Question

The molar conductivity of a solution of a weak acid $HX\left(0.01\text{M}\right)$ is 10 times smaller than the molar conductivity of a solution of a weak acid $HY\left(0.10\text{M}\right)$. If ${\lambda }_{X^{-}}^{\circ }\approx {\lambda }_{Y^{-}}^{\circ }$, the difference in their $p{K}_a$ values, $p{K}_a\left(HX\right)-p{K}_a\left(HY\right)$, is (consider degree of ionizaiton of both acids to be <<1)

Answer 1

${\lambda }_{X^{-}}^{\circ }\approx {\lambda }_{Y^{-}}^{\circ }$
$\Rightarrow {\lambda }_{H^{+}}^{\circ }+{\lambda }_{X^{-}}^{\circ }\approx {\lambda }_{H^{+}}^{\circ }+{\lambda }_{Y^{-}}^{\circ }\Rightarrow {\lambda }_{HX}^{\circ }\approx {\lambda }_{HY}^{\circ }\dots \dots \left(1\right)$
Also $\frac{{\lambda }_m}{{\lambda }_m^{\circ }}$= $\alpha$, So ${\lambda }_m\left(HX\right)={\lambda }_m^{\circ }{\alpha }_1$ and ${\lambda }_m\left(HY\right)={\lambda }_m^{\circ }{\alpha }_2$
Now,
${\lambda }_m\left(HY\right)=10{\lambda }_m\left(HX\right)$
$\Rightarrow {\lambda }_m^{\circ }{\alpha }_2=10\times {\lambda }_m^{\circ }{\alpha }_1$
${\alpha }_2=10{\alpha }_1\dots \dots \left(2\right)$
$K_a=\frac{C{\alpha }^2}{1-\alpha }$, but $\alpha <<1$, therefore $K_a=C{\alpha }^2$
$\Rightarrow \frac{K_a\left(HX\right)}{K_a\left(HY\right)}=\frac{0.01{\alpha }_1^2}{0.1{\alpha }_2^2}=\frac{0.01}{0.1}\times {\left(\frac{1}{10}\right)}^2=\frac{1}{1000}$.
$\Rightarrow log\left(K_a\right(HX\left)\right)-log\left(K_a\right(HY\left)\right)=-3$
$\Rightarrow p{K}_a\left(HX\right)-p{K}_a\left(HY\right)=3$