Q23 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
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)
Correct answer: 100
Solution
For a weak diprotic acid $H_2X$, the ionization steps are:
$H_2X \rightleftharpoons H^+ + HX^-$ with $K_{a1} = 2.5 \times 10^{-8}$
$HX^- \rightleftharpoons H^+ + X^{2-}$ with $K_{a2} = 1.0 \times 10^{-13}$
Since $K_{a1} \gg K_{a2}$, the concentration of $H^+$ is primarily determined by the first ionization.
$[H^+] \approx \sqrt{K_{a1} \cdot C} = \sqrt{2.5 \times 10^{-8} \times 0.1} = \sqrt{2.5 \times 10^{-9}} = 5 \times 10^{-5} M$
Also, from the first ionization, $[HX^-] \approx [H^+] = 5 \times 10^{-5} M$
For the second ionization:
$K_{a2} = \frac{[H^+][X^{2-}]}{[HX^-]}$
Substituting the values: $1.0 \times 10^{-13} = \frac{(5 \times 10^{-5})[X^{2-}]}{5 \times 10^{-5}}$
Thus, $[X^{2-}] = K_{a2} = 1.0 \times 10^{-13} M$
To express this in the form $x \times 10^{-15} M$:
$1.0 \times 10^{-13} = 100 \times 10^{-15} M$
The value is 100.
$H_2X \rightleftharpoons H^+ + HX^-$ with $K_{a1} = 2.5 \times 10^{-8}$
$HX^- \rightleftharpoons H^+ + X^{2-}$ with $K_{a2} = 1.0 \times 10^{-13}$
Since $K_{a1} \gg K_{a2}$, the concentration of $H^+$ is primarily determined by the first ionization.
$[H^+] \approx \sqrt{K_{a1} \cdot C} = \sqrt{2.5 \times 10^{-8} \times 0.1} = \sqrt{2.5 \times 10^{-9}} = 5 \times 10^{-5} M$
Also, from the first ionization, $[HX^-] \approx [H^+] = 5 \times 10^{-5} M$
For the second ionization:
$K_{a2} = \frac{[H^+][X^{2-}]}{[HX^-]}$
Substituting the values: $1.0 \times 10^{-13} = \frac{(5 \times 10^{-5})[X^{2-}]}{5 \times 10^{-5}}$
Thus, $[X^{2-}] = K_{a2} = 1.0 \times 10^{-13} M$
To express this in the form $x \times 10^{-15} M$:
$1.0 \times 10^{-13} = 100 \times 10^{-15} M$
The value is 100.
Hence, the correct answer is 100.