Chemistry for Majors: Atoms First

Module 14: Acid-Based Equilibria

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, K_a. 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 K_a 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.

Show Answer

The percent ionization for an acid is:

\frac{{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[{\text{HNO}}_{2}\right]}_{0}}\times 100

The chemical equation for the dissociation of the nitrous acid is:

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

. Since 10^−pH =

\left[\text{H}_{3}\text{O}^{+}\right]

, we find that 10^−2.09 = 8.1 × 10⁻³M, so that percent ionization is:

\frac{8.1\times {10}^{-3}}{0.125}\times 100=6.5\%

Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.

Check Your Learning

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

Show Answer

1.3% ionized

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.

Watch this simulation of strong and weak acids and bases at the molecular level.

As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant, (K_b) 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 K_w:

{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).

The diagram shows two horizontal bars. The first, labeled, “Relative acid strength,” at the top is red on the left and gradually changes to purple on the right. The red end at the left is labeled, “Stronger acids.” The purple end at the right is labeled, “Weaker acids.” Just outside the bar to the lower left is the label, “K subscript a.” The bar is marked off in increments with a specific acid listed above each increment. The first mark is at 1.0 with H subscript 3 O superscript positive sign. The second is ten raised to the negative two with H C l O subscript 2. The third is ten raised to the negative 4 with H F. The fourth is ten raised to the negative 6 with H subscript 2 C O subscript 3. The fifth is ten raised to a negative 8 with C H subscript 3 C O O H. The sixth is ten raised to the negative ten with N H subscript 4 superscript positive sign. The seventh is ten raised to a negative 12 with H P O subscript 4 superscript 2 negative sign. The eighth is ten raised to the negative 14 with H subscript 2 O. Similarly the second bar, which is labeled “Relative conjugate base strength,” is purple at the left end and gradually becomes blue at the right end. Outside the bar to the left is the label, “Weaker bases.” Outside the bar to the right is the label, “Stronger bases.” Below and to the left of the bar is the label, “K subscript b.” The bar is similarly marked at increments with bases listed above each increment. The first is at ten raised to the negative 14 with H subscript 2 O above it. The second is ten raised to the negative 12 C l O subscript 2 superscript negative sign. The third is ten raised to the negative ten with F superscript negative sign. The fourth is ten raised to a negative eight with H C O subscript 3 superscript negative sign. The fifth is ten raised to the negative 6 with C H subscript 3 C O O superscript negative sign. The sixth is ten raised to the negative 4 with N H subscript 3. The seventh is ten raised to the negative 2 with P O subscript 4 superscript three negative sign. The eighth is 1.0 with O H superscript negative sign.

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.

