Question

How much $AgBr$ could dissolve in $1.0L$ of $0.4MN{H}_3$? Assume that $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ is the only complex formed given, $K_f\left[Ag\right(N{H}_3{)}_2^{+}]=1.0\times {10}^8,Ksp(AgBr)=5.0\times {10}^{-13}$.

Answer 1

$AgBr\rightleftharpoons A{G}^{+}+B{r}^{-}$......(1)

$a{g}^{+}2N{H}_3\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}$......(2)

Let $x=$solubility.

from (2)

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

$\Rightarrow 1\times {10}^8=\frac{\left[Ag\right(N{H}_3{)}_2{]}^{+}}{\left(Ag{)}^{+}\right(0.4{)}^2}$

$\left(A{g}^{+}\right)=\frac{\left[Ag\right(N{H}_3{)}_2{]}^{+}}{1.6\times {10}^7}$

since, the majority of $Ag$ is in the form of complex

$x=\left[Br\right]=\left[Ag\right(N{H}_3{)}_2^{+}]$

$AgBr\rightleftharpoons \left[A{g}^{+}\right]+[-B{r}^{-}]$

$1$ $0$ $0$

$1-x$ $x$ $x$

$ksp=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$

$5\times {10}^{-13}=\frac{\left[Ag\right(N{H}_3{)}_2{]}^{+}\left[B{r}^{-}\right]}{1.6\times {10}^7}$

$5\times {10}^{-13}=\frac{x^2}{1.6\times {10}^7}$

$x^2=8.0\times {10}^{-6}$

$x=2.8\times {10}^{-3}$