JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Equilibrium

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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✓ Correct
(C)1.1, 2.2
(D)3.0, 2.2

Answer: (B)

Solution

At the start the solution is

0.2

M weak acid, for which

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.

Adding

10

mL of

0.2

M NaOH delivers

2

mmol of base to the

4

mmol of acid, neutralising half of it and leaving

2

mmol of HX alongside

2

mmol of its salt NaX. This is a buffer at half-neutralisation, so

pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]} = 3.3 + lo g 1 = 3.3.

The two pH values are

2.0

and

3.3

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Equilibrium

Solution

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