Question

Determine the silver ion concentration in a $0.2M$ solution of $\left[Ag{\left({NH}_3\right)}_2\right]{NO}_3$.
$Ag{\left({NH}_3\right)}_2^{+}\rightleftharpoons {Ag}^{+}+2{NH}_3$
$K_{diss.}=\frac{\left[{Ag}^{+}\right]{\left[{NH}_3\right]}^2}{\left[Ag{\left({NH}_3\right)}_2^{+}\right]}=6.8\times {10}^{-8}$

Answer 1

Let the concentration of ${Ag}^{+}$ at equilibrium be '$C$' M

$\left[{NH}_3\right]=2CM$

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

$6.8\times {10}^{-8}=\frac{C\times 4C^2}{0.2}$

$C=0.0015M$