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

$$NH_4Cl(s)⇌NH_4^+(aq)+Cl^-(aq)$$

The ammonium ion is the conjugate acid of the base ammonia, $NH_3$; its acid ionization (or acid hydrolysis) reaction is represented by

$$NH_4^+(aq)+H_2O(l)⇌H_3O^+(aq)+NH_3(aq)\qquad K_a=K_w/K_b$$

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

$$Cl^-(aq)+H_2O(l)⇌HCl(aq)+OH^-(aq)\qquad K_b=K_w/K_a$$

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 ($NH_4^+$) and inert anions (Cl⁻), resulting in an acidic solution.

Calculating the pH of an Acidic Salt Solution
Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, $[C_6H_5NH_3]Cl$, 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

$$C_6H_5NH_3^+(aq)+H_2O(l)⇌H_3O^+(aq)+C_6H_5NH_2(aq)$$

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

$$K_a=\frac{K_w}{K_b}=\frac{1.0×10^{-14}}{4.3×10^{-10}}=2.3×10^{-5}$$

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

Substituting these equilibrium concentration terms into the K_a expression gives

$$K_a=\frac{[C_6H_5NH_2][H_3O^+]}{[C_6H_5NH_3^+]}$$ $$2.3×10^{-5}=\frac{(x)(x)}{(0.233-x)}$$

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

$$2.3×10^{-5}=\frac{x^2}{0.233}$$ $$x=0.0023\;M$$

The ICE table defines x as the hydronium ion molarity, and so the pH is computed as $$\text{pH}=-log[H_3O^+]=-log(0.0023)=2.64$$