) + H2O(l) ⟶ A− (aq) + H3O+ (aq)
−
+H
H A AH O+ + O
H
HH
Water molecules surround the proton to form H-bonded species with the general 
formula (H2O)nH+. The proton’s charge density is so high that it attracts water mol-
ecules especially strongly, covalently bonding to one of the lone electron pairs of a 
water molecule’s O atom to form a hydronium ion, H3O+, which forms H bonds to 
several other water molecules (see Figure 4.11). To emphasize the active role of water 
and the proton-water interaction, the hydrated proton is usually shown in the text as 
H3O+(aq) [although, for simplicity, we sometimes show it as H+(aq)].
Release of H+ or OH− and the Arrhenius Acid-Base Definition
The earliest definition that highlighted the molecular nature of acids and bases is the 
Arrhenius acid-base definition, which classifies these substances in terms of their 
formulas and behavior in water:
∙ An acid is a substance with H in its formula that dissociates in water to yield H3O+.
∙ A base is a substance with OH in its formula that dissociates in water to yield OH−.
Some typical Arrhenius acids are HCl, HNO3, and HCN, and some typical Arrhe-
nius bases are NaOH, KOH, and Ba(OH)2. Because they are ionic compounds, Arrhe-
nius bases contain discrete OH− ions in their structures, but Arrhenius acids never 
contain discrete H+ ions. Instead, they contain covalently bonded H atoms that ionize 
when molecules of the acid dissolve in water.
When an acid and a base react, they undergo neutralization. The meaning of this 
term has changed, as we’ll see, but in the Arrhenius sense, neutralization occurs when 
the H+ from the acid and the OH− from the base form H2O. A key point about neu-
tralization that Arrhenius was able to explain is that no matter which strong acid and 
strong base react, and no matter which salt results, ΔH°rxn is −55.9 kJ per mole of water 
formed. Arrhenius suggested that the enthalpy change is always the same because the 
reaction is always the same—a hydrogen ion and a hydroxide ion form water:
H+ (aq) + OH− (aq) ⟶ H2O(l) ΔH°rxn = −55.9 kJ
The dissolved salt that is present, for example, NaCl in the reaction of sodium hydrox-
ide withhydrochloric acid, 
Na+ (aq) + OH− (aq) + H+ (aq) + Cl− (aq) ⟶ Na+ (aq) + Cl− (aq) + H2O(l)
exists as hydrated spectator ions and does not affect ΔH°rxn.
18.1 • Acids and Bases in Water 795
Despite its importance at the time, limitations in the Arrhenius definition soon 
became apparent. Arrhenius and many others realized that even though some sub-
stances do not have discrete OH− ions, they still behave as bases. For example, NH3 
and K2CO3 also yield OH− in water. As you’ll see shortly, broader acid-base defini-
tions are required to include species that do not have OH– ions in their structures.
Variation in Acid Strength: The Acid-Dissociation Constant (Ka)
Acids (and bases) are classified by their strength, the amount of H3O+ (or OH−) 
produced per mole of substance dissolved, in other words, by the extent of their dis-
sociation into ions (see Table 4.2). Because acids and bases are electrolytes, their 
strength correlates with electrolyte strength: strong electrolytes dissociate completely, 
and weak electrolytes dissociate slightly.
1. Strong acids dissociate completely into ions in water (Figure 18.1A):
HA(g or l) + H2O(l) ⟶ H3O+ (aq) + A− (aq)
In a dilute solution of a strong acid, HA molecules are no longer present: [H3O+] = 
[A−] ≈ [HA]init. In other words, [HA]eq ≈ 0, so the value of Kc is extremely large:
Qc =
[H3O+][A−]
[HA][H2O]
 (at equilibrium, Qc = Kc >> 1)
Because the reaction is essentially complete, we usually don’t express it as an equi-
librium process. In dilute aqueous nitric acid, for example, there are virtually no 
undissociated nitric acid molecules:
HNO3(l) + H2O(l) ⟶ H3O+ (aq) + NO3− (aq)
2. Weak acids dissociate slightly into ions in water (Figure 18.1B):
HA(aq) + H2O(l) �⥫⥬ H3O+ (aq) + A−(aq)
Figure 18.1 The extent of dissociation 
for strong acids and weak acids. The 
bar graphs show the relative numbers 
of moles of species before (left) and after 
(right) acid dissociation.
A Strong acid: 
HA HA H3O
+
H3O
+
A–
A–
R
el
at
iv
e 
nu
m
be
r o
f m
ol
es
Before
dissociation
After
dissociation
 HA(g or l ) + H2O(l ) H3O
+(aq) + A–(aq)
 
Complete dissociation:
virtually no HA molecules
are present.
R
el
at
iv
e 
nu
m
be
r o
f m
ol
es
HA HA
HA
H3O
+ A–
B Weak acid: 
 HA(aq) + H2O(l ) H3O
+(aq) + A–(aq)
A–
Before
dissociation
After
dissociation
H3O
+
Very little dissociation:
most HA molecules
remain intact.
796 Chapter 18 • Acid-Base Equilibria
In a dilute solution of a weak acid, the great majority of HA molecules are undissoci-
ated. Thus, [H3O+] = [A−] << [HA]init, and [HA]eq ≈ [HA]init, so Kc is very small. 
Hydrocyanic acid is an example of a weak acid:
HCN(aq) + H2O(l) ⥫⥬ H3O+ (aq) + CN− (aq)
Qc =
[H3O+][CN−]
[HCN][H2O]
 (at equilibrium, Qc = Kc << 1)
(As in Chapter 17, brackets with no subscript mean molar concentration at equilib-
rium; that is, [X] means [X]eq. In this chapter, we are dealing with systems at equi-
librium, so instead of writing Q and stating that Q equals K at equilibrium, we’ll 
express K directly as a collection of equilibrium concentration terms.)
The difference in [H3O+] causes a much higher rate for the reaction of a strong acid 
with an active metal like zinc than for the same reaction of a weak acid (Figure 18.2):
Zn(s) + 2H3O + (aq) ⟶ Zn2+(aq) + 2H2O(l) + H2(g)
In a strong acid, with its much higher [H3O+], zinc reacts rapidly, forming bubbles 
of H2 vigorously. In a weak acid, [H3O+] is much lower, so zinc reacts slowly.
The Meaning of Ka We write a specific equilibrium constant for acid dissociation 
that includes only the species whose concentrations change to any significant extent. 
For the dissociation of a general weak acid, HA,
HA(aq) + H2O(l ) ⥫⥬ H3O+(aq) + A−(aq)
the equilibrium expression is
Kc =
[H3O+][A−]
[HA][H2O]
The concentration of water ([H2O] =
1000 g
18.02 g/mol
= 55.5 M ) is typically several
orders of magnitude larger than [HA]. Therefore, [H2O] is essentially constant when 
HA dissociates, and so H2O can be considered a pure liquid. As we discussed in Sec-
tion 17.2, the concentration terms for pure liquids and solids are equal to 1 and do 
not appear in the equilibrium expression. Thus, we can define a new equilibrium 
constant, the acid-dissociation constant (or acid-ionization constant), Ka:
 
Kc × 1 = Ka =
[H3O+][A−]
[HA]
 (18.1)
Like any equilibrium constant, Ka is a number whose magnitude is temperature dependent 
and tells how far to the right the reaction has proceeded to reach equilibrium. Thus, the 
stronger the acid, the higher [H3O
+] is at equilibrium, and the larger the value of Ka:
Stronger acid ⟹ higher [H3O+] ⟹ larger Ka
The Range of Ka Values Acid-dissociation constants of weak acids range over many 
orders of magnitude. Some benchmark Ka values for typical weak acids in Table 18.2 
give an idea of the fraction of HA molecules that dissociate into ions.
 
Strong acid
(1 M HCl)
Weak acid
(1 M CH3COOH)
Figure 18.2 Reaction of zinc with a 
strong acid (left) and a weak acid (right).
Source: © McGraw-Hill Education/Stephen 
Frisch, photographer
 % Dissociation in a 
Magnitude of Ka 1 M Solution of HA Specific Example (Ka Value, % Dissociation)
Relatively high Ka (~10−2) ~10% 1 M chlorous acid (HClO2) (Ka = 1.1×10−2, 10.%)
Moderate Ka (~10−5) ~0.3% 1 M acetic acid (CH3COOH) (Ka = 1.8×10−5, 0.42%)
Relatively low Ka (~10−10) ~0.001% 1 M hydrocyanic acid (HCN) (Ka = 6.2×10−10, 0.0025%)
Table 18.2 Magnitude of Ka Values and Percent Dissociation for Weak Acids
Thus, for solutions of the same initial HA concentration, the smaller the Ka, the lower 
the percent dissociation of HA:
Weaker acid ⟹ lower % dissociation of HA ⟹ smaller Ka
18.1 • Acids and Bases in Water 797
Table 18.3 lists Ka values of some weak monoprotic acids, those with one ionizable 
proton. (A more extensive list is in Appendix C.) Note that the ionizable proton in 
organic acids is bound to the O in COOH; H atoms bonded to C do not ionize.
Classifying the Relative Strengths of Acids and Bases
Using a table of Ka values is the surest way to quantify strengths of weak acids, but 
you can classify acids and bases qualitatively as strong or weak from their formulas:
∙ Strong acids. Two types of strong acids, with examples you should memorize, are
 1. The hydrohalic acids HCl, HBr, and HI
 2. Oxoacids in which the number of O atoms exceeds the number of ionizable 
protons by two or more, such as HNO3, H2SO4, and HClO4; for example, in the 
case of H2SO4, 4 O’s − 2 H’s = 2
∙ Weak acids. There are many more weak acids than strong ones. Four types are
 1. The hydrohalic acid HF
 2. Acids in which H is not bonded to O or to a halogen, such as HCN and H2S
 3. Oxoacids in which the number of O atoms equals or exceeds by one the number 
of ionizable protons, such as HClO, HNO2, and H3PO4
 4. Carboxylic acids (general formula RCOOH, with the ionizable proton shown in 
red), such as CH3COOH and C6H5COOH
∙ Strong bases. Water-soluble compounds containing O2− or OH− ions are strong 
bases. The cations are usually those of the most active metals:
 1. M2O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs)
 2. MO or M(OH)2, where M = Group 2A(2) metal (Ca, Sr, Ba)
 [MgO2 and Mg(OH)2 are only slightly soluble in water, but the soluble portion 
dissociates completely.]
