JEE Main 2026 — Chemistry Physical Chemistry

Given:

Weak dibasic acid $H_2A$H2​A with $K_{a1} = 8.1 \times 10^{-8}$Ka1​=8.1×10−8 and $K_{a2} = 1.0 \times 10^{-13}$Ka2​=1.0×10−13

Initial concentration: $[H_2A]_0 = 0.1\text{ M}$[H2​A]0​=0.1 M

The solution also contains $0.1\text{ M HCl}$0.1 M HCl, which dissociates completely to give $[H^+]_0 = 0.1\text{ M}$[H+]0​=0.1 M.

The main source of $HA^-$HA− is the first dissociation:

$$H_2A \rightleftharpoons H^+ + HA^-$$H2​A⇌H++HA−

Let $x$x be the amount of $H_2A$H2​A that dissociates. Then at equilibrium:

$$[H_2A] = 0.1 - x$$[H2​A]=0.1−x

$$[H^+] = 0.1 + x$$[H+]=0.1+x

$$[HA^-] = x$$[HA−]=x

Applying the equilibrium expression:

$$K_{a1} = \dfrac{[H^+][HA^-]}{[H_2A]}$$Ka1​=[H2​A][H+][HA−]​

$$8.1 \times 10^{-8} = \dfrac{(0.1 + x)(x)}{0.1 - x}$$8.1×10−8=0.1−x(0.1+x)(x)​

Since $K_{a1}$Ka1​ is very small and the common $H^+$H+ ion from HCl suppresses dissociation, $x \ll 0.1$x≪0.1. Hence:

$0.1 - x \approx 0.1$0.1−x≈0.1 and $0.1 + x \approx 0.1$0.1+x≈0.1

$$8.1 \times 10^{-8} \approx \dfrac{(0.1)(x)}{0.1} = x$$8.1×10−8≈0.1(0.1)(x)​=x

Therefore:

$$[HA^-] = x = 8.1 \times 10^{-8}\text{ M}$$[HA−]=x=8.1×10−8 M

The second dissociation ($K_{a2} = 1.0 \times 10^{-13}$Ka2​=1.0×10−13) is negligible compared to the first, so the $HA^-$HA− consumed in that step is insignificant.

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