Question

The solubility product of $AgCl$ at ${25}^{\circ }C$ is $1\times {10}^{-10}$. A solution of $A{g}^{+}$ at a concentration of $4\times {10}^{-3}M$ just fails to yield a precipitate of $AgCl$ with a concentration of $1\times {10}^{-3}M$ of $C{l}^{-}$ when the concentration of $N{H}_3$ in the solution is $2\times {10}^{-2}M$. Calculate the equilibrium constant for
$\left[Ag\right(N{H}_3{)}_2{]}^{+}\rightleftharpoons A{g}^{+}+2N{H}_3$

• A. ${10}^{-8}$
• B. ${10}^{-10}$
• C. ${10}^{-12}$
• D. ${10}^{-2}$

Answer 1

The correct option is A ${10}^{-8}$

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]{10}^{-10}=\left[A{g}^{+}\right]\times {10}^{-3}\left[A{g}^{+}\right]={10}^{-7}M$

$\left[Ag\right(N{H}_3{)}_2{]}^{+}\rightleftharpoons A{g}^{+}+2N{H}_{3}Initially:4\times {10}^{-3}{10}^{-7}2\times {10}^{-2}Equilibrium:4\times {10}^{-3}-xx+{10}^{-7}2x+2\times {10}^{-2}$
Since,the above forward reaction is the dissociation of a complex, it will hardly occur
$4\times {10}^{-3}-x\approx 4\times {10}^{-3}x+{10}^{-7}\approx {10}^{-7}2x+2\times {10}^{-2}\approx 2\times {10}^{-2}$

Thus the equilibrium constant for the above dissociation equilibria of complex is given as:
$K=\frac{\left[A{g}^{+}\right][N{H}_3{]}^2}{\left[Ag\right(N{H}_3{)}_2{]}^{+}]}K=\frac{{10}^{-7}\times (2\times {10}^{-2}{)}^2}{4\times {10}^{-3}}K={10}^{-8}$

Since K has a very low value, our assumption is verified that forward reaction will hardly occur.