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Enthalpy of neutralization of H(3)PO(3) ...

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

`50.84kJ//mol`

B

`5kJ//mol`

C

`2.5kJ//mol`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`H_(3)PO_(3)rarr2H^(+)+HPO_(3)^(2-),Delta_(r)H=?`
`2H^(+)+2OH^(-)rarr2H_(2)O,`
`Delta_(r)H=-55.84xx2=-111.68`
`-106.68 = Delta_("iron")H-55.84 xx2`
`Delta_("iron")H=5` kJ/mol
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