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 lithium hydroxide (LiOH), we can use the given data about the heat of neutralization of LiOH with hydrochloric acid (HCl) and the standard heat of neutralization for strong acids and bases.

### Step-by-Step Solution:

1. **Understand the Heat of Neutralization**:
The heat of neutralization for a strong acid (HCl) and a strong base (LiOH) is typically around -57.1 kJ/mol. This value represents the heat released when one mole of water is formed from the neutralization reaction.

2. **Write the Neutralization Reaction**:
The neutralization reaction can be written as:
\[ \text{LiOH} + \text{HCl} \rightarrow \text{LiCl} + \text{H}_2\text{O} \]
The heat of neutralization for this reaction is given as -34.868 kJ/mol.

3. **Write the Ionization Reaction**:
The ionization of LiOH can be represented as:
\[ \text{LiOH} \rightarrow \text{Li}^+ + \text{OH}^- \]
We need to find the heat of this ionization reaction, which we will denote as \(\Delta H_{ionization}\).

4. **Use Hess's Law**:
According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can set up the following equations:
- For the formation of water from H+ and OH-:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \quad \Delta H = -57.1 \text{ kJ/mol} \quad \text{(Equation 1)} \]
- For the neutralization of LiOH with HCl:
\[ \text{LiOH} + \text{H}^+ \rightarrow \text{Li}^+ + \text{H}_2\text{O} \quad \Delta H = -34.868 \text{ kJ/mol} \quad \text{(Equation 2)} \]

5. **Combine the Equations**:
By combining Equation 1 and Equation 2, we can eliminate H2O:
\[ \text{H}^+ + \text{OH}^- + \text{LiOH} + \text{H}^+ \rightarrow \text{Li}^+ + \text{H}_2\text{O} + \text{H}^+ \]
Rearranging gives us:
\[ \text{LiOH} \rightarrow \text{Li}^+ + \text{OH}^- \]
The overall enthalpy change for this reaction will be:
\[ \Delta H_{ionization} = \Delta H_{neutralization} + \Delta H_{formation} \]
Thus:
\[ \Delta H_{ionization} = -34.868 \text{ kJ/mol} + 57.1 \text{ kJ/mol} \]

6. **Calculate the Heat of Ionization**:
\[ \Delta H_{ionization} = 22.232 \text{ kJ/mol} \]

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