100 mL of 0.20 M NH3 is mixed with 100 mL of 0.10 M HCl to form a buffer. Take Kb(NH3)=1.8×10^-5 and Ksp[Mg(OH)2]=1.0×10^-11. Neglecting any volume…

After mixing, $\mathrm{HCl}$ is limiting: $\mathrm{NH_3}$ initially $20$ mmol, $\mathrm{HCl}$ $10$ mmol, so $10$ mmol $\mathrm{NH_3}$ remain and $10$ mmol $\mathrm{NH_4^+}$ form, in $200\ \mathrm{mL}$: each is $0.050\ M$. For the basic buffer $[\mathrm{OH^-}]=K_b\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]}=1.8\times10^{-5}\times1=1.8\times10^{-5}\ M$. Precipitation of $\mathrm{Mg(OH)_2}$ begins when $[\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2=K_{sp}$, so $[\mathrm{Mg^{2+}}]_{max}=\frac{1.0\times10^{-11}}{(1.8\times10^{-5})^2}=\frac{10^{-11}}{3.24\times10^{-10}}=3.086\times10^{-2}\ M$. Hence $x\approx3.09$. The key idea is that the buffer fixes $[\mathrm{OH^-}]$ through the ratio $[\mathrm{NH_3}]/[\mathrm{NH_4^+}]=1$, independent of dilution, so the common-ion effect of $\mathrm{NH_4^+}$ on the ionisation of ammonia sets the hydroxide ion concentration. Using the ratio of moles directly is valid since both species share one volume.