This figure includes a table separated into a left half which is labeled “Acids” and a right half labeled “Bases.” A red arrow points up the left side, which is labeled “Increasing acid strength.” Similarly, a blue arrow points downward along the right side, which is labeled “Increasing base strength.” Names of acids and bases are listed next to each arrow toward the center of the table, followed by chemical formulas. Acids listed top to bottom are sulfuric acid, H subscript 2 S O subscript 4, hydrogen iodide, H I, hydrogen bromide, H B r, hydrogen chloride, H C l, nitric acid, H N O subscript 3, hydronium ion ( in pink text) H subscript 3 O superscript plus, hydrogen sulfate ion, H S O subscript 4 superscript negative, phosphoric acid, H subscript 3 P O subscript 4, hydrogen fluoride, H F, nitrous acid, H N O subscript 2, acetic acid, C H subscript 3 C O subscript 2 H, carbonic acid H subscript 2 C O subscript 3, hydrogen sulfide, H subscript 2 S, ammonium ion, N H subscript 4 superscript +, hydrogen cyanide, H C N, hydrogen carbonate ion, H C O subscript 3 superscript negative, water (shaded in beige) H subscript 2 O, hydrogen sulfide ion, H S superscript negative, ethanol, C subscript 2 H subscript 5 O H, ammonia, N H subscript 3, hydrogen, H subscript 2, methane, and C H subscript 4. The acids at the top of the listing from sulfuric acid through nitric acid are grouped with a bracket to the right labeled “Undergo complete acid ionization in water.” Similarly, the acids at the bottom from hydrogen sulfide ion through methane are grouped with a bracket and labeled, “Do not undergo acid ionization in water.” The right half of the figure lists bases and formulas. From top to bottom the bases listed are hydrogen sulfate ion, H S O subscript 4 superscript negative, iodide ion, I superscript negative, bromide ion, B r superscript negative, chloride ion, C l superscript negative, nitrate ion, N O subscript 3 superscript negative, water (shaded in beige), H subscript 2 O, sulfate ion, S O subscript 4 superscript 2 negative, dihydrogen phosphate ion, H subscript 2 P O subscript 4 superscript negative, fluoride ion, F superscript negative, nitrite ion, N O subscript 2 superscript negative, acetate ion, C H subscript 3 C O subscript 2 superscript negative, hydrogen carbonate ion, H C O subscript 3 superscript negative, hydrogen sulfide ion, H S superscript negative, ammonia, N H subscript 3, cyanide ion, C N superscript negative, carbonate ion, C O subscript 3 superscript 2 negative, hydroxide ion (in blue), O H superscript negative, sulfide ion, S superscript 2 negative, ethoxide ion, C subscript 2 H subscript 5 O superscript negative, amide ion N H subscript 2 superscript negative, hydride ion, H superscript negative, and methide ion C H subscript 3 superscript negative. The bases at the top, from perchlorate ion through nitrate ion are group with a bracket which is labeled “Do not undergo base ionization in water.” Similarly, the lower 5 in the listing, from sulfide ion through methide ion are grouped and labeled “Undergo complete base ionization in water.”

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 K_a ×K_b = K_w

Use the K_b for the nitrite ion,

{\text{NO}}_{2}^{-}

, to calculate the K_a for its conjugate acid.

Show Answer

K_b for

{\text{NO}}_{2}^{-}

is given in this section as 2.22 × 10⁻¹¹. The conjugate acid of

{\text{NO}}_{2}^{-}

is HNO₂; K_a for HNO₂ can be calculated using the relationship:

{K}_{\text{a}}\times {K}_{\text{b}}=1.0\times {10}^{-14}={K}_{\text{w}}

Solving for K_a, we get:

{K}_{\text{a}}=\frac{{K}_{\text{w}}}{{K}_{\text{b}}}=\frac{1.0\times {10}^{-14}}{2.22\times {10}^{-11}}=4.5\times {10}^{-4}

This answer can be verified by finding the K_a for HNO₂ in Ionization Constants of Weak Acids.

Check Your Learning

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}^{+}

Show Answer

{\text{NH}}_{4}^{+}

is the slightly stronger acid (K_a for

{\text{NH}}_{4}^{+}

= 5.6 × 10⁻¹⁰).

Exercises

Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
Use this list of important industrial compounds (and Figure 2) to answer the following questions regarding: CaO, Ca(OH)₂, CH₃CO₂H, CO_2, 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⁻.
Explain why the ionization constant, K_a, for H₂SO₄ is larger than the ionization constant for H₂SO₃.
Explain why the ionization constant, K_a, for HI is larger than the ionization constant for HF.
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, K_b = 4.4 × 10⁻⁴.
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, K_b = 7.4 × 10⁻⁴?

Show Selected Answers

2. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH⁻, which causes the solution to be basic. An example is NaCN. The CN⁻ reacts with water as follows:

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

4. The oxidation state of the sulfur in H₂SO₄ is greater than the oxidation state of the sulfur in H₂SO₃.

6. K_w = K_a × K_b; thus,

\begin{array}{rcl}{K}_{\text{a}}&=&\frac{{K}_{\text{w}}}{{K}_{\text{b}}}\\{K}_{\text{a}}&=&\frac{1.0\times {10}^{-14}}{4.4\times {10}^{-4}}=2.3\times {10}^{-11}\end{array}

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}

This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover.

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	K_a 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, K_b, 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]}

This photo shows two glass containers filled with a transparent liquid. In between the containers is a p H strip indicator guide. There are p H strips placed in front of each glass container. The liquid in the container on the left appears to have a p H of 10 or 11. The liquid in the container on the right appears to have a p H of about 13 or 14.

