Question

Enthalpy of neutralization of $H_{3}P{O}_3$ acid is -106.68 kJ/mol using $NaOH$. If enthalpy of neutralization of $HCl$ by $NaOH$ is -55.84 kJ/mol.
Calculate $△H_{ionization}of{H}_{3}P{O}_3$ into its ions:

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

Answer 1

The correct option is B 5 kJ/mol

$H_{3}P{O}_3\to 2H^{+}+HP{O}_3^{2-};{△}_{r}H=?$
$2H^{+}+2O{H}^{-}\to 2H_{2}O;$
${△}_{r}H=-55.84\times 2=-111.68$
$-106.68={△}_{ion}H-55.84\times 2$
${△}_{ion}H=5kJ/mol$