# Relative Strengths of Acids and Bases

### Learning Objectives

By the end of this module, you will be able to:

• Assess the relative strengths of acids and bases according to their ionization constants
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems

We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:

$\text{HA}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

Water is the base that reacts with the acid HA, A is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of
${\text{H}}_{3}{\text{O}}^{+}$
and A when the acid ionizes in water; Table 1 lists several strong acids. A weak acid gives small amounts of
${\text{H}}_{3}{\text{O}}^{+}$
and A.

Table 1. Some Common Strong acids and  Strong Bases
Strong Acids Strong Bases
HClO4 perchloric acid LiOH lithium hydroxide
HCl hydrochloric acid NaOH sodium hydroxide
HBr hydrobromic acid KOH potassium hydroxide
HI hydroiodic acid Ca(OH)2 calcium hydroxide
HNO3 nitric acid Sr(OH)2 strontium hydroxide
H2SO4 sulfuric acid Ba(OH)2 barium hydroxide
The relative strengths of acids may be determined by measuring their equlibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid HA:

$\text{HA}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$
,

we write the equation for the ionization constant as:

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\text{[HA]}}$

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger the concentration of
${\text{H}}_{3}{\text{O}}^{+}$
and A relative to the concentration of the nonionized acid, HA. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. (A table of ionization constants of weak acids appears in Ionization Constants of Weak Acids, with a partial listing in Table 1.)

The following data on acid-ionization constants indicate the order of acid strength CH3CO2H < HNO2 <
${\text{HSO}}_{4}^{-}:$

${\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}^{-}\left(aq\right){K}_{\text{a}}=1.8\times {10}^{-5}$

${\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{NO}}_{2}^{-}\left(aq\right){K}_{\text{a}}=4.6\times {10}^{-4}$

${\text{HSO}}_{4}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(aq\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{SO}}_{4}^{2-}\left(aq\right){K}_{\text{a}}=1.2\times {10}^{-2}$

Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:

$\text{\% ionization}=\frac{{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[\text{HA}\right]}_{0}}\times 100$

Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.

### Example 1: Calculation of Percent Ionization from pH

Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.

Calculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.

We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:

$\text{B}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HB}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Water is the acid that reacts with the base, HB+ is the conjugate acid of the base B, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH and HB+ when it reacts with water; Table 1 lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.

As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant, (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:

$\text{B}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HB}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$
,

we write the equation for the ionization constant as:

${K}_{\text{b}}=\frac{\left[{\text{HB}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$

where the concentrations are those at equilibrium. Again, we do not include [H2O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are:

${\text{NO}}_{2}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{HNO}}_{2}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=2.22\times {10}^{-11}$

${\text{CH}}_{3}{\text{CO}}_{2}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=5.6\times {10}^{-10}$

${\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{NH}}_{4}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=1.8\times {10}^{-5}$

A table of ionization constants of weak bases appears in Ionization Constants of Weak Bases (with a partial list in Table 2). As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution.

Consider the ionization reactions for a conjugate acid-base pair, HA − A:

$\text{HA}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{A}}^{-}\left(aq\right){K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

${\text{A}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+\text{HA}\left(aq\right){K}_{\text{b}}=\frac{\left[\text{HA}\right]\left[\text{OH}\right]}{\left[{\text{A}}^{-}\right]}$

Adding these two chemical equations yields the equation for the autoionization for water:

$\cancel{\text{HA}\left(aq\right)}+\text{H}_{2}\text{O}\left(l\right)+\cancel{\text{A}^{-}\left(aq\right)}+\text{H}_{2}\text{O}\left(l\right)\rightleftharpoons\text{H}_{3}\text{O}^{+}\left(aq\right)+\cancel{\text{A}^{-}\left(aq\right)}+\text{OH}^{-}\left(aq\right)+\cancel{\text{HA}\left(aq\right)}$

${\text{2H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ K expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:

${K}_{\text{a}}\times {K}_{\text{b}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\text{[HA]}}\times \frac{\text{[HA]}\left[{\text{OH}}^{-}\right]}{\left[{\text{A}}^{-}\right]}=\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]={K}_{\text{w}}$

For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion (CH3COO), is 5.6 × 10−10. The product of these two constants is indeed equal to Kw:

${K}_{\text{a}}\times {K}_{\text{b}}=\left(1.8\times {10}^{-5}\right)\times \left(5.6\times {10}^{-10}\right)=1.0\times {10}^{-14}={K}_{\text{w}}$

The extent to which an acid, HA, donates protons to water molecules depends on the strength of the conjugate base, A, of the acid. If A is a strong base, any protons that are donated to water molecules are recaptured by A. Thus there is relatively little A and
${\text{H}}_{3}{\text{O}}^{+}$
in solution, and the acid, HA, is weak. If A is a weak base, water binds the protons more strongly, and the solution contains primarily A and
${\text{H}}_{3}{\text{O}}^{+}$
—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure 1).

