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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.A. 50.84 kJ/molB. 5 kJ/molC. 2.5 kJ/molD. None of these |
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Answer» Correct Answer - B `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`. |
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