Question

How many moles of ${NH}_3$ must be added to $1$ litre of a $0.1M$ $Ag{NO}_3$ to reduce ${Ag}^{+}$ concentration to $2\times {10}^{-7}M$?
${\left[Ag{\left({NH}_3\right)}_2\right]}^{+}\rightleftharpoons {Ag}^{+}+2{NH}_3$
$K_{diss.}=\frac{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}{\left[Ag{\left({NH}_3\right)}_2^{+}\right]}=6.8\times {10}^{-8}$

Answer 1

Almost all the $A{g}^{+}$ is converted to $Ag\left[\right(N{H}_3{)}_2{]}^{+}$.

So $\left[Ag{\left({NH}_3\right)}_2^{+}\right]=0.1M$

$\left[{Ag}^{+}\right]=2\times {10}^{-7}M$
$K=\frac{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}{\left[Ag{\left({NH}_3\right)}_2^{+}\right]}$
$6.8\times {10}^{-8}=\frac{2\times {10}^{-7}{\left[{NH}_3\right]}^2}{0.1}$
$\left[{NH}_3\right]=0.184M$
It is the concentration of free ${NH}_3$
${\left[{NH}_3\right]}_{Total}={\left[{NH}_3\right]}_{Free}+{\left[{NH}_3\right]}_{Complexed}$
$=0.184+2\times 0.1=0.384M$