JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

At $2 5^{\circ}$ C, $20.0$ mL of $0.2$ M weak monoprotic acid HX is titrated against $0.2$ M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when $10$ mL of NaOH is added respectively, are :

Given : $K_{a} = 5 \times 1 0^{- 4}$ , p $K_{a} = 3.3$ , $α << 1$

A
$0.7$ , $2.0$
B
$2.0$ , $3.3$
C
$1.1$ , $2.2$
D
$3.0$ , $2.2$

Solution

(a) For a weak acid with $α << 1$ : $pH = \frac{1}{2} (p K_{a} - lo g C) = \frac{1}{2} (3.3 - lo g 0.2) = \frac{1}{2} (3.3 + 0.7) = 2.0$ .

(b) After adding $10$ mL of $0.2$ M NaOH to $20$ mL of $0.2$ M HX: initial HX $= 4$ mmol, NaOH $= 2$ mmol. After reaction, $2$ mmol HX remains and $2$ mmol of NaX is formed — a buffer at the half-equivalence point.

By Henderson-Hasselbalch: $pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]} = 3.3 + lo g 1 = 3.3$ .

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