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

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water