`HA + OH rightarrow H_(2)O + A^(-+) q_(1) kJ
H^(+) + OH ^(-) rightarrow H_(2)O + q_(2)kJ` the enthalpy of ionisation of HA is

Options

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

Answer

To determine the enthalpy of ionization of the acid HA, we can follow these steps:

### Step 1: Write the Ionization Reaction
The ionization of the acid HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
This reaction indicates that when HA ionizes, it produces hydrogen ions (H⁺) and the conjugate base (A⁻).

### Step 2: Write the Given Reactions
We have two reactions provided:
1. \( HA + OH^- \rightarrow H_2O + A^- \) with enthalpy change \( q_1 \) kJ
2. \( H^+ + OH^- \rightarrow H_2O \) with enthalpy change \( q_2 \) kJ

### Step 3: Reverse the Second Reaction
To relate the ionization of HA to the two given reactions, we need to reverse the second reaction:
\[ H_2O \rightarrow H^+ + OH^- \]
When we reverse a reaction, the sign of the enthalpy change also changes:
\[ \text{Enthalpy change becomes } -q_2 \text{ kJ} \]

### Step 4: Combine the Reactions
Now, we can add the two reactions together:
1. \( HA + OH^- \rightarrow H_2O + A^- \) (enthalpy change = \( q_1 \))
2. \( H_2O \rightarrow H^+ + OH^- \) (enthalpy change = \( -q_2 \))

When we add these two reactions, the \( H_2O \) cancels out:
\[ HA + OH^- + H_2O \rightarrow H_2O + A^- + H^+ \]

This simplifies to:
\[ HA \rightarrow H^+ + A^- \]

### Step 5: Write the Enthalpy Change for the Overall Reaction
The overall enthalpy change for the ionization of HA is:
\[ \Delta H = q_1 - q_2 \]

### Conclusion
Thus, the enthalpy of ionization of the acid HA is given by:
\[ \Delta H = q_1 - q_2 \]