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

Enthalpy of neutralisation of `H_(3)PO_(3)` acid is `-106.68kJ//mol` using NaOH. If enthaly of neutralisation of HCl by NaOH is `-55.84` kJ/mol. Calculate `DeltaH_("ionization") " of " H_(3)PO_(3)` into its ions

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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