Equivalent conductivity of BaCl₂, H₂SO₄and HCI, are x₁,x₂ and x₃scm^–1eq^–1at infinite dilution. If conductivity of saturated BaSO₄ solution is x Scm^–1, then Ksp of BaSO₄ is

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

$\frac{0 .25 x^{2}}{{(x_{1} + x_{2} - x_{3})}^{2}}$

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} {BaSO}_{4} = ({BaCl}_{2} + H_{2} {SO}_{4} - 2 HCl) \\ \land_{{BaSO}_{4}} = 2 (x_{1} + x_{2} - x_{3}) \\ S = M = \frac{K \times 1000}{\land_{M}} = \frac{x \times 1000}{2 (x_{1} + x_{2} - x_{3})} \\ K_{sp} ({BaSO}_{4}) = S^{2} = \frac{x^{2} \times (500)^{2}}{{(x_{1} + x_{2} - x_{3})}^{2}} \\ K_{sp} = \frac{x^{2} \times 2.5 \times 10^{5}}{{(x_{1} + x_{2} - x_{3})}^{2}} \end{array}$