Equivalent conductivity of ${BaCl}_{2}, H_{2} {SO}_{4}$ and HCl, are $x_{1}, x_{2}$ and $x_{3} {Scm}^{- 1} {eq}^{- 1}$ at infinite dilution. If conductivity of saturated ${BaSO}_{4}$ solution is $x S {cm}^{- 1},$ then $K_{s p}$ of $B a S O_{4}$ is

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$\frac{0.25 x^{2}}{{(x_{1} + x_{2} - x_{3})}^{2}}$

$\frac{500 x}{(x_{1} + x_{2} - 2 x_{3})}$

$\frac{2.5 \times 10^{5} x^{2}}{{(x_{1} + x_{2} - x_{3})}^{2}}$

$\frac{10^{6} x^{2}}{{(x_{1} + x_{2} - 2 x_{3})}^{3}}$

answer is C.

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Detailed Solution

$\begin{array}{l} Λ_{eq}^{\circ} ({Baso}_{4}) = Λ_{eq}^{\circ} ({Bacl}_{2}) + Λ_{eq}^{\circ} (H_{2} {So}_{4}) - Λ_{eq}^{\circ} (HCl) = x_{1} + x_{2} - x_{3} \\ Λ_{m}^{\circ} ({Baso}_{4}) = 2 (x_{1} + x_{2} - x_{3}) \\ S = \frac{K \times 10^{3}}{Λ_{m}^{\circ}} = \frac{x \times 10^{3}}{2 (x_{1} + x_{2} - x_{3})} \\ K_{sp} = s^{2} = \frac{x^{2} \times 10^{6}}{4 {(x_{1} + x_{2} - x_{3})}^{2}} \end{array}$