when Fe(OH)2 (S) ↔ Fe2+ + 2OH- is the reaction equation
let the molar solubility = X
when mol/L Fe(OH)2 gives X mol / L Fe2+ and 2X mol / L OH-
when the Ksp expression is:
Ksp = [Fe2+] [OH-]^2
when we assume [Fe2+] = X
and [OH-] = 2X
and when we have Ksp = 4.87 x 10^-17
so by substitution:
∴ 4.87 x 10^-17 = X*(2X)^2
∴ 4X^3 = 4.87 x 10^-17
∴X = 2.3 x 10^-6
∴ the molar solubility = X = 2.3 x 10^-6 mol/L
