If theta of reaction for the given acid-base reaction:
`HA+NaOHrarrNaA+H_(2)O , Delta H= -4.7 kcal`
The heat of dissocitation of HA is ________.

Answer

To find the heat of dissociation of the weak acid HA in the given acid-base reaction, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ \text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O} \]
where HA is a weak acid and NaOH is a strong base.

### Step 2: Identify Given Data
We are given:
- The enthalpy change (\( \Delta H \)) of the reaction:
\[ \Delta H = -4.7 \, \text{kcal} \]
- The enthalpy change of neutralization for a weak acid with a strong base is generally:
\[ \Delta H_{\text{neutralization}} = -13.7 \, \text{kcal} \]

### Step 3: Set Up the Relationship
The enthalpy change of the reaction can be expressed in terms of the heat of dissociation of the weak acid HA:
\[ \Delta H_{\text{neutralization}} = \Delta H_{\text{dissociation}} + \Delta H_{\text{reaction}} \]

### Step 4: Rearrange the Equation
We can rearrange the equation to find the heat of dissociation (\( \Delta H_{\text{dissociation}} \)):
\[ \Delta H_{\text{dissociation}} = \Delta H_{\text{neutralization}} - \Delta H_{\text{reaction}} \]

### Step 5: Substitute the Values
Now we substitute the known values into the equation:
\[ \Delta H_{\text{dissociation}} = -13.7 \, \text{kcal} - (-4.7 \, \text{kcal}) \]
This simplifies to:
\[ \Delta H_{\text{dissociation}} = -13.7 + 4.7 \]
\[ \Delta H_{\text{dissociation}} = -9.0 \, \text{kcal} \]

### Step 6: Conclusion
Thus, the heat of dissociation of HA is:
\[ \Delta H_{\text{dissociation}} = -9.0 \, \text{kcal} \]

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