Calculate the solubility of magnesium fluoride, MgF2, in water by knowing the Ksp = 5.2 x 10-11?

_              MgF₂(s) ⇌ Mg²⁺(aq) + 2F⁻(aq)  Ksp = 5.2 × 10⁻¹¹
_Initial:                        0 M            0 M
_Change:                   +s M         +2s M
_Eqm:    　　　           s M           2s M

At equilibrium:
Ksp = [Mg²⁺] [F⁻]²
5.2 × 10⁻¹¹ = s (2s)²
4s³ = 5.2 × 10⁻¹¹
s = ³√(5.2 × 10⁻¹¹ / 4)
s = 2.4 × 10⁻⁴
Solubility of MgF₂ in water = 2.4 × 10⁻⁴ M

MgF2 = Mg2+ + 2OH-

Ksp =  [Mg2+] * [OH-]^2

if x mole of the hydroxide dissolves we have

x * (2x)^2 =  5.2 x 10-11

4 x^3 = 5.2 x 10-11

x^3 = 1.3 * 10^-11 

x = 2.35 * 10 ^-4 = moles of MgF2 that dissolved/ liter 

molar solubility of MgF2 = 2.35 * 10 ^-4