Calculate the solubility (in grams per 1.00×102ml of solution) of magnesium hydroxide in a solution buffered at ph = 12
how does this compare to the solubility of mg(oh)2 in pure water?

The Ksp of Mg(OH)2 in water is 1.8 x 10-11. This means that in pure water, Mg(OH)2 has a solubility of:
∛[(1.8 x 10-11) / 4] = 1.65 x 10-4 mol/L
which is equal to
1.65 x 10-4 mol x (58.32) / 10 x 100 mL =  9.62 x 10-4g / 1x102 mL

If the pH is 12, the hydroxide concentration in the solvent is
10^-(14-12) = 0.01 mol/L
The solubility is solve using the formula
1.8 x 10-11 = x (2(0.01 + x))^2
x = 4.5x10-8 mol/L
which is equal to
4.5x10-8 mol x (58.32) / 10 x 100 mL =  2.62 x 10-7g / 1x102 mL

Is there a picture or answer choices?