Question

The solubility of a sparingly soluble a salt \(\text{AB}_2\) is \(1\times 10^{-6} \text{mol/dm}^3\). Calculate its solubility product?

• A. \(1\times 10^{-12}\)
• B. \(2\times 10^{-12}\)
• C. \(3\times 10^{-12}\)
• D. \(4\times 10^{-18}\)

Hint:

Write the dissociation of AB2 and express Ksp as 4s cubed.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a salt \(\text{AB}_2\) the dissolution equilibrium is \(\text{AB}_2 \rightleftharpoons \text{A}^{2+} + 2\text{B}^-\). The solubility product is the product of the ion concentrations, each raised to its coefficient.

Step 2: Key Formula or Approach:
If the solubility is \(s\), then \([\text{A}^{2+}] = s\) and \([\text{B}^-] = 2s\), so \(K_{sp} = s(2s)^2 = 4s^3\).

Step 3: Detailed Explanation:
Here \(s = 1 \times 10^{-6}\) mol/dm\(^3\).
\[ K_{sp} = 4s^3 = 4 \times (10^{-6})^3 = 4 \times 10^{-18} \]
Options A, B and C are of order \(10^{-12}\). They come from treating the salt like an \(\text{AB}\) type salt (\(s^2\)) or from squaring instead of cubing, and so they are not correct for \(\text{AB}_2\).

Final Answer:
\(K_{sp} = 4 \times 10^{-18}\), option (D). \[ \boxed{4 \times 10^{-18}} \]