Question

How much ${NH}_3$ must be added to $0.004M$ ${Ag}^{+}$ solution to prevent the precipitation of $AgCl$ when (${Cl}^{-}$) reaches $0;001M$? $K_{sp}$ for $AgCl$ is $1.8\times {10}^{-10}$ and $K$ for $Ag{\left({NH}_3\right)}_2^{+}$ is $5.9\times {10}^{-8}$.

Answer 1

$\left[{Ag}^{+}\right]=\frac{K_{sp}}{\left[{Cl}^{-}\right]}=\frac{1.8\times {10}^{-10}}{0.001}=1.8\times {10}^{-7}M$

${NH}_3$ is added to keep the conc. of ${Ag}^{+}$ below $1.8\times {10}^{-7}M$ to prevent precipitation

$\left[Ag{\left({NH}_3\right)}_2^{+}\right]$ at above limiting condition $=0.004-1.8\times {10}^{-7}=0.004M$

${\left[Ag{\left({NH}_3\right)}_2\right]}^{+}\rightleftharpoons {Ag}^{+}+2{NH}_3$

$K_d=\frac{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}{{\left[Ag{\left({NH}_3\right)}_2\right]}^{+}}$

$5.9\times {10}^{-8}=\frac{[1.8\times {10}^{-7}]{\left[{NH}_3\right]}^2}{{\left[0.004\right]}^{}}$

$\left[{NH}_3\right]=0.036M$

${\left[{NH}_3\right]}_{Total}+{\left[{NH}_3\right]}_{Free}+{\left[{NH}_3\right]}_{Complexed}=0.036+2\times 0.004=0.44mol/litre$