Q15 / 150 · MHT CET 2026 (17 April Shift 1) · Chemistry
Calculate the solubility in mol $\text{dm}^{-3}$ of sparingly soluble salt BA at 293 K if its solubility product is $8.56 \times 10^{-5}$ at same temperature.
- A$8.123 \times 10^{-3}$
- B$8.780 \times 10^{-3}$
- Correct answerC$9.252 \times 10^{-3}$
- D$7.756 \times 10^{-3}$
Solution
The dissociation of the sparingly soluble salt BA is given by:
$BA \rightleftharpoons B^+ + A^-$
Let the solubility of the salt be $s$ mol $\text{dm}^{-3}$.
The solubility product $K_{sp}$ is given by:
$K_{sp} = [B^+][A^-] = s \times s = s^2$
Given $K_{sp} = 8.56 \times 10^{-5}$, we have:
$s^2 = 8.56 \times 10^{-5} = 85.6 \times 10^{-6}$
$s = \sqrt{85.6 \times 10^{-6}} = 9.252 \times 10^{-3} \text{ mol dm}^{-3}$
Answer: $9.252 \times 10^{-3}$
$BA \rightleftharpoons B^+ + A^-$
Let the solubility of the salt be $s$ mol $\text{dm}^{-3}$.
The solubility product $K_{sp}$ is given by:
$K_{sp} = [B^+][A^-] = s \times s = s^2$
Given $K_{sp} = 8.56 \times 10^{-5}$, we have:
$s^2 = 8.56 \times 10^{-5} = 85.6 \times 10^{-6}$
$s = \sqrt{85.6 \times 10^{-6}} = 9.252 \times 10^{-3} \text{ mol dm}^{-3}$
Answer: $9.252 \times 10^{-3}$
Hence, the correct answer is option: 'C'.