The enthalpy of neutralisation of a strong acid by a string base is `-57.32 kJ mol^(-1)`. The enthalpy of formation of water is `-285.84 kJ mol^(-1)`. The enthalpy of formation of hydroxyl ion is

Answer

`-228 kJ//"mole"`

Explanation

`H^(+)(aq)+OH^(-)(aq)rarrH_(2)O(l)DeltaH= -57`
`DeltaH_("reaction")=Sigma Delta H_(f'"product")-Sigma Delta H_(f'"reactant")`
`= -57= -285-(DeltaH_(f'H^(+))+Delta H_(f'OH^(-)))`
`= Delta H_(f'OH^(-))= -228`