Q20 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry
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$3 \times 10^{-12}$
- Correct answerD$4 \times 10^{-18}$
Solution
The dissociation of the salt $AB_2$ is given by:
$AB_2 \rightleftharpoons A^{2+} + 2B^-$
Let the solubility of the salt be $S$. The equilibrium concentrations are:
$[A^{2+}] = S$
$[B^-] = 2S$
The solubility product $K_{sp}$ is:
$K_{sp} = [A^{2+}][B^-]^2$
$K_{sp} = (S)(2S)^2 = 4S^3$
Given $S = 1 \times 10^{-6}\ \text{mol/dm}^3$, substituting this value:
$K_{sp} = 4(1 \times 10^{-6})^3$
$K_{sp} = 4 \times 10^{-18}$
Answer: $4 \times 10^{-18}$
$AB_2 \rightleftharpoons A^{2+} + 2B^-$
Let the solubility of the salt be $S$. The equilibrium concentrations are:
$[A^{2+}] = S$
$[B^-] = 2S$
The solubility product $K_{sp}$ is:
$K_{sp} = [A^{2+}][B^-]^2$
$K_{sp} = (S)(2S)^2 = 4S^3$
Given $S = 1 \times 10^{-6}\ \text{mol/dm}^3$, substituting this value:
$K_{sp} = 4(1 \times 10^{-6})^3$
$K_{sp} = 4 \times 10^{-18}$
Answer: $4 \times 10^{-18}$
Hence, the correct answer is option: 'D'.