JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

The first and second ionization constants of a weak dibasic acid H $_{2}$ A are $8.1 \times 1 0^{- 8}$ and $1.0 \times 1 0^{- 13}$ respectively. $0.1$ mol of H $_{2}$ A was dissolved in $1$ L of $0.1$ M HCl solution. The concentration of HA $^{-}$ in the resultant solution is :

A
$0.1$ M
B
$9.53 \times 1 0^{- 6}$ M
C
$8.1 \times 1 0^{- 8}$ M
D
$1.0 \times 1 0^{- 13}$ M

Solution

HCl is a strong acid, fully dissociating to give $[H^{+}] = 0.1$ M.

By the common-ion effect, the ionisation of H $_{2}$ A is heavily suppressed, so $[H_{2} A] \approx 0.1$ M and $[H^{+}] \approx 0.1$ M.

From the first ionisation equilibrium, $K_{a_{1}} = \frac{[ H ^{+} ] [ HA ^{-} ]}{[ H _{2} A ]} \Rightarrow [HA^{-}] = \frac{K _{a_{1}} \cdot [ H _{2} A ]}{[ H ^{+} ]} = K_{a_{1}} = 8.1 \times 1 0^{- 8}$ M.

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