The standard heat of formation of sodium ions in aqueous solution from the following data :
Heat of formation of `NaOH(aq)` at `25^(@)C = - 470.7 KJ` :
Heat of formation of `OH^-` at `25^(@)C = -228.8 KJ` is :

Options

A. `-251.9 KJ`
B. 241.9 KJ
C. `-241.9 KJ`
D. `300 Kjmol^(-1)`

Answer

To find the standard heat of formation of sodium ions in aqueous solution, we can use the given data about the heat of formation of sodium hydroxide (NaOH) and hydroxide ions (OH⁻).

### Step-by-Step Solution:

1. **Write the Reaction**:
The dissociation of sodium hydroxide in aqueous solution can be represented as:
\[ \text{NaOH (aq)} \rightarrow \text{Na}^+ (aq) + \text{OH}^- (aq) \]

2. **Identify Given Values**:
- Heat of formation of NaOH (aq): \( \Delta H_f (\text{NaOH}) = -470.7 \, \text{kJ} \)
- Heat of formation of OH⁻: \( \Delta H_f (\text{OH}^-) = -228.8 \, \text{kJ} \)

3. **Apply Hess's Law**:
According to Hess's law, the heat of formation of a compound can be expressed as the sum of the heats of formation of its constituent ions. Therefore, we can write:
\[ \Delta H_f (\text{NaOH}) = \Delta H_f (\text{Na}^+) + \Delta H_f (\text{OH}^-) \]

4. **Rearranging the Equation**:
To find the heat of formation of sodium ions (\( \Delta H_f (\text{Na}^+) \)), we can rearrange the equation:
\[ \Delta H_f (\text{Na}^+) = \Delta H_f (\text{NaOH}) - \Delta H_f (\text{OH}^-) \]

5. **Substituting the Values**:
Now, substitute the known values into the equation:
\[ \Delta H_f (\text{Na}^+) = -470.7 \, \text{kJ} - (-228.8 \, \text{kJ}) \]
Simplifying this gives:
\[ \Delta H_f (\text{Na}^+) = -470.7 + 228.8 \]

6. **Calculating the Result**:
Performing the arithmetic:
\[ \Delta H_f (\text{Na}^+) = -241.9 \, \text{kJ} \]

### Final Answer:
The standard heat of formation of sodium ions in aqueous solution is:
\[ \Delta H_f (\text{Na}^+) = -241.9 \, \text{kJ} \]