A solution is 0.010 M in Mg^2+ and contains 0.10 M NH3. Solid NH4Cl is added (no volume change) so that Mg(OH)2 just fails to precipitate. Given…

Precipitation is just avoided when the ionic product equals $K_{sp}$: $[\mathrm{OH^-}]^2=K_{sp}/[\mathrm{Mg^{2+}}]=1.8\times10^{-11}/0.010=1.8\times10^{-9}$, so $[\mathrm{OH^-}]=4.243\times10^{-5}$ M. The buffer must hold this hydroxide concentration: $K_b=[\mathrm{NH_4^+}][\mathrm{OH^-}]/[\mathrm{NH_3}]$, giving $[\mathrm{NH_4^+}]=K_b[\mathrm{NH_3}]/[\mathrm{OH^-}]=(1.8\times10^{-5})(0.10)/(4.243\times10^{-5})=0.04243$ M. Hence $x=4.24$. Any larger ammonium concentration lowers $[\mathrm{OH^-}]$ further and keeps the hydroxide in solution. We neglected the small ammonium contribution of ionised ammonia itself because the buffer swamps it. The answer is expressed in units of $10^{-2}$ M, so 0.04243 M corresponds to 4.24. A larger ammonium concentration is the safe direction, so this is a minimum requirement.