Question

Given $Ag{\left({NH}_3\right)}_2^{+}\rightleftharpoons {Ag}^{+}+2{NH}_3$, $K_c=6.2\times {10}^{-8}$ and $K_{sp}$ of $AgCl$ is $1.8\times {10}^{-10}$ at $298K$. Calculate concentration of the complex in $1M$ aqueous ammonia.

Answer 1

$AgCl\rightleftharpoons {Ag}^{+}+{Cl}^{-}$
On adding ammonia solution, complex formation takes place
${Ag}^{+}+2{NH}_3\rightleftharpoons {\left[Ag{\left({NH}_3\right)}_2\right]}^{+}...\left(i\right)$
Where, $x=$ solubility of $AgCl$ in ${NH}_3$
$y=$ amount of complex formed
$K_{sp}$ of $AgCl=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]$
$1.8\times {10}^{-10}=(x-y)\times x...\left(ii\right)$
$K_c$ for equation (i) $=\frac{{\left[Ag{\left({NH}_3\right)}_2\right]}^{+}}{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}$
$\frac{1}{6.2\times {10}^{-8}}=\frac{y}{(x-y)1}....\left(iii\right)$
On solving euations (ii) and (iii) we get
$y=0.0539M$