The heat of neutralisation of acetic and NaOH is 55.9 kJ. If the heat of neutralisation of all strong acids and strong bases is 57.3 kJ, then calculate the heat of ionisation of acetic acid.

Explanation

Neutralisation reaction of strong acid-strong base in an aqueous solution is: `H^(+)(aq)+OH^(-)(aq)toH_(2)O(l)`
The change in enthalpy in this reaction=-57.3 kJ
The heat of neutralisation in the reaction of weak acid (acetic acid) and strong bse=heat of ionisation of acetic acid+heat of raction for [`H^(+)(aq)+OH^(-)(aq)toH_(2)O(l)`]
Given, `-55.9kJ`=heat of ionisation of acetic acid `-57.3kJ`. therefore, heat of ionisation of acetic acid `=57.3-55.9=1.4kJ*mol^(-1)`.