14.4 Hydrolysis of Salt Solutions

Acid-Base Neutralization

A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:

1. A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the conjugate partners produced are of negligible strength (see Figure 14.8):
   
   $$\text{HCl}(aq)+\text{NaOH}(aq)\rightleftharpoons \text{NaCl}(aq)+{\text{H}}_2\text{O}(l)$$

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2. A strong acid and a weak base yield a weakly acidic solution. The products of this neutralization reaction are the conjugate base of the strong acid (Kb ≈ 0, so it doesn’t affect pH) and the conjugate acid of the weak base (Ka > 0, so it ionizes to make the solution acidic).
3. A weak acid and a strong base yield a weakly basic solution. The products of this neutralization reaction are the conjugate acid of the strong base (Ka ≈ 0, so it doesn’t affect pH) and the conjugate base of the weak acid (Kb > 0, so it ionizes to make the solution basic).
4. A weak acid plus a weak base can yield either an acidic or basic solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. To predict whether a particular combination will be acidic or basic, tabulated K values of the conjugates must be compared. (Note: occasionally the weak acid and the weak base can have the same strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral.)

Chemistry in Everyday Life

Culinary Aspects of Chemistry

Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO₃ is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 14.15). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.

Figure 14.15 A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.

Salts of Weak Bases and Strong Acids

When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH₄Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:

A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:

The equilibrium equation for this reaction is simply the ionization constant. K_a, for the acid ${\text{NH}}_4{}^{\text{+}}:$

We will not find a value of K_a for the ammonium ion in Appendix H. However, it is not difficult to determine K_a for ${\text{NH}}_4{}^{\text{+}}$ from the value of the ionization constant of water, K_w, and K_b, the ionization constant of its conjugate base, NH₃, using the following relationship:

This relation holds for any base and its conjugate acid or for any acid and its conjugate base.

Example 14.15

The pH of a Solution of a Salt of a Weak Base and a Strong Acid

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, $[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}]\text{Cl},$ 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 aniline hydrochloride?

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

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Solution

The new step in this example is to determine K_a for the ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ ion. The ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ ion is the conjugate acid of a weak base. The value of K_a for this acid is not listed in Appendix H, but we can determine it from the value of K_b for aniline, C₆H₅NH₂, which is given as 4.3 $\times$ 10⁻¹⁰ (Table 14.3 and Appendix I):

$$K_{\text{a}}(\text{for}{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}})\times K_{\text{b}}(\text{for}{\text{C}}_6{\text{H}}_5{\text{NH}}_2)=K_{\text{w}}=1.0\times {10}^{\mathrm{-14}}$$

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$$K_{\text{a}}(\text{for}{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}})=\frac{K_{\text{w}}}{K_{\text{b}}(\text{for}{\text{C}}_6{\text{H}}_5{\text{NH}}_2)}=\frac{1.0\times {10}^{\mathrm{-14}}}{4.3\times {10}^{\mathrm{-10}}}=2.3\times {10}^{\mathrm{-5}}$$

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Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of H₃O⁺, and the pH:

With these steps we find [H₃O⁺] = 2.3 $\times$ 10⁻³ M and pH = 2.64

Check Your Learning

(a) Do the calculations and show that the hydronium ion concentration for a 0.233-M solution of ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ is 2.3 $\times$ 10⁻³ and the pH is 2.64.

(b) 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 ${\text{NO}}_3{}^{\text{−}}.$ Use the data in Table 14.3 to determine K_b for the ammonium ion. Which is the stronger acid ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ or ${\text{NH}}_4{}^{\text{+}}?$

Answer:

(a) $K_a(\text{for}{\text{NH}}_4{}^{\text{+}})=5.6\times {10}^{\mathrm{-10}},$ [H₃O⁺] = 7.5 $\times$ 10⁻⁶ M; (b) ${\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}$ is the stronger acid.

Salts of Weak Acids and Strong Bases

When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH₃CO₂, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:

A solution of this salt contains sodium ions and acetate ions. The sodium ion has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:

The equilibrium equation for this reaction is the ionization constant, K_b, for the base ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}.$ The value of K_b can be calculated from the value of the ionization constant of water, K_w, and K_a, the ionization constant of the conjugate acid of the anion using the equation:

For the acetate ion and its conjugate acid we have:

Some handbooks do not report values of K_b. They only report ionization constants for acids. If we want to determine a K_b value using one of these handbooks, we must look up the value of K_a for the conjugate acid and convert it to a K_b value.

Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base

In a solution of a salt formed by the reaction of a weak acid and a weak base, the salt’s cation will be a weak acid (the conjugate acid of the weak base reactant) and its anion will be a weak base (the conjugate base of the weak acid reactant). To predict the pH of the salt solution, we must know both the K_a of the acidic cation and the K_b of the basic anion. If K_a > K_b, the solution is acidic, and if K_b > K_a, the solution is basic.

The Ionization of Hydrated Metal Ions

If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, $\text{Al}{({\text{H}}_2\text{O})}_6{}^{\mathrm{3+}},$ dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this $\text{Al}{({\text{H}}_2\text{O})}_6{}^{\mathrm{3+}}$ cluster as well:

We frequently see the formula of this ion written simply as “Al³⁺(aq)”, without explicitly noting that six water molecules are covalently bonded to the aluminum ion. This is similar to the simplification of the formula of the hydronium ion, H₃O⁺ to H⁺. In this case, a bonded water molecule acts as a weak acid (Figure 14.16) and donates a proton to a water molecule.

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 the equations below:

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

Figure 14.16 When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.

Additional examples of the first stage in the ionization of hydrated metal ions are:

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

In spite of the unusual appearance of the acid, this is a typical acid ionization problem.

1. Step 1.
   Determine the direction of change. The equation for the reaction and K_a are:
   

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

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   The reaction shifts to the right to reach equilibrium.
2. Step 2. Determine x and equilibrium concentrations. Use the table:
   
3. Step 3.
   Solve for x and the equilibrium concentrations. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:
   

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

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

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   Solving this equation gives:
   

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

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   From this we find:
   

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

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

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4. Step 4. Check the work. The arithmetic checks; when 1.2 $\times$ 10⁻³ M is substituted for x, the result = K_a.

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

The constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.