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

`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 `eq. (2)-(1) HF rarr H_((aq.))^(+)+F_((aq.))^(-),`
`DeltaH^(@)=-2.7kcal`
The negative value is due to high heat of hydration of `F^(-)` ion on account of smaller size.