Question

$2.5\, mL$ of $\frac{2}{5}\, M$ weak monoacidic base $\left( K _{ b }=1 \times 10^{-12}\right.$ at $\left.25^{\circ} C \right)$ is titrated with $\frac{2}{15} \,M\, HCl$ in water at $25^{\circ} C$. The concentration of $H ^{+}$at equivalence point is $\left( K _{ w }=1 \times 10^{-14}\right.$ at $\left.25^{\circ} C \right)$

• 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)

$BOH + HC1 \longrightarrow \underset{C}{BCl} + H _{2} O$
$\underset{C (1- h )}{B ^{+}}+ H _{2} O \rightleftharpoons \underset{Ch}{BOH} + \underset{Ch}{H ^{+}}$
Volume of HCl used $=\frac{2.5 \times \frac{2}{5}}{2 / 15}=7.5\, m 1$
Concentration of Salt, $C =\frac{2.5 \times \frac{2}{5}}{10}=0.1\, M$
$\therefore \frac{ Ch ^{2}}{1- h }=\frac{ K _{ w }}{ K _{ b }}$
Solving $h =0.27$
$\left[ H ^{+}\right]= Ch =0.1 \times 0.27=2.7 \times 10^{-2} \,M$