Question

What is the solubility product of binary sparingly soluble salt BA if 100 mL saturated solution of salt consist \(10^{-4}\) moles at room temperature?

• A. \(1\times 10^{-4}\)
• B. \(1\times 10^{-6}\)
• C. \(1\times 10^{-8}\)
• D. \(1\times 10^{-10}\)

Hint:

Find solubility in mol/L from 1e-4 mol in 100 mL, then Ksp = S squared.

Answer 1

The Correct Option is B

Step 1: Concentration
Dissolved salt per litre: $10^{-4} \times 10 = 10^{-3}$ mol, so $S = 10^{-3}$ M.

Step 2: Ion product
Each formula unit gives one B+ and one A-, so $K_{sp} = [\text{B}^+][\text{A}^-] = 10^{-3} \times 10^{-3}$.

Step 3: Result
$K_{sp} = 10^{-6}$. Option (B).

Final Answer:
Option (B). \[ \boxed{1 \times 10^{-6}} \]