Chemistry - Ionic Equilibrium Question with Solution | TestHub

Due to higher value of K_f, mostly Ag⁺converted into complex.

$\text{AgI}\left(\text{s}\right)\rightleftharpoons {Ag}^{+}\left(\text{aq.}\right)+{\text{I}}^{-}\left(\text{aq.}\right){;\; K}_{\text{sp}}$

${\text{Ag}}^{+}\left(\text{aq.}\right)+2{\text{NH}}_3\left(\text{aq.}\right)\rightleftharpoons \text{Ag}{\left({\text{NH}}_3\right)}_2^{+}\left(\text{aq.}\right);K_{\text{f}}$
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overal reaction is

$\text{AgI}\left(\text{s}\right)+2{\text{NH}}_3\left(\text{aq.}\right)\rightleftharpoons Ag{\left({\text{NH}}_3\right)}_2^{+}\left(\text{aq.}\right)+{\text{I}}^{-}\left(\text{aq.}\right)$

${\text{K}}_{\text{eq}}={\text{K}}_{\text{sp}}·{\text{K}}_{\text{f}}$

${\text{K}}_{\text{sp}}·{\text{K}}_{\text{f}}=\frac{{\text{x}}^2}{1^2}$

$\therefore x=\text{4.9}\times 10^{-5}mol/litre$

So, mass of AgI required is

$=\text{4.9}\times 10^{-5}\times 235=\text{0.0115 g/L}$

$\mathrm{I}\mathrm{n}\mathrm{ }\mathrm{0.1 }\mathrm{L},\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{A}\mathrm{g}\mathrm{I}\mathrm{ }\mathrm{ }\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{w}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{b}\mathrm{e}=\frac{0.0115}{10}\mathrm{ }\mathrm{g}=1.15\times 10{-\mathrm{3 }}^{}\mathrm{g}\mathrm{ }\mathrm{ }$