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For a concentrated solution of a weak electrolyte ($$K_{eq}$$ = equilibrium constant) A$$_2$$B$$_3$$ of concentration 'C', the degree of dissociation '$$\alpha$$' is
For the weak electrolyte A$$_2$$B$$_3$$ dissociating as
$$\text{A}_2\text{B}_3 \rightleftharpoons 2\text{A} + 3\text{B}$$If the initial concentration is $$C$$ and degree of dissociation is $$\alpha$$, then at equilibrium:
$$[\text{A}_2\text{B}_3] = C(1 - \alpha) \approx C$$ (since $$\alpha$$ is small for a concentrated solution), $$[\text{A}] = 2C\alpha$$, and $$[\text{B}] = 3C\alpha$$.
The equilibrium constant is
$$K_{eq} = \frac{[\text{A}]^2[\text{B}]^3}{[\text{A}_2\text{B}_3]} = \frac{(2C\alpha)^2(3C\alpha)^3}{C}$$ $$K_{eq} = \frac{4C^2\alpha^2 \times 27C^3\alpha^3}{C} = 108C^4\alpha^5$$Solving for $$\alpha$$,
$$\alpha^5 = \frac{K_{eq}}{108C^4}$$ $$\alpha = \left(\frac{K_{eq}}{108C^4}\right)^{1/5}$$Hence, the correct answer is $$\alpha = \left(\dfrac{K_{eq}}{108c^4}\right)^{1/5}$$.
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