When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous solution 108kJ of heat is liberated ,then the enthalpy of ionization of the oxalic acid is(Given `H^+(aq)+OH^-(aq)rarrH_2O(l)Delta_neulH=-57.3kJ`)

Options

A. `4.6kJ mol^(-1)`
B. `-4.6kJ mol^(-1)`
C. `-6.6kJ mol^(-1)`
D. `6.6kJ mol^(-1)`

Answer

To solve the problem, we need to determine the enthalpy of ionization of oxalic acid when it reacts with sodium hydroxide (NaOH). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
Oxalic acid (C₂H₂O₄) is a diprotic acid, meaning it can donate two protons (H⁺ ions). When it reacts with NaOH, it will neutralize to form water and the corresponding salt.

### Step 2: Write the Reaction
The reaction can be represented as:
\[ \text{C}_2\text{H}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]

### Step 3: Enthalpy of Neutralization
The problem states that when 1 mole of oxalic acid is treated with excess NaOH, 108 kJ of heat is liberated. This is the enthalpy change for the neutralization reaction:
\[ \Delta H_{\text{neutralization}} = -108 \text{ kJ} \]

### Step 4: Enthalpy of Ionization of H⁺ and OH⁻
The enthalpy change for the neutralization of a strong acid (H⁺) with a strong base (OH⁻) is given as:
\[ \Delta H = -57.3 \text{ kJ/mol} \]
Since oxalic acid is diprotic, it will release 2 moles of H⁺ ions, so for 2 moles:
\[ \Delta H_{\text{neutralization}} = 2 \times (-57.3) = -114.6 \text{ kJ} \]

### Step 5: Calculate the Enthalpy of Ionization
The enthalpy of ionization (\( \Delta H_{\text{ionization}} \)) can be calculated using the relationship:
\[ \Delta H_{\text{ionization}} = \Delta H_{\text{neutralization}} + \Delta H_{\text{neutralization of strong acid}} \]

Substituting the values:
\[ \Delta H_{\text{ionization}} = -108 \text{ kJ} - (-114.6 \text{ kJ}) \]
\[ \Delta H_{\text{ionization}} = -108 + 114.6 \]
\[ \Delta H_{\text{ionization}} = 6.6 \text{ kJ} \]

### Step 6: Conclusion
The enthalpy of ionization of oxalic acid is:
\[ \Delta H_{\text{ionization}} = +6.6 \text{ kJ/mol} \]

### Final Answer:
The enthalpy of ionization of oxalic acid is \( 6.6 \text{ kJ/mol} \).

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