Question

Calculate the solubility of AgCl in 0.2 M - $N{H}_3$ solution.
Given : Ksp of AgCl = 2 x 10, Kf of $Ag\left(N{H}_3\right)2^{+}$ = 8 x ${10}^6$.

Answer 1

$AgCl⟶A{g}^{+}+C{l}^{-}$ $\Rightarrow K_{sp}=2\times {10}^{-10}$

$Ag+2N{H}_3⟶Ag\left(N{H}_3\right)2$ $\Rightarrow K_f=8\times {10}^6$

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$AgCl+2N{H}_3⟶Ag(N{H}_3{)}_2+C{l}^{-}$ net equilibrium constant= $2\times 8\times {10}^{-4}=0.00016$

$K_f=\frac{\left[Ag\right(N{H}_3{)}_2]}{\left[Ag\right][N{H}_3{]}^2}\Rightarrow 0.00016=\frac{x^2}{0.1-2x}$

$\Rightarrow x^2=0.000016-0.00032x$

$\Rightarrow x^2+0.00032x-0.000016=0$ .