What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?

               M(OH)₂(s) ⇌ M⁺(aq) + 2OH⁻(aq)    Ksp = 8.05 × 10⁻¹²
Initial (M):                      0.202         0

Change (M):                    +s         +2s
Eqm (M):                  (0.202 + s)     2s
                                   ≈ 0.202

At equilibrium:
Ksp = [M⁺] [OH⁻]
8.05 × 10⁻¹² = (0.202)(2s)²
0.808s² = 8.05 × 10⁻¹²
s = √{(8.05 × 10⁻¹²)/0.808}
s = 3.16 × 10⁻⁶

Solubility = 3.16 × 10⁻⁶ M