14.4 Hydrolysis of Salts

Salts with Acidic Ions

Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt's constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation

The ammonium ion is the conjugate acid of the base ammonia, NH₃; its acid ionization (or acid hydrolysis) reaction is represented by

Since ammonia is a weak base, K_b is measurable and K_a > 0 (ammonium ion is a weak acid).

The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis) reaction is represented by

Since HCl is a strong acid, K_a is immeasurably large and K_b ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).

Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (${\text{NH}}_4{}^{+}$) and inert anions (Cl⁻), resulting in an acidic solution.

Example 14.15

Calculating the pH of an Acidic Salt Solution

Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, $[{\text{C}}_6{\text{H}}_5{\text{NH}}_3]\text{Cl},$ a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M solution of anilinium chloride

$${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{C}}_6{\text{H}}_5{\text{NH}}_2(aq)$$

Solution

The K_a for anilinium ion is derived from the K_b for its conjugate base, aniline (see Appendix I):

$$K_{\text{a}}=\frac{K_{\text{w}}}{K_{\text{b}}}=\frac{1.0\times {10}^{\mathrm{-14}}}{4.3\times {10}^{\mathrm{-10}}}=2.3\times {10}^{\mathrm{-5}}$$

Using the provided information, an ICE table for this system is prepared:

Substituting these equilibrium concentration terms into the K_a expression gives

$$\begin{array}{l}{\text{K}}_a=[{\text{C}}_6{\text{H}}_5{\text{NH}}_2][{\text{H}}_3{\text{O}}^{+}]\text{/}[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{+}]\\ 2.3\times {10}^{\text{−}5}=(x)(x)\text{/}0.233-x)\end{array}$$

Assuming x << 0.233, the equation is simplified and solved for x:

$$\begin{array}{}\\ \\ 2.3\times {10}^{\text{−}5}=x^2\text{/}0.233\\ x=0.0023M\end{array}$$

The ICE table defines x as the hydronium ion molarity, and so the pH is computed as

$$\text{pH}=-\text{log}[{\text{H}}_3{\text{O}}^{+}]=\text{−}\text{log}(0.0023)=2.64$$

Check Your Learning

What is the hydronium ion concentration in a 0.100-M solution of ammonium nitrate, NH₄NO₃, a salt composed of the ions ${\text{NH}}_4{}^{\text{+}}$ and ${\text{NO}}_3{}^{\text{−}}.$ Which is the stronger acid ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ or ${\text{NH}}_4{}^{\text{+}}?$

Answer:

[H₃O⁺] = 7.5 $\times$ 10⁻⁶ M; ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ is the stronger acid.

Salts with Basic Ions

As another example, consider dissolving sodium acetate in water:

The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion's formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.

The acetate ion, ${\text{CH}}_3{\text{CO}}_2{}^{-}\text{,}$ is the conjugate base of acetic acid, CH₃CO₂H, and so its base ionization (or base hydrolysis) reaction is represented by

Because acetic acid is a weak acid, its K_a is measurable and K_b > 0 (acetate ion is a weak base).

Dissolving sodium acetate in water yields a solution of inert cations (Na⁺) and weak base anions ${\text{(CH}}_3{\text{CO}}_2{}^{-}),$ resulting in a basic solution.

Salts with Acidic and Basic Ions

Some salts are composed of both acidic and basic ions, and so the pH of their solutions will depend on the relative strengths of these two species. Likewise, some salts contain a single ion that is amphiprotic, and so the relative strengths of this ion’s acid and base character will determine its effect on solution pH. For both types of salts, a comparison of the K_a and K_b values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.

The Ionization of Hydrated Metal Ions

Unlike the group 1 and 2 metal ions of the preceding examples (Na⁺, Ca²⁺, etc.), some metal ions function as acids in aqueous solutions. These ions are not just loosely solvated by water molecules when dissolved, instead they are covalently bonded to a fixed number of water molecules to yield a complex ion (see chapter on coordination chemistry). As an example, the dissolution of aluminum nitrate in water is typically represented as

However, the aluminum(III) ion actually reacts with six water molecules to form a stable complex ion, and so the more explicit representation of the dissolution process is

As shown in Figure 14.13, the $\text{Al}{({\text{H}}_{\text{2}}\text{O})}_6{}^{\mathrm{3+}}$ ions involve bonds between a central Al atom and the O atoms of the six water molecules. Consequently, the bonded water molecules' O–H bonds are more polar than in nonbonded water molecules, making the bonded molecules more prone to donation of a hydrogen ion:

The conjugate base produced by this process contains five other bonded water molecules capable of acting as acids, and so the sequential or step-wise transfer of protons is possible as depicted in few equations below:

This is an example of a polyprotic acid, the topic of discussion in a later section of this chapter.

Figure 14.13 When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

Aside from the alkali metals (group 1) and some alkaline earth metals (group 2), most other metal ions will undergo acid ionization to some extent when dissolved in water. The acid strength of these complex ions typically increases with increasing charge and decreasing size of the metal ions. The first-step acid ionization equations for a few other acidic metal ions are shown below:

Example 14.18

Hydrolysis of [Al(H₂O)₆]³⁺

Calculate the pH of a 0.10-M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion ${[\text{Al}{({\text{H}}_2\text{O})}_6]}^{\mathrm{3+}}$ in solution.

Solution

The equation for the reaction and K_a are:

$$\text{Al}{({\text{H}}_2\text{O})}_6{}^{\mathrm{3+}}(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+\text{Al}{({\text{H}}_2\text{O})}_5({\text{OH})}^{\text{2+}}(aq)K_{\text{a}}=1.4\times {10}^{\mathrm{-5}}$$

An ICE table with the provided information is

Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:

$$K_{\text{a}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][\text{Al}{({\text{H}}_2\text{O})}_5({\text{OH})}^{\mathrm{2+}}]}{[\text{Al}{({\text{H}}_2\text{O})}_6{}^{\mathrm{3+}}]}$$

$$=\frac{(x)(x)}{0.10-x}=1.4\times {10}^{\mathrm{-5}}$$

Assuming x << 0.10 and solving the simplified equation gives:

$$x=1.2\times {10}^{\mathrm{-3}}M$$

The ICE table defined x as equal to the hydronium ion concentration, and so the pH is calculated to be

$$[{\text{H}}_3{\text{O}}^{\text{+}}]=0+x=1.2\times {10}^{\mathrm{-3}}M$$

$$\text{pH}=\text{−log}[{\text{H}}_3{\text{O}}^{\text{+}}]=2.92(\text{an acidic solution})$$

Check Your Learning

What is $[\text{Al}{({\text{H}}_2\text{O})}_5({\text{OH})}^{\mathrm{2+}}]$ in a 0.15-M solution of Al(NO₃)₃ that contains enough of the strong acid HNO₃ to bring [H₃O⁺] to 0.10 M?

Answer:

2.1 $\times$ 10⁻⁵ M