JEE Advanced 2025 — Paper 1 — Question 38

Because concentration of $\mathrm{H}+$ from weak acid is less we need to consider self ionization of $\mathrm{H}_{2} \mathrm{O}$ also.

$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-}(\mathrm{aq})$ $10^{-3}-x \quad x+y \quad x$ $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$

$x+y \quad y$

Approximation : $\left(10^{-3}-\mathrm{x}\right) \simeq 10^{-3}$

$\Rightarrow \quad \frac{\mathrm{x}(\mathrm{x}+\mathrm{y})}{10^{-3}}=\mathrm{K}_{\mathrm{a}}=4 \times 10^{-11}$

$\Rightarrow \quad \mathrm{y}(\mathrm{x}+\mathrm{y})=\mathrm{Kw}=10^{-14}$

Add (1) + (2)

$\Rightarrow \quad(\mathrm{x}+\mathrm{y})^{2}=5 \times 10^{-14}$ $\Rightarrow \quad x+y=\left[H^{+}\right]=\sqrt{5} \times 10^{-7}$

$\Rightarrow \quad x=\sqrt{5}=2.236$ Answer 2.23 or 2.24