JEE Main 2026 — Ionic Equilibrium Question with Solution

For complete neutralization of $20$ mL of acetic acid, the millimoles of NaOH required is:

$28.4\times 0.1=2.84$ mmol

Thus, $20$ mL of the acetic acid solution contains $2.84$ mmol of $C{H}_{3}COOH$.

For the preparation of solution (X), the millimoles of NaOH added is:

$14.2\times 0.1=1.42$ mmol

The reaction between acetic acid and sodium hydroxide is:

$C{H}_{3}COOH+NaOH\to C{H}_{3}COONa+H_{2}O$

Millimoles of $C{H}_{3}COOH$ remaining after the reaction = $2.84-1.42=1.42$ mmol

Millimoles of $C{H}_{3}COONa$ formed = $1.42$ mmol

Since the solution contains a weak acid and its conjugate base, it forms an acidic buffer. Using the Henderson-Hasselbalch equation:

$pH=p{K}_a+\log \left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$

Substituting the values:

$pH=4.75+\log \left(\frac{1.42}{1.42}\right)$

$pH=4.75+\log (1)$

$pH=4.75$

Answer: $4.75$