What is the solubility-product constant of magnesium hydroxide, Mg(OH)2? Its solubility is 9.0 x 10–4 g/100 g

Chemistry help Please!

I am assuming that the solubility of Mg(OH)2 is 9.0 x 10^-4 g of Mg(OH)2 to 100 g water.

Since the molar mass of Mg(OH)2 is 24.31 + 2(16 + 1) = 58.31 g/mol, we see that we have:
(9.0 x 10^-4 g Mg(OH)2) * (1 mol Mg(OH)2)/(58.31 g Mg(OH)2)
= 1.54 x 10^-5 mol Mg(OH)2.

Then, we see that Mg(OH)2 dissolves as follows:
Mg(OH)2 <=> Mg2+ + 2OH-.

Its K(sp) expression is:
K(sp) = [Mg2+][OH-]^2.

Since [Mg2+] = (1.54 x 10^-5)/0.100 = 1.54 x 10^-4 and [OH-] = 2(1.54 x 10^-4) = 3.08 x 10^-4:
K(sp) = [Mg2+][OH-]^2
= (1.54 x 10^-4)(3.08 x 10^-4)^2
= 1.5 x 10^-11.

I hope this helps!

Ksp = [activity(Mg2+) * activity(OH-)^2]/[activity (Mg(OH)2)]

Mg(OH)2 is a pure solid and as such the activity of this is 1
so:
Ksp = [activity(Mg2+) * activity(OH-)^2]/[activity (Mg(OH)2)]

when we have a substance with low solubilit the activity of the ions can be approximated as the concentration/solubility of the ions so:

Ksp = [Mg2+] * [OH-]^2

we have to convert the solubility into mol/L from g/100g, we can do this using the molecular weight of magnesium hydroxide = 58.32 g/mol

so in 100g we have:
number of moles = [9.0 x 10^-4]/[58.32] = 1.5432 x 10^-5 mol

Because we have a dilute solution (i.e. mostly water we can approximate that the volume of 100g is 100mL)
so:
number of moles in 100mL = 1.5432 x 10^-5 mol
number of moles per litre = 1.5432 x 10^-4 mol

then we can substitute this into the equation derived previously:\
note: The concentration of the hydroxide ions is twice that of the magnesium ions due to the stoichiometry of the dissociation reaction
Mg(OH)2 --> Mg2+ + 2OH-
Ksp = [Mg2+] * [OH-]^2
Ksp = [1.5432 x 10^-4] * [2*1.5432 x 10^-4]^2
Ksp = 1.47 x 10^-11