Chemical And Ionic Equilibrium Result Question 49
49. The molar conductivity of a solution of a weak acid $\mathrm{H} X(0.01$ $\mathrm{M})$ 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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Answer:
Correct Answer: 49. $( 3 )$
Solution:
- Degree of ionisation $(\alpha)=\dfrac{\lambda_{\mathrm{m}}}{\lambda_{\infty}}$
Let $\lambda _\mathrm{m}(\mathrm{H} Y)=x \Rightarrow \lambda _\mathrm{m}(\mathrm{H} X)=\dfrac{X}{10}$
$ \Rightarrow \quad \dfrac{\lambda \mathrm{m}(\mathrm{H} X)}{\lambda{\mathrm{m}}(X Y)}=\dfrac{1}{10}=\dfrac{\alpha(\mathrm{H} X)}{\alpha(\mathrm{HY})} \quad\left[\because {\lambda _\infty}(\mathrm{H} X)={\lambda _\infty}(\mathrm{HY})\right] $
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 $
$ \begin{array}{lr} \Rightarrow & \dfrac{K_a(\mathrm{H} X)}{K_a(\mathrm{H} Y)}=\dfrac{0.01}{10}=\dfrac{1}{1000} \\ \Rightarrow & \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 & \mathrm{p} K_a(\mathrm{H} X)-\mathrm{p} K_a(\mathrm{H} Y)=3 \end{array} $
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