The enthalpy of neutralization of a strong acid with a strong base is -57.32 kJ/mole. The enthalpy of formation of water is -285.84 kJ/mole. The enthalpy of formation of hydroxyl ions 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`