JEE Main 2026 · 2 Apr · Shift 2 · Chemistry Q52

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ChemistrySingle correct+4 / -1Question 52 of 75

The first and second ionization constants of a weak dibasic acid $\ce{H2A}$ are $8.1\times10^{-8}$ and $1.0\times10^{-13}$ respectively. 0.1 mol of $\ce{H2A}$ was dissolved in 1 L of 0.1 M HCl solution. The concentration of $\ce{HA-}$ in the resultant solution is :

0.1 M

9.53\times10^{-6}

8.1\times10^{-8}

1.0\times10^{-13}

Answer and worked solution

Official answer: C

$[\ce{H+}]$ is fixed by HCl: 0.1 M (the weak acid adds negligibly).

First ionisation.
$[\ce{HA-}]=\frac{K_{a1}[\ce{H2A}]}{[\ce{H+}]}=\frac{8.1\times10^{-8}\times0.1}{0.1}=8.1\times10^{-8}\text{ M}.$

(C)

Topics:Common ion effect pH and ionisation

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JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 52