What is the pH of a neutral solution at physiological temperature, 37 °C?
A neutral solution is defined when [H3O+] = [OH–]
Use thermodynamic data to find ΔG° for the autoionization of water at 37 °C. Then find Kw at 37 °C. Finally, this allows determination of [H3O+] = [OH–] and then the pH.
Reaction:
2 H2O(l) → ← H3O+(aq) + OH–(aq)
ΔH°
ΔHf°(H2O(l)) = –285.8 kJ/mol
ΔHf°(H3O+(aq)) = –285.8 kJ/mol
ΔHf°(OH–(aq)) = –230.0 kJ/mol
ΔH° = [–285.8 + –230.0] – [2(–285.8)] = 55.8 kJ/mol
ΔS°
S°(H2O(l)) = 69.91 J/mol·K
S°(H3O+(aq)) = 69.91 J/mol·K
S°(OH–(aq)) = –10.75 J/mol·K
ΔS° = [69.91 + –10.75] – [2(69.91)] = –80.66 J/mol·K
Then, ΔG° = ΔH° – TΔS°
At T= 37 °C = 310 K, ΔG° = 55.8 kJ/mol – (310 K)(–0.08066 kJ/mol·K) = 80.8 kJ/mol
Kw = exp(–ΔG°/RT) = exp(–80800/(8.314)(310)) = 2.43×10–14