$2.5 \mathrm{~mL}$ of $\frac{2}{5} \mathrm{M}$ weak monoacidic base $\left(\mathrm{K}_{\mathrm{b}}=1 \times…
$\left(\mathrm{K}_{\mathrm{b}}=1 \times 10^{-12}ight.$ at $\left.25^{\circ} \mathrm{C}ight)$ 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 $\left(\mathrm{K}_{\mathrm{w}}=1 \times 10^{-14}ight.$ at $\left.25^{\circ} \mathrm{C}ight)$
- $3.7 \times 10^{-13} \mathrm{M}$
- $3.2 \times 10^{-7} \mathrm{M}$
- $3.2 \times 10^{-2} \mathrm{M}$
- $2.7 \times 10^{-2} \mathrm{M}$
Solution
$=$ equivalents of base
$\frac{2}{15} \times \mathrm{V}=2.5 \times \frac{2}{5}$
$\mathrm{~V}=7.5 \mathrm{ml}$
Total volume $=2.5+7.5=10 \mathrm{ml}$ concentration of salt at equivalence point
$=\frac{2.5 \times \frac{2}{5}}{10}$
$\mathrm{C}=0.1 \mathrm{M}$
$\mathrm{BOH}+\mathrm{HCl} \longrightarrow \mathrm{BCl}+\mathrm{H}_{2} \mathrm{O}$
Now BCl undergoes hydrolysis
$\mathrm{B}^{+}+\mathrm{H}_{2} \mathrm{O} ightleftharpoons \mathrm{BOH}+\mathrm{H}^{+}$
$\left[\mathrm{H}^{+}ight]=\mathrm{Ch}=\mathrm{C} \times \sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}} \times \mathrm{C}}}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \times \mathrm{C}}$
$=3.33 \times 10^{-2} \mathrm{M}$ ^
Asked in: JEE-TOPICTESTS-CHEMISTRY