`HA+OH^(-) to H_(2)O+A^(-)+q_(1) kJ`
`H^(+)+OH^(-) to H_(2)O+ q_(2) kJ`
The enthalpy of dissociation of `HA` is

Options

A. `(q_(1)+q_(2))`
B. `(q_(1)-q_(2))`
C. `(q_(2)-q_(1))`
D. `-(q_(1)+q_(2))`

Explanation

To find the enthalpy of dissociation of HA, we can analyze the given reactions step by step.

### Step 1: Write down the reactions
We have two reactions provided:
1. \( \text{HA} + \text{OH}^- \rightarrow \text{H}_2\text{O} + \text{A}^- + q_1 \, \text{kJ} \)
2. \( \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} + q_2 \, \text{kJ} \)

### Step 2: Identify the dissociation of HA
The dissociation of HA can be represented as:
\[ \text{HA} \rightarrow \text{H}^+ + \text{A}^- \]

### Step 3: Relate the reactions to the dissociation of HA
To find the enthalpy of dissociation of HA, we can manipulate the two given reactions. We can subtract the second reaction from the first reaction.

### Step 4: Subtract the second reaction from the first
Subtracting the second reaction from the first gives:
\[ (\text{HA} + \text{OH}^- \rightarrow \text{H}_2\text{O} + \text{A}^- + q_1) - (\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} + q_2) \]

### Step 5: Cancel out common terms
When we perform the subtraction, the \( \text{OH}^- \) and \( \text{H}_2\text{O} \) terms cancel out:
\[ \text{HA} - \text{H}^+ \rightarrow \text{A}^- + (q_1 - q_2) \]

### Step 6: Rearranging the equation
This simplifies to:
\[ \text{HA} \rightarrow \text{H}^+ + \text{A}^- + (q_1 - q_2) \]

### Step 7: Conclusion
The enthalpy of dissociation of HA is given by the heat change associated with this reaction, which is:
\[ \Delta H_{\text{dissociation}} = q_1 - q_2 \]

Thus, the enthalpy of dissociation of HA is \( q_1 - q_2 \).

### Final Answer
The enthalpy of dissociation of HA is \( q_1 - q_2 \).

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