Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}\left(\mathrm{s}\right)+\underset{0-2\mathrm{x}}{2\mathrm{N}{\mathrm{H}}_3}\left(\mathrm{a}\mathrm{q}\right)\rightleftharpoons \underset{\mathrm{x}}{\mathrm{A}\mathrm{g}(\mathrm{N}{\mathrm{H}}_3{)}_2^{+}}+\underset{\mathrm{x}}{\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{a}\mathrm{q}\right)}$
${\mathrm{K}}_1=1.6\times 10^{-10}\times 10^7=1.6\times 10^{-3}$
${\mathrm{K}}_1=\frac{\left[\mathrm{A}\mathrm{g}\right(\mathrm{N}{\mathrm{H}}_3{)}_2^{+}\left[\mathrm{C}{\mathrm{l}}^{-}\right]}{[\mathrm{N}{\mathrm{H}}_3{]}^2}$
$\frac{\left[\mathrm{C}{\mathrm{l}}^{-}\right]}{\left[\mathrm{C}{\mathrm{N}}^{-}\right]}=\frac{1.6\times 10^{-3}}{2.5\times 10^{-9}}=6.4\times 10^5$
${\mathrm{K}}_1=\frac{{\mathrm{x}}^2}{(1-2\mathrm{x}{)}^2}=1.6\times 10^{-3}$
$\frac{\mathrm{x}}{(1-2\mathrm{x})}=0.04$
$\mathrm{x}=0.037$
$\left[\mathrm{C}{\mathrm{N}}^{-}\right]=\frac{0.037}{6.4\times 10^5}=5.78\times 10^{-8}$