The first and second ionization constants of H 2 X are 2.5 × 10 -8 and 1.0 × 10 -13 respectively. The concentration of X 2- in 0.1 M H 2 X solution is \ \ \ \ × 10 -15 M . (Nearest Integer)

Chemistry · Ionic Equilibrium

JEE Main 2026 — 21 January, Evening Shift — Question 64

The first and second ionization constants of $\mathrm{H}_{2} \mathrm{X}$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $\mathrm{X}^{2-}$ in $0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{X}$ solution is $\_\_\_\_$ $\times 10^{-15} \mathrm{M}$ . (Nearest Integer)

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{H}_{2} \mathrm{X} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HX}^{-}$ , $0.1-x x+y x-y$ $2.5 \times 10^{-8}=\frac{(\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})}{0.1-\mathrm{x}}$ $\mathrm{HX}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{2-}$ , $x-y x+y y$ $1 \times 10^{-13}=\frac{(\mathrm{x}+\mathrm{y})(\mathrm{y})}{\mathrm{x}-\mathrm{y}}$ Approximate : $\mathrm{K}_{\mathrm{a}_{1}} \gg \mathrm{~K}_{\mathrm{a}_{2}} \Rightarrow$ So $\mathrm{x} \gg \mathrm{y}$ . $\mathrm{x}+\mathrm{y} \approx \mathrm{x}, \mathrm{x}-\mathrm{y} \approx \mathrm{x}$ $10^{-13}=\frac{x \cdot y}{x}$ $\mathrm{y}=10^{-13}$ $\left[\mathrm{X}^{2-}\right]=10^{-13}$ $\left[\mathrm{X}^{2-}\right]=100 \times 10^{-15}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Solutions with mixture of acids or bases