For a concentrated solution of a weak electro | vedclass

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Question

(A) The dissociation reaction is: $A_2B_3 (aq.) \rightleftharpoons 2 A^{3+} (aq.) + 3 B^{2-} (aq.)$ At equilibrium,the concentrations are: $[A_2B_3] =

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$\left(\frac{K_{eq}}{108 c^4}\right)^{\frac{1}{5}}$

Vedclass · Answer

(A) The dissociation reaction is: $A_2B_3 (aq.) \rightleftharpoons 2 A^{3+} (aq.) + 3 B^{2-} (aq.)$ At equilibrium,the concentrations are: $[A_2B_3] = c(1 - \alpha)$,$[A^{3+}] = 2c\alpha$,and $[B^{2-}] = 3c\alpha$. The equilibrium constant expression is: $K_{eq} = \frac{[A^{3+}]^2 [B^{2-}]^3}{[A_2B_3]} = \frac{(2c\alpha)^2 (3c\alpha)^3}{c(1 - \alpha)}$. For a weak electrolyte,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$. Thus,$K_{eq} = \frac{(4c^2\alpha^2)(27c^3\alpha^3)}{c} = 108c^4\alpha^5$. Solving for $\alpha$: $\alpha^5 = \frac{K_{eq}}{108c^4} \implies \alpha = \left(\frac{K_{eq}}{108 c^4}\right)^{\frac{1}{5}}$.

For a concentrated solution of a weak electrolyte ($K_{eq} = $ equilibrium constant) $A_2B_3$ of concentration '$c$',the degree of dissociation '$\alpha$' is

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