Figure 1. This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.

Figure 2 lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.

Figure 2. The chart shows the relative strengths of conjugate acid-base pairs.

The first six acids in Figure 2 are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.

Those acids that lie between the hydronium ion and water in Figure 2 form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure 3 exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.

The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure 2. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.

### Example 2: The Product Ka ×Kb = Kw

Use the Kb for the nitrite ion,
${\text{NO}}_{2}^{-}$
, to calculate the Ka for its conjugate acid.

We can determine the relative acid strengths of
${\text{NH}}_{4}^{+}$
and HCN by comparing their ionization constants. The ionization constant of HCN is given in Ionization Constants of Weak Acids as 4 × 10−10. The ionization constant of
${\text{NH}}_{4}^{+}$
is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 × 10−5. Determine the ionization constant of
${\text{NH}}_{4}^{+}$
, and decide which is the stronger acid, HCN or
${\text{NH}}_{4}^{+}$
.

### Exercises

1. Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
2. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
3. Use this list of important industrial compounds (and Figure 2) to answer the following questions regarding: CaO, Ca(OH)2, CH3CO2H, CO2, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.

1. Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
2. List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying between those of
${\text{H}}_{3}{\text{O}}^{+}$
and H2O.
3. List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lying between those of H2O and OH.

4. Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant for H2SO3.
5. Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.
6. What is the ionization constant at 25 °C for the weak acid
${\text{CH}}_{3}{\text{NH}}_{3}^{+}$
, the conjugate acid of the weak base CH3NH2, Kb = 4.4 × 10−4.
7. What is the ionization constant at 25 °C for the weak acid
${\left({\text{CH}}_{3}\right)}_{2}{\text{NH}}_{2}^{+}$
, the conjugate acid of the weak base (CH3)2NH, Kb = 7.4 × 10−4?

## The Ionization of Weak Acids and Weak Bases

Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).

Acetic acid, CH3CO2H, is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:

${\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}^{-}\left(aq\right)$
,

giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure 3). The remaining weak acid is present in the nonionized form.

For acetic acid, at equilibrium:

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=1.8\times {10}^{-5}$

Figure 3. pH paper indicates that a 0.l-M solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [H3O+] = 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) is has a pH of 3 ([H3O+] = 0.001 M) because the weak acid CH3CO2H is only partially ionized. In this solution, [H3O+] < [CH3CO2H]. (credit: modification of work by Sahar Atwa)

Table 2. Ionization Constants of Some Weak Acids
Ionization Reaction Ka at 25 °C
${\text{HSO}}_{4}^{-}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{SO}}_{4}^{2-}$
1.2 × 10−2
$\text{HF}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{F}}^{-}$
7.2 × 10−4
${\text{HNO}}_{2}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{NO}}_{2}^{-}$
4.5 × 10−4
$\text{HNCO}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{NCO}}^{-}$
3.46 × 10−4
${\text{HCO}}_{2}\text{H}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{HCO}}_{2}^{-}$
1.8 × 10−4
${\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{\text{CO}}_{2}^{-}$
1.8 × 10−5
$\text{HCIO}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{CIO}}^{-}$
3.5 × 10−8
$\text{HBrO}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{BrO}}^{-}$
2 × 10−9
$\text{HCN}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}+{\text{CN}}^{-}$
4 × 10−10
Table 2 gives the ionization constants for several weak acids; additional ionization constants can be found in Ionization Constants of Weak Acids.

At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).

For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation

${\left({\text{CH}}_{3}\right)}_{3}\text{N}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$
,

giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.

We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 4). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, Kb, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:

${K}_{\text{b}}=\frac{\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{CH}}_{3}\right)}_{3}\text{N}\right]}$

Figure 4. pH paper indicates that a 0.1-M solution of NH3 (left) is weakly basic. The solution has a pOH of 3 ([OH] = 0.001 M) because the weak base NH3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)
The ionization constants of several weak bases are given in Table 3 and in Ionization Constants of Weak Bases.

Table 3. Ionization Constants of Some Weak Bases
Ionization Reaction Kb at 25 °C
${\left({\text{CH}}_{3}\right)}_{2}\text{NH}+{\text{H}}_{2}\text{O}\rightleftharpoons {\left({\text{CH}}_{3}\right)}_{2}{\text{NH}}_{2}^{+}+{\text{OH}}^{-}$
7.4 × 10−4
${\text{CH}}_{3}{\text{NH}}_{2}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{CH}}_{3}{\text{NH}}_{3}^{+}+{\text{OH}}^{-}$
4.4 × 10−4
${\left({\text{CH}}_{3}\right)}_{3}\text{N}+{\text{H}}_{2}\text{O}\rightleftharpoons {\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}+{\text{OH}}^{-}$
7.4 × 10−5
${\text{NH}}_{3}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{NH}}_{4}^{+}+{\text{OH}}^{-}$
1.8 × 10−5
${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{C}}_{6}{\text{N}}_{5}{\text{NH}}_{3}^{+}+{\text{OH}}^{-}$
4.6 × 10−10

Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN, is the stronger base? See Table 3.

