Chemical And Ionic Equilibrium Result Question 14
14. 2.5 mL of $\dfrac{2}{5} \mathrm{M}$ weak monoacidic base $\left(K_b=1 \times 10^{-12}\right.$ at $25^{\circ} \mathrm{C}$ ) is titrated with $\dfrac{2}{15} \mathrm{M} \mathrm{HCl}$ 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, 3M)
(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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Answer:
Correct Answer: 14. ( d )
Solution:
- $\mathrm{mmol}$ of base $=2.5 = \dfrac{2}{5} = 1$
$\mathrm{mmol}$ of acid required to reach the end point $=1$
Volume of acid required to reach the end point $=\dfrac{15}{2} \mathrm{~mL}$
Total volume at the end point $=\dfrac{15}{2}+2.5=10 \mathrm{~mL}$
Molarity of salt at the end point $=\dfrac{1}{10}=0.10$
$\begin{gathered} \underset{C (1-\alpha)}{B^{+}}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{C \alpha}{B \mathrm{OH}}+\underset{C \alpha}{\mathrm{H}^{+}} \\ K_h=\dfrac{K_w}{K_b}=10^{-2} \\ K_h=10^{-2}=\dfrac{C \alpha^2}{1-\alpha}=\dfrac{0.1 \alpha^2}{1-\alpha} \\ \Rightarrow \quad 10 \alpha^2+\alpha-1=0 \\ \Rightarrow \quad \alpha=\dfrac{-1+\sqrt{1+40}}{20}=0.27 \\ \Rightarrow \quad\left[\mathrm{H}^{+}\right]=C \alpha=0.1 \times 0.27=0.027 \mathrm{M} \end{gathered}$
BETA
