Question

Heat of neutralization `(DeltaH)` of `NH_(4)OH` and `HF` are `-51.5` and `-68.6 kJ` respectively. Calculate their heat of dissociation?
(i) `HCl (aq)+NaOH(aq) rarr NaCl (aq)+H_(2)O, " "DeltaH= -57.3 kJ`
(ii) `HCl (aq)+underset("(weak base)")(NH_(4)OH (aq)) rarr NH_(4)Cl (aq)+H_(2)O, " "DeltaH= -51.5 kJ`

Answer 1

`:.` The heat of dissociation of `NH_(4)OH`,
`DeltaH=-51.5-(-57.3)=5.8 kJ`
Similarly we have
`HF(aq) +NaOH(aq) rarr NaF(aq)+H_(2)O, DeltaH=-68.6 kJ`
`:.` The heat of diffociation of `HF`,
`DeltaH =-68.6-(-57.3)=-11.3 kJ`