In group III the analyst adds ammonium hydroxide in excess to a solution that holds magnesium ions…

Step 1. Ammonia alone. $[\text{OH}^-] = \sqrt{1.8 \times 10^{-5} \times 0.10} = 1.34 \times 10^{-3}\ \text{mol/L}$. $\text{pOH} = 2.87$, so $\text{pH} = 11.13$.

Step 2. With ammonium chloride. $[\text{OH}^-] = 1.8 \times 10^{-5} \times \dfrac{0.10}{0.50} = 3.6 \times 10^{-6}\ \text{mol/L}$. $\text{pOH} = 5.44$, so $\text{pH} = 8.56$.

Step 3. Ionic product, first case. $Q = 0.0050 \times (1.34 \times 10^{-3})^2 = 9.0 \times 10^{-9}$. $\dfrac{Q}{K_{sp}} = \dfrac{9.0 \times 10^{-9}}{5.6 \times 10^{-12}} = 1607$. The hydroxide precipitates.

Step 4. Second case. $Q = 0.0050 \times (3.6 \times 10^{-6})^2 = 6.5 \times 10^{-14}$. $\dfrac{Q}{K_{sp}} = 0.0116$. Nothing precipitates.