∙ Weak bases. Many compounds with an electron-rich nitrogen atom are weak bases 
(none is an Arrhenius base). The common structural feature is an N atom with a 
lone electron pair (shown in blue):
 1. Ammonia (· ·NH3)
 2. Amines (general formula R · ·NH2, R2· ·NH, or R3· ·N), such as CH3CH2· ·NH2, 
(CH3)2· ·NH, and (C3H7)3· ·N
Table 18.3 Ka Values for Some Monoprotic Acidsat 25°C
Name (Formula)* Lewis Structure* Dissociation Reaction Ka
*Red type indicates the ionizable proton; all atoms have zero formal charge.
ClH O O
NH O O
H F
H HC O
O
C H
H
H
CH O
O
C H
H
H
CC
H
H
H O
O
ClH O
NH C
Nitrous acid (HNO2)
Chlorous acid (HClO2)
Hydrofluoric acid (HF)
Formic acid (HCOOH)
Acetic acid (CH3COOH)
Propanoic acid (CH3CH2COOH)
Hypochlorous acid (HClO)
Hydrocyanic acid (HCN) 
A
C
ID
 S
TR
EN
G
TH
1.1 × 10−2
7.1 × 10−4
6.8 × 10−4
1.8 × 10−4
1.8 × 10−5
1.3 × 10−5
2.9 × 10−8
6.2 × 10−10
HClO2(aq) + H2O(l) H3O+(aq) + ClO2−(aq)
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
HF(aq) + H2O(l) H3O+(aq) + F−(aq)
CH3COOH(aq) + H2O(l) 
 H3O+(aq) + CH3COO−(aq)
CH3CH2COOH(aq) + H2O(l) 
 H3O+(aq) + CH3CH2COO−(aq)
HClO(aq) + H2O(l) H3O+(aq) + ClO−(aq)
HCN(aq) + H2O(l) H3O+(aq) + CN−(aq)
HCOOH(aq) + H2O(l) 
 H3O+(aq) + HCOO−(aq)
ClH O O
NH O O
H F
H HC O
O
C H
H
H
CH O
O
C H
H
H
CC
H
H
H O
O
ClH O
NH C
Nitrous acid (HNO2)
Chlorous acid (HClO2)
Hydrofluoric acid (HF)
Formic acid (HCOOH)
Acetic acid (CH3COOH)
Propanoic acid (CH3CH2COOH)
Hypochlorous acid (HClO)
Hydrocyanic acid (HCN) 
A
C
ID
 S
TR
EN
G
TH
1.1 × 10−2
7.1 × 10−4
6.8 × 10−4
1.8 × 10−4
1.8 × 10−5
1.3 × 10−5
2.9 × 10−8
6.2 × 10−10
HClO2(aq) + H2O(l) H3O+(aq) + ClO2−(aq)
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
HF(aq) + H2O(l) H3O+(aq) + F−(aq)
CH3COOH(aq) + H2O(l) 
 H3O+(aq) + CH3COO−(aq)
CH3CH2COOH(aq) + H2O(l) 
 H3O+(aq) + CH3CH2COO−(aq)
HClO(aq) + H2O(l) H3O+(aq) + ClO−(aq)
HCN(aq) + H2O(l) H3O+(aq) + CN−(aq)
HCOOH(aq) + H2O(l) 
 H3O+(aq) + HCOO−(aq)
ClH O O
NH O O
H F
H HC O
O
C H
H
H
CH O
O
C H
H
H
CC
H
H
H O
O
ClH O
NH C
Nitrous acid (HNO2)
Chlorous acid (HClO2)
Hydrofluoric acid (HF)
Formic acid (HCOOH)
Acetic acid (CH3COOH)
Propanoic acid (CH3CH2COOH)
Hypochlorous acid (HClO)
Hydrocyanic acid (HCN) 
A
C
ID
 S
TR
EN
G
TH
1.1 × 10−2
7.1 × 10−4
6.8 × 10−4
1.8 × 10−4
1.8 × 10−5
1.3 × 10−5
2.9 × 10−8
6.2 × 10−10
HClO2(aq) + H2O(l) H3O+(aq) + ClO2−(aq)
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
HF(aq) + H2O(l) H3O+(aq) + F−(aq)
CH3COOH(aq) + H2O(l) 
 H3O+(aq) + CH3COO−(aq)
CH3CH2COOH(aq) + H2O(l) 
 H3O+(aq) + CH3CH2COO−(aq)
HClO(aq) + H2O(l) H3O+(aq) + ClO−(aq)
HCN(aq) + H2O(l) H3O+(aq) + CN−(aq)
HCOOH(aq) + H2O(l) 
 H3O+(aq) + HCOO−(aq)
ClH O O
NH O O
H F
H HC O
O
C H
H
H
CH O
O
C H
H
H
CC
H
H
H O
O
ClH O
NH C
Nitrous acid (HNO2)
Chlorous acid (HClO2)
Hydrofluoric acid (HF)
Formic acid (HCOOH)
Acetic acid (CH3COOH)
Propanoic acid (CH3CH2COOH)
Hypochlorous acid (HClO)
Hydrocyanic acid (HCN) 
A
C
ID
 S
TR
EN
G
TH
1.1 × 10−2
7.1 × 10−4
6.8 × 10−4
1.8 × 10−4
1.8 × 10−5
1.3 × 10−5
2.9 × 10−8
6.2 × 10−10
HClO2(aq) + H2O(l) H3O+(aq) + ClO2−(aq)
HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq)
HF(aq) + H2O(l) H3O+(aq) + F−(aq)
CH3COOH(aq) + H2O(l) 
 H3O+(aq) + CH3COO−(aq)
CH3CH2COOH(aq) + H2O(l) 
 H3O+(aq) + CH3CH2COO−(aq)
HClO(aq) + H2O(l) H3O+(aq) + ClO−(aq)
HCN(aq) + H2O(l) H3O+(aq) + CN−(aq)
HCOOH(aq) + H2O(l) 
 H3O+(aq) + HCOO−(aq)
1.1×10−2
7.1×10−4
6.8×10−4
1.8×10−4
1.8×10−5
1.3×10−5
2.9×10−8
6.2×10−10
798 Chapter 18 • Acid-Base Equilibria
Summary of Section 18.1
› In aqueous solution, water binds the proton released from an acid to form a hydrated 
species represented by H3O
+(aq).
› By the Arrhenius definition, acids contain H and yield H3O
+ in water, bases contain OH and yield 
OH− in water, and an acid-base reaction (neutralization) is the reaction of H+ and OH− to form H2O.
› Acid strength depends on [H3O
+] relative to [HA] in aqueous solution. Strong acids dissociate 
completely and weak acids slightly.
› The extent of dissociation is expressed by the acid-dissociation constant, Ka. Most weak acids 
have Ka values ranging from about 10
−2 to 10−10.
› Many acids and bases can be classified as strong or weak based on their formulas.
›
Problem Classify each of the following compounds as a strong acid, weak acid, strong 
base, or weak base, and write the Ka expression for any weak acid:
(a) KOH (b) (CH3)2CHCOOH (c) H2SeO4 (d) (CH3)2CHNH2
Plan We examine the formula and classify each compound as acid or base, using the 
text descriptions. Particular points to note for acids are the numbers of O atoms relative 
to ionizable H atoms and the presence of the COOH group. For bases, we note the 
nature of the cation or the presence of an N atom that has a lone pair.
Solution (a) Strong base: KOH is one of the Group 1A(1) hydroxides.
(b) Weak acid: (CH3)2CHCOOH is a carboxylic acid, as indicated by the COOH 
group; the dissociation reaction is:
(CH3)2CHCOOH(aq) + H2O(l) ⥫⥬ H3O+ (aq) + (CH3)2CHCOO− (aq)
and
Ka =
[H3O+][(CH3)2CHCOO−]
[(CH3)2CHCOOH]
(c) Strong acid: H2SeO4 is an oxoacid in which the number of O atoms exceeds the 
number of ionizable protons by 2.
(d) Weak base: (CH3)2CHNH2 has a lone electron pair on the N and is an amine.
FOLLOW-UP PROBLEMS
Brief Solutions for all Follow-up Problems appear at the end of the chapter.
18.1A Which member of each pair is the stronger acid or base?
(a) HClO or HClO3 (b) HCl or CH3COOH (c) NaOH or CH3NH2
18.1B Classify each of the following compounds as a strong acid, weak acid, strong 
base, or weak base, and write the Ka expression for any weak acid:
(a) (CH3)3N (b) HI (c) HBrO (d) Ca(OH)2
SOME SIMILAR PROBLEMS 18.15–18.18
SAMPLE PROBLEM 18.1
Classifying Acid and Base Strength 
from the Chemical Formula
 18.2 AUTOIONIZATION OF WATER 
AND THE pH SCALE
Before we discuss the next major acid-base definition, let’s examine a crucial property 
of water that enables us to quantify [H3O+]: water dissociates very slightly into ions 
in an equilibrium process known as autoionization (or self-ionization):
H2O(l ) H2O(l ) H3O
+(aq) OH–(aq) 
+ + 
+ + 
Lone pair of 
O binds H+. 
18.2 • Autoionization of Water and the pH Scale 799
The Equilibrium Nature of Autoionization: The Ion-Product 
Constant for Water (Kw)
Like any equilibrium process, the autoionization of water is described quantitatively 
by an equilibrium constant:
Kc =
[H3O+][OH−]
[H2O]2
Because the concentration of H2O (55.5 M) remains essentially constant, it is consid-
ered a pure liquid and is eliminated from the equilibrium expression. Thus, we obtain 
the ion-product constant for water, Kw:
 Kc × (1)2 = Kw = [H3O+][OH] = 1.0×1014 (at 25°C) (18.2)
Notice that one H3O+ ion and one OH− ion form for each H2O molecule that dissociates. 
Therefore, in pure water, we find that
[H3O+] = [OH−] = √1.0×10−14 = 1.0×10−7 M (at 25°C)
Since pure water has a concentration of about 55.5 M, these equilibrium concentra-
tions are attained when only 1 in 555 million water molecules dissociates reversibly 
into ions!
Autoionization of water affects aqueous acid-base chemistry in two major ways:
1. A change in [H3O+] causes an inverse change in [OH−], and vice versa:
Higher [H3O+] ⟹ lower [OH−] and Higher [OH−] ⟹ lower [H3O+]
Recall from Le Châtelier’s principle (Section 17.6) that a change in concentration 
shifts the equilibrium position but does not change the equilibrium constant.There-
fore, if some acid is added, [H3O+] increases and [OH−] decreases as the autoionization 
reaction proceeds to the left and the ions react to form water; similarly, if some base 
is added, [OH−] increases and [H3O+] decreases. In both cases, as long as the tem-
perature is constant, the value of Kw is constant.
