A solution contains 0.010 M Mg^2+ and 1.5 M NH4Cl. Aqueous ammonia is added gradually. Given Ksp[Mg(OH)2]=4.0×10^-12 and Kb(NH3)=1.0×10^-5, find the…

Precipitation of $\mathrm{Mg(OH)_2}$ begins when the ionic product equals $K_{sp}$: $[\mathrm{OH^-}]^2=\dfrac{K_{sp}}{[\mathrm{Mg^{2+}}]}=\dfrac{4.0\times10^{-12}}{0.010}=4.0\times10^{-10}$, so $[\mathrm{OH^-}]=2.0\times10^{-5}\ \mathrm{M}$. The hydroxide concentration is fixed by the $\mathrm{NH_3/NH_4^+}$ buffer through $K_b=\dfrac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}$. Therefore $[\mathrm{NH_3}]=\dfrac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{K_b}=\dfrac{1.5\times2.0\times10^{-5}}{1.0\times10^{-5}}=3.0\ \mathrm{M}$. Any lower ammonia concentration keeps the ionic product below $K_{sp}$ and no precipitate forms; this is why ammonium chloride is added to prevent precipitation of the magnesium group in qualitative analysis. The answer is $3$. Note that using $K_b$ with the correct buffer ratio is essential: using the ammonium concentration alone would not give the required free ammonia.