Chapter 18, Problem 56E

A solution is 0.10 M in free ${\text{NH}}_{\text{2}}$ ,0.10M in ${\text{NH}}_{\text{4}}\text{Cl}$ , and 0.015 M in ${\left[\text{Cu}{\left({\text{NH}}_{\text{2}}\right)}_{\text{4}}\right]}^{\text{2+}}$ precipitate from this solution? $K_{ap}$ of $\text{Cu}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{2}\text{.2}\times {\text{10}}^{-20}$ .