. . . (ii) H++OH−⟶H2O . . . (iii) ∆H=−55.90kJmol−1 (from neutralisation of strong acid and strong base) From equation (i), (ii) and (iii) NH4OH+HCl⟶NH4++Cl−+H2O ∆H=−51.46kJmol−1 ∴x+(−55.90)=−51.46 x=−51.46+55.90 =4.44kJmol−1 ∴ Enthalpy of ionisation of NH4OH=4.44kJmol−1