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