Enthalpy of neutralization of H3PO3 acid is -106.68 kJ/mol using NaOH. If enthalpy of neutralization of HCl by NaOH is -55.84 kJ/mol. Calculate △HionizationofH3PO3 into its ions:
A
50.84 kJ/mol
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B
5 kJ/mol
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C
2.5 kJ/mol
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D
None of these
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Solution
The correct option is B 5 kJ/mol H3PO3→2H++HPO2−3;△rH=? 2H++2OH−→2H2O; △rH=−55.84×2=−111.68 −106.68=△ionH−55.84×2 △ionH=5kJ/mol