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

medium CBSE JEE-MAIN NEET 4 min read

Question

Predict the pH of salt solutions based on the strength of the parent acid and base. Derive the pH formula for a salt of a weak acid and strong base.

(JEE Main, NEET, CBSE 11 — salt hydrolysis pH calculation is a high-frequency numerical)


Solution — Step by Step

Salt TypeParent AcidParent BaseSolution pHExample
Strong acid + Strong baseStrongStrongNeutral (pH = 7)NaCl, KNO3
Weak acid + Strong baseWeakStrongBasic (pH > 7)CH3COONa, Na2CO3
Strong acid + Weak baseStrongWeakAcidic (pH < 7)NH4Cl, FeCl3
Weak acid + Weak baseWeakWeakDepends on Ka vs KbCH3COONH4

The rule: the “stronger parent wins.” If the base is stronger (weak acid + strong base), the solution is basic.

For a salt like CH3COONa (concentration cc):

The acetate ion hydrolyses: CH3COO+H2OCH3COOH+OHCH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-

Kh=KwKaK_h = \frac{K_w}{K_a} pH=7+12logKa+12logc\text{pH} = 7 + \frac{1}{2}\log K_a + \frac{1}{2}\log c

Or equivalently: pH=7+12(pKa+logc)\text{pH} = 7 + \frac{1}{2}(pK_a + \log c)… wait, let us write it correctly:

pH=7+12pKa+12logc\text{pH} = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log c

For a salt like NH4Cl (concentration cc):

pH=712pKb12logc\text{pH} = 7 - \frac{1}{2}pK_b - \frac{1}{2}\log c

For weak acid + weak base salt (like CH3COONH4):

pH=7+12pKa12pKb\text{pH} = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b

Notice: the pH of a weak acid + weak base salt is independent of concentration.

graph TD
    A["Salt Solution"] --> B{"Parent acid and base?"}
    B -->|"Strong + Strong"| C["pH = 7 Neutral"]
    B -->|"Weak acid + Strong base"| D["pH > 7 Basic"]
    B -->|"Strong acid + Weak base"| E["pH < 7 Acidic"]
    B -->|"Weak + Weak"| F{"Compare Ka and Kb"}
    F -->|"Ka > Kb"| G["Acidic"]
    F -->|"Ka < Kb"| H["Basic"]
    F -->|"Ka = Kb"| I["Neutral"]

Why This Works

When a salt dissolves, it dissociates completely. If one of the ions comes from a weak parent (weak acid or weak base), that ion reacts with water (hydrolyses) to partially regenerate the weak parent. This reaction produces either OH- (making the solution basic) or H+ (making it acidic).

For CH3COONa: the Na+ ion does not hydrolyse (comes from strong base). But CH3COO- hydrolyses — it grabs a proton from water, producing CH3COOH and OH-. The excess OH- makes the solution basic.


Alternative Method

For JEE numericals, the quick formula approach:

  • Weak acid + strong base: pH=7+12pKa+12logc\text{pH} = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log c
  • Strong acid + weak base: pH=712pKb12logc\text{pH} = 7 - \frac{1}{2}pK_b - \frac{1}{2}\log c
  • Weak + weak: pH=7+12(pKapKb)\text{pH} = 7 + \frac{1}{2}(pK_a - pK_b)

Just plug in values. Remember pKa+pKb=pKw=14pK_a + pK_b = pK_w = 14 if you need to convert.


Common Mistake

The most common error: saying “NaCl solution is always neutral.” NaCl in pure water gives pH 7. But NaCl in a buffer or in the presence of other solutes may not be neutral. The neutrality applies only to the salt’s own hydrolysis behaviour.

Also, students forget that the pH of a weak acid + weak base salt does not depend on concentration. The formula pH=7+12(pKapKb)\text{pH} = 7 + \frac{1}{2}(pK_a - pK_b) has no concentration term. Diluting CH3COONH4 does not change its pH significantly. This is a common JEE trap.

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