JEE Main 2026 — Ionic Equilibrium Question with Solution

Given:
Weak dibasic acid $H_{2}A$ with $K_{a1}=8.1\times 10^{-8}$ and $K_{a2}=1.0\times 10^{-13}$
Initial concentration: $[H_{2}A{]}_0=0.1\text{ M}$
The solution also contains $0.1\text{ M HCl}$, which dissociates completely to give $[H^{+}{]}_0=0.1\text{ M}$.

The main source of $H{A}^{-}$ is the first dissociation:
$H_{2}A\rightleftharpoons H^{+}+H{A}^{-}$

Let $x$ be the amount of $H_{2}A$ that dissociates. Then at equilibrium:
$[H_{2}A]=0.1-x$
$[H^{+}]=0.1+x$
$[H{A}^{-}]=x$

Applying the equilibrium expression:
$K_{a1}=\frac{[H^{+}][H{A}^{-}]}{[H_{2}A]}$

$8.1\times 10^{-8}=\frac{(0.1+x)(x)}{0.1-x}$

Since $K_{a1}$ is very small and the common $H^{+}$ ion from HCl suppresses dissociation, $x\ll 0.1$. Hence:
$0.1-x\approx 0.1$ and $0.1+x\approx 0.1$

$8.1\times 10^{-8}\approx \frac{(0.1)(x)}{0.1}=x$

Therefore:
$[H{A}^{-}]=x=8.1\times 10^{-8}\text{ M}$

The second dissociation ($K_{a2}=1.0\times 10^{-13}$) is negligible compared to the first, so the $H{A}^{-}$ consumed in that step is insignificant.

Hence, the correct option is $(3)8.1\times 10^{-8}\text{ M}$.