2.20 g of solid NaOH are added to 250 mL of a 0.10 M FeCl2 solution. Given that Ksp for Fe(OH)2 is 1.6 × 10-14, calculate the following?

a.
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Initial number of moles of NaOH = (2.20 g) / (40.0 g/mol) = 0.0550 mol
Initial number of moles of FeCl₂ = (0.10 mol/L) × (250/1000 L) = 0.0250 mol

FeCl₂(aq) + 2NaOH(aq) → Fe(OH)₂(s) + 2NaCl(aq)
Mole ratio FeCl₂ : NaOH = 1 : 2

If FeCl₂ completely reacts, NaOH needed = (0.0250 mol) × 2 = 0.0500 mol < 0.0550 mol
Hence, NaOH is in excess, and FeCl₂ completely react (limiting reactant).

As Ksp is very small, it can be assumed that the dissolving of Fe(OH)₂ is negligible.
Assume that the reaction between FeCl₂ and NaOH is complete, and there is no loss in the reaction.
Mole ratio FeCl₂ : Fe(OH)₂ = 1 : 1
No. of moles of FeCl₂ reacted = 0.0250 mol
Then, no. of moles of Fe(OH)₂ = 0.0250 mol

Molar mass Fe(OH)₂ = (55.8 + 16.0×2 + 1.0×2) = 89.8 g/mol
Mass of Fe(OH)₂ formed = (0.0250 g) × (89.8 g/mol) = 2.25 g (to 3 sig. fig.)


b.
In the final solution, no. of moles of unreacted NaOH = (0.0550 - 0.0500) mol = 0.0050 mol
Volume of the final solution = 250 mL = 0.250 L
Before dissociation of Fe(OH)₂, concentration of unreacted NaOH = (0.0050 mol) / (0.250 L) = 0.020 M

________ Fe(OH)₂(s) ⇌ Fe²⁺(aq) + 2OH⁻(aq) …. Ksp = 1.6 × 10⁻¹⁴
Initial: _____________ 0 M ____ 0.020 M
Change: ___________ +s M ______ +s M
At eqm: ____________ s M ___ (0.020 + s) M

As Ksp is very small, we can assume that 0.020 ≫ s M
We can assume that [OH⁻] at eqm = (0.020 + s) M ≈ 0.020 M

Ksp = [Fe²⁺] [OH⁻]²
[Fe²⁺] [OH⁻]² = 1.6 × 10⁻¹⁴
[Fe²⁺] (0.02)² = 1.6 × 10⁻¹⁴
[Fe²⁺] = 1.6 × 10⁻¹⁴ / (0.02)² M
[Fe²⁺] at the final solution = 4.0 × 10⁻¹¹ M