Why is enthalpy of neutralization of strong acids and strong bases always - 57.1 kj ?

Explanation

This is due to the fact that strong acids and strong bases are completely ionised in dilute aqueous solutions. The reaction between any strong acid and strong base may be written as,
`H^(+)(aq)=Cl(aq)+Na^(+)(aq)OH^(-)(aq)toNa^(+)(aq)^(+)Cl(aq)^(+)H_(2)O,DeltaH^(@)=-57.1kj/mol` By cancelling the common ions, we get
`H^(+)(aq) + OH^(-)(aq)to H_(2)O(l) Delta H^(@) " "_(n) = - 57.1 kj`
So, neutralization is simply a reaction between `H^(+)(aq)` ions given by acid and `OH^(-)` (aq) of base. These ions combine to from `H_(2)O`. THis suggests that the net chemical reaction in all neutralization reactions is same `H^(+) (aq) + OH^(-) (aq) to H_(2)O(l) , Delta H^(@) = - 57.1 kj`.