Question

How much $N{H}_3$ should be added to a 0.004 M $A{g}^{+}$ solution to prevent the precipitation of AgCl when $C{l}^{-}$ ion reaches 0.001? $K_{sp}AgCl=1.8\times {10}^{-10},K_{d}Ag(N{H}_3{)}_2^{+}=6\times {10}^{-8}$

• A. 0.0011 molar
• B. 0.36 molar
• C. 0.0036 molar
• D. 1.6 molar

Answer 1

The correct option is A 0.0011 molar

For AgCl,
$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$ or $1.8\times {10}^{-10}=\left[A{g}^{+}\right]\times 0.001$
Hence, $\left[A{g}^{+}\right]=0.00018$ M.
Thus when silver ion concentration exceeds 0.00018 M, precipitation will occur.
Thus to prevent precipitation, the silver ion concentration should be reduced by $0.004-0.00018=0.00382$ M.Hence the concentration of ammonia required for it in $[Ag(N{H}_3{)}_2{]}^{2+}$ is given by,

, $K_d=\frac{\left[A{g}^{+}\right][N{H}_3{]}^2}{\left[Ag\right(N{H}_3{)}_2{]}^{2+}}$

Substitute values in the above expression.
$6\times {10}^{-8}=\frac{0.00018[N{H}_3{]}^2}{0.00382}$
Hence, $\left[N{H}_3\right]=0.0011$ M.