Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\begin{array}{ccccccccc}{\text{AgCl}}_{\left(\text{s}\right)}& +& \text{aq}& \rightleftharpoons & {\text{Ag}}_{\left(\text{aq}\right)}^{+}& +& {\text{Cl}}_{\text{aq}}^{-}& \text{;}& {\text{K}}_{\text{sp}}\\ {\text{Ag}}_{\text{aq}}^{+}& +& 2{\text{NH}}_{3_{\text{aq}}}& \rightleftharpoons & \left[\text{Ag}{\left({\text{NH}}_3\right)}_{2_{\text{aq}}}^{+}\right]& & & \text{;}& {\text{K}}_{\text{f}}\end{array}$

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$\begin{array}{ccccccccccc}{\text{AgCl}}_{\left(\text{s}\right)}& +& 2{\text{NH}}_3& \rightleftharpoons & \left[\text{Ag}{\left({\text{NH}}_3\right)}_{2_{\text{aq}}}^{+}\right]& +& {\text{Cl}}_{\text{aq}}^{-}& & \text{ K}={\text{K}}_{\text{sp}}& \times & {\text{K}}_{\text{f}}\end{array}$

$\left(\mathrm{C}-0.2\right)\mathrm{M} \; \; \; \; 0.1\mathrm{ }\mathrm{M} \; \; \; \; \; \; \; 0.1\mathrm{ }\mathrm{M}$Equilibrium concentration

$\text{K}=10^{-10}\times 10^8=10^{-2}=\frac{\left[\text{Ag}[{\text{NH}}_3{)}_2^{+}\right] \left[{\text{Cl}}^{-}\right]}{{\left[{\text{NH}}_3\right]}^2}$

$10^{-2}=\frac{10^{-1}\times 10^{-1}}{{\left(\text{M}-0\text{.}2\right)}^2}$

$\therefore \mathrm{M}-0.2 =1$

$\therefore \mathrm{M}=1.2\mathrm{M}$