2. Both ions are present in all aqueous systems. Thus, all acidic solutions contain 
a low [OH−], and all basic solutions contain a low [H3O+]. The equilibrium nature of 
autoionization allows us to define “acidic” and “basic” solutions in terms of relative 
magnitudes of [H3O+] and [OH−]:
 In an acidic solution, [H3O+] > [OH− ]
 In a neutral solution, [H3O+ ] = [OH− ]
 In a basic solution, [H3O+ ] < [OH− ]
Figure 18.3 summarizes these relationships. Moreover, if you know the value of Kw 
at a particular temperature and the concentration of one of the two ions, you can find 
the concentration of the other:
[H3O+] =
Kw
[OH−]
 or [OH−] =
Kw
[H3O+]
Figure 18.3 The relationship between 
[H3O
+] and [OH−] and the relative acidity 
of solutions.
ACIDIC
SOLUTION
[H3O+]
[H3O+] > [OH
–]
Divide into Kw
NEUTRAL
SOLUTION
[H3O+] = [OH
–]
BASIC
SOLUTION
[OH–]
[H3O+] < [OH
–]
800 Chapter 18 • Acid-Base Equilibria
Problem A research chemist adds a measured amount of HCl gas to pure water at 25°C 
and obtains a solution with [H3O+] = 3.0×10−4 M. Calculate [OH−]. Is the solution 
neutral, acidic, or basic?
Plan We use the known value of Kw at 25°C (1.0×10−14) and the given [H3O+] 
(3.0×10−4 M) to solve for [OH−]. Then we compare [H3O+] with [OH−] to determine 
whether the solution is acidic, basic, or neutral (see Figure 18.3).
Solution Calculating [OH−]:
 [OH−] =
Kw
[H3O+]
=
1.0×10−14
3.0×10−4
= 3.3×10−11 M
Because [H3O+] > [OH−], the solution is acidic.
Check It makes sense that adding an acid to water results in an acidic solution. 
Also,  since [H3O+] is greater than 10−7 M, [OH−] must be less than 10−7 M to give 
a constant Kw.
FOLLOW-UP PROBLEMS
18.2A Calculate [H3O+] in a solution that has [OH−] = 6.7×10–2 M at 25°C. Is the 
solution neutral, acidic, or basic?
18.2B An aqueous solution of window cleaner has [H3O+] = 1.8×10–10 M at 25°C. 
Calculate [OH−]. Is the solution neutral, acidic, or basic?
SOME SIMILAR PROBLEMS 18.27–18.30
SAMPLE PROBLEM 18.2
Calculating [H3O
+] or [OH−] in Aqueous 
Solution
Expressing the Hydronium Ion Concentration: The pH Scale
In aqueous solutions, [H3O+] can vary from about 10 M to 10−15 M. To handle num-
bers with negative exponents more conveniently in calculations, we convert them to 
positive numbers using a numerical system called a p-scale, the negative of the com-
mon (base-10) logarithm of the number. Applying this numerical system to [H3O+] 
gives pH, the negative of the common logarithm of [H+] (or [H3O+]):
 pH = −log [H3O+] (18.3)
What is the pH of a 10−12 M H3O+ solution?
pH = −log [H3O+] = −log 10−12 = (−1)(−12) = 12
Similarly, a 10−3 M H3O+ solution has a pH of 3, and a 5.4×10–4 M H3O+ solution 
has a pH of 3.27:
pH = −log [H3O+] = (−1)(log 5.4 + log 10−4) = 3.27
As with any measurement, the number of significant figures in a pH value reflects 
the precision with which the concentration is known. However, a pH value is a loga-
rithm, so the number of significant figures in the concentration equals the number of 
digits to the right of the decimal point in the pH value (see Appendix A). In the 
preceding example, 5.4×10–4 M has two significant figures, so its negative logarithm, 
3.27, has two digits to the right of the decimal point.
Note in particular that the higher the pH, the lower the [H3O+]. Therefore, an 
acidic solution has a lower pH (higher [H3O
+]) than a basic solution. At 25°C in 
pure water, [H3O+] is 1.0×10−7 M, so
 pH of an acidic solution < 7.00
 pH of a neutral solution = 7.00
 pH of a basic solution > 7.00
Figure 18.4 shows that the pH values of some familiar aqueous solutions fall within 
a range of 0 to 14.
M
O
R
E 
B
A
S
IC
M
O
R
E 
A
C
ID
IC
pH
1 M NaOH 
(14.0)
Lye (13.0)
Household
ammonia (11.9)
Milk of magnesia
(10.5)
Detergent
solution (~10)
Seawater (7.0–8.3)
Blood (7.4)
Milk (6.4)
Urine (4.8–7.5)
Unpolluted
rainwater
(5.6)
Beer
(4.0–4.5)
Vinegar
(2.4–3.4)
Lemon juice
(2.2–2.4)
Stomach acid
(1.0–3.0)
1 M HCl (0.0)
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
NEUTRAL
Figure 18.4 The pH values of some 
 familiar aqueous solutions.
18.2 • Autoionization of Water and the pH Scale 801
Because the pH scale is logarithmic, a solution of pH 1.0 has an [H3O+] that is 10 
times higher than that of a pH 2.0 solution, 100 times higher than that of a pH 3.0 
solution, and so forth. To find the [H3O+] from the pH, you perform the opposite 
arithmetic process; that is, you find the negative antilog of pH:
[H3O+] = 10−pH
A p-scale is used to express other quantities as well:
1. Hydroxide ion concentration can be expressed as pOH:
pOH = −log [OH−]
Acidic solutions have a higher pOH (lower [OH−]) than basic solutions.
2. Equilibrium constants can be expressed as pK:
pK = –log K
Specifically for weak acids:
 pKa = −log Ka and Ka = 10−pKa
A low pK corresponds to a high K. So
∙ a reaction that reaches equilibrium with mostly products present (proceeds far 
to the right) has a low pK (high K);
∙ a reaction that has mostly reactants present at equilibrium has a high pK (low K).
Table 18.4 shows this relationship for aqueous equilibria of some weak acids.
The Relationships Among pH, pOH, and pKw Taking the negative log of both 
sides of the Kw expression gives a useful relationship among pKw, pH, and pOH:
 Kw = [H3O+][OH−] = 1.0×10−14 (at 25°C)
 −log Kw = (−log [H3O+]) + (−log [OH−]) = −log (1.0×10−14)
 pKw = pH + pOH = 14.00 (at 25°C) (18.4)
Note these important points:
1. The sum of pH and pOH is pKw for any aqueous solution at any temperature, and 
pKw equals 14.00 at 25°C.
2. Because Kw is constant, pH, pOH, [H3O+], and [OH−] are interrelated:
∙ [H3O+] and [OH−] change in opposite directions.
∙ pH and pOH also change in opposite directions.
∙ At 25°C, the product of [H3O+] and [OH−] is 1.0×10−14, and the sum of pH 
and pOH is 14.00 (Figure 18.5, next page).
Calculating pH for Strong Acids and Bases Since strong acids and bases dissoci-
ate completely, calculating pH, [H3O+], [OH−], and pOH for these substances is 
straightforward:
∙ The equilibrium concentration of H3O+ is equal to the initial concentration of a 
strong acid: 0.200 M HCl dissociates to produce 0.200 M H3O+.
∙ The equilibrium concentration of OH− is equal to the initial concentration of a 
Group 1A(1) hydroxide: 0.200 M KOH dissociates to produce 0.200 M OH−.
 
Table 18.4 The Relationship Between Ka and pKa
Acid Name (Formula) Ka at 25°C pKa
Hydrogen sulfate ion (HSO4−) 1.0×10−2 1.99
Nitrous acid (HNO2) 7.1×10−4 3.15
Acetic acid (CH3COOH) 1.8×10−5 4.75
Hypobromous acid (HBrO) 2.3×10−9 8.64
Phenol (C6H5OH) 1.0×10−10 10.00 A
C
ID
 S
TR
EN
G
TH
802 Chapter 18 • Acid-Base Equilibria
∙ The equilibrium concentration of OH− is equal to twice the initial concentration of 
a Group 2A(2) hydroxide: 0.200 M Ba(OH)2 dissociates to produce 2(0.200 M) = 
0.400 M OH− since there are two moles of OH− in every mole of Ba(OH)2.
Figure 18.5 The relationships among 
[H3O
+], pH, [OH−], and pOH.
M
O
R
E 
B
A
SI
C
M
O
R
E 
A
C
ID
IC
[H3O+] [OH–]pH pOH
15.00
14.00
13.00
12.00
1 1 .00
10.00
9.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.00
–1.00
BASIC
NEUTRAL
ACIDIC
11.00
–1 .00
15.00
14.00
13.00
12.00
10.009.00
8.00
7.00
6.00
5.00
4.00
3.00
2.00
1.00
0.001.0x10–14
1.0x10–13
1.0x10–12
1.0x10–11
1.0x10–10
1.0x10–9
1.0x10–8
1.0x10–7
1.0x10–6
1.0x10–5
1.0x10–4
1.0x10–3
1.0x10–2
1.0x10–1
1.0x100
1.0x101
1.0x10–15
1.0x10–14
1.0x10–13
1.0x10–12
1.0x10–11
1.0x10–10
1.0x10–9
1.0x10–8
1.0x10–7
1.0x10–6
1.0x10–5
1.0x10–4
1.0x10–3
1.0x10–2
1.0x10–1
1.0x100
1.0x101
1.0x10–15
Problem Calculate [H3O+], pH, [OH−], and pOH for each solution at 25°C:
(a) 0.30 M HNO3, used for etching copper metal 
(b) 0.0042 M Ca(OH)2, used in leather tanning to remove hair from hides
Plan We know that HNO3 is a strong acid and dissociates completely; thus, [H3O+] = 
[HNO3]init. Similarly, Ca(OH)2 is a strong base and dissociates completely; the molar ratio 
is 1 mol Ca(OH)2/2 mol OH− so [OH−] = 2[Ca(OH)2]init. We use these concentrations 
and the value of Kw at 25°C (1.0×10−14) to find [OH−] or [H3O+], which we then use to 
calculate pH and pOH.
