Question

Number of moles of $AgCl$ that can be dissolved in 10 liter of 2.7 M of $N{H}_3$:
$\left(K_{sp}ofAgClandKof\right[Ag(N{H}_3{)}_2{]}^{+}are1.0\times {10}^{-10}{M}^{2}and1.6\times {10}^7{M}^{-2}respectively)$

Answer 1

$\begin{array}{ccccc}AgCl\left(s\right)& \stackrel{K_{SP}={10}^{-10}}{\rightleftharpoons }& A{g}_{aq}^{+}& +& C{l}_{aq}^{-}\\ & & s-x& & s\end{array}$
$\begin{array}{ccccc}A{g}_{aq}^{+}& +& 2N{H}_3& \stackrel{K_f=1.6\times {10}^7}{\rightleftharpoons }& \left[Ag\right(N{H}_3{)}_2{]}^{+}\\ s-x& & 2.7-2x& & x\end{array}$
So, $K_{sp}={10}^{-10}=(s-x)\times s$
Again $K=1.6\times {10}^7=\frac{x}{(s-x)\times (2.7-2x{)}^2}$
$\Rightarrow 1.6\times {10}^{-3}=\frac{sx}{(2.7-2x{)}^2}$
Since $K$ is very high, so $s\approx x$
$\Rightarrow \frac{x}{2.7-2x}=0.04\Rightarrow x=0.1mol/L$
So, in 10 L, moles of AgCl that can be dissolved = 1 mol