Calculate the molar solubility of fe (oh) 2 in pure water. (the value of ksp for fe (oh) 2 is 4.87*10-17.)

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