Solution (a) For 0.30 M HNO3:
 [H3O+] = 0.30 M
 pH = −log [H3O+] = −log 0.30 = 0.52
 [OH−] =
Kw
[H3O+]
=
1.0×10−14
0.30
= 3.3×10−14 M
 pOH = −log [OH−] = −log (3.3×10−14) = 13.48
(b) For 0.0042 M Ca(OH)2: Ca(OH)2(aq) ⟶ Ca2+(aq) + 2OH−(aq)
 [OH−] = 2(0.0042 M) = 0.0084 M
 pOH = −log [OH−] = −log (0.0084) = 2.08
[H3O+] =
Kw
[OH−]
=
1.0×10−14
0.0084
= 1.2×10−12 M
 pH = −log [H3O+] = −log (1.2×10−12) = 11.92
Check The strong acid has pH < 7 and the strong base has a pH > 7, as expected. In 
each case, pH + pOH = 14, so the arithmetic seems correct.
SAMPLE PROBLEM 18.3
Calculating [H3O
+], pH, [OH−], and pOH for 
Strong Acids and Bases
18.3 • Proton Transfer and the Brønsted-Lowry Acid-Base Definition 803
Measuring pH In the laboratory, pH values are usually obtained in two ways:
1. Acid-base indicators are organic molecules whose colors depend on the acidity 
of the solution in which they are dissolved. A pH can be estimated quickly with pH 
paper, a paper strip impregnated with one or a mixture of indicators. A drop of solu-
tion is placed on the strip, and the color is compared with a chart (Figure 18.6A).
2. A pH meter measures [H3O+] by means of two electrodes immersed in the test 
solution. One electrode supplies a reference system; the other consists of a very thin 
glass membrane that separates a known internal [H3O+] from the unknown external 
[H3O+]. The difference in [H3O+] creates a voltage difference across the membrane, 
which is displayed in pH units (Figure 18.6B). We examine this device in Chapter 21.
FOLLOW-UP PROBLEMS
18.3A Sodium hydroxide is used to clear clogged drains. A solution of NaOH has a 
pOH of 4.48 at 25°C. What are its pH, [OH−], and [H3O+]?
18.3B Hydrochloric acid is sold commercially under the name muriatic acid and is 
used to clean concrete masonry. A dilute solution of HCl has a pH of 2.28 at 25°C. 
Calculate its pOH, [H3O+], and [OH−].
SOME SIMILAR PROBLEMS 18.23–18.30
Figure 18.6 Methods for measuring the 
pH of an aqueous solution. A, pH paper. 
B, pH meter. Solution is 10−4 M NaOH.
Source: © McGraw-Hill Education/Charles 
Winters/Timeframe Photography, Inc.
A
B
NaOHNaOH
Summary of Section 18.2
› Pure water has a low conductivity because it autoionizes to a small extent in a process 
whose equilibrium constant is the ion-product constant for water, Kw (1.0×10−14 at 25°C).
› [H3O
+] and [OH−] are inversely related: in acidic solution, [H3O
+] is greater than [OH−]; the 
reverse is true in basic solution; and the two are equal in neutral solution.
› To express small values of [H3O
+], we use the pH scale: pH = −log [H3O+]. Similarly, 
pOH = −log [OH−], and pK = −log K.
› A high pH corresponds to a low [H3O
+]. In acidic solutions, pH < 7.00; in basic solutions, 
pH > 7.00; and in neutral solutions, pH = 7.00. The sum of pH and pOH equals 
pKw (14.00 at 25°C).
› A pH is typically measured with either an acid-base indicator or a pH meter.
›
 18.3 PROTON TRANSFER AND THE BRØNSTED-
LOWRY ACID-BASE DEFINITION
Earlier we noted a key limitation of the Arrhenius definition: many substances that 
yield OH− ions in water do not contain OH in their formulas. Examples include 
ammonia, the amines, and many salts of weak acids, such as NaF. Another limitation 
is that water had to be the solvent for acid-base reactions. In the early 20th century, 
J. N. Brønsted and T. M. Lowry suggested definitions that remove these limitations. 
(We introduced some of these ideas in Section 4.4.)
According to the Brønsted-Lowry acid-base definition,
∙ An acid is a proton donor, any species that donates an H+ ion. An acid must 
contain H in its formula; HNO3 and H2PO4− are two of many examples. All Arrhenius 
acids are Brønsted-Lowry acids.
∙ A base is a proton acceptor, any species that accepts an H+ ion. A base must 
contain a lone pair of electrons to bind H+; a few examples are NH3, CO32−, and 
F−, as well as OH− itself. Brønsted-Lowry bases are not Arrhenius bases, but all 
Arrhenius bases contain the Brønsted-Lowry base OH−.
From this perspective, an acid-base reaction occurs when one species donates a proton 
and another species simultaneously accepts it: an acid-base reaction is thus a proton-
transfer process. Acid-base reactions can occur between gases, in nonaqueous solu-
tions, and in heterogeneous mixtures, as well as in aqueous solutions.
804 Chapter 18 • Acid-Base Equilibria
According to this definition, an acid-base reaction occurs even when an acid (or 
a base) just dissolves in water, because water acts as the proton acceptor (or donor):
1. Acid donates a proton to water (Figure 18.7A). When HCl dissolves in water, an 
H+ ion (a proton) is transferred from HCl to H2O, where it becomes attached to a 
lone pair of electrons on the O atom, forming H3O+. Thus, HCl (the acid) has 
donated the H+, and H2O (the base) has accepted it:
HCl(g) + H2O(l) Cl−(aq) + H3O+(aq)
2. Base accepts a proton from water (Figure 18.7B). When ammonia dissolves in 
water, an H+ from H2O is transferred to the lone pair of N, forming NH4+, and the 
H2O becomes an OH− ion:
NH3(aq) + H2O(l) NH4+(aq) + OH−(aq)
In this case, H2O (the acid) has donated the H+, and NH3 (the base) has accepted it.
Note that H2O is amphiprotic: it acts as a base (accepts an H+) in one case and as 
an acid (donates an H+) in the other. Many other species are amphiprotic as well.
Conjugate Acid-Base Pairs
The Brønsted-Lowry definition provides a new way to look at acid-base reactions 
because it focuses on the reactants and the products as acids and bases. For example, 
let’s examine the reaction between hydrogen sulfide and ammonia:
H2S + NH3 �⥫⥬ HS− + NH4+
In the forward reaction, H2S acts as an acid by donating an H+ to NH3, which acts as 
a base by accepting it. In the reverse reaction, the ammonium ion, NH4+, acts as an 
acid by donating an H+ to the hydrogen sulfide ion, HS−, which acts as a base. Notice 
that the acid, H2S, becomes a base, HS−, and the base, NH3, becomes an acid, NH4+.
In Brønsted-Lowry terminology, H2S and HS− are a conjugate acid-base pair: 
HS− is the conjugate base of the acid H2S. Similarly, NH3 and NH4+ are a conjugate 
acid-base pair: NH4+ is the conjugate acid of the base NH3. Every acid has a conjugate 
base, and every base has a conjugate acid. For any conjugate acid-base pair,
∙ The conjugate base has one fewer H and one more negative charge than the acid.
∙ The conjugate acid has one more H and one fewer negative charge than the base.
A Brønsted-Lowry acid-base reaction occurs when an acid and a base react to 
form their conjugate base and conjugate acid, respectively:
acid1 + base2 �⥫⥬ base1 + acid2
Table 18.5 shows some Brønsted-Lowry acid-base reactions. Note these points:
∙ Each reaction has an acid and a base as reactants andas products, comprising two 
conjugate acid-base pairs.
∙ Acids and bases can be neutral molecules, cations, or anions.
∙ The same species can be an acid or a base (amphiprotic), depending on the other 
species reacting. Water behaves this way in reactions 1 and 4 in Table 18.5 and 
HPO42− does so in reactions 4 and 6.
Figure 18.7 Dissolving of an acid or 
base in water as a Brønsted-Lowry 
 acid-base reaction. A, The acid HCl dis-
solving in the base water. B, The base 
NH3 dissolving in the acid water. HCl
(acid, H+ donor)
Cl–
H2O
(base, H+ acceptor)
Lone pair of 
O binds H+ .
H3O
+
A
B
+ +
NH3
(base, H+ acceptor)
 H2O 
 (acid, H+ donor) OH
–NH4
+
+ +
Lone pair of
N binds H+ .
18.3 • Proton Transfer and the Brønsted-Lowry Acid-Base Definition 805
Table 18.5 The Conjugate Pairs in Some Acid-Base Reactions 
Reaction 1 HF + H2O ⥫⥬ F− + H3O+
Reaction 2 HCOOH + CN− ⥫⥬ HCOO− + HCN
Reaction 3 NH4+ + CO32− ⥫⥬ NH3 + HCO3
−
Reaction 4 H2PO4− + OH− ⥫⥬ HPO42− + H2O
Reaction 5 H2SO4 + N2H5
+ ⥫⥬ HSO4− + N2H62+
Reaction 6 HPO42− + SO32− ⥫⥬ PO43− + HSO3−
Acid + + AcidBase Base
Conjugate Pair
Conjugate Pair
Problem The following reactions are important environmental processes. Identify the 
conjugate acid-base pairs.
(a) H2PO4− (aq) + CO32− (aq) ⥫⥬ HCO3−(aq) + HPO42− (aq)
(b) H2O(l) + SO32− (aq) ⥫⥬ OH− (aq) + HSO3− (aq)
Plan To find the conjugate pairs, we find the species that donated an H+ (acid) and the 
species that accepted it (base). The acid (or base) on the left becomes its conjugate base 
(or conjugate acid) on the right. Remember, the conjugate acid has one more H and one 
fewer negative charge than its conjugate base.
Solution (a) H2PO4− has one more H+ than HPO42−; CO32− has one fewer H+ than 
HCO3−. Therefore, H2PO4− and HCO3− are the acids, and HPO42− and CO32− are the bases. 
The conjugate acid-base pairs are H2PO4−/HPO42− and HCO3−/CO32−.
(b) H2O has one more H+ than OH−; SO32− has one fewer H+ than HSO3−. The acids 
are H2O and HSO3−; the bases are OH− and SO32−. The conjugate acid-base pairs are 
H2O/OH− and HSO3−/SO32−.
