`CsOH +HCI rarr CsCI +H_(2)O, DeltaH =- 13.4 kcal mol^(-1)……….(i)`
`CsOH +HF rarr CsF +H_(2)O, DeltaH =- 16.4 kcal mol^(-1) ………(ii)`
Calculate `DeltaH` for the ionisation of `HF` in `H_(2)O`.

Explanation

`HF rarr H^(o+) +F^(Theta) DeltaH = ?`
From equation (i), we get
`CsOH +HcI rarr CsCI +H_(2)O`
or
`overset(Theta)OH (Strong base) +H^(o+)(acid) rarr H_(2)O DeltaH_(2) =- 13.4 kcal`
From equation (ii), we get
`overset(Theta)OH (Strong base) +HF (Weak acid) rarr H_(2)O DeltaH_(2) =- 16.4 kcal`
`DeltaH = DeltaH_(2) - DeltaH_(1)`
`=- 16.4 -(-13.4) == 3.0 kcal`