Question

The solubility of sparingly soluble salt \(\text{AX}_2\) is \(1 \times 10^{-4} \text{ mol dm}^{-3}\) at 298 K . Calculate its solubility product.

• A. \(2 \times 10^{-12}\)
• B. \(4 \times 10^{-12}\)
• C. \(2 \times 10^{-10}\)
• D. \(4 \times 10^{-10}\)

Hint:

For a salt of type: \[ \text{AX}_2 \rightleftharpoons \text{A}^{2+} + 2\text{X}^- \] if solubility is \(s\), then: \[ K_{sp}=s(2s)^2=4s^3 \]

Answer 1

The Correct Option is B

Concept:
If the solubility of a salt is \(s\), then we first write its dissociation and express ion concentrations in terms of \(s\).

Step 1: Write the dissociation of \(\text{AX}_2\).
\[ \text{AX}_2 \rightleftharpoons \text{A}^{2+} + 2\text{X}^- \]

Step 2: Express ion concentrations in terms of solubility.
Given solubility: \[ s = 1\times10^{-4}\ \text{mol dm}^{-3} \] So, \[ [\text{A}^{2+}] = s = 1\times10^{-4} \] \[ [\text{X}^-] = 2s = 2\times10^{-4} \]

Step 3: Write the expression of solubility product.
\[ K_{sp} = [\text{A}^{2+}][\text{X}^-]^2 \] Substitute the values: \[ K_{sp} = (1\times10^{-4})(2\times10^{-4})^2 \] \[ K_{sp} = (1\times10^{-4})(4\times10^{-8}) \] \[ K_{sp} = 4\times10^{-12} \] Hence, the correct answer is:
\[ \boxed{(B)\ 4\times10^{-12}} \]