Equivalent conductivity of BaC𝑙₂, H₂SO₄ and HC𝑙, are 𝑥₁, 𝑥₂ and 𝑥₃scm^-1eq^-1 at infinite dilution. If conductivity of saturated BaSO₄ solution is𝑥 ScM^-1, then K_sp of BaSO₄ is

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

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

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

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

answer is C.

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Detailed Solution

$λ_{{BaSO}_{4}} = λ_{{BaCl}_{2}} + λ_{H_{2} {SO}_{4}} - λ_{HCl}$

$= x_{1} + x_{2} - x_{3}$

k=x

$N = \frac{kx 1000}{λ} = \frac{x x 10^{3}}{x_{1} + x_{2} - x_{3}}$

$S = M = \frac{N}{2} = \frac{0.5 x x 10^{3}}{(x_{1} + x_{2} - x_{3})}$

$K_{SP} = S^{2} = \frac{0.25 x^{2} X 10^{6}}{{(x_{1} + x_{2} - x_{3})}^{2}}$

$K_{SP} = \frac{2.5 X 10^{5}}{{(x_{1} + x_{2} - x_{3})}^{2}}$

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