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: Concentrations of ions:
One mole of dissolved $\text{AB}_2$ gives one mole of $\text{A}^{2+}$ and two moles of $\text{B}^-$.
So $[\text{A}^{2+}] = 10^{-6}$ M and $[\text{B}^-] = 2 \times 10^{-6}$ M.

Step 2: Multiply:
$K_{sp} = [\text{A}^{2+}][\text{B}^-]^2 = (10^{-6})(2 \times 10^{-6})^2 = 10^{-6} \times 4 \times 10^{-12} = 4 \times 10^{-18}$.

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