Question

Determine the number of moles of $AgI$ which may be dissolved in $1$ litre of $1M$ of ${CN}^{-}$ solution $K_{sp}$ for $AgI$ and $K_c$ for $Ag{\left(CN\right)}_2^{-}$ $1.2\times {10}^{-17}{M}^2$ and $7.1\times {10}^{-19}{M}^{-2}$ respectively.

Answer 1

$AgI\left(s\right)\rightleftharpoons {Ag}^{+}+I^{-}$
$K_{sp}=\left[{Ag}^{+}\right]\left[I^{-}\right]=1.2\times {10}^{-17}$
$K_f=\frac{\left[Ag{\left(CN\right)}_2^{-}\right]}{\left[{Ag}^{+}\right]{\left[{CN}^{-}\right]}^2}=7.1\times {10}^{-19}$
$K_{eq}=K_{sp}\times K_f=\frac{\left[Ag{\left(CN\right)}_2^{-}\right]\left[I^{-}\right]}{{\left[{CN}^{-}\right]}^2}=8.52\times {10}^2$
Let us consider the following equilibrium
$AgI\left(s\right)+2{CN}^{-}\rightleftharpoons \left[Ag{\left(CN\right)}_2^{-}\right]+I^{-}$
$t=0$ $1$ $0$ $0$
$t_{eq}$ $1-2x$ $x$ $x$
Let ${}^{\prime }{x}^{\prime }$ moles of $AgI$ be dissolved in ${CN}^{-}$
Then,
$AgI\left(s\right)+2{CN}^{-}\rightleftharpoons \left[Ag{\left(CN\right)}_2^{-}\right]+I^{-}$
$K_{eq}=8.52\times {10}^2=\frac{x\times x}{{(1-2x)}^2}$
$x=0.49mole$