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(B) For a weak electrolyte,the degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at concentration $C$ $(\Lambda_m)$ to the

Vedclass · Accepted Answer

$K_a = \frac{C\Lambda_m^2}{\Lambda_m^\infty (\Lambda_m^\infty - \Lambda_m)}$

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(B) For a weak electrolyte,the degree of dissociation $(\alpha)$ is given by the ratio of molar conductivity at concentration $C$ $(\Lambda_m)$ to the molar conductivity at infinite dilution $(\Lambda_m^\infty)$: $\alpha = \frac{\Lambda_m}{\Lambda_m^\infty}$ The ionisation constant $(K_a)$ for a weak acid $HA$ is given by the Ostwald dilution law: $K_a = \frac{C\alpha^2}{1 - \alpha}$ Substituting the value of $\alpha$: $K_a = \frac{C(\frac{\Lambda_m}{\Lambda_m^\infty})^2}{1 - \frac{\Lambda_m}{\Lambda_m^\infty}}$ $K_a = \frac{C \cdot \frac{\Lambda_m^2}{(\Lambda_m^\infty)^2}}{\frac{\Lambda_m^\infty - \Lambda_m}{\Lambda_m^\infty}}$ $K_a = \frac{C\Lambda_m^2}{\Lambda_m^\infty (\Lambda_m^\infty - \Lambda_m)}$

What is the expression for the ionisation constant $(K_a)$ of a weak acid $(HA)$ in terms of molar conductivity at infinite dilution $(\Lambda_m^\infty)$ and molar conductivity at a given concentration $(\Lambda_m)$?

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