Question
Consider the following data :
| Electrolyte | $$\mathbf{\Lambda}_{\mathbf{m}}^{\mathbf{o}}\left( \mathbf{Scm}^{\mathbf{2}}{\text{ }\mathbf{mol}}^{\mathbf{- 1}} \right)$$ |
|---|---|
| $${BaCl}_{2}$$ | $$x_{1}$$ |
| $$H_{2}{SO}_{4}$$ | $$x_{2}$$ |
| HCl | $$x_{3}$$ |
${BaSO}_{4}$ is sparingly soluble in water. If the conductivity of the saturated ${BaSO}_{4}$ solution is $\times S{cm}^{- 1}$ then the solubility product of ${BaSO}_{4}$ can be given as.
(Here $\land_{m} = \land_{m}^{o}$ )
Options
Solution
Sol. [Key Concept: ${BaSO}_{4}$ is sparingly soluble: get $\Lambda_{m}^{\circ} = (x_{1} + x_{2} - 2x_{3})$ by Kohlrausch, then $S = \frac{\kappa \times 1000}{\Lambda_{m}}$ and $K_{sp} = S^{2}$.]
$${BaSO}_{4}(\text{ }s) \rightleftharpoons {Ba}_{(aq)}^{+ 2} + {SO}_{4(aq)}^{2 -}$$
$$\begin{matrix} & K_{sp} = S^{2} \\ & k\text{~}\text{of}\text{~}{BaSO}_{4} = x\text{ }S{\text{ }cm}^{- 1} \\ & \Lambda_{m}^{o}\text{~}\text{of}\text{~}{BaSO}_{4} = \left( x_{1} + x_{2} - 2x_{3} \right) \\ & \Lambda_{m} = \alpha\Lambda_{M}^{o} = \frac{k \times 1000}{\text{ }S} \\ & \text{ }S = \frac{x \times 1000}{\alpha\left( x_{1} + x_{2} - 2x_{3} \right)} \\ & k_{sp} = S^{2} = \frac{1}{\alpha^{2}}\left\lbrack \frac{x \times 1000}{x_{1} + x_{2} - 2x_{3}} \right\rbrack^{2} \end{matrix}$$
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