Chapter 14 Acid Base Equilibria
14.5 Hydrolysis of Salts
Learning Objectives
By the end of this section, you will be able to:
- Predict whether a salt solution will be acidic, basic, or neutral
- Calculate the concentrations of the various species in a salt solution
- Describe the acid ionization of hydrated metal ions
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, Kb is measurable and Ka > 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, Ka is immeasurably large and Kb ≈ 0 (chloride ions don’t undergo appreciable hydrolysis).
Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (NH4+) and inert anions (Cl−), resulting in an acidic solution.
Example 14.18 – Calculating the pH of an Acidic Salt Solution
Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride,
Solution
The Ka for anilinium ion is derived from the Kb for its conjugate base, aniline (see Appendix I):
| C6H5NH3+ | + | H2O | ⇌ | C6H6NH2 | + | H3O+ | |
| I (M) | 0.233 | — | 0 | ~0 | |||
| C (M) | −x | (−x) | +x | +x | |||
| E (M) | 0.233 − x | — | x | x |
Substituting these equilibrium concentration terms into the Ka expression gives
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NO3– is the conjugate base of a strong acid (HNO3), so it will not affect the hydronium ion concentration.
The Ka for ammonium (NH4+) is
[latex]K_a=\frac{K_w}{K_b}=\frac{1\times10^{-14}}{1.8\ \times10^{-5}}=5.6\ \times10^{-10}[/latex]
| NH4+ | + | H2O | ⇌ | NH3 | + | H3O+ | |
| I (M) | 0.100 | — | 0 | ~0 | |||
| C (M) | −x | (−x) | +x | +x | |||
| E (M) | 0.100 − x | — | x | x |
Substituting these equilibrium concentration terms into the Ka expression gives
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, CH3CO2−, 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 Ka is measurable and Kb > 0 (acetate ion is a weak base).
Dissolving sodium acetate in water yields a solution of inert cations (Na+) and weak base anions (CH3CO2−), resulting in a basic solution.
Example 14.19 – Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base
Determine the acetic acid ([latex]CH_3CO_2H[/latex]) concentration in a solution with [CH3CO2−] = 0.050 M and [OH−] = 2.5 × 10−6 M at equilibrium. The reaction is:
Solution
The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which
Solving the above equation for the acetic acid molarity yields [CH3CO2H] = 1.1 × 10−5 M.
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According to Appendix H, the Ka of HCN is 4.9 x 10-10. The Kb of CN– is therefore
[latex]K_b=\frac{K_w}{K_a}=\frac{1\times10^{-14}}{4.9\times10^{-10}}=2.0\times10^{-5}[/latex]
The ICE table for the base ionization of CN– is:
| CN– | + | H2O | ⇌ | HCN | + | OH– | |
| I (M) | 0.083 | — | 0 | ~0 | |||
| C (M) | −x | (−x) | +x | +x | |||
| E (M) | 0.083 − x | — | x | x |
Substituting these equilibrium concentration terms into the Kb expression gives
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 Ka and Kb values allows prediction of the solution’s acid-base status, as illustrated in the following example exercise.
Example 14.20 – Determining the Acidic or Basic Nature of Salts
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
(a) KBr
(b) NaHCO3
(c) Na2HPO4
(d) NH4F
Solution
Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:
(a) The K+ cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.
(b) The Na+ cation is inert and will not affect the pH of the solution. The HCO3− anion is amphiprotic. The Ka of HCO3− is 4.7 × 10−11, and its Kb is [latex]\frac {1.0 \times 10^{-14}}{4.3 \times 10^{-7}} = 2.3 \times 10^{-8}[/latex]. Since Kb >> Ka, the solution is basic.
(c) The Na+ cation is inert and will not affect the pH of the solution, while the HPO42− anion is amphiprotic. The Ka of HPO42− is 4.2 × 10−13, and its Kb is [latex]\frac {1.0 \times 10^{-14}}{6.2 \times 10^{-8}} = 1.6 \times 10^{-7}[/latex]. Because Kb >> Ka, the solution is basic.
(d) The NH4+ ion is acidic (see above discussion) and the F− ion is basic (conjugate base of the weak acid HF). Consider their ionization constants: Ka of NH4+ is 5.6 × 10−10 and the Kb of F− is 1.6 × 10−11. Because Ka > Kb, the solution is acidic.
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