Consider the following data. Electrolyte ^ _m (S cm ^2 mol ^ -1 )...

Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of $\text{BaSO}_4$ is given by:

$\Lambda^{\circ}_m(\text{BaSO}_4) = \Lambda^{\circ}_m(\text{BaCl}_2) + \Lambda^{\circ}_m(\text{H}_2\text{SO}_4) - 2\Lambda^{\circ}_m(\text{HCl})$

Substituting the given values:

$\Lambda^{\circ}_m(\text{BaSO}_4) = x_1 + x_2 - 2x_3$

For a sparingly soluble salt, the solubility $S$ (in mol L$^{-1}$) is related to its conductivity $\kappa$ (in S cm$^{-1}$) and molar conductivity $\Lambda_m$ (in S cm$^2$ mol$^{-1}$) by the relation:

$\Lambda_m = \dfrac{\kappa \times 1000}{S}$

Given that $\kappa = x$ and $\Lambda_m = \Lambda^{\circ}_m$ (which implies complete dissociation, meaning the degree of dissociation $\alpha = 1$), we get:

$S = \dfrac{1000 x}{x_1 + x_2 - 2x_3}$

The solubility product $K_{sp}$ for $\text{BaSO}_4$ (which dissociates into $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$) is:

$K_{sp} = S^2 = \left( \dfrac{1000 x}{x_1 + x_2 - 2x_3} \right)^2 = \dfrac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$

Since $\alpha = 1$, this expression is equivalent to $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$.

Answer: $\dfrac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$