Use ksp=4.87*10-17 to calculate the solubility of iron (ii) hydroxide in pure water in grams per 100.0 ml of solution. g

The ksp = 4.87 * 10^-17

Therefore; 4.87 * 10^-17 = [Fe2+][OH-]^2 = (X) (2X) ^2 = 4X^3

Hence;

4x^3 = 4.87 * 10^-17

x = 2.30 * 10^-6 M

therefore the molarity is 2.3 * 10^-6 M

Thus;

The mass of Fe (OH) 2 in 100 ml of water, will be given by:

= 2.30 * 10^-6 mol/L = 2.3 * * 10^-5 mole/100 mL

= 89.86 g/mol * 2.3 * 10^-6 * 0.100 = 2.07 * 10^-5 g

= 2.07 * 10^-5 g

= 2.07 * 10^-5 g