A solution contains 0.10 M MgCl2 and 0.70 M NH4Cl. Ammonia gas is passed in slowly at 298 K. Given Ksp[Mg(OH)2]=1.0×10^-11 and Kb(NH3)=1.0×10^-5, the…

Precipitation of $\mathrm{Mg(OH)_2}$ just begins when $[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2=K_{sp}$. With $[\mathrm{Mg^{2+}}]=0.10\ \mathrm{M}$ this gives $[\mathrm{OH^-}]^2=10^{-11}/0.10=10^{-10}$, so $[\mathrm{OH^-}]_{max}=1.0\times10^{-5}\ \mathrm{M}$ (the solution can be at most $\mathrm{pH}=9$). The ammonia buffer sets the hydroxide ion concentration through $K_b=\dfrac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}$. Because the ionisation of $\mathrm{NH_3}$ is negligible, $[\mathrm{NH_4^+}]\approx0.70\ \mathrm{M}$ from the salt. Rearranging, $[\mathrm{NH_3}]=\dfrac{[\mathrm{OH^-}][\mathrm{NH_4^+}]}{K_b}=\dfrac{10^{-5}\times0.70}{10^{-5}}=0.70\ \mathrm{M}$. In units of $0.10\ \mathrm{M}$ this is $7$. Any more ammonia would push $[\mathrm{OH^-}]$ above $10^{-5}$ and precipitate magnesium hydroxide. Notice that the large ammonium ion concentration from the salt suppresses the ionisation of ammonia, which is why the common-ion buffer lets a fairly concentrated ammonia solution coexist with magnesium ions. The same logic is used in qualitative analysis, where ammonium chloride is added before ammonia so that group III hydroxides precipitate but magnesium stays in solution.