Question

What is the solubility of $AgCl$ in $0.20M$ ${NH}_3$?

Given: $K_{sp}\left(AgCl\right)=1.7\times {10}^{-10}$ $M^2$, $K_1=\left[Ag{\left({NH}_3\right)}^{+}\right]/\left[{Ag}^{+}\right]\left[{NH}_3\right]=2.33\times {10}^3{M}^{-1}$ and $K_2=\left[Ag{\left({NH}_3\right)}_2^{+}\right]/\left[Ag{\left({NH}_3\right)}^{+}\right]\left[{NH}_3\right]=7.14\times {10}^3{M}^{-1}$

• A. $9.6\times {10}^{-3}$
• B. $19.2\times {10}^{-3}$
• C. $15.6\times {10}^{-4}$
• D. $31\times {10}^{-4}$

Answer 1

The correct option is A $9.6\times {10}^{-3}$

$AgC{l}_{\left(s\right)}\rightleftharpoons {Ag}^{+}+{Cl}^{-}{K}_{sp}$
${Ag}^{+}+2{NH}_3\rightleftharpoons {Ag{\left({NH}_3\right)}_2}^{+}{K}_1\times K_2$
$AgC{l}_{\left(s\right)}+2{NH}_3\rightleftharpoons {Ag{\left({NH}_3\right)}_2}^{+}+{Cl}^{-}$
$0.2$
$0.2-2x$ $x$ $x$
$K=\frac{x^2}{(0.2-2x)}=K_{sp}{K}_1{K}_2=0.002828$
$\frac{x}{0.2-2x}=0.05318$
$x=0.009613$
Solubility $=9.6\times {10}^{-3}M$