A solution contains 0.100 M NH3 and 0.010 M Mg^2+. Given Kb(NH3)=1.8×10^-5 and Ksp[Mg(OH)2]=1.2×10^-11, what mass of solid NH4Cl (molar mass 53.5 g…

Precipitation of $\mathrm{Mg(OH)_2}$ is just avoided when $[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2=K_{sp}$. So $[\mathrm{OH^-}]_{max}=\sqrt{1.2\times10^{-11}/0.010}=3.46\times10^{-5}\ \mathrm{M}$. The ammonia equilibrium $\mathrm{NH_3+H_2O\rightleftharpoons NH_4^++OH^-}$ gives $[\mathrm{OH^-}]=K_b\,[\mathrm{NH_3}]/[\mathrm{NH_4^+}]$, so the ammonium concentration needed is $[\mathrm{NH_4^+}]=\dfrac{1.8\times10^{-5}\times0.100}{3.46\times10^{-5}}=0.0520\ \mathrm{M}$. In $0.500\ \mathrm{L}$ this is $0.0260$ mol, and the mass is $0.0260\times53.5=1.39\ \mathrm{g}$. Forgetting that the volume is $500\ \mathrm{mL}$ gives $2.78\ \mathrm{g}$, and $0.70\ \mathrm{g}$ results from halving the ammonium concentration. The common-ion effect of $\mathrm{NH_4^+}$ suppresses ionisation of the base and lowers $[\mathrm{OH^-}]$ below the limit, which is why ammonium salts keep magnesium ions in solution in the presence of aqueous ammonia, a fact used in qualitative analysis to separate the groups of cations.