Q9 / 48 · JEE Advanced 2025 (Paper 1) · Chemistry
At $25^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}^{+}$ions in $1.00 \times 10^{-3} \mathrm{M}$ aqueous solution of a weak monobasic acid having acid dissociation constant $\left(K_a\right)=4.00 \times 10^{-11}$ is $\boldsymbol{X} \times 10^{-7} \mathrm{M}$. The value of $\boldsymbol{X}$ is _____ .
Use: Ionic product of water $\left(K_w\right)=1.00 \times 10^{-14}$ at $25^{\circ} \mathrm{C}$
Correct answer: 2.23
Solution
Because concentration of $\mathrm{H}+$ from weak acid is less we need to consider self ionization of $\mathrm{H}_2 \mathrm{O}$ also.
$\begin{array}{rl}\mathrm{HX}(\mathrm{aq}) \rightleftharpoons & \mathrm{H}^{+}+\mathrm{X}^{-}(\mathrm{aq}) \\ 10^{-3}-\mathrm{x} \quad & \mathrm{x}+\mathrm{y} \quad \mathrm{x} \\ \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons & \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ & \mathrm{x}+\mathrm{y} \quad \quad \mathrm{y}\end{array}$
Approximation : $\left(10^{-3}-\mathrm{x}\right) \simeq 10^{-3}$
$\begin{aligned}
& \Rightarrow \quad \frac{x(x+y)}{10^{-3}}=K_a=4 \times 10^{-11}....(1) \\
& \Rightarrow \quad y(x+y)=K w=10^{-14}.....(2)
\end{aligned}$
$\begin{array}{ll}
\text { Add }(1)+(2) \\
\end{array}$
$\begin{array}{ll}\Rightarrow & (\mathrm{x}+\mathrm{y})^2=5 \times 10^{-14} \\ \Rightarrow & \mathrm{x}+\mathrm{y}=\left[\mathrm{H}^{+}\right]=\sqrt{5} \times 10^{-7} \\ \Rightarrow & \mathrm{x}=\sqrt{5}=2.236\end{array}$
Answer 2.23 or 2.24.
Hence, the correct answer is 2.23.