Find the molar solubility of Fe(OH)3 in water at 25 °C.
Ksp for Fe(OH)3 is not listed in the Table of Solubility Product Constants
This means we need to find Ksp from thermodynamic data
Write the balanced reaction:
Fe(OH)3(s) → ← Fe3+(aq) + 3 OH–(aq)
ΔGf°(Fe(OH)3(s)) = –696.6 kJ/mol
ΔGf°(Fe3+(aq)) = –4.7 kJ/mol
ΔGf°(OH–(aq)) = –157.2 kJ/mol
ΔG° = [–4.7 + 3(–157.2)] – [–696.6] = 220.3 kJ/mol
Then, Ksp = exp(–(220300 J)/(8.314 J·mol–1K–1 × 298 K) = 2.42×10–39
Now, the solubility can be done determined using the equilibrium constant:
Fe(OH)3(s) → ← Fe3+(aq) + 3 OH–(aq)
Ksp = [Fe3+]e[OH–]e3 = 2.42×10–39
Initial00
Change+x+3x
Equilibriumx3x
2.42×10–39 = (x)(3x)3 = 27x4
x = 9.73×10–11
The solubility of Fe(OH)3 in water at 25 °C is 9.73×10–11 M