Q13 / 155 · WBJEE 2021 · Chemistry
$2.5 \mathrm{ml}$ of $0.4(\mathrm{M})$ weak monoacidic base $\left(\mathrm{k}_{\mathrm{b}}=1 \times 10^{-12}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ is titrated with $\frac{2}{15}(\mathrm{M}) \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(\mathrm{k}_{\mathrm{w}}=1 \times 10^{-14}\right.$, at $\left.25^{\circ} \mathrm{C}\right)$,
- A$3.7 \times 10^{-13}$ (M)
- B$3.2 \times 10^{-7}(\mathrm{M})$
- Correct answerC$3.2 \times 10^{-2}(\mathrm{M})$
- D$2.7 \times 10^{-2}(\mathrm{M})$
Solution
$\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$ (At equivalence point)
$0.4 \times 2.5=\frac{2}{15} \times \mathrm{V}_{2}$
$\mathrm{BOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{B} \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(I)$
$\therefore \mathrm{V}_{2}=\frac{0.4 \times 2.5}{\frac{2}{15}}=\frac{0.4 \times 2.5}{2} \times 15=7.5 \mathrm{ml}$
$\therefore$ Concentration of salt $=\frac{0.4 \times 2.5}{2.5+7.5}$
or, Salt of weak base and strong acid = $0.1$
$\therefore\left[\mathrm{H}^{+}\right]=\mathrm{Ch}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}}=\sqrt{10^{-3}}=3.2 \times 10^{-2} \mathrm{M}$
as $C h^{2}=\frac{K_{w}}{K_{b}} \quad \therefore h=\sqrt{\frac{K_{w}}{K_{b} C}}$
Hence, the correct answer is option: 'C'.