Q18 / 150 · MHT CET 2026 (11 April Shift 2) · Chemistry
What is the solubility of $\text{BaSO}_4$ in g /$\text{dm}^3$ if its solubility product is $1.0 \times 10^{-10}$ at $25^\circ$C. [Molar mass of $\text{BaSO}_4 = 233$ g/mol]
- Correct answerA$2.33 \times 10^{-3}$ g/$\text{dm}^3$
- B$4.66 \times 10^{-3}$ g/$\text{dm}^3$
- C$3.48$ g/$\text{dm}^3$
- D$1.16 \times 10^{-3}$ g/$\text{dm}^3$
Solution
Let the solubility of $\text{BaSO}_4$ be $s$ mol/$\text{dm}^3$.
The dissociation reaction is:
$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2$
Substituting the given value of $K_{sp}$:
$s^2 = 1.0 \times 10^{-10}$
$s = 1.0 \times 10^{-5}$ mol/$\text{dm}^3$
To find the solubility in g/$\text{dm}^3$, multiply the molar solubility by the molar mass of $\text{BaSO}_4$:
Solubility = $s \times \text{Molar mass}$
Solubility = $1.0 \times 10^{-5} \times 233$
Solubility = $2.33 \times 10^{-3}$ g/$\text{dm}^3$
Answer: $2.33 \times 10^{-3}$ g/$\text{dm}^3$
The dissociation reaction is:
$\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2$
Substituting the given value of $K_{sp}$:
$s^2 = 1.0 \times 10^{-10}$
$s = 1.0 \times 10^{-5}$ mol/$\text{dm}^3$
To find the solubility in g/$\text{dm}^3$, multiply the molar solubility by the molar mass of $\text{BaSO}_4$:
Solubility = $s \times \text{Molar mass}$
Solubility = $1.0 \times 10^{-5} \times 233$
Solubility = $2.33 \times 10^{-3}$ g/$\text{dm}^3$
Answer: $2.33 \times 10^{-3}$ g/$\text{dm}^3$
Hence, the correct answer is option: 'A'.