The ionisation constant of a weak acid (HA) i | vedclass

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Question

(B) For a weak acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$At equilibrium,the concentrations

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a weak acid $HA$,the dissociation equilibrium is: $HA_{(aq)} ightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$At equilibrium,the concentrations are: $C(1 - \alpha)$,$C\alpha$,and $C\alpha$ respectively.The expression for the ionisation constant is $K_a = \frac{C\alpha^2}{1 - \alpha}$.The degree of dissociation $\alpha$ is given by the ratio of molar conductivity at concentration $C$ to molar conductivity at infinite dilution: $\alpha = \frac{\Lambda _m}{\Lambda _m^\infty}$.Substituting $\alpha$ into the $K_a$ expression:$K_a = \frac{C(\frac{\Lambda _m}{\Lambda _m^\infty})^2}{1 - \frac{\Lambda _m}{\Lambda _m^\infty}}$Simplifying the expression:$K_a = \frac{C\Lambda _m^2}{(\Lambda _m^\infty)^2} 	imes \frac{\Lambda _m^\infty}{(\Lambda _m^\infty - \Lambda _m)}$$K_a = \frac{C\Lambda _m^2}{\Lambda _m^\infty (\Lambda _m^\infty - \Lambda _m)}$

The ionisation constant of a weak acid $(HA)$ in terms of molar conductivity at infinite dilution $(\Lambda _m^\infty)$ and molar conductivity at concentration $C$ $(\Lambda _m)$ is given by:

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