[practice-area rows="4"][/practice-area]

### Exercises

1. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO2H and 0.10 M in HClO.
2. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO2 and 0.120 M in HBrO.
3. Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH3NH2 and 0.10 M in C5H5N (Kb = 1.7 × 10−9).
4. Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH3 and 0.100 M in C6H5NH2.
5. Using the Ka values in Ionization Constants of Weak Acids, place
$\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}^{3+}$
in the correct location in Figure 3.

### Example 3: Determination of Ka from Equilibrium Concentrations

Acetic acid is the principal ingredient in vinegar (Figure 5); that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and
$\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right]=0.00118M$
. What is the value of Ka for acetic acid?

Figure 5. Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

What is the equilibrium constant for the ionization of the
${\text{HSO}}_{4}^{-}$
ion, the weak acid used in some household cleansers:

${\text{HSO}}_{4}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{SO}}_{4}^{2-}\left(aq\right)$

In one mixture of NaHSO4 and Na2SO4 at equilibrium,
$\left[{\text{H}}_{3}{\text{O}}^{+}\right]$
= 0.027 M;
$\left[{\text{HSO}}_{4}^{-}\right]=0.29M$
; and
$\left[{\text{SO}}_{4}^{2-}\right]=0.13M$
.

### Example 4: Determination of Kb from Equilibrium Concentrations

Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M,
$\left[{\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}{\text{H}}^{+}\right]$
= 5.0 × 10−3M, and [OH] = 2.5 × 10−3M?

What is the equilibrium constant for the ionization of the
${\text{HPO}}_{4}^{2-}$
ion, a weak base:

${\text{HPO}}_{4}^{2-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{2}{\text{PO}}_{4}^{-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

In a solution containing a mixture of NaH2PO4 and Na2HPO4 at equilibrium, [OH] = 1.3 × 10−6M;
$\left[{\text{H}}_{2}{\text{PO}}_{4}^{-}\right]=0.042M$
; and
$\left[{\text{HPO}}_{4}^{2-}\right]=0.341M$
.

### Example 5: Determination of Ka or Kb from pH

The pH of a 0.0516-M solution of nitrous acid, HNO2, is 2.34. What is its Ka?

${\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{NO}}_{2}^{-}\left(aq\right)$

The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.

### Example 6: Equilibrium Concentrations in a Solution of a Weak Acid

Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings (Figure 6).

Figure 6. The pain of an ant’s sting is caused by formic acid. (credit: John Tann)

What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?

${\text{HCO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{HCO}}_{2}^{-}\left(aq\right){K}_{\text{a}}=1.8\times {10}^{-4}$

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?

${\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}^{-}\left(aq\right){K}_{\text{a}}=1.8\times {10}^{-5}$

(Hint: Determine
$\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right]$
at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or
$\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right]}{{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}_{\text{initial}}}\times 100$
.

Example 7 shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.

### Example 7: Equilibrium Concentrations in a Solution of a Weak Base

Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:

${\left({\text{CH}}_{3}\right)}_{3}\text{N}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=7.4\times {10}^{-5}$

1. Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 7.4 × 10−5.
2. Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or
$\frac{\left[{\text{NH}}_{4}^{+}\right]}{\left[{\text{NH}}_{3}\right]}\times 100$

Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.

### Example 8: Equilibrium Concentrations in a Solution of a Weak Acid

Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the
${\text{HSO}}_{4}^{-}$
ion, a weak acid. What is the pH of a 0.50-M solution of
${\text{HSO}}_{4}^{-}?$

${\text{HSO}}_{4}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{SO}}_{4}^{2-}\left(aq\right){K}_{\text{a}}=1.2\times {10}^{-2}$

1. Show that the quadratic formula gives x = 7.2 × 10−2.
2. Calculate the pH in a 0.010-M solution of caffeine, a weak base:
${\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}{\text{H}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right){K}_{\text{b}}=2.5\times {10}^{-4}$

(Hint: It will be necessary to convert [OH] to
$\left[{\text{H}}_{3}{\text{O}}^{+}\right]$
or pOH to pH toward the end of the calculation.)

### Exercises

Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Ionization Constants of Weak Acids and Ionization Constants of Weak Bases.

#### Solution 1

0.0092 M HClO, a weak acid

#### Solution 2

0.0784 M C6H5NH2, a weak base

#### Solution 3

0.0810 M HCN, a weak acid

#### Solution 4

0.11 M (CH3)3N, a weak base