MCQ
$2.5 \mathrm{~mL}$ of $\frac{2}{5} \mathrm{M}$ 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 $\left(\mathrm{K}_W=1 \times 10^{-14}\right.$ at $\left.25^0 \mathrm{C}\right)$
  • A
    $3.7 \times 10^{-13} \mathrm{M}$
  • B
    $3.2 \times 10^{-7} \mathrm{M}$
  • C
    $3.2 \times 10^{-2} \mathrm{M}$
  • $2.7 \times 10^{-2} \mathrm{M}$

Answer

Correct option: D.
$2.7 \times 10^{-2} \mathrm{M}$
d
$\begin{aligned} & \mathrm{BOH}+\mathrm{HCl} \longrightarrow \mathrm{BCl}+\mathrm{H}_2 \mathrm{O} \\ & \text { C }  \end{aligned}$

$\begin{aligned} & \mathrm{B}^{+}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{BOH}+\mathrm{H}^{+} \\ & C(1-h) \\ & \text { Ch } \quad \text { Ch } \\ & \end{aligned}$

Volume of $\mathrm{HCl}$ used $=\frac{2.5 \times \frac{2}{5}}{2 / 15}=7.5 \mathrm{ml}$

Concentration of Salt, $\mathrm{C}=\frac{2.5 \times \frac{2}{5}}{10}=0.1 \mathrm{M}$

$\therefore \frac{\mathrm{Ch}^2}{1-\mathrm{h}}=\frac{\mathrm{K}_\pi}{\mathrm{K}_{\mathrm{b}}}$

Solving $\mathrm{h}=0.27$

$\left[\mathrm{H}^{+}\right]=\mathrm{Ch}=0.1 \times 0.27=2.7 \times 10^{-2} \mathrm{M}$

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