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Question

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 Hionization of H3PO3 into its ions: 
  1. 5 kJ/mol
  2. 2.5 kJ/mol
  3. 50.84 kJ/mol
  4. None of these


Solution

The correct option is A 5 kJ/mol
H3PO32H++HPO23;     rH=?
2H++2OH2H2O;
rH=55.84×2=111.68
106.68=ionH55.84×2
ionH=5 kJ/mol

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