The solubility product expression for iron(II) hydroxide, Fe(OH)2, is:
Ksp = [Fe2+][OH-]^2 = 4.87×10^-17
Let's assume that the initial concentration of Fe2+ and OH- ions in pure water is x M. Then, the equilibrium concentration of Fe2+ and OH- ions will also be x M.
Therefore, the solubility product expression becomes:
Ksp = x * (2x)^2 = 4x^3
Solving for x:
4x^3 = 4.87×10^-17
x^3 = 1.2175×10^-17
x = (1.2175×10^-17)^(1/3)
x = 2.312×10^-6 M
The solubility of Fe(OH)2 is equal to the concentration of Fe2+ ions, which is x.
To convert this to grams per 100.0 ml of solution, we need to multiply by the molar mass of Fe(OH)2 and the volume of the solution:
solubility = x * molar mass * 100 / volume
Assuming the molar mass of Fe(OH)2 is 89.86 g/mol and the volume of the solution is 100.0 ml, we get:
solubility = (2.312×10^-6 M) * (89.86 g/mol) * 100 / 100.0 ml
solubility = 0.00208 g/100.0 ml
Therefore, the solubility of iron(II) hydroxide in pure water is 0.00208 g/100.0 ml of solution.
*IG:whis.sama_ent*