K_c for the reaction: ${\left[\mathrm{Ag}(\mathrm{CN}{)}_2\right]}^{-}\rightleftharpoons {\mathrm{Ag}}^{+}+2{\mathrm{CN}}^{-}$ the equilibrium constant at 25°C is 4 .0 x 10^-19, then the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar in AgNO₃ is:

$\underset{\left(0. 1-0.06\right)}{\underset{0.1}{2\mathrm{KCN}}}+\underset{0}{\underset{0.03}{{\mathrm{AgNO}}_3}}⟶\underset{0.03}{\underset{0}{\mathrm{Ag}(\mathrm{CN}{)}_2^{-}}}+\underset{0.03}{\underset{0}{{\mathrm{KNO}}_3}}+\underset{0.03}{\underset{0}{{\mathrm{K}}^{+}}}$
$\therefore \left[\mathrm{Ag}(\mathrm{CN}{)}_2^{-}\right]=0.03$
Now, $\underset{\left(0..03-\mathrm{a}\right)}{\underset{0.03}{\mathrm{Ag}(\mathrm{CN}{)}_2^{-}}}\rightleftharpoons \underset{\mathrm{a}}{\underset{0}{{\mathrm{Ag}}^{+}}}+\underset{\left(0.04+\mathrm{a}\right)}{\underset{0.04}{2{\mathrm{CN}}^{-}}}\underset{ }{\underset{(\mathrm{left} \mathrm{from} \mathrm{KCN})}{ }}$
K_c is small $\therefore$0.03- a = 0.03and 0.04 + a =0. 04
$\therefore {\mathrm{K}}_{\mathrm{c}}=4\times {10}^{-19}=\frac{(0.04{)}^2\times \mathrm{a}}{0.03};\therefore \mathrm{a}=7.5\times {10}^{-18}$