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 $\Delta$Hionization of H3PO3 into its ions :

• A. 50.84 kJ/mol
• B. 5 kJ/mol
• C. 2.5 kJ/mol
• D. None

Answer 1

Answer: (B)

5 kJ/mol

Answer 2

H3PO3 $\rightarrow$ 2H+ + HPO ; $\Delta$rH = ?

2H+ + 2OH– $\rightarrow$ 2H2O;

$\Delta$rH = – 55.84 × 2 = – 111.68

–106.68 = $\Delta$ionH – 55.84 × 2

Dion H = 5 kJ/mol.