Enthalpy of neutralization of `H_(3)PO_(3) "with " NaOH " is" -106.68 "kJ"//"mol"`. If enthalpy of neutralization of HCL with NaOH is -55.84`"kJ"//"mole"`, then calculate enthalpy of ionization of `H_(3)PO_(3)` in to its ions in kJ.

Answer

To find the enthalpy of ionization of \( H_3PO_3 \) into its ions, we can use the enthalpy of neutralization values provided in the question. We will set up the problem using Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- The neutralization of \( H_3PO_3 \) with \( NaOH \):
\[ H_3PO_3 + 2NaOH \rightarrow Na_2HPO_3 + 2H_2O \quad (\Delta H = -106.68 \, \text{kJ/mol}) \]
- The neutralization of \( HCl \) with \( NaOH \):
\[ HCl + NaOH \rightarrow NaCl + H_2O \quad (\Delta H = -55.84 \, \text{kJ/mol}) \]

2. **Calculate the Enthalpy Change for 2 Moles of \( HCl \) Neutralization:**
- Since we need to consider the enthalpy change for 2 moles of \( HCl \):
\[ \Delta H_{neutralization \, of \, HCl} = 2 \times (-55.84 \, \text{kJ/mol}) = -111.68 \, \text{kJ} \]

3. **Set Up the Equation Using Hess's Law:**
- According to Hess's law, we can express the enthalpy change for the ionization of \( H_3PO_3 \) as:
\[ \Delta H_{ionization} + \Delta H_{neutralization \, of \, HCl} = \Delta H_{neutralization \, of \, H_3PO_3} \]
- Plugging in the values we have:
\[ \Delta H_{ionization} - 111.68 \, \text{kJ} = -106.68 \, \text{kJ} \]

4. **Solve for the Enthalpy of Ionization:**
- Rearranging the equation to solve for \( \Delta H_{ionization} \):
\[ \Delta H_{ionization} = -106.68 \, \text{kJ} + 111.68 \, \text{kJ} \]
\[ \Delta H_{ionization} = 5.00 \, \text{kJ} \]

### Final Answer:
The enthalpy of ionization of \( H_3PO_3 \) into its ions is \( 5.00 \, \text{kJ} \).

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