We're used to the standard molar enthalpy of formation of a compound. Ions in solution also have a heat of formation: when we add a substance such as HCl to water it breaks apart into ions.

HCl(g) -> H⁺(aq) + Cl^-(aq)

We can't measure the enthalpy of just one ion, however: they come in pairs. To get a standard value, we assume that the enthalpy of formation of the hydrogen ion, H⁺(aq) =0. Knowing that, we can get the other enthalpies of formation: in the example above, we can compute DH_f^oCl^-(aq), for example.

These enthalpies can be used exactly like the standard molar enthalpies of formation of molecules in reactions that involve ions.

Example: A solution containing barium ions is mixed with a solution containing chloride ions. What is the change in enthalpy in the reaction?

Solution: First, write down the balanced equation for the reaction

Ba⁺²(aq) + 2Cl^-(aq) -> BaCl₂(s)

Next, look up the enthalpies of formation of the products and reactants

Substance	DHfo
Ba+2(aq)	-537.6 kJ/mol
Cl-(aq)	-167.2 kJ/mol
BaCl2(s)	-858.6 kJ/mol

S DH_f^o reactants = DH_f^oBa⁺²(aq)+ 2*DH_f^oCl^-(aq) = -537.6 + 2*(-167.2) = -872.0 kJ/mol
S DH_f^o products = DH_f^oBaCl₂(s) = -858.6 kJ/mol

DH^o = S DH_f^o products - S DH_f^o reactants = -858.6 - (-872.0) = 13.4 kJ/mol

The final thermochemical reaction is

Ba⁺²(aq) + 2Cl^-(aq) -> BaCl₂(s)      DH = 13.4 kJ

Information provided by: http://learn.chem.vt.edu