The ksp of iron(ii hydroxide, fe(oh2, is 4.87 × 10-17. calculate the molar solubility of this compound.

1- getting the molar solubility:
We have the compound Fe(OH)2. It dissociates based on the following equation:
Fe(OH)₂ <> Fe2+ + 2OH-
we can note that the concentration of hydroxide is twice that of iron
Therefore:
Assuming that the molar solubility is x,
Ksp = x * (2x)²
4.87 * 10^-17 = 4x³
x³ = 1.2175 * 10^-17
x = 2.3 * 10^-6 mole / liter

In case you want the solubility in grams / liter:
2- getting the molar mass of Fe(OH)2:
From the periodic table:
mass of Fe = 55.8 grams
mass of O = 16 grams
mass of H = 1 gram
Therefore:
molar mass of Fe(OH)2 = 55.8 + 2(16+1) = 89.8 grams

3- getting the solubility in grams / liter:
solubility i grams / liter = solubility in moles / liter * molar mass
solubility in grams / liter = 2.3 * 10^-6 * 89.8
solubility = 2.0654 * 10^-4 gram / liter

Hope this helps :)

Ireally don’t know this