The heat released on neutralisation of a strong alkali say `NaOH` with strong acid is `13.7kcal//mol.` The heat released on neutralisation of `NaOH` with `HF` if `-16.4 kcal//mol.` Calculate `DeltaH^(@)` of ionisation of `HF` in water.

Explanation

The required equation is `:`
`HFrarr H_((aq.))^(+)+F_((aq.))^(-), DeltaH^(@)=?`
Given that
`OH_((aq.))^(-)+H_((aq.))^(+)rarr H_(2)O_((aq.)), ...(1)`
`DeltaH^(@)=-13.7kcal`
`OH_((aq.))^(-)+HF_((l))rarr H_(2)O+F_((aq.))^(-), ....(2)`
`DeltaH^(@)=-16.4kcal`
By subtracting eq.` (1)` from `(2)`
The high exothermic heat of hydration of`F^(-)` ion due to its smaller size.