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	K_b 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

Think about It

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

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.
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.
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 (K_b = 1.7 × 10⁻⁹).
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.
Using the K_a 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.

Show Answer to Question 2

Question 2

The reactions and equilibrium constants are:

\begin{array}{l}{\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}}=4.5\times {10}^{-4}\\ \text{HBrO}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{BrO}}^{-}\left(aq\right){K}_{\text{a}}=2\times {10}^{-9}\end{array}

As K_a is much larger for HNO₂ than for HBrO, the first equilibrium will dominate. The equilibrium expression is

{K}_{\text{a}}=\frac{\left[{\text{NO}}_{2}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HNO}}_{2}\right]}=4.5\times {10}^{-5}

The initial and equilibrium concentrations for this system can be written as follows:

	[HNO₂]	[H₃O⁺]	[NO₂⁻]
Initial concentration (M)	0.134	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.134 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.134 − x) ≈ 0.134 gives:

\frac{\left[{\text{NO}}_{2}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{HNO}}_{2}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.134-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.134}=4.5\times {10}^{-4}

Solving for x gives 7.77 × 10⁻³M. Because this value is 5.8% of 0.134, our assumption is incorrect. Therefore, we must use the quadratic formula. Using the data gives the following equation: x² + 4.5 × 10⁻⁷x − 6.03 × 10⁻⁵ = 0

Using the quadratic formula gives (a = 1, b = 4.5 × 10⁻⁴, and c = −6.03 × 10⁻⁵)

\begin{array}{rcl}{ }x&=&\frac{-b\pm \sqrt{{b}^{\text{2+}}-4ac}}{2a}=\frac{-\left(4.5\times {10}^{-4}\right)\pm \sqrt{{\left(4.5\times {10}^{-4}\right)}^{\text{2+}}-4\left(1\right)\left(-6.03\times {10}^{-5}\right)}}{2\left(1\right)}\\{}&=&\frac{-\left(4.5\times {10}^{-4}\right)\pm \left(1.55\times {10}^{-2}\right)}{2}=7.54\times {10}^{-3}M\left(\text{positive root}\right)\end{array}

The equilibrium concentrations are therefore:

$\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\left[{\text{NO}}_{2}^{-}\right]=7.54\times {10}^{-3}=7.5\times {10}^{-3}M$
[HNO₂] = 0.134 − 7.54 × 10⁻³ = 0.1264 = 0.126

[OH⁻] can be calculated using K_w:

\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{7.54\times {10}^{-3}}=1.33\times {10}^{-12}=1.3\times {10}^{-12}M

Finally, use the other equilibrium to find the other concentrations. Assume for [HBrO] that (0.120 − x) ≈ 0.124 M.

{K}_{\text{a}}=\frac{\left[{\text{BrO}}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[\text{HBrO}\right]}=2\times {10}^{-9}\approx \frac{\left[{\text{BrO}}^{-}\right]\left(7.54\times {10}^{-3}\right)}{0.120}

Solving for [BrO⁻] gives:

[BrO⁻] = 3.2 × 10⁻⁸ = 3.2 × 10⁻⁸M
[HBrO] = 0.120 − 3.2 × 10⁻⁸ = 0.120 M

Show Answer to Question 4

Question 4

The reactions and equilibrium constants are:

\begin{array}{l}{}{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+{\text{NH}}_{4}^{+}\left(aq\right){K}_{\text{b}}=1.8\times {10}^{-5}\\ {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{OH}}^{-}\left(aq\right)+{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\left(aq\right){K}_{\text{b}}=4.6\times {10}^{-10}\end{array}

As K_b is much larger for NH₃ than for C₆H₅NH_2, the first equilibrium will dominate. The equilibrium expression is

{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=1.8\times {10}^{-5}

The initial and equilibrium concentrations for this system can be written as follows:

	[NH₃]	[OH⁻]	[NH₄⁺]
Initial concentration (M)	0.115	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.115 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.115 − x) ≈ 0.115 gives:

\frac{\left[{\text{NH}}_{4}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.115-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.115}=1.8\times {10}^{-5}

