For a fairly concentrated solution of a weak electrolyte $A_{x} B_{y}$ , the degree of dissociation is given as

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$α = \sqrt{K_{e q} c / x y}$

$α = {(K_{eq} / c^{x + y - 1} x^{x} y^{y})}^{1 / (x + y)}$

$α = \sqrt{K_{eq} x y / c}$

$α = \sqrt{K_{eq} / c (x + y)}$

answer is B.

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Detailed Solution

$\begin{array}{l} A_{x} B_{y} ⇌ x A^{y +} + y B^{x -} \\ c (1 - α) c (x α) c (y α) \\ K_{eq} = \frac{(c x α)^{x} (c y α)^{y}}{c (1 - α)} ≃ c^{x + y - 1} x^{x} y^{y} α^{x + y} \end{array}$

or $α = {(\frac{K_{eq}}{c^{x + y - 1} x^{x} y^{y}})}^{1 / (x + y)}$

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