Ionic EquilibriumHard
Question
What concentration of free CN– must be maintained in a solution that is 1.8 M-AgNO3 and 0.16 M-NaCl to prevent AgCl from precipitating? The value of Kf for Ag(CN)2− = 6.4 × 1017 and Ksp for AgCl = 1.8 × 10–10.
Options
A.2.5 × 10−9 M
B.5 × 10–5 M
C.2.5 × 10−5 M
D.1 × 10−4 M
Solution
To prevent precipitation of AgCl, the concentration of Ag+ needed in solution
$= \frac{K_{sp}(AgCl)}{\left( Cl^{-} \right)} = \frac{1.8 \times 10^{- 10}}{0.16} < < 1.8\text{ M}$
Hence, almost all Ag+ ion must form complete with CN– ions.
$Ag^{+} + 2CN^{-} \rightleftharpoons Ag(CN)_{2}^{-};K_{f} = 6.4 \times 10^{17} $$$\frac{1.8 \times 10^{- 10}}{0.16}M\text{CM1.8 M}$$
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