Question

$2.5mL$ of $\frac{2}{5}M$ weak monoacidic base $(K_b=1\times {10}^{-12}$ at ${25}^{0}C)$ is titrated with $\frac{2}{15}MHCl$ in water at ${25}^{o}C$. The concentration of $H^{+}$ at equivalence point is $(K_w=1\times {10}^{-14}$ at ${25}^{0}C)$ :

• A. $3.7\times {10}^{-13}M$
• B. $3.2\times {10}^{-7}M$
• C. $3.2\times {10}^{-2}M$
• D. $2.7\times {10}^{-2}M$

Answer 1

Answer: (D)

$2.7\times {10}^{-2}M$

$BOH+HCl\to BCl+H_{2}O$
At equilibrium, all $BOH$ is converted into salt.

Salt hydrolysis at equivalence point: $\underset{C(1-h)}{B^{+}}+H_{2}O\rightleftharpoons \underset{Ch}{BOH}+\underset{Ch}{H^{+}}$
Volume of $HCl$ used $=\frac{2.5\times \frac{2}{5}}{2/15}=7.5ml$
Concentration of Salt, $C=\frac{2.5\times \frac{2}{5}}{10}=0.1M$
$K_h$ = $\frac{C{h}^2}{1-h}=\frac{K_w}{K_b}$
Solving $h=0.27$
$\left[H^{+}\right]=Ch=0.1\times 0.27=2.7\times {10}^{-2}M$

The correct option is 'D'