Q19 / 150 · MHT CET 2026 (16 April Shift 1) · Chemistry
The solubility product of a sparingly soluble salt BA is $6.4 \times 10^{-13}$. Calculate it's solubility in $\text{g dm}^{-3}$ .
Molar mass of salt is $190\ \text{g mol}^{-1}$ .
Molar mass of salt is $190\ \text{g mol}^{-1}$ .
- Correct answerA$1.52 \times 10^{-4}$
- B$1.25 \times 10^{-4}$
- C$2.1 \times 10^{-4}$
- D$1.9 \times 10^{-4}$
Solution
For a sparingly soluble salt $BA$, the dissociation equilibrium is given by $BA \rightleftharpoons B^+ + A^-$.
Let the solubility of the salt be $s$ in $\text{mol dm}^{-3}$.
The solubility product $K_{sp}$ is given by:
$K_{sp} = [B^+][A^-] = s \times s = s^2$
Substituting the given value of $K_{sp}$:
$s^2 = 6.4 \times 10^{-13} = 64 \times 10^{-14}$
$s = \sqrt{64 \times 10^{-14}} = 8 \times 10^{-7}\ \text{mol dm}^{-3}$
To convert the solubility from $\text{mol dm}^{-3}$ to $\text{g dm}^{-3}$, multiply by the molar mass of the salt:
Solubility in $\text{g dm}^{-3} = s \times \text{Molar mass}$
Solubility $= 8 \times 10^{-7}\ \text{mol dm}^{-3} \times 190\ \text{g mol}^{-1}$
Solubility $= 1520 \times 10^{-7}\ \text{g dm}^{-3} = 1.52 \times 10^{-4}\ \text{g dm}^{-3}$
Answer: $1.52 \times 10^{-4}$
Let the solubility of the salt be $s$ in $\text{mol dm}^{-3}$.
The solubility product $K_{sp}$ is given by:
$K_{sp} = [B^+][A^-] = s \times s = s^2$
Substituting the given value of $K_{sp}$:
$s^2 = 6.4 \times 10^{-13} = 64 \times 10^{-14}$
$s = \sqrt{64 \times 10^{-14}} = 8 \times 10^{-7}\ \text{mol dm}^{-3}$
To convert the solubility from $\text{mol dm}^{-3}$ to $\text{g dm}^{-3}$, multiply by the molar mass of the salt:
Solubility in $\text{g dm}^{-3} = s \times \text{Molar mass}$
Solubility $= 8 \times 10^{-7}\ \text{mol dm}^{-3} \times 190\ \text{g mol}^{-1}$
Solubility $= 1520 \times 10^{-7}\ \text{g dm}^{-3} = 1.52 \times 10^{-4}\ \text{g dm}^{-3}$
Answer: $1.52 \times 10^{-4}$
Hence, the correct answer is option: 'A'.