Salt hydrolysis — pH of salt solutions from strong/weak acid-base combinations

Question

How do we predict whether a salt solution is acidic, basic, or neutral? What formulas give us the exact pH for salts of different strong/weak acid-base combinations?

(JEE Main, NEET, CBSE 11 — salt hydrolysis pH calculations appear in every competitive exam paper)

Solution — Step by Step

When a salt dissolves in water, it dissociates completely into ions. If any of those ions come from a weak acid or weak base, they react with water (hydrolyse) to produce $H^+$ or $OH^-$ , making the solution acidic or basic.

Strong acid + Strong base salts (e.g., NaCl, $KNO_3$ ): No hydrolysis. pH = 7.

Why? $Na^+$ comes from strong base NaOH (no tendency to react with water). $Cl^-$ comes from strong acid HCl (no tendency to accept $H^+$ ). Neither ion disturbs the water equilibrium.

Case 1: Strong acid + Strong base (NaCl, $K_2SO_4$ )

No hydrolysis. pH = 7 (neutral)

Case 2: Strong acid + Weak base ( $NH_4Cl$ , $CuSO_4$ )

Cation hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$
Solution is acidic (pH < 7)

pH = 7 - \frac{1}{2}pK_b - \frac{1}{2}\log c

Case 3: Weak acid + Strong base ( $CH_3COONa$ , $Na_2CO_3$ )

Anion hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Solution is basic (pH > 7)

pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log c

Case 4: Weak acid + Weak base ( $CH_3COONH_4$ )

Both ions hydrolyse. pH depends on relative strengths.

pH = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b

Note: pH is independent of concentration in this case.

Find the pH of 0.1 M $CH_3COONa$ solution. ( $K_a$ of $CH_3COOH = 1.8 \times 10^{-5}$ )

This is a weak acid + strong base salt (Case 3).

$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$

pH = 7 + \frac{1}{2}(4.74) + \frac{1}{2}\log(0.1)

pH = 7 + 2.37 + \frac{1}{2}(-1) = 7 + 2.37 - 0.5 = \mathbf{8.87}

The solution is basic, as expected — the acetate ion hydrolyses to produce $OH^-$ .

flowchart TD
    A["Identify the salt"] --> B{"From strong acid + strong base?"}
    B -->|"Yes"| C["pH = 7 (neutral)"]
    B -->|"No"| D{"From strong acid + weak base?"}
    D -->|"Yes"| E["pH < 7 (acidic)<br/>pH = 7 - ½pKb - ½log c"]
    D -->|"No"| F{"From weak acid + strong base?"}
    F -->|"Yes"| G["pH > 7 (basic)<br/>pH = 7 + ½pKa + ½log c"]
    F -->|"No"| H["Weak acid + Weak base<br/>pH = 7 + ½pKa - ½pKb<br/>(independent of c)"]

Why This Works

Salt hydrolysis is just the reverse of neutralization. When a weak acid was neutralized by a strong base, the reaction went almost to completion. But in dilute aqueous solution, the reverse reaction (hydrolysis) proceeds to a small extent, regenerating the weak acid/base and releasing $OH^-$ or $H^+$ .

The formulas come from combining the hydrolysis equilibrium constant ( $K_h$ ) with $K_w$ , $K_a$ , and $K_b$ . For Case 3: $K_h = K_w / K_a$ , and the standard weak-base-like calculation gives the pH formula.

Common Mistake

In Case 4 (weak acid + weak base), students often try to use concentration in the pH formula. The pH of $CH_3COONH_4$ is $7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b$ regardless of whether the solution is 0.01 M or 1 M. The concentration cancels out in the derivation. JEE Main has set traps by giving concentration data that is completely irrelevant to the answer.

Quick classification: NaCl = neutral, $NH_4Cl$ = acidic, $CH_3COONa$ = basic, $CH_3COONH_4$ = depends on $K_a$ vs $K_b$ . If $K_a > K_b$ , acidic. If $K_a < K_b$ , basic. If equal, neutral.

Question

Solution — Step by Step

Why This Works

Common Mistake

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