FOLLOW-UP PROBLEMS
18.4A Identify the conjugate acid-base pairs:
(a) CH3COOH(aq) + H2O(l) �⥫⥬ CH3COO− (aq) + H3O+ (aq)
(b) H2O(l) + F − (aq) ⥫⥬ OH− (aq) + HF(aq)
18.4B Give the formula of each of the following:
(a) The conjugate acid of HSO3− (b) The conjugate base of C5H5NH+
(c) The conjugate acid of CO32− (d) The conjugate base of HCN
SOME SIMILAR PROBLEMS 18.43–18.52
SAMPLE PROBLEM 18.4 Identifying Conjugate Acid-Base Pairs
Student data indicate that you may struggle with 
the identification of conjugate acid-base pairs. 
Access the Smartbook to view additional 
Learning Resources on this topic.
Student Hot Spot
Relative Acid-Base Strength and the Net Direction of Reaction
The net direction of an acid-base reaction depends on relative acid and base strengths: 
A reaction proceeds to the greater extent in the direction in which a stronger acid 
and stronger base form a weaker acid and weaker base.
Competition for the Proton The net direction of the reaction of H2S and NH3 is 
to the right (Kc > 1) because H2S is a stronger acid than NH4+, the other acid present, 
and NH3 is a stronger base than HS−, the other base:
 H2S + NH3 ⥫⥬ HS− + NH4+
 stronger acid + stronger base ⟶ weaker base + weaker acid
You might think of the process as a competition for the proton between the two bases, 
NH3 and HS−, in which NH3 wins.
806 Chapter 18 • Acid-Base Equilibria
In effect, the extent of acid (HA) dissociation in water can be viewed as the result 
of a competition for the proton between the two bases, A− and H2O. Strong and weak 
acids give different results:
1. Strong acids. When the strong acid HNO3 dissolves, it completely transfers an 
H+ to the base, H2O, forming the conjugate base NO3− and the conjugate acid H3O+:
 HNO3 + H2O ⥫⥬ NO3− + H3O+
 stronger acid + stronger base ⟶ weaker base + weaker acid
Even though we show an equilibrium arrow here, the net direction is so far to the 
right that Kc >> 1 and the reaction is essentially complete. HNO3 is a stronger acid 
than H3O+, and H2O is a stronger base than NO3−. Thus, with a strong HA, H2O wins 
the competition for the proton because A− is a much weaker base. In fact, the only 
acidic species that remains in strong-acid solutions is H3O+.
2. Weak acids. On the other hand, with weak acids such as HF, the A− (F− for this 
example) wins the competition because it is a stronger base than H2O and the net 
direction is to the left (Kc < 1), with the net result that only a small percentage of 
HF molecules dissociate:
 HF + H2O ⥫⥬ F− + H3O+
 weaker acid + weaker base ⟵ stronger base + stronger acid 
Ranking Conjugate Pairs Based on evidence from many such reactions, we can 
rank conjugate pairs in terms of the ability of the acid to transfer its proton (Figure 18.8). 
Note that a weaker acid has a stronger conjugate base: the acid can’t give up its 
proton very readily because its conjugate base attracts the proton too strongly.
Figure 18.8 Strengths of conjugate 
acid-base pairs. The stronger the acid is, 
the weaker its conjugate base. The stron-
gest acid is at top left and the strongest 
base at bottom right. When an acid reacts 
with a base farther down the list, the 
 reaction proceeds to the right (Kc > 1).
HCN
A
C
ID
 S
TR
EN
G
TH
B
A
S
E 
S
TR
EN
G
TH
ACID BASE
Strong
Weak
Negligible
Weak
Negligible
Strong
HCl
H2SO4
HNO3
H3O+
HSO4
–
H2SO3
H3PO4
HF
CH3COOH
H2CO3
H2S
HSO3
–
H2PO4
–
NH4
+
HCO3
–
HPO4
2–
H2O
HS–
OH–
Cl –
HSO4
–
NO3
–
H2O
SO4
2–
HSO3
–
H2PO4
–
F–
CH3COO–
HCO3
–
HS–
SO3
2–
HPO4
2–
NH3
CO3
2–
PO4
3–
OH–
S2–
O2–
CN–
18.3 • Proton Transfer and the Brønsted-Lowry Acid-Base Definition 807
We use Figure 18.8 to predict the net direction of a reaction between any two 
pairs, that is, whether the equilibrium position lies to the right (Kc > 1, mostly prod-
ucts) or to the left (Kc < 1, mostly reactants). A reaction proceeds to the right if an 
acid reacts with a base lower on the list. The following two sample problems dem-
onstrate this key idea.
Problem Predict the net direction and indicate whether Kc is greater or less than 1 for 
each of the following reactions (assume equal initial concentrations of all species):
(a) H2PO4− (aq) + NH3(aq) ⥫⥬ NH4+ (aq) + HPO42− (aq)
(b) H2O(l) + HS− (aq) ⥫⥬ OH− (aq) + H2S(aq)
Plan We identify the conjugate acid-base pairs and consult Figure 18.8 to see which 
acid and base are stronger. The reaction proceeds in the direction in which the stronger 
acid and base form the weaker acid and base. If the reaction as written proceeds to the 
right, then [products] is higher than [reactants], so Kc > 1.
Solution (a) The conjugate pairs are H2PO4−/HPO42− and NH4+/NH3. Since H2PO4− is 
higher on the list of acids, it is stronger than NH4+; since NH3 is lower on the list of 
bases, it is stronger than HPO42−. Therefore,
H2PO4−(aq) + NH3(aq) ⥫⥬ HPO42−(aq) + NH4+(aq)
 stronger acid + stronger base ⟶ weaker base + weaker acid 
The net direction is to the right, so Kc > 1.
(b) The conjugate pairs are H2O/OH− and H2S/HS−. Since H2S is higher on the list of 
acids, and OH− is lower on the list of bases, we have
 H2O(l) + HS−(aq) ⥫⥬ OH−(aq) + H2S(aq)
 weaker acid + weaker base ⟵ stronger base + stronger acid
The net direction is to the left, so Kc < 1.
FOLLOW-UP PROBLEMS
18.5A Use the following conjugate acid-base pairs and Figure 18.8 to write acid-base 
reactions with Kc values as specified:
(a) H2SO3/HSO3− and HCO3−/CO32−, Kc > 1 (b) HF/F− and HCN/CN−, Kc < 1
18.5B Use balanced equations that show the net direction of thereaction to explain 
each of the following observations:
(a) You smell ammonia when NH3 dissolves in water. 
(b) The odor goes away when you add an excess of HCl to the solution in part (a). 
(c) The odor returns when you add an excess of NaOH to the solution in part (b).
SOME SIMILAR PROBLEMS 18.53–18.58
SAMPLE PROBLEM 18.5
Predicting the Net Direction of 
an Acid-Base Reaction
Problem Given that 0.10 M HX (blue and green) has a pH of 2.88, and 0.10 M HY 
(blue and orange) has a pH of 3.52, which scene best represents the final mixture after 
equimolar solutions of HX and Y− are mixed?
1 2 3
SAMPLE PROBLEM 18.6
Using Molecular Scenes to Predict the Net 
Direction of an Acid-Base Reaction
808 Chapter 18 • Acid-Base Equilibria
Summary of Section 18.3
› The Brønsted-Lowry acid-base definition does not require that bases contain OH in their 
formula or that acid-base reactions occur in aqueous solution.
› An acid is a species that donates a proton and a base is one that accepts it, so an acid-base 
reaction is a proton-transfer process.
› When an acid donates a proton, it becomes the conjugate base; when a base accepts 
a proton, it becomes the conjugate acid. In an acid-base reaction, acids and bases form 
their conjugates. A stronger acid has a weaker conjugate base, and vice versa.
› An acid-base reaction proceeds to the greater extent (K > 1) in the direction in which 
a stronger acid and base form a weaker base and acid.
›
Plan A stronger acid and base yield a weaker acid and base, so we have to determine the 
relative acid strengths of HX and HY in order to choose the correct molecular scene. The 
concentrations of the acid solutions are equal, so we can pick the stronger acid directly 
from the pH values of the two acid solutions. Because the stronger acid reacts to a greater 
extent, fewer molecules of it will be in the scene than molecules of the weaker acid.
Solution The HX solution has a lower pH (2.88) than the HY solution (3.52), so we 
know right away that HX is the stronger acid and X− is the weaker base and that HY is 
the weaker acid and Y− is the stronger base. Therefore, the reaction of HX and Y− has 
Kc > 1, which means the equilibrium mixture will have more HY than HX. Scene 1 has 
equal numbers of HX and HY, which would occur if the acids were of equal strength, 
and scene 2 shows fewer HY than HX, which would occur if HY were stronger. 
Therefore, only scene 3 is consistent with the relative acid strengths.
FOLLOW-UP PROBLEMS
18.6A The left-hand scene in the margin represents the equilibrium mixture after 0.10 M 
solutions of HA (blue and red) and B− (black) react: Does this reaction have a Kc greater 
or less than 1? Which acid is stronger, HA or HB?
18.6B The right-hand scene depicts an aqueous solution of two conjugate acid-base pairs: 
HC/C− and HD/D−. HD is a stronger acid than HC. What colors represent the base C− and 
the base D−? Does the reaction between HC and D− have a Kc greater or less than 1?
A SIMILAR PROBLEM 18.3918.6A
 18.4 SOLVING PROBLEMS INVOLVING 
WEAK-ACID EQUILIBRIA
Just as you saw in Chapter 17 for equilibrium problems in general, there are two types 
of equilibrium problems involving weak acids and their conjugate bases:
1. Given equilibrium concentrations, find Ka.
2. Given Ka and some concentrations, find other equilibrium concentrations.
For all of these problems, we’ll apply the same problem-solving approach, notation 
system, and assumptions:
∙ The problem-solving approach. Start with what is given in the problem and move 
toward what you want to find. Make a habit of applying the following steps:
1. Write the balanced equation and Ka expression; these tell you what to find.
2. Define x as the unknown change in concentration that occurs during the reac-
tion. Frequently, x = [HA]dissoc, the concentration of HA that dissociates, which, 
based on certain assumptions, also equals [H3O+] and [A−] at equilibrium.
3. Construct a reaction table (for most problems) that incorporates x.
4. Make assumptions (usually that x is very small relative to the initial concentra-
tion) that simplify the calculations.
5. Substitute the values into the Ka expression, and solve for x.
6. Check that the assumptions are justified with the 5% test first used in Sample 
Problem 17.9. If they are not justified, use the quadratic formula to find x.
