JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Equilibrium

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20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (pKₐ value of acetic acid is 4.75).

(A)7.0
(B)4.75✓ Correct
(C)3.5
(D)4.82

Answer: (B)

Solution

The

20 mL

of acetic acid needs

28.4 mL

0.1 M

NaOH for complete neutralisation, so its equivalence point is at

28.4 mL

of base. Adding only

14.2 mL

of the same NaOH delivers exactly half that amount, leaving the acid half-neutralised with

[A^{-}] = [HA]

. At this half-equivalence point the Henderson–Hasselbalch equation gives

pH = p K_{a} + lo g \frac{[ A ^{-} ]}{[ HA ]} = p K_{a}

. With

p K_{a} = 4.75

for acetic acid, the pH of solution

X

4.75

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option

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