JEE Chemistry 2026 Question with Solution

Given: The first and second ionization constants of $H_{2}X$ are $K_{a1} = 2.5 \times 10^{-8}$ and $K_{a2} = 1.0 \times 10^{-13}$. The concentration of the acid is $0.1 \, \text{M}$.

Find: The value of $[X^{2-}]$ in the form _____ $\times 10^{-13} \, \text{M}$.

For the diprotic acid, the dissociation steps are

$$H_{2}X \rightleftharpoons H^{+} + HX^{-}$$

and

$$HX^{-} \rightleftharpoons H^{+} + X^{2-}$$

Since $K_{a1} \gg K_{a2}$, the first ionization produces most of $H^{+}$ and $HX^{-}$, so $[H^{+}] \approx [HX^{-}]$.

Using the second ionization expression,

$$K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]}$$

Therefore,

$$[X^{2-}] = \frac{K_{a2}[HX^{-}]}{[H^{+}]}$$

Now using $[H^{+}] \approx [HX^{-}]$,

$$[X^{2-}] \approx K_{a2} = 1.0 \times 10^{-13} \, \text{M}$$

Therefore, the concentration of $X^{2-}$ is $1 \times 10^{-13} \, \text{M}$ and the value to be filled in the blank is 1.

Given: $K_{a1} = 2.5 \times 10^{-8}$, $K_{a2} = 1.0 \times 10^{-13}$, and concentration of $H_{2}X$ is $0.1 \, \text{M}$.

Find: $[X^{2-}]$.

From the first dissociation of a weak acid,

$$[H^{+}] \approx [HX^{-}] = \sqrt{K_{a1}C}$$

For the second dissociation,

$$K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]}$$

So,

$$[X^{2-}] = \frac{K_{a2}[HX^{-}]}{[H^{+}]}$$

Substituting $[H^{+}] \approx [HX^{-}]$,

$$[X^{2-}] \approx K_{a2}$$

Hence,

$$[X^{2-}] = 1.0 \times 10^{-13} \, \text{M}$$

Therefore, the required nearest integer is 1.