18.6A 18.6B
18.4 • Solving Problems Involving Weak-Acid Equilibria 809
∙ The notation system. As always, molar concentration is indicated with brackets. A 
subscript indicates where the species comes from or when it occurs in the reaction 
process. For example, [H3O+]from HA is the molar concentration of H3O+ that comes 
from the dissociation of HA; [HA]init is the initial molar concentration of HA, that 
is, before dissociation; [HA]dissoc is the molar concentration of HA that dissociates; 
and so forth. A bracketed formula with no subscript represents the molar concentra-
tion of the species at equilibrium.
∙ The assumptions. We make two assumptions to simplify the arithmetic:
1. [H3O+] from the autoionization of water is negligible. It is so much smaller than 
the [H3O+] from the dissociation of HA that we can neglect it in these problems:
[H3O+] = [H3O+]from HA + [H3O+]from H2O ≈ [H3O
+]from HA
 Note that each molecule of HA that dissociates forms one H3O+ and one A−:
[HA]dissoc = [H3O+] = [A−]
2. A weak acid has a small Ka. Therefore, it dissociates to such a small extent that 
we can neglect the change in its concentration to find its equilibrium concentration:
[HA] = [HA]init − [HA]dissoc ≈ [HA]init
Finding Ka Given Concentrations
This type of problem involves finding Ka of a weak acid from the concentration of 
one of the species in solution, usually [H3O+] from a given pH:
HA(aq) + H2O(l) ⥫⥬ H3O+(aq) + A−(aq)
Ka =
[H3O+][A−]
[HA]
You prepare an aqueous solution of HA and measure its pH. Thus, you know 
[HA]init, can calculate [H3O+] from the pH, and then determine [A−] and [HA] at 
equilibrium. You substitute these values into the Ka expression and solve for Ka. Let’s 
go through the approach in a sample problem.
Problem Phenylacetic acid (C6H5CH2COOH, simplified here to HPAc; see model) 
builds up in the blood of persons with phenylketonuria, an inherited disorder that, if 
untreated, causes mental retardation and death. A study of the acid shows that the pH of 
0.12 M HPAc is 2.62. What is the Ka of phenylacetic acid?
Plan We are given [HPAc]init (0.12 M) and the pH (2.62) and must find Ka. As always, 
we first write the equation for HPAc dissociation and the expression for Ka to see 
which values we need. We set up a reaction table and use the given pH to find [H3O+], 
which equals [PAc−] and [HPAc]dissoc (we assume that [H3O+]from H2O is negligible). To 
find [HPAc], we assume that, because it is a weak acid, very little dissociates, so 
[HPAc]init − [HPAc]dissoc = [HPAc] ≈ [HPAc]init. We make these assumptions, substitute 
the equilibrium values, solve for Ka, and then check the assumptions using the 5% rule 
(Sample Problem 17.9).
Solution Writing the dissociation equation and Ka expression:
HPAc(aq) + H2O(l) ⥫⥬ H3O+(aq) + PAc−(aq) Ka =
[H3O+][PAc−]
[HPAc]
Setting up a reaction table, with x = [HPAc]dissoc = [H3O+]from HPAc = [PAc−]:
Concentration (M) HPAc(aq) + H2O(l) ⥫⥬ H3O+(aq) + PAc−(aq)
Initial 0.12 — 0 0
Change −x — +x +x
Equilibrium 0.12 − x — x x
SAMPLE PROBLEM 18.7 Finding Ka of a Weak Acid from the Solution pH
Phenylalanine, one of the amino acids 
that make up aspartame, is metabolized 
to phenylacetic acid (model). 
810 Chapter 18 • Acid-Base Equilibria
Finding Concentrations Given Ka
The second type of equilibrium problem gives some concentration data and Ka and 
asks for the equilibrium concentration of some component. Such problems are very 
similarto those we solved in Chapter 17 in which a substance with a given initial 
concentration reacted to an unknown extent (see Sample Problems 17.8 to 17.10). We 
will use a reaction table in these problems to find the values, and, as we just found, 
[H3O+]from H2O is so small relative to [H3O
+]from HA that we will neglect it and enter 
the initial [H3O+] in all reaction tables as zero.
Calculating [H3O+]:
[H3O+] = 10−pH = 10−2.62 = 2.4×10−3 M
Making the assumptions:
1. The calculated [H3O+] (2.4×10−3 M) >> [H3O+]from H2O (1.0×10
−7 M), so we assume that 
[H3O+] ≈ [H3O+]from HPAc = [PAc−] = x (the change in [HPAc], or [HPAc]dissoc).
2. HPAc is a weak acid, so we assume that [HPAc] = 0.12 M − x ≈ 0.12 M.
Solving for the equilibrium concentrations:
x ≈ [H3O+] = [PAc−] = 2.4×10−3 M
[HPAc] = 0.12 M − x = 0.12 M − (2.4×10−3 M) ≈ 0.12 M (to 2 sf)
Substituting these values into Ka:
Ka =
[H3O+][PAc−]
[HPAc]
≈
(2.4×10−3)(2.4×10−3)
0.12
= 4.8×10−5
Checking the assumptions by finding the percent error in concentration:
1. For [H3O+]from H2O: 
1×10−7 M
2.4×10−3 M
× 100 = 4×10−3% (<5%; assumption is justified.)
2. For [HPAc]dissoc: 
2.4×10−3 M
0.12 M
× 100 = 2.0% (<5%; assumption is justified.)
Check The [H3O+] makes sense: pH 2.62 should give [H3O+] between 10−2 and 10−3 
M. The Ka calculation also seems in the correct range: (10−3)2/10−1 = 10−5, and this 
value seems reasonable for a weak acid.
FOLLOW-UP PROBLEMS
18.7A The conjugate acid of ammonia is the weak acid NH4+. If a 0.2 M NH4Cl 
solution has a pH of 5.0, what is the Ka of NH4+?
18.7B Over a million tons of acrylic acid (H2CCHCOOH) are produced each year for 
manufacturing plastics, adhesives, and paint. A 0.30 M solution of the acid has a pH of 
2.43. What is Ka for this acid?
SOME SIMILAR PROBLEMS 18.64, 18.65, 18.72, and 18.73
Problem Propanoic acid (CH3CH2COOH, which we simplify to HPr) is a carboxylic 
acid whose salts are used to retard mold growth in foods. What are the [H3O+] and the 
pH of 0.10 M HPr (Ka = 1.3×10−5)?
Plan We know the initial concentration (0.10 M) and Ka (1.3×10−5) of HPr, and we 
need to find [H3O+] and pH. First, we write the balanced equation and the expression 
for Ka. We know [HPr]init but not [HPr] (that is, the concentration at equilibrium). If we 
let x = [HPr]dissoc, x is also [H3O+]from HPr and [Pr−] because each HPr dissociates into 
one H3O+ and one Pr−. With this information, we set up a reaction table. We assume 
SAMPLE PROBLEM 18.8
Determining Concentration and pH from Ka 
and Initial [HA]
The Effect of Concentration on the Extent 
of Acid Dissociation
If we repeat the calculation in Sample Problem 18.8, but start with a lower [HPr], 
we observe a very interesting fact about the extent of dissociation of a weak acid. 
Suppose the initial concentration of HPr is one-tenth as much, 0.010 M, rather than 
Student data indicate that you may struggle with 
calculating the pH of a weak acid. Access the 
Smartbook to view additional Learning 
Resources on this topic.
Student Hot Spot
that, because HPr has a small Ka, it dissociates very little. After solving for x, which is 
[H3O+], we check the assumption. Then we use the value for [H3O+] to find the pH.
Solution Writing the balanced equation and expression for Ka:
HPr(aq) + H2O(l) ⥫⥬ H3O+(aq) + Pr−(aq) Ka =
[H3O+][Pr−]
[HPr]
= 1.3×10−5
Setting up a reaction table, with x = [HPr]dissoc = [H3O+]from HPr = [Pr−] = [H3O+]:
Concentration (M) HPr(aq) + H2O(l) ⥫⥬ H3O+(aq) + Pr−(aq)
Initial 0.10 — 0 0
Change −x — +x +x
Equilibrium 0.10 − x — x x
Making the assumption: Ka is small, so x is small compared with [HPr]init; therefore, 
[HPr]init − x = [HPr] ≈ [HPr]init, or 0.10 M − x ≈ 0.10 M. Substituting into the Ka 
expression and solving for x:
 Ka =
[H3O+][Pr−]
[HPr]
= 1.3×10−5 ≈
(x)(x)
0.10
 x ≈ √(0.10)(1.3×10−5) = 1.1×10−3 M = [H3O+]
Checking the assumption for [HPr]dissoc:
[H3O+]
[HPr]init
× 100 =
1.1×10−3 M
0.10 M
× 100 = 1.1% (<5%; assumption is justified.)
Finding the pH:
pH = −log [H3O+] = −log (1.1×10−3) = 2.96
Check The [H3O+] and pH seem reasonable for a dilute solution of a weak acid with 
a moderate Ka. By reversing the calculation, we can check the math: (1.1×10−3)2/0.10 = 
1.2×10−5, which is within rounding of the given Ka.
Comment In Chapter 17 we introduced a benchmark, aside from the 5% rule, to see if 
the assumption is justified (see the discussion following Sample Problem 17.9):
· If 
[HA]init
Ka
 > 400, the assumption is justified: neglecting x introduces an error <5%.
· If 
[HA]init
Ka
 < 400, the assumption is not justified; neglecting x introduces an error >5%, 
so we solve a quadratic equation to find x.
In this sample problem, we have 
0.10
1.3×10−5
= 7.7 × 103, which is greater than 400. The 
alternative situation occurs in the next follow-up problem.
FOLLOW-UP PROBLEMS
18.8A Cyanic acid (HOCN) is an extremely acrid, unstable substance. What are the 
[H3O+] and the pH of 0.10 M HOCN (Ka = 3.5×10−4)?
18.8B Benzoic acid (C6H5COOH) is used as a food preservative. What are the [H3O+] 
and the pH of 0.25 M C6H5COOH (pKa = 4.20)?
