Equivalent conductance of BaCl₂, H₂SO₄ and HCl are x₁, x₂ and x₃ S cm² equiv^–1 at infinite dilution. If specific conductance of saturated BaSO₄ solution is y S cm^–1, then Ksp of BaSO₄ is

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

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

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

$\frac{x_{1} + x_{2} - 2 x_{3}}{10^{6} y^{2}}$

answer is C.

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Detailed Solution

equivalent conductance = 2(x₁+x₂-x₃)

Molarity = $\frac{1000 \times y}{2 (x_{1} + x_{2} - x_{3})}$

K_sp = $\frac{10^{6} y^{2}}{4 {(x_{1} + x_{2} - x_{3})}^{2}}$

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