Ionic EquilibriumHard

Question

The solubility of AgCN in a buffer solution of pH = 3.0 is (Ksp of AgCN = 1.2 × 10−16; Ka of HCN = 4.8 × 10−10)

Options

A.1.58 × 10−5 M
B.2.0 × 10−5 M
C.1.58 × 10−4 M
D.2.5 × 10−9 M

Solution

AgCN (s) $\rightleftharpoons$ Ag+ + CN–; Ksp = 1.2 × 10–16

$CN^{-} + H^{+} \rightleftharpoons HCN;K_{eq} = \frac{1}{K_{a}} = \frac{1}{4.8 \times 10^{- 10}}$

Adding both reactions,

AgCN (s) + H+ $\rightleftharpoons$ Ag+ + HCN; $K_{eq} = \frac{K_{sp}}{K_{a}} = \frac{1}{4 \times 10^{6}}$

10–3 SM SM

Now, $\frac{1}{4 \times 10^{6}} = \frac{S.S}{10^{- 3}} \Rightarrow 1.58 \times 10^{- 5}\text{ M}$

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