Question

Consider the following data.

$$\begin{array}{|c|c|} \hline \text{Electrolyte} & \Lambda_m^\circ (\text{S cm}^2 \text{ mol}^{-1}) \\ \hline \text{BaCl}_2 & x_1 \\ \hline \text{H}_2\text{SO}_4 & x_2 \\ \hline \text{HCl} & x_3 \\ \hline \end{array}$$

$\text{BaSO}_4$ is sparingly soluble in water. If the conductivity of the saturated $\text{BaSO}_4$ solution is $x$ S cm$^{-1}$ then the solubility product of $\text{BaSO}_4$ can be given as
(Here $\Lambda_m = \Lambda_m^\circ$)

• A. $\frac{10^6 x^2}{\alpha^2(x_1 + x_2 - 2x_3)^2}$
• B. $\frac{x^2}{(x_1 + x_2 - 2x_3)^2}$
• C. $\frac{\alpha^2(x_1 + x_2 - 2x_3)^2}{10^6 x^2}$
• D. $\frac{x^2}{(x_1 + x_2 + 2x_3)^2}$

Hint:

Find the molar conductivity of BaSO4 using Kohlrausch's law. Then use the formula for solubility $S = 1000\kappa / \Lambda_m$ and find $K_{sp} = S^2$.

Answer 1

The Correct Option is A

This problem utilizes Kohlrausch's law of independent migration of ions to find the molar conductivity at infinite dilution ($\Lambda_m^\circ$) of $\text{BaSO}_4$ from other electrolytes.
According to the law:
$\Lambda_m^\circ(\text{BaSO}_4) = \lambda^\circ(\text{Ba}^{2+}) + \lambda^\circ(\text{SO}_4^{2-})$
From the given data:
1) $\Lambda_m^\circ(\text{BaCl}_2) = \lambda^\circ(\text{Ba}^{2+}) + 2\lambda^\circ(\text{Cl}^-) = x_1$
2) $\Lambda_m^\circ(\text{H}_2\text{SO}_4) = 2\lambda^\circ(\text{H}^+) + \lambda^\circ(\text{SO}_4^{2-}) = x_2$
3) $\Lambda_m^\circ(\text{HCl}) = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{Cl}^-) = x_3$

To get $\text{BaSO}_4$, we perform: $(1) + (2) - 2 \times (3)$:
$$\Lambda_m^\circ(\text{BaSO}_4) = x_1 + x_2 - 2x_3$$

For a sparingly soluble salt, the solution is so dilute that $\Lambda_m \approx \Lambda_m^\circ$. The relationship between conductivity ($\kappa$), molar conductivity ($\Lambda_m$), and solubility ($S$ in mol/L) is:
$$\Lambda_m = \frac{\kappa \times 1000}{S} \implies S = \frac{x \times 1000}{\Lambda_m^\circ} = \frac{1000x}{x_1 + x_2 - 2x_3}$$
The solubility product ($K_{sp}$) for $\text{BaSO}_4 \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$ is:
$$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S \times S = S^2$$
$$K_{sp} = \left( \frac{1000x}{x_1 + x_2 - 2x_3} \right)^2 = \frac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$$
If we account for the degree of dissociation $\alpha$ (where $\Lambda_m = \alpha \Lambda_m^\circ$), the formula becomes $\frac{10^6 x^2}{\alpha^2 (x_1 + x_2 - 2x_3)^2}$. Since the question implies $\Lambda_m = \Lambda_m^\circ$, we take $\alpha=1$.