Solving for x gives 1.44 × 10⁻³M. Because this value is less than 5% of 0.115 M, our assumption is correct. The equilibrium concentrations are therefore:

[OH⁻] =
$\left[{\text{NO}}_{4}^{+}\right]$
= 1.44 × 10⁻³ = 0.0014 M
[NH₃] = 0.115 − 0.00144 = 0.1136 = 0.144 M

\left[{\text{H}}_{3}{\text{O}}^{+}\right]

can be calculated using K_w:

\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{0.00144}=6.94\times {10}^{-12}=6.9\times {10}^{-12}M

Finally, use the other equilibrium to find the other concentrations. Assume for [C₆H₅NH₂] that (0.100 − x) ≈ 0.100 M:

\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{0.00144}=6.94\times {10}^{-12}=6.9\times {10}^{-12}M

Solving for

\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right]

gives:

$\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right]$
= 3.9 × 10⁻⁸M
[C₆H₅NH₂] = 0.100 − 3.19 × 10⁻⁸ = 0.100 M

Example 3: Determination of K_a 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 K_a for acetic acid?

An image shows the label of a bottle of distilled white vinegar. The label states that the contents have been reduced with water to 5 percent acidity.

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

Show Answer

We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

{\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}}=\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]}=\frac{\left(0.00118\right)\left(0.00118\right)}{0.0787}=1.77\times {10}^{-5}

Check Your Learning

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

Show Answer

K_a for

{\text{HSO}}_{4}^{-}

= 1.2 × 10⁻²

Example 4: Determination of K_b from Equilibrium Concentrations

Caffeine, C₈H₁₀N₄O₂ is a weak base. What is the value of K_b 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?

Show Answer

At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

{\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}}=\frac{\left[{\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}\right]}=\frac{\left(5.0\times {10}^{-3}\right)\left(2.5\times {10}^{-3}\right)}{0.050}=2.5\times {10}^{-4}

Check Your Learning

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

Show Answer

K_b for

{\text{HPO}}_{4}^{2-}=1.6\times {10}^{-7}

Example 5: Determination of K_a or K_b from pH

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

{\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)

Show Answer

We determine an equilibrium constant starting with the initial concentrations of HNO₂,

{\text{H}}_{3}{\text{O}}^{+}

, and

{\text{NO}}_{2}^{-}

as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)

We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:

A diagram is shown that includes six rust-red rectangles. The first rectangle is labeled, “p H,” and there is an arrow that points to the right to the second rectangle. The second rectangle is labeled, “[ H subscript 3 O superscript plus ].” There is an arrow that points right to the third rectangle. The third rectangle is labeled, “Calculate x subscript [ H subscript 3 O superscript plus ].” There is an arrow that points right to the fourth rectangle. The fourth rectangle is labeled, “Calculate x subscript [ H N O subscript 2 ] and x subscript [ N O subscript 2 superscript negative sign ].” There is an arrow that points down to the fifth rectangle. The fifth rectangle is labeled, “Calculate equilibrium concentrations.” There is an arrow that points down to the sixth rectangle. The sixth rectangle is labeled, “Calculate K subscript a.”

We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, “H N O subscript 2 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign N O subscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.0516, negative x, [ H N O subscript 2 ] subscript i plus ( negative x ) equals 0.0516 plus sign ( negative x ). The second column is blank in the first row, positive sign, blank for the third row. The third column has the following: approximately 0, x, [ H subscript 3 O ] superscript positive sign plus sign x [ N O subscript 2 ] superscript negative sign plus sign x plus sign 0 plus sign x. The fourth column has the following: 0, x, 0.0046.

To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate

\left[{\text{H}}_{3}{\text{O}}^{+}\right]

, the equilibrium concentration of

{\text{H}}_{3}{\text{O}}^{+}

, from the pH:

\left[{\text{H}}_{3}{\text{O}}^{+}\right]={10}^{-2.34}=0.0046M

The change in concentration of

{\text{H}}_{3}{\text{O}}^{+}

{x}_{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}

, is the difference between the equilibrium concentration of

{\text{H}}_{3}{\text{O}}^{+}

, which we determined from the pH, and the initial concentration,

{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}_{\text{i}}

. The initial concentration of

{\text{H}}_{3}{\text{O}}^{+}

is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).

