Chemical and Ionic Equilibrium 2 Question 14
14. $2.5 \mathrm{~mL}$ of $\frac{2}{5} \mathrm{M}$ weak monoacidic base $\left(K_{b}=1 \times 10^{-12}\right.$ at $25^{\circ} \mathrm{C}$ ) is titrated with $\frac{2}{15} \mathrm{M} \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(K_{w}=1 \times 10^{-14}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$
$(2008,3 M)$
(a) $3.7 \times 10^{-13} \mathrm{M}$
(b) $3.2 \times 10^{-7} \mathrm{M}$
(c) $3.2 \times 10^{-2} \mathrm{M}$
(d) $2.7 \times 10^{-2} \mathrm{M}$
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Solution:
- $\mathrm{mmol}$ 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 $=\frac{15}{2} \mathrm{~mL}$
Total volume at the end point $=\frac{15}{2}+2.5=10 \mathrm{~mL}$
Molarity of salt at the end point $=\frac{1}{10}=0.10$
$$ \begin{array}{cc} \underset{C(1-\alpha)}{B^{+}}+\mathrm{H}{2} \mathrm{O} \rightleftharpoons \underset{C \alpha}{B \mathrm{OH}}+\underset{C \alpha}{\mathrm{H}^{+}} \ K{h}=\frac{K_{w}}{K_{b}}=10^{-2} \ \Rightarrow \quad K_{h}=10^{-2}=\frac{C \alpha^{2}}{1-\alpha}=\frac{0.1 \alpha^{2}}{1-\alpha} \ \Rightarrow \quad 10 \alpha^{2}+\alpha-1=0 \ \Rightarrow \quad \alpha=\frac{-1+\sqrt{1+40}}{20}=0.27 \ \quad\left[\mathrm{H}^{+}\right]=C \alpha=0.1 \times 0.27=0.027 \mathrm{M} \end{array} $$
