Derive the relation between the dissociation constant $K_a$ and the limiting molar conductivity $\Lambda_m^o$ for a solution of a weak electrolyte.

(N/A) For a weak electrolyte,the degree of dissociation $\alpha$ is related to molar conductivity $\Lambda_m$ and limiting molar conductivity $\Lambda_m^o$ by the equation: $\alpha = \frac{\Lambda_m}{\Lambda_m^o}$.
For a weak electrolyte $AB$ dissociating as $AB \rightleftharpoons A^+ + B^-$,the dissociation constant $K_a$ is given by: $K_a = \frac{c \alpha^2}{1 - \alpha}$.
Substituting $\alpha = \frac{\Lambda_m}{\Lambda_m^o}$ into the expression for $K_a$:
$K_a = \frac{c (\frac{\Lambda_m}{\Lambda_m^o})^2}{1 - \frac{\Lambda_m}{\Lambda_m^o}}$.
Simplifying the expression:
$K_a = \frac{c \Lambda_m^2}{(\Lambda_m^o)^2 (\frac{\Lambda_m^o - \Lambda_m}{\Lambda_m^o})}$.
Thus,the final relation is: $K_a = \frac{c \Lambda_m^2}{\Lambda_m^o (\Lambda_m^o - \Lambda_m)}$.