The change in concentration of

{\text{NO}}_{2}^{-}

is equal to the change in concentration of

\left[{\text{H}}_{3}{\text{O}}^{+}\right]

. For each 1 mol of

{\text{H}}_{3}{\text{O}}^{+}

that forms, 1 mol of

{\text{NO}}_{2}^{-}

forms. The equilibrium concentration of HNO₂ is equal to its initial concentration plus the change in its concentration.

Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:

Finally, we calculate the value of the equilibrium constant using the data in the table:

{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{NO}}_{2}^{-}\right]}{\left[{\text{HNO}}_{2}\right]}=\frac{\left(0.0046\right)\left(0.0046\right)}{\left(0.0470\right)}=4.5\times {10}^{-4}

Check Your Learning

The pH of a solution of household ammonia, a 0.950-M solution of NH_3, is 11.612. What is K_b for NH₃.

Show Answer

K_b = 1.8 × 10⁻⁵

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).

A photograph is shown of a large black ant on the end of a human finger.

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}

Show Answer

Step 1: Determine x and equilibrium concentrations

The equilibrium expression is

{\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)

The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):

Step 2: Solve for x and the equilibrium concentrations

At equilibrium:

\begin{array}{rcl}{K}_{\text{a}}&=&1.8\times {10}^{-4}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{HCO}}_{2}^{-}\right]}{\left[{\text{HCO}}_{2}\text{H}\right]}\\{}&=&\frac{\left(x\right)\left(x\right)}{0.534-x}=1.8\times {10}^{-4}\end{array}

Now solve for x. Because the initial concentration of acid is reasonably large and K_a is very small, we assume that x << 0.534, which permits us to simplify the denominator term as (0.534 − x) = 0.534. This gives:

{K}_{\text{a}}=1.8\times {10}^{-4}=\frac{{x}^{\text{2+}}}{0.534}

Solve for x as follows:

\begin{array}{rcl}{x}^{\text{2+}}&=&0.534\times \left(1.8\times {10}^{-4}\right)=9.6\times {10}^{-5}\\x&=&\sqrt{9.6\times {10}^{-5}}\\&=&9.8\times {10}^{-3}\end{array}

To check the assumption that x is small compared to 0.534, we calculate:

\frac{x}{0.534}=\frac{9.8\times{10}^{-3}}{0.534}=1.8\times{10}^{-2}\left(1.8\%\text{ of 0.534}\right)

x is less than 5% of the initial concentration; the assumption is valid.

We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:

\begin{array}{rcl}\left[{\text{H}}_{3}{\text{O}}^{+}\right]&=&~0+x=0+9.8\times {10}^{-3}M\\{}&=&9.8\times {10}^{-3}M\end{array}

The pH of the solution can be found by taking the negative log of the

\left[{\text{H}}_{3}{\text{O}}^{+}\right]

, so:

-\text{log}\left(9.8\times {10}^{-3}\right)=2.01

Check Your Learning

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

Show Answer

percent ionization = 1.3%

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}

Show Answer

This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example 6. The reactants and products will be different and the numbers will be different, but the logic will be the same:

A diagram is shown with 4 tan rectangles connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The final rectangle is labeled “Check the math.”

A diagram is shown with 4 tan rectangles connected with right pointing arrows. The first is labeled “Determine the direction of change.” The second is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The final rectangle is labeled “Check the math.”

Step 1: Determine x and equilibrium concentrations

The table shows the changes and concentrations:

Step 2: Solve for x and the equilibrium concentrations

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]}=\frac{\left(x\right)\left(x\right)}{0.25-x}=7.4\times {10}^{-5}

If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives

x=4.3\times {10}^{-3}

.

This change is less than 5% of the initial concentration (0.25), so the assumption is justified.Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):

\left[{\text{OH}}^{-}\right]=~0+x=x=4.3\times {10}^{-3}M

=4.3\times {10}^{-3}M

Then calculate pOH as follows:

\text{pOH}=-\text{log}\left(4.3\times {10}^{-3}\right)=2.37

.

