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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, then the ΔHionization of H3PO3 into its ions is

Options

A.50.84 kJ/mol
B.5 kJ/mol
C.10 kJ/mol
D.2.5 kJ/mol

Solution

$\Delta H = ( - 106.68) - 2 \times ( - 55.84) = 5\text{ kJ/mol}$

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