Enthalpy of neutralisation of acetic acid by `NaOH` is `-50.6 kJ mol^(-1)`. Calculate `DeltaH` for ionisation of `CH_(3)COOH`. Given. The heat of neutralisation of a strong acid with a strong base is `-55.9 kJ mol^(-1)`.

Explanation

We have, `DeltaH` (neutralisation)
`=DeltaH ( "ionisation of" CH_(3)COOH)+DeltaH(H^(+)+OH^(-))`
`:.DeltaH("ionisation of" CH_(3)COOH)=-50.6-(-55.9)=5.3 "kJ mol"^(-1)`
[DeltaH (ionisation of NaOH)=0 as NaOH is a strong base]