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.

Options

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

Explanation

To calculate the enthalpy of ionization of \( H_3PO_3 \) (phosphorous acid) into its ions, we can use the given enthalpy values for the neutralization reactions and apply Hess's law. Here’s a step-by-step solution:

### Step 1: Understand the Neutralization Reactions
1. **Neutralization of \( H_3PO_3 \) with NaOH**:
The reaction can be represented as:
\[ H_3PO_3 + 2 NaOH \rightarrow Na_2HPO_3 + 2 H_2O \]
The enthalpy of this reaction is given as:
\[ \Delta H_{neutralization} = -106.68 \, \text{kJ/mol} \]

2. **Neutralization of HCl with NaOH**:
The reaction can be represented as:
\[ HCl + NaOH \rightarrow NaCl + H_2O \]
The enthalpy of this reaction is given as:
\[ \Delta H_{neutralization} = -55.84 \, \text{kJ/mol} \]

### Step 2: Determine the Enthalpy of Ionization
Since \( H_3PO_3 \) is a weak acid, it does not dissociate completely. We denote the enthalpy of ionization of \( H_3PO_3 \) as \( X \).

The overall process for the ionization of \( H_3PO_3 \) can be broken down into two steps:
1. The ionization of \( H_3PO_3 \) to produce \( 2 H^+ \) ions:
\[ H_3PO_3 \rightarrow 2 H^+ + HPO_3^{2-} \]
This requires energy \( X \).

2. The neutralization of the \( H^+ \) ions with \( NaOH \):
\[ 2 H^+ + 2 NaOH \rightarrow 2 H_2O + 2 Na^+ \]
The enthalpy change for this step is:
\[ -2 \times (-55.84) = -111.68 \, \text{kJ} \]

### Step 3: Apply Hess's Law
According to Hess's law, the total enthalpy change for the reaction can be expressed as:
\[ X + (-111.68) = -106.68 \]

### Step 4: Solve for \( X \)
Rearranging the equation gives:
\[ X = -106.68 + 111.68 \]
\[ X = 5 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of ionization of \( H_3PO_3 \) is \( 5 \, \text{kJ/mol} \).

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