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

We use the concept that for a sparingly soluble salt like $\text{BaSO}_4$, the molar conductivity at saturation is essentially equal to its molar conductivity at infinite dilution. First, find $\Lambda_m^\circ$ for $\text{BaSO}_4$ using the provided electrolytes:
$\Lambda_m^\circ(\text{BaSO}_4) = \Lambda_m^\circ(\text{BaCl}_2) + \Lambda_m^\circ(\text{H}_2\text{SO}_4) - 2\Lambda_m^\circ(\text{HCl})$
$\Lambda_m^\circ(\text{BaSO}_4) = x_1 + x_2 - 2x_3$.

The molar conductivity is defined as $\Lambda_m = \frac{\kappa}{C}$, where $\kappa$ is conductivity and $C$ is concentration (solubility $S$). To use units of S cm$^2$ mol$^{-1}$ with conductivity in S cm$^{-1}$, $C$ must be in mol/cm$^3$.
$S (\text{mol/cm}^3) = \frac{\kappa}{\Lambda_m^\circ} = \frac{x}{x_1 + x_2 - 2x_3}$.

To convert solubility to molarity (mol/L), we multiply by 1000:
$S (\text{mol/L}) = \frac{1000x}{x_1 + x_2 - 2x_3}$.

For $\text{BaSO}_4$, $K_{sp} = S^2$. Substituting the value of $S$:
$$K_{sp} = \left(\frac{1000x}{x_1 + x_2 - 2x_3}\right)^2 = \frac{10^6 x^2}{(x_1 + x_2 - 2x_3)^2}$$
In cases where dissociation is not complete, $K_{sp} = (\alpha S)^2$. Given the options provided, the correct form matches Option 1.