Question

The solubility of BaSO₄ in water is 2·42 × 10^-3 gL^-1 at 298K. The value of its solubility product (K_sp) will be (Given molar mass of BaSO₄ = 233 g mol^-1)

• A. 1.08 × 10^-10 mol² L^-2
• B. 1.08 × 10^-14 mol² L^-2
• C. 1.08 × 10^-12 mol² L^-2
• D. 1.08 × 10^-8 mol² L^-2

Answer 1

The Correct Option is A

To determine the solubility product (K_{sp}) of \text{BaSO}_4, we can use the solubility information provided and follow these steps:

1. First, we need to convert the solubility of \text{BaSO}_4 from grams per liter to moles per liter (mol/L). We use the formula:
   \text{Solubility (mol/L)} = \frac{\text{Solubility (g/L)}}{\text{Molar Mass (g/mol)}}
   Substituting the given values:
   \text{Solubility (mol/L)} = \frac{2.42 \times 10^{-3}}{233}
   \text{Solubility (mol/L)} \approx 1.039 \times 10^{-5} \text{ mol/L}
2. \text{BaSO}_4 dissociates in water as follows:
   \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq)
   Hence, at equilibrium, the concentration of \text{Ba}^{2+} and \text{SO}_4^{2-} will be equal to the solubility in mol/L calculated above.
3. The solubility product, K_{sp}, is calculated using:
   K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]
   Since the concentrations are equal, we have:
   K_{sp} = (1.039 \times 10^{-5})^2
   K_{sp} \approx 1.08 \times 10^{-10} \text{ mol}^2 \text{ L}^{-2}

Thus, the solubility product of \text{BaSO}_4 is 1.08 × 10^-10 mol² L^-2, which matches the first option given.