Heat of neutralisation of `NaOH` and `HCl` is `-57.46 kJ //` equivalent. The heat of ionisation of water in `kJ //mol` is `:`

Options

A. `-57.46`
B. `+57.46`
C. `-114.92`
D. `+114.92`

Answer

To find the heat of ionization of water based on the given heat of neutralization of NaOH and HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The neutralization reaction between NaOH (a strong base) and HCl (a strong acid) can be represented as:
\[ \text{NaOH (aq)} + \text{HCl (aq)} \rightarrow \text{NaCl (aq)} + \text{H}_2\text{O (l)} \]
The heat of this reaction is given as -57.46 kJ per equivalent.

2. **Identify the Ionization of Water**: The ionization of water can be represented as:
\[ \text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ (\text{aq}) + \text{OH}^- (\text{aq}) \]
This reaction is the reverse of the neutralization reaction.

3. **Relate the Heat of Neutralization to Ionization**: The heat released during the neutralization reaction corresponds to the formation of water from its ions. Therefore, the heat of ionization of water is the reverse of the heat of neutralization:
\[ \Delta H_{\text{ionization}} = -(\Delta H_{\text{neutralization}}) \]
Thus:
\[ \Delta H_{\text{ionization}} = -(-57.46 \text{ kJ}) = +57.46 \text{ kJ} \]

4. **Final Answer**: The heat of ionization of water is:
\[ \Delta H_{\text{ionization}} = +57.46 \text{ kJ/mol} \]