SOME SIMILAR PROBLEMS 18.66–18.69 and 18.74–18.77
18.4 • Solving Problems Involving Weak-Acid Equilibria 811
812 Chapter 18 • Acid-Base Equilibria
0.10 M. After filling in the reaction table and making the same assumptions, we 
find that
x = [H3O+] = [HPr]dissoc = 3.6×10−4 M
Now, let’s compare the percentages of HPr molecules dissociated at the two different 
initial acid concentrations, using the relationship
 Percent HA dissociated = 
[HA]dissoc
[HA]init
× 100 (18.5)
Case 1: [HPr]init = 0.10 M
Percent HPr dissociated =
1.1×10−3 M
1.0×10−1 M
× 100 = 1.1%
Case 2: [HPr]init = 0.010 M
Percent HPr dissociated =
3.6×10−4 M
1.0×10−2 M
× 100 = 3.6%
As the initial acid concentration decreases, the percent dissociation of the acid 
increases. Don’t confuse the concentration of HA dissociated with the percent HA 
dissociated. The concentration, [HA]dissoc, is lower in the diluted HA solution because 
the actual number of dissociated HA molecules is smaller. It is the fraction (or the 
percent) of dissociated HA molecules that increases with dilution.
This phenomenon is analogous to a change in container volume (pressure) for a 
reaction involving gases at equilibrium (Section 17.6). In the case of gases, an increase 
in volume shifts the equilibrium position to favor more moles of gas. In the case of 
HA dissociation, a lower HA concentration, which is the same as an increase in vol-
ume, shifts the equilibrium position to favor more moles of ions.
Problem In 2011, researchers showed that hypochlorous acid (HClO) generated by 
white blood cells kills bacteria. Calculate the percent dissociation of (a) 0.40 M HClO; 
(b) 0.035 M HClO (Ka = 2.9×10−8).
Plan We know the Ka of HClO and need [HClO]dissoc to find the percent dissociation 
at two different initial concentrations. We write the balanced equation and the 
expression for Ka and then set up a reaction table, with x = [HClO]dissoc = [ClO−] = 
[H3O+]. We assume that because HClO has a small Ka, it dissociates very little. Once 
[HClO]dissoc is known, we use Equation 18.5 to find the percent dissociation and check 
the assumption.
Solution (a) Writing the balanced equation and the expression for Ka:
HClO(aq) + H2O(l) ⥫⥬ H3O+(aq) + ClO−(aq) Ka =
[ClO−][H3O+]
[HClO]
= 2.9×10−8 
Setting up a reaction table with x = [HClO]dissoc = [ClO−] = [H3O+]:
Concentration (M) HClO(aq) + H2O(l) ⥫⥬ H3O+(aq) + ClO−(aq)
Initial 0.40 — 0 0
Change −x — +x +x
Equilibrium 0.40 − x — x x
Making the assumption: Ka is small, so x is small compared with [HClO]init; therefore, 
[HClO]init − x ≈ [HClO]init, or 0.40 M − x ≈ 0.40 M. Substituting into the Ka 
expression andsolving for x:
Ka =
[ClO−][H3O+]
[HClO]
= 2.9×10−8 ≈
(x)(x)
0.40
SAMPLE PROBLEM 18.9 Finding the Percent Dissociation of a Weak Acid
Thus,
x2 = (0.40)(2.9×10−8)
x = √(0.40)(2.9×10−8) = 1.1×10−4 M = [HClO]dissoc
Finding the percent dissociation:
Percent dissociation =
[HClO]dissoc
[HClO]init
× 100 =
1.1×10−4 M
0.40 M
× 100 = 0.028%
Since the percent dissociation is <5%, the assumption is justified.
(b) Performing the same calculations using [HClO]init = 0.035 M:
 Ka =
[ClO−][H3O+]
[HClO]
= 2.9×10−8 ≈
(x)(x)
0.035
 x = √(0.035)(2.9×10−8) = 3.2×10−5 M = [HClO]dissoc
Finding the percent dissociation:
Percent dissociation =
[HClO]dissoc
[HClO]init
× 100 =
3.2×10−5 M
0.035 M
× 100 = 0.091%
Since the percent dissociation is <5%, the assumption is justified.
Check The percent dissociation is very small, as we expect for an acid with such a low 
Ka. Note, however, that the percent dissociation is larger for the lower initial 
concentration, as we also expect.
FOLLOW-UP PROBLEMS
18.9A Calculate the percent dissociation of 0.75 M HCN, an extremely poisonous acid 
that can be obtained from the pits of fruits such as cherries and apples (Ka = 6.2×10−10).
18.9B A weak acid is 3.16% dissociated in a 1.5 M solution. What is the Ka of the acid?
SOME SIMILAR PROBLEMS 18.70, 18.71, 18.78, and 18.79
The Behavior of Polyprotic Acids
Acids with more than one ionizable proton are polyprotic acids. In solution, each 
dissociation step has a different Ka. For example, phosphoric acid is a triprotic acid 
(three ionizable protons), so it has three Ka values:
H3PO4(aq) + H2O(l) ⥫⥬ H2PO4−(aq) + H3O+(aq)
 Ka1 =
[H2PO4 −][H3O+]
[H3PO4]
= 7.2×10−3
H2PO4−(aq) + H2O(l) ⥫⥬ HPO42− (aq) + H3O+(aq)
 Ka2 =
[HPO42−][H3O+]
[H2PO4−]
= 6.3×10−8
HPO42− (aq) + H2O(l) ⥫⥬ PO43− (aq) + H3O+(aq)
 Ka3 =
[PO43−][H3O+]
[HPO42−]
= 4.2×10−13
The relative Ka values show that H3PO4 is a much stronger acid than H2PO4−, which 
is much stronger than HPO42−.
Table 18.6 (next page) lists some common polyprotic acids and their Ka values. 
(More are listed in Appendix C.) Note that the general pattern seen for H3PO4 occurs 
for all polyprotic acids:
Ka1 >> Ka2 >> Ka3
This trend occurs because it is more difficult for the positively charged H+ ion to leave 
a singly charged anion (such as H2PO4−) than to leave a neutral molecule (such as H3PO4), 
and more difficult still for it to leave a doubly charged anion (such as HPO42−). Successive 
Ka values typically differ by several orders of magnitude. This fact simplifies calculations 
because we usually neglect the H3O+coming from the subsequent dissociations.
18.4 • Solving Problems Involving Weak-Acid Equilibria 813
814 Chapter 18 • Acid-Base Equilibria
*Red type indicates the ionizable protons.
Table 18.6 Successive Ka Values for Some Polyprotic Acids at 25°C 
Problem Ascorbic acid (H2C6H6O6; represented as H2Asc for this problem), known as 
vitamin C, is a diprotic acid (Ka1 = 1.0×10−5 and Ka2 = 5×10−12) found in citrus fruit. 
Calculate [HAsc−], [Asc2−], and the pH of 0.050 M H2Asc.
Plan We know the initial concentration (0.050 M) and both Ka’s for H2Asc, and we have 
to calculate the equilibrium concentrations of all species and convert [H3O+] to pH. We 
first write the equations and Ka expressions. Because Ka1 >> Ka2, we can assume that the 
first dissociation produces almost all the H3O+: [H3O+]from H2Asc >> [H3O
+]from HAsc−. Also, 
because Ka1 is small, the amount of H2Asc that dissociates can be neglected. We set up a 
reaction table for the first dissociation, with x = [H2Asc]dissoc, and then we solve for 
[H3O+] and [HAsc−]. Because the second dissociation occurs to a much lesser extent, we 
can substitute values from the first dissociation directly to find [Asc2−] from the second.
Solution Writing the equations and Ka expressions:
H2Asc(aq) + H2O(l) ⥫⥬ HAsc−(aq) + H3O+(aq)
 Ka1 =
[HAsc−][H3O+]
[H2Asc]
= 1.0×10−5
HAsc−(aq) + H2O(l) ⥫⥬ Asc2− (aq) + H3O+(aq)
 Ka2 =
[Asc2−][H3O+]
[HAsc−]
= 5×10−12
Setting up a reaction table with x = [H2Asc]dissoc = [HAsc−] ≈ [H3O+]:
Concentration (M) H2Asc(aq) + H2O(l) ⥫⥬ H3O+(aq) + HAsc−(aq)
Initial 0.050 — 0 0
Change −x — +x +x
Equilibrium 0.050 − x — x x
SAMPLE PROBLEM 18.10
Calculating Equilibrium Concentrations 
for a Polyprotic Acid
C HCH OO
OO
S HH OO
O
P H
H
H OO
O
O
As H
H
H OO
O
O
C HH OO
O
H HS
A
C
ID
 S
TR
EN
G
TH
Oxalic acid (H2C2O4)
Sulfurous acid (H2SO3)
Phosphoric acid (H3PO4)
Arsenic acid (H3AsO4)
Carbonic acid (H2CO3)
Hydrosulfuric acid (H2S) 9×10−8
4.5×10−7
1×10−17
4.7×10−11
6×10−3 1.1×10−7 3×10−12
7.2×10−3 6.3×10−8 4.2×10−13
1.4×10−2 6.5×10−8
5.6×10−2 5.4×10−5
Name (Formula) Lewis Structure* Ka1 Ka2 Ka3
Summary of Section 18.4
› Two types of weak-acid equilibrium problems involve finding Ka from a given concentration 
and finding a concentration from a given Ka.
› We simplify the arithmetic by assuming (1) that [H3O
+]from H2O is much smaller than [H3O
+]from HA 
and can be neglected and (2) that weak acids dissociate so little that [HA]init ≈ [HA] at equilibrium.
› The fraction of weak acid molecules that dissociate is greater in a more dilute solution, even 
though the total [H3O
+] is lower.
› Polyprotic acids have more than one ionizable proton, but we assume that the first 
dissociation provides virtually all the H3O
+.
›
Making the assumptions:
1. Because Ka2 << Ka1, [H3O+]from HAsc− << [H3O+]from H2Asc. Therefore,
[H3O+]from H2Asc ≈ [H3O
+]
2. Because Ka1 is small, [H2Asc]init − x = [H2Asc] ≈ [H2Asc]init. Thus,
[H2Asc] = 0.050 M − x ≈ 0.050 M
Substituting into the expression for Ka1 and solving for x:
 Ka1 =
[H3O+][HAsc−]
[H2Asc]
= 1.0×10−5 =
x2
0.050 − x
≈
x2
0.050
 x = [HAsc−] ≈ [H3O+] ≈ 7.1×10−4 M
 pH = −log [H3O+] = −log (7.1×10−4) = 3.15
Checking the assumptions:
1. [H3O+]from HAsc− << [H3O+]from H2Asc: For any second dissociation that does occur,
Ka2 =
[H3O+][Asc2−]
[HAsc−]
= 5×10−12 =
(x)(x)
7.1×10−4
x = [H3O+]from HAsc− = 6×10−8 M
This is even less than [H3O+]from H2O, so the assumption is justified.
