Question

The enthalpy of neutralization of one mole of $H_{3}P{O}_3$ acid using two moles of $NaOH$ is $-106.68{\text{kJ.mol}}^{-1}$. If the enthalpy of neutralization of $HCl$ by $NaOH$ is $-55.84{\text{kJ.mol}}^{-1}$, calculate the $\Delta H_{\text{ionization}}$ of $H_{3}P{O}_3$ into its ions:

• A. $50.84{\text{kJ.mol}}^{-1}$
• B. $5{\text{kJ.mol}}^{-1}$
• C. $2.5{\text{kJ.mol}}^{-1}$
• D. None of these

Answer 1

The correct option is B $5{\text{kJ.mol}}^{-1}$

$H_{3}P{O}_3\to 2H^{+}+HP{O}_3^{2-};{\Delta }_{r}H=?$
$2H^{+}+2O{H}^{-}\to 2H_{2}O$

If it is a strong acid, the enthalpy of neutralization will be
${\Delta }_{r}H=-55.84\times 2=-111.68$, since 2 protons are replaced.

If we substract this value from the actual value, the enthalpy of dissociation can be obtained.
$-106.68={\Delta }_{\text{ion}}H-55.84\times 2$
$\Rightarrow \Delta H_{\text{ionization}}=5{\text{kJ.mol}}^{-1}$