Question

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

• 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

The correct option is C $3.2\times {10}^{-2}M$

Given, $[BOH=\frac{2}{5}M,K_b=1\times {10}^{-12},[HCl]=\frac{2}{15}M,K_w=1\times {10}^{-14}at{25}^{o}C$
$\underset{Base}{N_1{V}_1}=\underset{Acid}{N_2{V}_2}$
$2.5\times \frac{2}{5}=\frac{2}{15}\times V_2$
or $V_2\left(acid\right)=\frac{15}{2}=7.5mL$
$BOH+HCl\rightleftharpoons BCl+H_{2}O$
$\text{millimoles of BOH}=2.5\times \frac{2}{5}=1mmol$
$\therefore$ salt $BCl$ formed = $1mmol$
$\text{Total volume of solution}=2.5+7.5=10mL=10\times {10}^{-3}L$
$\therefore \left[BCl\right]\text{in solution}=\frac{1\times {10}^{-3}}{10\times {10}^{-3}}=0.1M$
For the salt of weak base and the strong acid
$\left[H^{+}\right]=\sqrt{\frac{K_{w}C}{K_b}}=\sqrt{\frac{{10}^{-14}\times 0.1}{{10}^{-12}}}$
$\left[H^{+}\right]=3.2\times {10}^{-2}M$