The heat of neutralization of LiOh and HCI at `25 ^(@)C` is `34.868 kJ mol^(-1)` . The heat of ionisation of LiOh will be

Options

A. `44.674 kJ`
B. `22.232 kJ`
C. `32.684 kJ`
D. `96.464 kJ`

Answer

To find the heat of ionization of LiOH, we can use the information given about the heat of neutralization and the known heat of neutralization for strong acids and bases. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When LiOH (a strong base) reacts with HCl (a strong acid), the reaction can be represented as:
\[ \text{LiOH} + \text{HCl} \rightarrow \text{Li}^+ + \text{Cl}^- + \text{H}_2\text{O} \]
- The heat of neutralization (\( \Delta H_{neut} \)) for this reaction is given as \( -34.868 \, \text{kJ/mol} \).

2. **Heat of Neutralization for Strong Acid and Base**:
- For a strong acid (HCl) and a strong base (NaOH, for example), the heat of neutralization is typically \( -57.1 \, \text{kJ/mol} \). This value represents the heat released when one mole of water is formed from the neutralization of a strong acid and a strong base.

3. **Setting Up the Equations**:
- The heat of ionization of LiOH can be represented by the following reaction:
\[ \text{LiOH} \rightarrow \text{Li}^+ + \text{OH}^- \]
- We denote the heat of ionization of LiOH as \( \Delta H_{ion} \).

4. **Combining the Reactions**:
- The overall process can be viewed as:
- Ionization of LiOH:
\[ \text{LiOH} \rightarrow \text{Li}^+ + \text{OH}^- \quad (\Delta H_{ion}) \]
- Neutralization of the OH⁻ ion with H⁺ from HCl:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \quad (\Delta H = -57.1 \, \text{kJ/mol}) \]
- The total reaction can be expressed as:
\[ \text{LiOH} + \text{HCl} \rightarrow \text{Li}^+ + \text{Cl}^- + \text{H}_2\text{O} \]
- Therefore, we can write:
\[ \Delta H_{neut} = \Delta H_{ion} + (-57.1 \, \text{kJ/mol}) \]

5. **Calculating the Heat of Ionization**:
- Rearranging the equation gives:
\[ \Delta H_{ion} = \Delta H_{neut} + 57.1 \, \text{kJ/mol} \]
- Substituting the values:
\[ \Delta H_{ion} = -34.868 \, \text{kJ/mol} + 57.1 \, \text{kJ/mol} \]
- Performing the calculation:
\[ \Delta H_{ion} = 22.232 \, \text{kJ/mol} \]

### Final Answer:
The heat of ionization of LiOH is \( 22.232 \, \text{kJ/mol} \).