When we have this balanced equation for a reaction: Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH- so we can assume [Fe+2] = X and [OH-] = 2 X when Ksp = [Fe+2][OH-]^2 and have Ksp = 4.87x10^-17 [Fe+2]= X [OH-] = 2X so by substitution 4.87x10^-17 = X*(2X)^2 ∴X^3 = 4.8x10^-17 / 4 ∴the molar solubility X = 2.3x10^-6 M
