The first and second ionization constants of H_2 X are 2.5… - JEE Main 2026 Chemistry PYQ

JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

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)

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

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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.

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JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

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.

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