To calculate the molar solubility of Fe(OH)2 at a pH of 8, we need to consider the equilibrium reaction and the hydrolysis of the compound.

1. Equilibrium Reaction:
The equilibrium reaction for the dissolution of Fe(OH)2 can be represented as:
Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH-(aq)

The solubility product constant (Ksp) expression for this equilibrium can be written as:
Ksp = [Fe2+][OH-]^2

2. Hydrolysis Reaction:
In an aqueous solution, the OH- ions can react with H+ ions from the water to form water molecules. This reaction is known as hydrolysis and can be represented as:
OH-(aq) + H+(aq) ⇌ H2O(l)

At pH 8, the concentration of H+ ions can be calculated using the equation:
pH = -log[H+]

3. Solubility Calculation:
To calculate the molar solubility of Fe(OH)2 at pH 8, we need to find the concentration of OH- ions and Fe2+ ions in the solution.

Step 1: Calculate the concentration of H+ ions at pH 8.
pH = -log[H+]
8 = -log[H+]
[H+] = 10^(-8) M

Step 2: Calculate the concentration of OH- ions at pH 8.
Since water undergoes autoionization, the concentration of OH- ions can be calculated using the equation:
[H+][OH-] = 1.0 x 10^(-14) M^2 (Kw)

[OH-] = (1.0 x 10^(-14)) / [H+]
[OH-] = (1.0 x 10^(-14)) / (10^(-8))
[OH-] = 10^(-6) M

Step 3: Calculate the molar solubility of Fe(OH)2.
Using the Ksp expression:
Ksp = [Fe2+][OH-]^2
1.6 x 10^(-14) = [Fe2+](10^(-6))^2
[Fe2+] = (1.6 x 10^(-14)) / (10^(-12))
[Fe2+] = 1.6 x 10^(-2) M

Therefore, the molar solubility of Fe(OH)2 at a pH of 8 is 1.6 x 10^(-2) M.

Answer: B. 0.016