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.

Options

A. 50.84 kJ/mol
B. 5 kJ/mol
C. 2.5 kJ/mol
D. None of these

Answer

B

Explanation

`H_(3)PC_(3)rarr2H^(+)+HPO_(3)^(2-) , Delta_(r )H=?`
`2H^(+)+2OH^(-)rarr2H_(2)O`,
`Delta_(r )H= -55.84xx2= -111.68`
`-106.68=Delta _("ion")-55.84xx2`
`Delta H_("ion")H=5 kJ//mol`.