pH problem?

(a)
Molar mass of Mg(OH)₂ = (24.3 + 16.0×2 + 1.0×2) g/mol = 58.3 g/mol
No. of moles of Mg(OH)₂ = (0.0233 g) / (58.3 g/mol) = 0.0004 mol
Molarity of the solution = (0.0004 mol) / (400/1000 L) = 0.001 M

(b)
1 mole of Mg(OH)₂ contains 1 mole of Mg²⁺ ions.
No. of moles of Mg²⁺ ions = 0.0004 mol

(c)
1 mole of Mg(OH)₂ contains 2 mole of OH⁻ ions.
No. of moles of OH⁻ ions = (0.0004 mol) × 2 = 0.0008 mol

(d)
pOH = -log[OH⁻] = -log(0.0008) = 3.1

(e)
pH = 14 - pOH = 14 - 3.1 = 10.9

this is the actual solution >>>>>
Mg(OH)2 = Mg2+ + 2OH-
Ksp Mg(OH)2 = 1.8 x 10-11
Ksp = [Mg2+] * [OH-]^2
we have 0.0233 g of the hydroxide or 0.233g / 58.3197 g/mol = 0.0040
molarity of the "solution" = 0.004 mol / 0.40 L = 0.0100 M however very little of the hydroxide dissolves
it appears as a suspension
we need to use the Ksp expression to find the Mg and OH- ions in solution
1.8 x 10^-11 = x * x^2
x^3 = 1.8 x 10^-11
x= 0.000262
[Mg2+] = 0.000262 M
[OH-] = 2*0.000262 = 0.000524 M
pOH = 3.28
pH= 10.72