The enthalpy of neutralization of ammonium hydroxide by hydrochloric acid is 51.46 k| mol-1. Calculate the enthalpy of ionization of ammonium hydroxide.

(i)  NH4OH(aq)+H+(aq)→NH4+(aq)+H2O(l)ΔH∘=−51.46kJmol−1(ii) H+(aq)+OH−(aq)→H2O(l); ΔH∘=−57.32kJmol−1 (iii) NH4OH(aq)→NH4+(aq)+OH−(aq); ΔH∘=ΔH∘ ionization On adding (ii) and (iii), we getNH4OH(aq)+H+(aq)NH4+→NH4+(aq)+H2O(l)orΔH∘=−57.32+ΔHionization ∘From equation (i), we have ΔH∘=−51.46kJmol−1 Hence −57.32+ΔHionization ∘kJmol−1=−51.46kJmol−1 or  ΔHIonization ∘=+5.86kJmol−1 .