Question

How much AgBr could dissolve in $1.0$L of $0.40$ M $N{H}_3$? Assume that $Ag(N{H}_3{)}_2^{+}$ is the only complex formed. $\left[K_f\right(Ag\left(N{H}_3{)}_2^{+}\right)=1\times {10}^8$; $K_{sp}\left(AgBr\right)=5\times {10}^{-13}]$
$[\sqrt{50}=7]$.

Answer 1

${Ag}^{+}+{2NH}_3\rightleftharpoons {\left[Ag{\left({NH}_3\right)}_2\right]}^{2+}$

$K_f=\frac{C_{{\left[Ag{\left({NH}_3\right)}_2\right]}^{+}}}{C_{A_5^{+}}.C_{{NH}_3}^2}$

now, $C_{{NH}_3}=0.4M$ $\therefore$ $\frac{C_{{\left[Ag{\left({NH}_3\right)}_2\right]}^{+}}}{C_{{Ag}^{+}}}=1\times {10}^8\times 0.16⟶\left(1\right)$

Let the solubility of $AgBr$ in $0.4M$ be $x$

$\therefore$ $C_{{\left[Ag{\left({NH}_3\right)}_2\right]}^{+}}\approx x$

now, $K_{sp}=\left[{Ag}^{+}\right]\left[{Br}^{-}\right]=\left[{Ag}^{+}\right]\times x$

$\therefore$ $\left[{Ag}^{+}\right]=\frac{5\times {10}^{-13}}{x}⟶\left(2\right)$

$\therefore$ From equation $\left(1\right)$ and $\left(2\right)$ we get

$\frac{5\times {10}^{-13}}{x}=\frac{x}{1\times {10}^8\times 0.16}\Rightarrow x^2=5\times 0.16\times {10}^{-5}$

$\therefore$ $x=2.83\times {10}^{-3}$

$\therefore$ Solubility of $AgBr$ will $2.83\times {10}^{-3}M$