Final answer:
The molar solubility of Fe(OH)2 in pure water, given the Ksp value of 4.87 x 10^-17, is calculated as 1.35 x 10^-6 M. The molar solubility of Fe(OH)2 in pure water is 1.35 × 10−6 M.
Step-by-step explanation:
To calculate the molar solubility of Fe(OH)2, we first establish the dissociation equilibrium:
Fe(OH)2 (s) ⇌ Fe2+ (aq) + 2OH− (aq)
For each mole of Fe(OH)2 that dissolves, it produces 1 mole of Fe2+ and 2 moles of OH−.
Thus, the solubility product expression based on the dissociation will be:
Ksp = [Fe2+] [OH−]2 = 4.87 × 10−17
Let s be the molar solubility of Fe(OH)2, then:
Ksp = (s)(2s)2 = 4s3
Substituting the given Ksp value and solving for s, we get:
4.87 × 10−17 = 4s3
s = ∓(4.87 × 10−17 / 4)
s = 1.35 × 10−6 M
The molar solubility of Fe(OH)2 in pure water is 1.35 × 10−6 M.