Using the relation introduced in the previous section of this chapter, we get

\text{pH}+\text{pOH}=\text{p}{K}_{\text{w}}=14.00

, which permits the computation of pH:

\text{pH}=14.00-\text{pOH}=14.00 - 2.37=11.63

Step 3: Check the work

A check of our arithmetic shows that K_b = 7.4 × 10⁻⁵.

Check Your Learning

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 K_b = 7.4 × 10⁻⁵.
Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a K_b of 1.76 × 10⁻⁵. Calculate the percent ionization of ammonia, the fraction ionized × 100, or
$\frac{\left[{\text{NH}}_{4}^{+}\right]}{\left[{\text{NH}}_{3}\right]}\times 100$

Show Answer

7.56 × 10⁻⁴M
2.33%

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}

Show Answer

We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of

{\text{HSO}}_{4}^{-}

so that we can use

\left[{\text{H}}_{3}{\text{O}}^{+}\right]

to determine the pH. As in the previous examples, we can approach the solution by the following steps:

Three rectangles are shown with right pointing arrows between them. The first is labeled “Determine x and the equilibrium concentrations.” The second is labeled “Solve for x and the equilibrium concentrations. The third is labeled “Check the math.”

Three rectangles are shown with right pointing arrows between them. The first is labeled “Determine x and the equilibrium concentrations.” The second is labeled “Solve for x and the equilibrium concentrations. The third is labeled “Check the math.”

Step 1: Determine x and equilibrium concentrations

This table shows the changes and concentrations:

Step 2: Solve for x and the concentrations

As we begin solving for x, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x. At equilibrium:

{K}_{\text{a}}=1.2\times {10}^{-2}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{SO}}_{4}^{2-}\right]}{\left[{\text{HSO}}_{4}^{-}\right]}=\frac{\left(x\right)\left(x\right)}{0.50-x}

If we assume that x is small and approximate (0.50 − x) as 0.50, we find

x=7.7\times {10}^{-2}

. When we check the assumption, we calculate:

\frac{x}{{\left[{\text{HSO}}_{4}^{-}\right]}_{\text{i}}}

\frac{x}{0.50}=\frac{7.7\times {10}^{-2}}{0.50}=0.15\left(15\%\right)

The value of x is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find x. The equation

{K}_{\text{a}}=1.2\times {10}^{-2}=\frac{\left(x\right)\left(x\right)}{0.50-x}

gives

6.0\times {10}^{-3}-1.2\times {10}^{-2}x={x}^{\text{2+}}

{x}^{\text{2+}}+1.2\times {10}^{-2}x - 6.0\times {10}^{-3}=0

.

This equation can be solved using the quadratic formula. For an equation of the form

a{x}^{\text{2+}}+bx+c=0

, x is given by the equation

x=\frac{-b\pm \sqrt{{b}^{\text{2+}}-4ac}}{2a}

.

In this problem, a = 1, b = 1.2 × 10⁻³, and c = −6.0 × 10⁻³.Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

x=7.2\times {10}^{-2}

.

Now determine the hydronium ion concentration and the pH:

\left[{\text{H}}_{3}{\text{O}}^{+}\right]=~0+x=0+7.2\times {10}^{-2}M=7.2\times {10}^{-2}M

. The pH of this solution is

\text{pH}=-\text{log}\left[{\text{H}}_{3}{\text{O}}^{+}\right]=-\text{log}7.2\times {10}^{-2}=1.14

Check Your Learning

Show that the quadratic formula gives x = 7.2 × 10⁻².
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.)

Show Answer

pH 11.16

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

Show Answer

The reaction is:

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

The equilibrium expression is

{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{ClO}}^{-}\right]}{\left[\text{HClO}\right]}=3.5\times {10}^{-8}

The initial and equilibrium concentrations for this system can be written as follows:

	[HClO]	[H₃O⁺]	[ClO⁻]
Initial concentration (M)	0.0092	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0092 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0092 − x) ≈ 0.0092 gives:

\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{ClO}}^{-}\right]}{\left[\text{HClO}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0092-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0092}=3.5\times {10}^{-8}

Solving for x gives 1.79 × 10⁻⁵M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

