JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Equilibrium

The electrolyte ionises as ${\mathrm{A}}_2{\mathrm{B}}_3\rightleftharpoons 2{\mathrm{A}}^{3+}+3{\mathrm{B}}^{2-}$. From an initial concentration $c$ with degree of dissociation $\alpha$, the equilibrium concentrations are $[{\mathrm{A}}^{3+}]=2c\alpha$, $[{\mathrm{B}}^{2-}]=3c\alpha$ and $[{\mathrm{A}}_2{\mathrm{B}}_3]=c(1-\alpha )$. Hence $K_{\mathrm{eq}}=\frac{(2c\alpha {)}^2(3c\alpha {)}^3}{c(1-\alpha )}=\frac{108c^4{\alpha }^5}{1-\alpha }$. For a concentrated solution of a weak electrolyte $\alpha$ is small, so $(1-\alpha )\approx 1$ and $K_{\mathrm{eq}}\approx 108c^4{\alpha }^5$. Solving for the degree of dissociation gives $\alpha ={\left(\frac{K_{\mathrm{eq}}}{108c^4}\right)}^{1/5}$.