Chemical and Ionic Equilibrium 2 Question 49
49. The molar conductivity of a solution of a weak acid $\mathrm{H} X(0.01$ M) is 10 times smaller than the molar conductivity of a solution of a weak acid $\mathrm{H} Y(0.10 \mathrm{M})$. If $\lambda_{X^{-}}^{0} \approx \lambda_{Y^{-}}^{0}$, the difference in their $\mathrm{p} K_{a}$ values, $\mathrm{p} K_{a}(\mathrm{H} X)-\mathrm{p} K_{a}(\mathrm{H} Y)$, is (consider degree of ionisation of both acids to be $«1$ ).
(2015 Adv.)
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Solution:
- Degree of ionisation $(\alpha)=\frac{\wedge_{\mathrm{m}}}{\wedge^{\infty}}$
$$ \begin{aligned} & \text { Let } \quad{ }^{\wedge} \mathrm{m}(\mathrm{H} Y)=x \Rightarrow{ }^{\wedge} \mathrm{m}(\mathrm{H} X)=\frac{X}{10} \ & \Rightarrow \quad \frac{{ }^{\wedge} \mathrm{m}(\mathrm{H} X)}{{ }^{\wedge} \mathrm{m}(X Y)}=\frac{1}{10}=\frac{\alpha(\mathrm{H} X)}{\alpha(\mathrm{H} Y)} \quad\left[\because{ }^{\wedge \infty}(\mathrm{H} X)=\wedge^{\wedge}(\mathrm{H} Y)\right] \end{aligned} $$
Also : $\quad K_{a}(\mathrm{H} X)=(0.01)[\alpha(\mathrm{H} X)]^{2}$
$K_{\mathrm{a}}(\mathrm{H} Y)=(0.10)[\alpha(\mathrm{H} Y)]^{2}$
$$ =0.10[10 \alpha(\mathrm{H} X)]^{2}=10[\alpha(\mathrm{H} X)]^{2} $$
$\Rightarrow \quad \frac{K_{a}(\mathrm{H} X)}{K_{a}(\mathrm{H} Y)}=\frac{0.01}{10}=\frac{1}{1000}$
$\Rightarrow \quad \log K_{a}(\mathrm{H} X)-\log K_{a}(\mathrm{H} Y)=-3$
$\Rightarrow-\log K_{a}(\mathrm{H} X)-\left[-\log K_{a}(\mathrm{H} Y)\right]=3$
$\Rightarrow \quad \mathrm{p} K_{a}(\mathrm{H} X)-\mathrm{p} K_{a}(\mathrm{H} Y)=3$
