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In my lab manual this is written but I can’t see how this is true

The heat of neutralization of weak acids by strong bases is less than 13.7 Cal/mole because the weak acid becomes completely ionized before the neutralization reaction by absorption of heat.

If the weak acid can absorb heat before the neutralization reaction to completely ionize, then why do we say that it doesn’t fully ionize in the first place? Before reading that text I thought that the reason was that less than 1 mole of H+ ions would be in the solution so less water will form therefore less heat will be released from the neutralization reaction, and that the difference in the heat released is the amount of heat that is needed to fully ionize the remaining weak acid. Where have I gone wrong?

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  • $\begingroup$ Could you be more specific about the reaction you are referring to? Remember solvent effects on both acid and base. Sometimes dissociation into ions can be energy absorbing. "Complete ionization" may be local (not all at once). $\endgroup$
    – Robert DiGiovanni
    Commented Mar 30, 2023 at 7:35
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    $\begingroup$ Note that effort to insert in the question title as much info as possible is contra-productive. $\endgroup$
    – Poutnik
    Commented Mar 30, 2023 at 7:38
  • $\begingroup$ @RobertDiGiovanni the weak acid neutralization reaction is that of acetic acid and sodium hydroxide $\endgroup$
    – Jay
    Commented Mar 31, 2023 at 12:26
  • $\begingroup$ It would appear that dissociation on acetic acid in water to acetate and H3O+ is "endothermic", or heat absorbing. $\endgroup$
    – Robert DiGiovanni
    Commented Mar 31, 2023 at 14:18
  • $\begingroup$ Interestingly, even dissolving common NaCl is slightly endothermic. Solid NaOH is quite the opposite. $\endgroup$
    – Robert DiGiovanni
    Commented Mar 31, 2023 at 14:28

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Neutralization of strong acids and strong bases is the exothermic reaction

$$\ce{H+(aq) + OH-(aq) -> H2O(l) \label{R1} \tag{R1}}$$

Neutralization of weaks acids and strong bases is formally the reaction chain of endothermic reaction

$$\ce{HA(aq) -> H+(aq) + A-(aq) \label{R2} \tag{R2}}$$

followed by the reaction $\eqref{R1}$

The dominant reaction is more probably the faster direct reaction

$$\ce{HA(aq) + OH-(aq) -> H2O(l) + A-(aq) \tag{R3}}$$

but with the same energy outcome as the chain $\eqref{R2}$+$\eqref{R1}$, due the Hess law that is consequence of the fundamental law of energy conservation.

The net reaction enthalpy is $\Delta H_{\mathrm{R3}} = \Delta H_{\mathrm{R2}}+\Delta H_{\mathrm{R1}} > \Delta H_{\mathrm{R2}}$

as some heat is absorbed by acid dissociation what diminishes the net heat release.

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