$\left[{\text{H}}_{3}{\text{O}}^{+}\right]$
= [ClO] = 1.8 × 10⁻⁵M
[HClO] = 0.0092 − 1.79 × 10⁻⁵ = 0.00918 = 0.00092 M
$\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{1.79\times {10}^{-5}}=5.6\times {10}^{-10}M$

Solution 2

0.0784 M C₆H₅NH₂, a weak base

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The reaction is

{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)

The equilibrium expression is

{K}_{\text{a}}=\frac{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]}=4.6\times {10}^{-10}

The initial and equilibrium concentrations for this system can be written as follows:

	[C₆H₅NH₂]	[C₅H₅NH₃⁺]	[OH⁻]
Initial concentration (M)	0.0784	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0784 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0784 − x) ≈ 0.0784 gives:

\frac{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0784-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0784}=4.6\times {10}^{-10}

Solving for x gives 6.01 × 10⁻⁶M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

$\left[{\text{CH}}_{3}{\text{CO}}_{2}^{-}\right]$
= [OH⁻] = 6.0 × 10⁻⁶M
[C₆H₅NH₂] = 0.0784 − 6.01 × 10⁻⁶ = 0.007839 = 0.00784
$\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{w}}}{\left[{\text{OH}}^{-}\right]}=\frac{1.0\times {10}^{-14}}{6.01\times {10}^{-6}}=1.7\times {10}^{-9}M$

Solution 3

0.0810 M HCN, a weak acid

Show Answer

The reaction is

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

.

The equilibrium expression is

{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}=4\times {10}^{-10}

The initial and equilibrium concentrations for this system can be written as follows:

	[HClO]	[H₃O⁺]	[CN⁻]
Initial concentration (M)	0.0810	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0810 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0810 − x) ≈ 0.0810 gives:

\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0810-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0810}=4\times {10}^{-10}

Solving for x gives 5.69 × 10⁻⁶M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

$\left[{\text{H}}_{3}{\text{O}}^{+}\right]$
= [CN⁻] = 5.7 × 10⁻⁶M
[HCN] = 0.0810 − 5.69 × 10⁻⁶ = 0.08099 = 0.0810 M
$\left[{\text{OH}}^{-}\right]=\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{1.0\times {10}^{-14}}{5.69\times {10}^{-6}}=1.8\times {10}^{-9}M$

Solution 4

0.11 M (CH₃)₃N, a weak base

Show Answer

The reaction is

{\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)

The equilibrium expression is

{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]}=7.4\times {10}^{-5}

The initial and equilibrium concentrations for this system can be written as follows:

	[(CH₃)₃N]	[(CH₃)₃NH⁺]	[OH⁻]
Initial concentration (M)	0.11	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.11 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.11 − x) ≈ 0.11 gives:

\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]}=\frac{\left(x\right)\left(x\right)}{\left(0.11-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.11}=7.4\times {10}^{-5}

Solving for x gives 2.85 × 10⁻³M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:

$\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{+}\right]$

Chemistry for Majors: Atoms First

Module 14: Acid-Based Equilibria

Relative Strengths of Acids and Bases

Learning Objectives

Example 1: Calculation of Percent Ionization from pH

Check Your Learning

Example 2: The Product Ka ×Kb = Kw

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Exercises

The Ionization of Weak Acids and Weak Bases

Think about It

Exercises

Question 2

Question 4

Example 3: Determination of Ka from Equilibrium Concentrations

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Example 4: Determination of Kb from Equilibrium Concentrations

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Example 5: Determination of Ka or Kb from pH

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

Step 1: Determine x and equilibrium concentrations

Step 2: Solve for x and the equilibrium concentrations

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

Step 1: Determine x and equilibrium concentrations

Step 2: Solve for x and the equilibrium concentrations

Step 3: Check the work

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

Step 1: Determine x and equilibrium concentrations

Step 2: Solve for x and the concentrations

Check Your Learning

Exercises

Solution 1

Solution 2

Solution 3

Solution 4

Example 2: The Product K_a ×K_b = K_w

Example 3: Determination of K_a from Equilibrium Concentrations

Example 4: Determination of K_b from Equilibrium Concentrations

Example 5: Determination of K_a or K_b from pH