Question

Determine solubility of \(\mathrm{AgBr}\) in a \(0.1\mathrm{\ M\ KCN}\) solution. \([K_f \text{ of } [\mathrm{Ag(CN)_2}]^- = 5.6 \times 10^8; K_{sp} \text{ of } \mathrm{AgBr} = 7.7 \times 10^{-13}]\)

• A. \(2 \times 10^{-3}\mathrm{\ M}\)
• B. \(4 \times 10^{-6}\mathrm{\ M}\)
• C. \(2 \times 10^{-6}\mathrm{\ M}\)
• D. \(4 \times 10^{-3}\mathrm{\ M}\)

Hint:

Overall equilibrium constant \(K = K_{sp} \times K_f\).

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \mathrm{AgBr}(s) + 2\mathrm{CN}^- \rightleftharpoons [\mathrm{Ag(CN)_2}]^- + \mathrm{Br}^- \]
Step 2: Calculation / Simplification}
\(K_{eq} = K_{sp} \times K_f = 7.7 \times 10^{-13} \times 5.6 \times 10^8 = 4.312 \times 10^{-4}\)
Let solubility \(= s\): \([\mathrm{CN}^-] = 0.1 - 2s \approx 0.1\)
\(K_{eq} = \frac{s \cdot s}{(0.1)^2} = 100s^2\)
\(4.312 \times 10^{-4} = 100s^2 \Rightarrow s^2 = 4.312 \times 10^{-6} \Rightarrow s = 2.08 \times 10^{-3} \approx 2 \times 10^{-3}\mathrm{\ M}\)
Step 3: Final Answer
\[ 2 \times 10^{-3}\mathrm{\ M} \]