Question

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 \(190 \text{g mol}^{-1}\) .

• A. \(1.52\times 10^{-4}\)
• B. \(1.25\times 10^{-4}\)
• C. \(2.1\times 10^{-4}\)
• D. \(1.9\times 10^{-4}\)

Hint:

For a 1:1 salt, S = sqrt(Ksp); multiply by molar mass for g per dm3.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
BA dissociates as \(\text{BA} \rightleftharpoons \text{B}^+ + \text{A}^-\). If solubility is \(S\) mol/L, both ion concentrations equal \(S\).

Step 2: Key Formula or Approach:
\[ K_{sp} = S \times S = S^2 \]

Step 3: Detailed Explanation:
\[ S = \sqrt{6.4\times10^{-13}} = 8\times10^{-7}\ \text{mol dm}^{-3} \]
Solubility in grams per dm\(^3\) = \(8\times10^{-7} \times 190 = 1.52\times10^{-4}\) g dm\(^{-3}\).
Options (B), (C) and (D) do not come from this calculation.

Step 4: Final Answer:
Solubility is \(1.52\times10^{-4}\) g dm\(^{-3}\), option (A).

Final Answer:
Solubility is 1.52e-4 g per dm3. \[ \boxed{\text{(A) }1.52\times10^{-4}\ \text{g dm}^{-3}} \]