Heat of neutralisation of `NH_(4)OH` and HCl is `-51.46 " kJ mol"^(-1)` .Calculate the ionization enthalpy of `NH_(4)OH` . [Enthalpy of neutralisation of strong acid-strong base is `-57.35 "kJ mol"^(-1)`]

Explanation

`NH_(4)OH(aq)+HCl (aq) to NH_(4)Cl(aq)+H_(2)O(l), Delta H^(0)= -51.46 " kJ mol"^(-1)` ..[1]
`NH_(4)OH(aq) to NH_(4)^(+)(aq)+OH^(-)(aq), Delta H^(0)= ? ` .. [2]
Enthalpies of [1] and [2] = Enthalpy of neutralisation of strong acid and strong base.
So, `- 51.46+DeltaH^(0)= - 57.35 therefore Delta H^(0)= 5.89 kJ`