Find out the ionisation constant of a weak ac | vedclass

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Question

(B) For a weak acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$. The ionisation constant $K_a$ is given by $K_a = \frac{C \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a weak acid $HA$,the dissociation equilibrium is $HA ightleftharpoons H^+ + A^-$. The ionisation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{(1 - \alpha)}$,where $\alpha$ is the degree of dissociation. The degree of dissociation $\alpha$ is related to molar conductivity $\Lambda_m^c$ and limiting molar conductivity $\Lambda_m^o$ as $\alpha = \frac{\Lambda_m^c}{\Lambda_m^o}$. Substituting $\alpha$ into the expression for $K_a$: $K_a = \frac{C 	imes (\frac{\Lambda_m^c}{\Lambda_m^o})^2}{1 - \frac{\Lambda_m^c}{\Lambda_m^o}} = \frac{C 	imes \frac{(\Lambda_m^c)^2}{(\Lambda_m^o)^2}}{\frac{\Lambda_m^o - \Lambda_m^c}{\Lambda_m^o}} = \frac{C (\Lambda_m^c)^2}{\Lambda_m^o(\Lambda_m^o - \Lambda_m^c)}$.

Find out the ionisation constant of a weak acid $(HA)$ in terms of $\Lambda_m^o$ and $\Lambda_m^c$. (Given $\alpha$ cannot be ignored with respect to $1$)

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