On the basis of the following thermochemical data ∆fG0H+aq=0H2Ol  → H+aq + OH-aq;         ∆H=57.32 kJ H2g+12O2g → H2Ol;             ∆H=-286.02 kJThe value of enthalpy of formation of OH-  ion  at  250C  is

Consider the heat of formation of H2g+12O2g → H2Og; ∆H=-286.20 kJ ∆Hr=∆HfH2O, l -∆HfH2, g-∆Hf(O2,g)  = -286.20 = ∆HfH2O, l-0-0 ∆HfH2O, l = -286.20 Now, consider the ionization of H2O H2Ol → H+aq+OH-aq ∆H=57.32 kJ ∆Hr=∆HfH+, aq +∆HfOH-, aq-∆HfH2O, l 57.32=0+∆HfOH-, aq--286.20 Thus, ∆HfOH-, aq = 57.32-286.20=-228.80 kJ