The enthalpy of neutralisation of a strong acid by a string base is `-57.32 kJ mol^(-1)`. The enthalpy of formation of water is `-285.84 kJ mol^(-1)`. The enthalpy of formation of hydroxyl ion is

Options

A. `+228.52" kJ/mole"`
B. `-114.26 " kJ/mole"`
C. `-228.52" kJ/mole"`
D. `+114.26" kJ/mole"`

Answer

To find the enthalpy of formation of the hydroxyl ion (OH⁻), we will use the given data about the enthalpy of neutralization and the enthalpy of formation of water.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The neutralization of a strong acid (H⁺) by a strong base (OH⁻) can be represented as:
\[ \text{H}^+ (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O} (l) \]
The enthalpy change for this reaction (enthalpy of neutralization) is given as:
\[ \Delta H_{\text{neutralization}} = -57.32 \, \text{kJ/mol} \]

2. **Reversing the Reaction**:
If we reverse the above reaction, we get:
\[ \text{H}_2\text{O} (l) \rightarrow \text{H}^+ (aq) + \text{OH}^- (aq) \]
The enthalpy change for this reverse reaction will be:
\[ \Delta H = +57.32 \, \text{kJ/mol} \]

3. **Enthalpy of Formation of Water**:
The enthalpy of formation of water (H₂O) from its elements (H₂ and O₂) is given as:
\[ \text{H}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{H}_2\text{O} (l) \]
The enthalpy change for this reaction is:
\[ \Delta H_{\text{formation of water}} = -285.84 \, \text{kJ/mol} \]

4. **Combining the Reactions**:
Now, we will combine the two reactions to find the enthalpy of formation of the hydroxyl ion (OH⁻):
- The first reaction (reversed) gives us:
\[ \text{H}_2\text{O} (l) \rightarrow \text{H}^+ (aq) + \text{OH}^- (aq) \quad \Delta H = +57.32 \, \text{kJ/mol} \]
- The second reaction is:
\[ \text{H}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{H}_2\text{O} (l) \quad \Delta H = -285.84 \, \text{kJ/mol} \]

5. **Adding the Enthalpy Changes**:
When we add these two reactions, we can cancel out H₂O:
\[ \text{H}_2 (g) + \frac{1}{2} \text{O}_2 (g) + \text{H}_2\text{O} (l) \rightarrow \text{H}^+ (aq) + \text{OH}^- (aq) \]
The total enthalpy change for this combined reaction is:
\[ \Delta H = +57.32 \, \text{kJ/mol} - 285.84 \, \text{kJ/mol} = -228.52 \, \text{kJ/mol} \]

6. **Conclusion**:
Therefore, the enthalpy of formation of the hydroxyl ion (OH⁻) is:
\[ \Delta H_{\text{formation of OH}^-} = -228.52 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of the hydroxyl ion (OH⁻) is \(-228.52 \, \text{kJ/mol}\).

---