Question

How much AgBr could dissolve in $1.0Lof0.4MN{H}_3$ ? Assume that ${\left[Ag\right(N{H}_3{)}_2]}^{+}$ is the only complex formed given , $K_f{\left[Ag\right(N{H}_3{)}_2]}^{+}=1.0\times {10}^8,K_{sp}\left(AgBr\right)=5.0\times {10}^{-13}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$AgBr\rightleftharpoons A{g}^{+}+B{r}^{-}$ ---------------(1)
$A{g}^{+}+2N{H}_3\rightleftharpoons {\left[Ag\right(N{H}_3{)}_2]}^{+}$ ---------------(2)
Let x = solubility
Then $x=\left[B{r}^{-}\right]=\left[A{g}^{+}\right]$ initially, however due to very high $K_f$ and presence of sufficient $N{H}_3$ , almost all of it will convert to ${\left[Ag\right(N{H}_3{)}_2]}^{+}$ causing more AgBr to dissolve.
$\therefore \frac{{\left[Ag\right(N{H}_3{)}_2]}^{+}}{\left[A{g}^{+}\right]{\left[N{H}_3\right]}^2}=1\times {10}^8$
$\frac{X}{{\left[A{g}^{+}\right[0.4-2X]}^2}=1.0\times {10}^8;$
Also $Br=x\Rightarrow A{g}^{+}\frac{ksp}{x}$
$\frac{x^2}{5\times {10}^{-13}[0.4-2x{]}^2}=1\times {10}^8$
Solving which
$x=2.8\times {10}^{-3}mols/lit$