JEE Main 2026 — Ionic Equilibrium Question with Solution

For part (a), at the start of the titration, the solution contains only the weak acid HX.

The concentration of HX is $C=0.2$ M.

Using the formula for the hydrogen ion concentration of a weak acid with $\alpha \ll 1$:

$[H^{+}]=\sqrt{K_a\times C}=\sqrt{5\times 10^{-4}\times 0.2}=\sqrt{10^{-4}}=10^{-2}$ M

$pH=-\log (10^{-2})=2.0$

For part (b), when $10$ mL of $0.2$ M NaOH is added to $20$ mL of $0.2$ M HX:

Initial millimoles of HX $=20\times 0.2=4$ mmol

Millimoles of NaOH added $=10\times 0.2=2$ mmol

The added NaOH neutralizes half of the HX to form NaX, creating an acidic buffer.

Millimoles of HX remaining $=4-2=2$ mmol

Millimoles of NaX formed $=2$ mmol

Using the Henderson-Hasselbalch equation:

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

Since the millimoles of salt and acid are equal, $\log (1)=0$.

$pH=p{K}_a=3.3$

Answer: $2.03.3$