Question

The solubility of $BaSO _{4}$ in water is $2.42 \times 10^{-3}\, gL ^{-1}$ at $298 \,K$. The value of its solubility product $\left( K _{ sp }\right)$ will be (Given molar mass of $BaSO _{4}=233 \,g\, mol ^{-1}$ )

• A. $1.08 \times 10^{-8} \, mol^{2} \, L^{-2}$
• B. $1.08 \times 10^{-10} \, mol^{2} \, L^{-2}$
• C. $1.08 \times 10^{-14} \, mol^{2} \, L^{-2}$
• D. $1.08 \times 10^{-12} \, mol^{2} \, L^{-2}$

Answer 1

The Correct Option is B

 To find the solubility product (\(K_{sp}\)) of \(BaSO_4\), we start by understanding the dissolution reaction:

\(BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq)\)

The solubility product expression for this reaction is:

\(K_{sp} = [Ba^{2+}][SO_4^{2-}]\)

Given the solubility of \(BaSO_4\) in water as \(2.42 \times 10^{-3}\, \text{g/L}\), we first convert this into moles per liter (mol/L) using the molar mass of \(BaSO_4\), which is \(233 \, \text{g/mol}\).

Calculating the molarity (solubility in mol/L):

\(S = \frac{2.42 \times 10^{-3}\, \text{g/L}}{233 \, \text{g/mol}}\)

Solving for \(S\):

\(S \approx 1.039 \times 10^{-5}\, \text{mol/L}\)

Since the stoichiometry of the dissolution is 1:1 for \(BaSO_4\), \(Ba^{2+}\) and \(SO_4^{2-}\) ions, the concentrations are the same:

\([Ba^{2+}] = [SO_4^{2-}] = S\\)

Substituting into the \(K_{sp}\) expression:

\(K_{sp} = S \times S = S^2\)

Calculate \(K_{sp}\):

\(K_{sp} = (1.039 \times 10^{-5})^2 = 1.08 \times 10^{-10} \, \text{mol}^2 \, \text{L}^{-2}\)

Thus, the solubility product of \(BaSO_4\) is \(1.08 \times 10^{-10} \, \text{mol}^2 \, \text{L}^{-2}\), matching the correct answer option.