Enthalpy of neutralization of H3PO3 by is -106.68 kJ/mol. If the enthalpy of neutralization of by is -55.84 kJ/mol. The δ H_ionization of H3PO3 into its ions is

Question

Enthalpy of neutralization of H3PO3 by $NaOH$ is -106.68 kJ/mol. If the enthalpy of neutralization of $HCl$ by $NaOH$ is -55.84 kJ/mol. The δ H_ionization of H3PO3 into its ions is

Prashant Kotian · Accepted Answer

Correct answer: B. H3PO3 is a dibasic acid. Complete neutralization of 1 mole of H3PO3 involves 2 equivalents of H+. If H3PO3 were strong, theoretical heat of neutralization would be 2 * (-55.84 kJ/mol) = -111.68 kJ/mol. The actual heat of neutralization is -106.68 kJ/mol. The difference is the enthalpy of ionization: δ H_ion = -106.68 - (-111.68) = +5.00 kJ/mol.

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