Find the pH where a metal hydroxide starts to precipitate

Answer: C ($$10.36$$)

After dilution, the metal ion concentration is $$[X^{2+}]$$$${}=(0.0150\ \mathrm{M})$$$${}(20.0\ \mathrm{mL}/50.0\ \mathrm{mL})$$$${}=0.00600\ \mathrm{M}$$. Precipitation begins when the ion product equals $$K_{\mathrm{sp}}$$, so $$K_{\mathrm{sp}}$$$${}=[X^{2+}][\mathrm{OH}^{-}]^{2}$$. Thus, $$[\mathrm{OH}^{-}]$$$${}=\sqrt{(3.20\times10^{-10})/(0.00600)}$$$${}=2.31\times10^{-4}\ \mathrm{M}$$. The corresponding $$\mathrm{pOH}$$ is $$3.64$$, and $$\mathrm{pH}$$$${}=14.00-3.64$$$${}=10.36$$.