# Enthalpy of neutralization of strong acid and strong base differing from enthalpy of formation of water from ions

If the neutralization between strong acid and bases has enthalpy of neutralization of around $\pu{-57.1 kJ mol^-1}$, why is it that when I try to calculate the enthalpy of formation of water from $\ce{H+}$ and $\ce{OH-}$ ions I get approximately $\pu{-55.8 kJ mol^-1}$ using the reference values?

Why is there this difference in enthalpy if the reaction between say $\ce{NaOH}$ and $\ce{HCl}$ is only between $\ce{H+}$ and $\ce{OH-}$ ions, while $\ce{Na}$ and $\ce{Cl}$ remain dissociated?

\begin{align} \ce{H2 (g) + 1/2 O2 (g) &-> H2O (l)}\tag{\Delta H_\mathrm{f,1}=\pu{-285.8 kJ/mol}}\\ \ce{1/2 H2 (g) &-> H+ (aq) + e-}\tag{\Delta H_\mathrm{f,2}=0}\\ \ce{1/2 H2 (g) + 1/2 O2 (g) + e- &-> OH- (aq)}\tag{\Delta H_\mathrm{f,3}=\pu{-230 kJ/mol}} \end{align}
Computing $\Delta H_\mathrm{f,1}-\Delta H_\mathrm{f,2}-\Delta H_\mathrm{f,3}$ gives us the reaction $\ce{H+ (aq) + OH- (aq) -> H2O (l)}$ with the associated enthalpy change $\Delta H_\mathrm{f, \ce{H2O}} = \pu{-55.8kJ/mol}$