Enthalpy of neutralisation fo acetic acid by `NaOH` is `-50.6 kJ mol^(-1)`. Calculate `Deltah` for ionisation of `CH_(3)COOH`. Given, the hat of neutralisation of a strong acid with a strong base is `-55.9 kJ mol^(-1)`.

Explanation

The neutralisation of a strong acid by a strong base is repersented as :
`H^(+)(aq)+OH^(-)(aq)rarr H_(2)O(l) " " , " " Delta_("Neutralisation")H^(Θ)=-57.3` kJ/mol
Now acetic acid is weak acid, so it neutralisation is represented in two stages : First complete ionisation of weak acid and then complete neutralisation of the fully ionised acid by the base.
1. `CH_(3)COOH(aq)rarr CH_(3)COO^(-)(aq)+H^(+)(aq) " " , " " Delta_("Ionisation")H^(Θ)= ?`
2. `H^(+)(aq)+OH^(-)(aq)rarr H_(2)O(l) " " , " " Delta_("Neutralisation")H^(Θ) =-57.3` kJ/mol
Now adding the two equations to get the equation of neutralisation ,
`CH_(3)COOH(aq)+OH^(-)(aq)rarr CH_(3)COO^(-)(aq)+H_(2)O(l) , " " Delta_("Neutralisation")H^(Θ)=-50.6` kJ/mol
From Hess's Law : `" " Delta_("ionisation")H^(Θ)+(-57.3)=-50.6 " " rArr " " Delta_("ionisation")H^(Θ)=+6.7` kJ/mol.