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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?

Calculations: From the Wikipedia page, the standard enthalpies of formation are:

$$\begin{align} \ce{H2(g) + 1/2 O2(g) &-> H2O(l)} &\quad \Delta_\mathrm{f,1}H &= \pu{-285.8 kJ mol-1}\tag{1}\\ \ce{1/2 H2(g) &-> H+(aq) + e-} &\quad \Delta_\mathrm{f,2}H &= 0 \tag{2}\\ \ce{1/2 H2(g) + 1/2 O2(g) + e- &-> OH-(aq)}&\quad \Delta_\mathrm{f,3}H &= \pu{-230 kJ mol-1} \tag{3} \end{align}$$

Computing $\Delta_\mathrm{f,1}H - \Delta_\mathrm{f,2}H - \Delta_\mathrm{f,3}H$ gives us the reaction $\ce{H+ (aq) + OH- (aq) -> H2O (l)}$ with the associated enthalpy change $\Delta_\mathrm{f}H(\ce{H2O(l)}) = \pu{-55.8 kJ mol-1}.$

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    $\begingroup$ Good first question, there are some things to consider. Firstly there might be the possibility that the different answers are due to differences in the values measured for a given thing by different authors. Secondly it looks like you are using a Hess's law method, the thing is that errors might build up there. Lastly what do you mean by "strong acids and strong bases", do you mean an infinitely dilute solution of both or do you mean something like 4 M nitric acid and 5 M sodium hydroxide ? $\endgroup$
    – Nuclear Chemist
    Jul 6, 2018 at 4:22
  • $\begingroup$ @NuclearChemist, regarding the Hess's law method, what are the possible errors that can build up in there? The strong acids and base I am referring to are dilute solution of strong HCl and NaOH, e.g. 2M HCl, but not infinitely dilute. (I am not sure what infinitely dilute means) $\endgroup$
    – LHC2012
    Jul 6, 2018 at 4:28
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    $\begingroup$ Well if you have A forming B, and the measured delta H is 100 +/- 1 kJ mol-1. If you measure A to C, C to D, D to E and E to B all with an error of +/- 1 kJ mol-1. Under some conditions the effect of the errors will build up so this Hess's law based method will give a worse answer than a more direct measurement of A to B. $\endgroup$
    – Nuclear Chemist
    Jul 6, 2018 at 4:48

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-55.8 is pretty close -57.1 kJ/mol. There could be other reasons, but remember that:

1) The enthalpy of neutralization was measured colorimetrically. Different authors reporting different results. I personally only trust articles from the late 80's and onwards since accurate solution calorimeters were not available prior to that era. It is typical to see +- 2kJ/mol difference.

2) many use a single concentration of acid and bases to be reacted and repeat the experiment multiple times to get and average value. What I do instead, I do multiple experiments in incremental amounts of acid to be titrated on excess of base. This way, you're changing one variable and probing against the change in that variable only (the amount). You will get a slope of J/mol in the best statistical way: linear regression.

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2) Many authors are not careful about using only dilute solutions and correcting for the enthalpy of dilution when titrating in a calorimeter

3) Hess's law method combines all the errors in the heats of formation

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