Enthalpy of neutralization of `H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol"`. If enthalpy of neutralization of HCL with NaOH is -55.84`"kJ"//"mole"`, then calculate enthalpy of ionization of `H_(3)PO_(3)` in to its ions in kJ.

Explanation

The heat of neutralization is defined as " The change in enthalpy when one equivalent of an acid is completely nentralised by one gram equivalent of a base or vice versa in dilute solution".
`H_((aq))^(+) +OH_((aq))H_(2)O_((l))rarrDeltaH=-57.32 KJ`