Question

$2.5 \,ml$ of $ \frac{2}{5} M$ weak monoacidic base $\left(K_{p}=1 \times 10^{-12}\right.$ at $25^{\circ} C$) 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)

$m\,mol$ of base $=2.5 \times \frac{2}{5}=1$
$m\,mol$ of acid required to reach the end point $=1$
Volume of acid required to reach the end point $=\frac{15}{2}$
Total volume at the end point $=\frac{15}{2}+2.5=10\, ml$.
Molarity of salt at the end point $=\frac{1}{10}= 0.10$

$K_{h}=\frac{K_{w}}{K_{b}}=10^{-2}$
$K_h = 10^{-2} = \frac{C\alpha^2}{- -\alpha} = \frac{0.1 \alpha^2}{1 - \alpha} $
$\Rightarrow 10\alpha^{2}+\alpha-1=0$
$\Rightarrow \alpha=\frac{-1+\sqrt{1+40}}{20}=0.27$
$\Rightarrow\left[H^{+}\right]=C \alpha=0.1 \times 0.27=0.027\, M$