the enthalpy of formation of H2O(l) is -280.70 kj/mol and enthalpy of neutralisation of strong acid and strong base is -56.70 kj/mol. What is the enthalpy of formation of OH- ions

1. Formation of H2O (l)  from H2 and O2 gases is as follows :
  
      H2(g) + 12O2(g) H2O                 (Hf=-280.70kJ/mol)

2. Neutralisation reaction of strong acid and strong base involves :
     
      H+ +OH-  H2O (l)             (Hn= -56.70kJ/mol)

3. Formation of OH- ions :
                   H2(g) + 12O2(g) H2O                 Hf=-280.70kJ/mol                    -           H+ (l) + OH-(l)  H2O                  Hn = -56.70kJ/mol                               H2(g) + 12O2(g)  H+ (l) +OH-(l)    H=-280.70+56.70  =-224kJ/mol

4. The enthalpy of formation of OH- ions is -224kJ/mol.

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H2O + 0.5 O2---> H2O
enthalpy of formation of water is -280.70
now,
H^+ + OH^- --> H^+ +OH^-
enthalpy of strong acid and base is -56.70
this is eq 2
now,
subtract 2 from 1
you get enthalpy of OH ion which is -224 kJ.
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