On the basis of the following thermochemical data: (ΔHofH+(aq) = 0) H₂O(l) → H+(aq) + OH⁻(aq); ΔH = 57.32 kJ H₂(g) + ½O₂(g) → H₂O(l); ΔH = -286.20 kJ The value of enthalpy of formation of OH⁻ ion at 25°C is:

In ionization of H₂O, H₂O(l) → H+(aq) + OH⁻(aq); ΔH = 57.32 kJ
ΔH = ΔHf(OH⁻(aq)) + ΔHf(H+(aq)) - ΔHf(H₂O(l))
57.32 = ΔHf(OH⁻(aq)) + ΔHf(H+(aq)) - ΔHf(H₂O(l))
Given that ΔHf(H+(aq)) = 0 and ΔHf(H₂O(l)) = -286.20 kJ
57.32 = ΔHf(OH⁻(aq)) + 0 - (-286.20)
ΔHf(OH⁻(aq)) = 57.32 - 286.20 = -228.88 kJ