The scenario is this. I have $50$ mL of $0.1$ M $\ce{NH4^+}$ at a certain temperature which gives it a $K_a=5.2\times 10^{-8}$. To this solution, I add $0.02$ moles of $\ce{Cd(NO_3)_2}$. It is known that cadmium ions undergo the following reaction: $\ce{Cd^2+ + 4NH3 <=> [Cd(NH_3)_4]^2+}$ which ha...