While performing a thermodynamics experiment, a student made the following observations, $\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}\Delta \mathrm{H}=-57.3 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

${\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}\to {\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$

$\mathrm{\Delta H}=-55.3 \mathrm{kJ}{ \mathrm{mol}}^{-1}$.

The enthalpy of ionization of ${\mathrm{CH}}_3\mathrm{COOH}$ as calculated by the student is $\mathrm{kJ} {\mathrm{mol}}^{-1}$. 

The enthalpy of neutralization (ΔH_n) is the change in enthalpy that occurs when one equivalent of an acid and a base undergo a neutralization reaction to form water and a salt. It is a special case of the enthalpy of reaction. It is defined as the energy released with the formation of 1 mole of water.

${△}_{\mathrm{H}} \mathrm{Ionization} \mathrm{of} \mathrm{weak} \mathrm{acid} ={∆}_{\mathrm{H}} \mathrm{Neutralization} \mathrm{of} \mathrm{W} \mathrm{A} + \mathrm{S} \mathrm{B} - {∆}_{\mathrm{H}} \mathrm{Neutralization} \mathrm{of} \mathrm{S} \mathrm{A} + \mathrm{S} \mathrm{B} =$

${\mathrm{\Delta H}}_{\mathrm{ionisation}}$ of ${\mathrm{CH}}_3\mathrm{COOH}=-55.3-(-57.3)$

$=2\mathrm{KJ}/\mathrm{mol}$

Hence, $2\mathrm{KJ}/\mathrm{mol}$ of heat is absorbed for the ionization of acetic acid.