Question

The molar conductivity of a solution of a weak acid $HX\left(0.01M\right)$ is $10$ times smaller than the molar conductivity of a solution of a weak acid $HY\left(0.1M\right)$. 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 ionization of both acids to be $<<1$).

Answer 1

Given $\frac{{\wedge }_m\left(HX\right)}{{\wedge }_m\left(HY\right)}=\frac{1}{10}$
$\frac{{\alpha }_1}{{\alpha }_2}=\frac{{\wedge }_m\left(HX\right)}{{\wedge }_m^{\circ }\left(HX\right)}/\frac{{\wedge }_m\left(HY\right)}{{\wedge }_m^{\circ }\left(HY\right)}=\frac{1}{10}$
$\frac{K_a\left(HX\right)}{K_a\left(HY\right)}=\frac{C_1{\alpha }_1^2}{C_2{\alpha }_2^2}$
$=\frac{0.01}{0.1}\times {\left(\frac{1}{10}\right)}^2=\frac{1}{1000}$
$\log K_a\left(HX\right)-\log K_a\left(HY\right)=-3$
$\therefore p{K}_a\left(HX\right)-p{K}_a\left(HY\right)=3$.