No of moles of solute =molarity×volumeofsolution
No. of moles of acid =0.2×400ml=80mmol
No. of moles of base =0.5×800ml=400mmol
CH3COOH+NaOH⟶CH3COONa+H2O
80mmol400mmol80mmol
(For CH3COOH and NaOH mole and equivalent are the same because both have value of n−factor=1)
No. of equivalents of acid or base neutralised completely =80milieq.1000
801000 equivalents give → 4.416 KJ
∴ 1 equivalent give →4.41680×1000=55.2kJ
∴ heat of ionisation of CH3COOH=–55.2+57.3=+2.1kJ/mole