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

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

Vedclass · Answer

(B) For a weak acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$At equilibrium,the concentrations are: $[HA] = c(1-\alpha)$,$[H^+] = c\alpha$,$[A^-] = c\alpha$.The dissociation constant is given by: $K_a = \frac{c\alpha^2}{1-\alpha}$.The degree of dissociation $\alpha$ is related to molar conductivity by: $\alpha = \frac{\Lambda _m}{\Lambda _m^\infty}$.Substituting $\alpha$ into the expression for $K_a$:$K_a = \frac{c(\frac{\Lambda _m}{\Lambda _m^\infty})^2}{1 - \frac{\Lambda _m}{\Lambda _m^\infty}}$$K_a = \frac{c\Lambda _m^2}{(\Lambda _m^\infty)^2 \left( \frac{\Lambda _m^\infty - \Lambda _m}{\Lambda _m^\infty} \right)}$$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 $\Lambda _m^\infty$ and $\Lambda _m$ is:

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