2. [H2Asc]dissoc << [H2Asc]init: 
7.1×10−4 M
0.050 M
 × 100 = 1.4% (<5%; assumption is justified). 
Also, note that [H2Asc]init
Ka1
=
0.050
1.0×10−5
= 5000 > 400
Using the equilibrium concentrations from the first dissociation to calculate [Asc2−]:
Ka2 =
[H3O+][Asc2−]
[HAsc−]
 and [Asc2−] =
(Ka2)[HAsc−]
[H3O+]
[Asc2−] =
(5×10−12)(7.1×10−4)
7.1×10−4
= 5×10−12 M
Check Ka1 >> Ka2, so it makes sense that [HAsc−] >> [Asc2−] because Asc2− is 
produced only in the second (much weaker) dissociation. Both Ka’s are small, so all 
concentrations except [H2Asc] should be much lower than the original 0.050 M.
FOLLOW-UP PROBLEMS
18.10A Oxalic acid (HOOCCOOH, or H2C2O4) is the simplest diprotic carboxylic 
acid. Its commercial uses include bleaching straw and leather and removing rust and ink 
stains. Calculate the equilibrium values of [H2C2O4], [HC2O4−], and [C2O42−], and find 
the pH of a 0.150 M H2C2O4 solution. Use Ka values from Appendix C. (Hint: First 
check whether you need the quadratic equation to find x.)
18.10B Carbonic acid (H2CO3) plays a role in blood chemistry, cave formation, and 
ocean acidification. Using its Ka values from Appendix C, calculate the equilibrium 
values of [H2CO3], [HCO3−], and [CO32−], and find the pH of a 0.075 M H2CO3 solution.
SOME SIMILAR PROBLEMS 18.80 and 18.81
18.4 • Solving Problems Involving Weak-Acid Equilibria 815
816 Chapter 18 • Acid-Base Equilibria
 18.5 MOLECULAR PROPERTIES 
AND ACID STRENGTH
The strength of an acid depends on its ability to donate a proton, which dependsin 
turn on the strength of the bond to the acidic proton. In this section, we apply trends 
in atomic and bond properties to determine the trends in acid strength of nonmetal 
hydrides and oxoacids and then discuss the acidity of hydrated metal ions.
Acid Strength of Nonmetal Hydrides
Two factors determine how easily a proton is released from a nonmetal hydride:
∙ The electronegativity of the central nonmetal (E)
∙ The strength of the EH bond
Figure 18.9 displays two periodic trends among the nonmetal hydrides:
1. Across a period, acid strength increases. The electronegativity of the nonmetal 
E determines this horizontal trend. From left to right, as E becomes more electro-
negative, it withdraws electron density from H, and the EH bond becomes more 
polar. As a result, H+ is pulled away more easily by an O atom of a water molecule. 
In aqueous solution, the hydrides of Groups 3A(13) to 5A(15) do not behave as acids, 
but an increase in acid strength is seen from Group 6A(16) to 7A(17).
2. Down a group, acid strength increases. EH bond strength determines this vertical 
trend. As E becomes larger, the EH bond becomes longer and weaker, so H+ comes 
off more easily.* For example, hydrohalic acid strength increases down Group 7A(17):
Acid strength: HF << HCl < HBr < HI
Bond length (pm): 92 127 141 161
Bond energy (kJ/mol): 565 427 363 295
(This trend is not seen in aqueous solution, where HCl, HBr, and HI are all equally 
strong; we discuss how it is observed in Section 18.8.)
Acid Strength of Oxoacids
All oxoacids have the acidic H atom bonded to an O atom, so bond length is not 
involved. As we mentioned for the halogen oxoacids (Section 14.9), two other factors 
determine the acid strength of oxoacids:
∙ The electronegativity of the central nonmetal (E)
∙ The number of O atoms around E (related to the oxidation number, O.N., of E)
Figure 18.10 summarizes these trends:
1. For oxoacids with the same number of O atoms, acid strength increases with 
the electronegativity of E. Consider the hypohalous acids (HOE, where E is a halogen 
atom). The more electronegative E is, the more polar the OH bond becomes and 
the more easily H+ is lost (Figure 18.10A). Electronegativity (EN) decreases down a 
group, as does acid strength:
Ka of HOC1 = 2.9×10–8 Ka of HOBr = 2.3×10–9 Ka of HOI = 2.3×10–11
 EN of Cl = 3.5 EN of Br = 2.8 EN of I = 2.5
Similarly, in Group 6A(16), H2SO4 (EN of S = 2.5) is stronger than H2SeO4 (EN of 
Se = 2.4); in Group 5A(15), H3PO4 (EN of P = 2.1) is stronger than H3AsO4 (EN of 
As = 2.0), and so forth.
2. For oxoacids with different numbers of O atoms, acid strength increases with the 
number of O atoms (or with the O.N. of the central nonmetal). The electronegative O 
atoms pull electron density away from E, which makes the OH bond more polar. The 
Electronegativity increases,
so acidity increases.
H2O
H2S
Ka1 = 9×10−8
H2Se
Ka1 = 1.3×10−4
H2Te
Ka1 = 2.3×10−3
HF
Ka = 6.8×10−4
HCl
Strong acid
HBr
Strong acid
HI
Strong acid
6A(16) 7A(17)
B
on
d 
st
re
ng
th
 d
ec
re
as
es
,
so
 a
ci
di
ty
 in
cr
ea
se
s.
Figure 18.9 The effect of atomic 
and molecular properties on nonmetal 
hydride acidity.
*Actually, bond energy refers to bond breakage that forms an H atom, whereas acidity refers to 
bond breakage that forms an H+ ion. Although these two types of bond breakage are not the 
same, the trends in bond energy and acid strength are opposites.
18.5 • Molecular Properties and Acid Strength 817
more O atoms present, the greater the shift in electron density, and the more easily the 
H+ ion comes off (Figure 18.10B). Therefore, the chlorine oxoacids (HOClOn, with n 
from 0 to 3) increase in strength with the number of O atoms (and the O.N. of Cl):
 +1 +3 +5 +7
Ka of HOCl = 2.9×10−8 Ka of HOClO = 1.12×10−2 Ka of HOClO2 ≈ 1 Ka of HOClO3 > 107
Similarly, HNO3 is stronger than HNO2, H2SO4 is stronger than H2SO3, and so forth.
Acidity of Hydrated Metal Ions
The aqueous solutions of certain metal ions are acidic because the hydrated metal ion 
transfers an H+ ion to water. Consider a general metal nitrate, M(NO3)n, as it dissolves 
in water. The ions separate and the metal ion becomes bonded to some number of 
H2O molecules. This equation shows the hydration of the cation (Mn+) using H2O 
molecules and “(aq)”; hydration of the anion (NO3−) is indicated with just “(aq)”:
M(NO3)n(s) + xH2O(l) ⟶ M(H2O)xn+ (aq) + nNO3− (aq)
If the metal ion, Mn+, is small and highly charged, its high charge density withdraws 
sufficient electron density from the OH bonds of the bound water molecules for an 
H+ to be released. Thus, the hydrated cation, M(H2O)xn+, is a typical Brønsted-Lowry 
acid. The bound H2O that releases the H+ becomes a bound OH− ion:
M(H2O)xn+ (aq) + H2O(l ) ⥫⥬ M(H2O)x−1OH(n−1)+ (aq) + H3O+ (aq)
Salts of most M2+ and M3+ ions yield acidic aqueous solutions. The Ka values for 
some acidic hydrated metal ions appear in Appendix C.
Consider the small, highly charged Al3+ ion. When an aluminum salt, such as 
Al(NO3)3, dissolves in water, the following steps occur:
Al(NO3)3(s) + 6H2O(l) ⟶ Al(H2O)63+ (aq) + 3NO3− (aq)
[dissolution and hydration]
 Al(H2O)63+ (aq) + H2O(l ) ⥫⥬ Al(H2O)5OH2+ (aq) + H3O+ (aq) Ka = 1×10−5
[dissociation of weak acid]
Note the formulas of the hydrated metal ions in the last step. When H+ is released, the 
number of bound H2O molecules decreases by 1 (from 6 to 5) and the number of bound 
OH− ions increases by 1 (from 0 to 1), which reduces the ion’s positive charge by 1 
(from 3 to 2) (Figure 18.11). This pattern of changes in the formula of the hydrated metal 
ion before and after it loses a proton occurs with any highly charged metal ion in water.
Figure 18.11 The acidic behavior of 
the hydrated Al3+ ion. The hydrated Al3+ 
ion is small and highly charged and pulls 
electron density from the OH bonds, so 
an H+ ion can be transferred to a nearby 
water molecule.
Al(
 
H2O)6
3+ Al(
 
H2O)5OH
2+
Al3+
Electron density
is drawn
toward Al3+.
 O—H bond 
 becomes 
 more polar.
Nearby H2O
acts as a base.
Al3+
H2O
+ +
–
H3O
+
A
B
< H O Br
δ+ δ–
<
H O Cl <<
δ+ δ–
H O Cl O
O
O
Electronegativity increases, so acidity increases.
H O I
δ+ δ–
Number of O atoms increases, so acidity increases.
δ+ δ–
H O Cl
δ+ δ–
Figure 18.10 The relative strengths of 
oxoacids. A, Cl withdraws electron density 
(thickness of green arrow) from the OH 
bond most effectively, making that bond 
most polar (relative size of δ symbols). 
B, Additional O atoms pull more electron 
density from the OH bond.
818 Chapter 18 • Acid-Base Equilibria
Summary of Section 18.5
› For nonmetal hydrides, acid strength increases across a period, with the electronegativity of 
the nonmetal (E), and down a group, with the length of the EH bond.
› For oxoacids with the same number of O atoms, acid strength increases with electro-
negativity of E; for oxoacids with the same E, acid strength increases with number of 
O atoms (or O.N. of E).
› Small, highly charged metal ions are acidic in water because they withdraw electron density 
from the OH bonds of bound H2O molecules, releasing an H+ ion to the solution.
›
 18.6 WEAK BASES AND THEIR RELATION 
TO WEAK ACIDS
The Brønsted-Lowry concept expands the definition of a base to encompass a host of 
species that the Arrhenius definition excludes: to accept a proton, a base needs only 
a lone electron pair.
Let’s examine the equilibrium system of a weak base (B) as it dissolves in water: 
B accepts a proton from H2O, which acts as an acid, leaving behind an OH− ion:
B(aq) + H2O(l ) ⥫⥬ BH+ (aq) + OH− (aq)
This general reaction for a base in water is described