For strong acid and strong base neutralisation,the net chemical change is $H^{+} + OH^{-} \longrightarrow H_2O_{(l)}$; $\Delta_r H^{\circ} = -55.84 \ kJ \ mol^{-1}$. If the enthalpy of neutralisation of $CH_3COOH$ by $NaOH$ is $-49.86 \ kJ \ mol^{-1}$,then the enthalpy of ionisation of $CH_3COOH$ is:
- A$5.98 \ kJ \ mol^{-1}$
- B$-5.98 \ kJ \ mol^{-1}$
- C$105.7 \ kJ \ mol^{-1}$
- D$-59.8 \ kJ \ mol^{-1}$
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