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: Understanding the Concept
A binary salt BA dissociates as \(\text{BA} \rightleftharpoons \text{B}^+ + \text{A}^-\). If the solubility is \(S\) mol/L, then \([\text{B}^+] = [\text{A}^-] = S\) and \(K_{sp} = S^2\).

Step 2: Detailed Explanation
100 mL of saturated solution contains \(10^{-4}\) mol. For one litre:
\[ S = \frac{10^{-4}}{0.1} = 10^{-3} \text{ mol/L} \]
\[ K_{sp} = S^2 = (10^{-3})^2 = 10^{-6} \]
Option (A) forgets to convert the volume to one litre; (C) and (D) come from using the wrong power.

Final Answer:
\(K_{sp} = 1 \times 10^{-6}\), option (B). \[ \boxed{1 \times 10^{-6}} \]