Question

The solubility of Fe(OH)₂ is 2.1 x 10^-4 g L^-1. What is the K_sp for Fe(OH)₂ at 25 °C?

Answer 1

Molar mass of Fe(OH)2,
MM = 1*MM(Fe) + 2*MM(O) + 2*MM(H)
= 1*55.85 + 2*16.0 + 2*1.008
= 89.866 g/mol


Molar mass of Fe(OH)2= 89.866 g/mol
s = 2.1*10^-4 g/L
To covert it to mol/L, divide it by molar mass
s = 2.1*10^-4 g/L / 89.866 g/mol
s = 2.337*10^-6 g/mol

At equilibrium:
Fe(OH)2   <---->     Fe2+     +         2 OH-   

                     s                  2s      

Ksp = [Fe2+][OH-]^2
Ksp = (s)*(2s)^2
Ksp = 4(s)^3
Ksp = 4(2.337*10^-6)^3
Ksp = 5.104*10^-17
Answer: 5.10*10^-17