Chem help?

Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq) ___ Ksp
Initial: _______ 0 M _____ 0 M
Change: _____ +s M ____ +2s M
eqm: ________ s M _____ 2s M

Molar solubility = s M = 6.8 × 10⁻³ M
Hence, s = 6.8 × 10⁻³

At equilibrium:
Ksp = [Ca²⁺] [OH⁻]²
Ksp = s × (2s)² = 4s³ = 4 × (6.8 × 10⁻³)³ = 1.26 × 10⁻⁶

Ksp = [Ca2+][OH-]^2

Because 6.8X10^-3 = molar solubility, this will be [Ca2+] in a saturated solution. From the subscript on OH- in the formula, [OH-] = 2[Ca2+] = 1.36X10^-2 M. So,

Ksp = 6.8X10^-3 (1.36X10^-2)^2 = 1.27X10^-6 or 1.3X10^-6

Another way to get to this is to let S = molar solubility = [Ca2+], and 2S = [OH-]. Then,

Ksp = S (2S)^2 = 4S^3 = 4(6.8X10^-3)^3 = 1.26X10^-6 or 1.3X10^-6

Ca(OH)2 = [Ca][OH]^2 Let solubility of Ca(OH)2 be x
x.x^2 = 6.8 x 10^-3
x^3 = 6.8 x 10^-3
x = cube root of 6.8 x 10^-3 = 0.1894M