Question

At temperature \(T(K)\), the solubility product of AgBr is \(4 \times 10^{-13}\). What is its solubility in 0.1 M KBr solution?

• A. \(2 \times 10^{-6} M\)
• B. \(4 \times 10^{-10} M\)
• C. \(4 \times 10^{-12} M\)
• D. \(4 \times 10^{-14} M\)

Hint:

Common ion effect reduces solubility; use adjusted ion concentrations.

Answer 1

The Correct Option is C

Step 1: Write solubility equilibrium
\[ AgBr \leftrightarrow Ag^+ + Br^- \] Let solubility be \(S\). In presence of 0.1 M \(KBr\), common ion effect applies: \[ K_{sp} = [Ag^+][Br^-] = S \times (0.1 + S) \approx 0.1 S \] Step 2: Calculate solubility
\[ S = \frac{K_{sp}}{0.1} = \frac{4 \times 10^{-13}}{0.1} = 4 \times 10^{-12} M \] Step 3: Conclusion
Solubility is \(4 \times 10^{-12} M\).