Question

2.5 mL 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 point 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 D

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

mmol of base $=2.5\times \frac{2}{5}=1$
mmol of acid required to reach the end point = 1
Volume of acid required to reach the end point = $\frac{15}{2}mL$
Total volume at the end point = $\frac{15}{2}$ + 2.5 = 10 mL
Molarity of salt at the end point = $\frac{1}{10}=0.10$
$\underset{C(1-\alpha )}{B^{+}}+H_{2}O\rightleftharpoons \underset{C\alpha }{BOH}+\underset{C\alpha }{H^{+}}$
$K_h=\frac{K_w}{K_b}={10}^{-2}$
$K_h={10}^{-2}=\frac{C\alpha }{1-\alpha }=\frac{0.1\alpha }{1-\alpha }$
$\Rightarrow 10{\alpha }^2+\alpha -1=0$
$\Rightarrow \alpha =\frac{-1+\sqrt{1+40}}{20}=0.27$
$\Rightarrow \left[H^{+}\right]=C\alpha =0.1\times 0.27=0.027M$