A student is given an aqueous mixture containing Fe^ 3+ and Mg^ 2+...

Ammonium hydroxide ($\mathrm{NH_4OH}$) is a weak base, and its ionization is suppressed by the addition of the strong electrolyte ammonium chloride ($\mathrm{NH_4Cl}$) due to the common ion effect of $\mathrm{NH_4^+}$ ions.

$\mathrm{NH_4OH} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH}^{-}$
$\mathrm{NH_4Cl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl}^{-}$

This suppression keeps the concentration of $\mathrm{OH}^{-}$ ions very low in the solution. The solubility product ($K_{\mathrm{sp}}$)$of \mathrm{Fe(OH)_3}$ is much lower (around $10^{-38}$) compared to that of $\mathrm{Mg(OH)_2}$ (around $10^{-11}$). The low $[\mathrm{OH}^{-}]$ is sufficient to make the ionic product of $\mathrm{Fe(OH)_3}$ exceed its $K_{\mathrm{sp}}$, leading to its precipitation. However, the ionic product of $\mathrm{Mg(OH)_2}$ does not exceed its relatively higher $K_{\mathrm{sp}}$, so $\mathrm{Mg}^{2+}$ remains dissolved in the solution.

Answer: the common ion effect of $\mathrm{NH_4^+}$ keeps $[\mathrm{OH}^{-}]$ low, which fails to exceed the higher $K_{\mathrm{sp}}$ of $\mathrm{Mg(OH)_2}$