The enthalpy of formation of `H_(2)O(l)` is `-285 KJ mol^(-1)` and enthalpy of neutralization of a stron acid and a strong bas is `-55 KJ mol^(-1)`. What is the enthalpy of formation of `OH^(-)` ions?

Options

A. `-228.5 KJ "mol"^(-1)`
B. `228.5 KJ "mol"^(-1)`
C. `114.5 KJ "mol"^(-1)`
D. `-114.5 KJ "mol"^(-1)`

Explanation

To find the enthalpy of formation of the hydroxide ion (OH⁻), we can use the given data about the enthalpy of formation of water (H₂O) and the enthalpy of neutralization of a strong acid and a strong base. Here’s how we can solve the problem step by step:

### Step 1: Write the reaction for the formation of water
The enthalpy of formation of liquid water (H₂O(l)) is given by the reaction:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \]
The enthalpy change (\(\Delta H\)) for this reaction is:
\[ \Delta H = -285 \, \text{kJ/mol} \]

### Step 2: Write the reaction for the neutralization of a strong acid and a strong base
The enthalpy of neutralization is given by the reaction:
\[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2O(l) \]
The enthalpy change for this reaction is:
\[ \Delta H = -55 \, \text{kJ/mol} \]

### Step 3: Reverse the neutralization reaction
If we reverse the neutralization reaction, we get:
\[ \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{OH}^-(aq) \]
The enthalpy change for this reversed reaction will be:
\[ \Delta H = +55 \, \text{kJ/mol} \]

### Step 4: Combine the two reactions
Now we can add the two reactions together:
1. Formation of water:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \quad (\Delta H = -285 \, \text{kJ/mol}) \]
2. Reversed neutralization:
\[ \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{OH}^-(aq) \quad (\Delta H = +55 \, \text{kJ/mol}) \]

Adding these reactions gives:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}^+(aq) + \text{OH}^-(aq) \]

### Step 5: Calculate the overall enthalpy change
Now we can calculate the overall enthalpy change for the combined reaction:
\[ \Delta H = -285 \, \text{kJ/mol} + 55 \, \text{kJ/mol} = -230 \, \text{kJ/mol} \]

### Step 6: Conclusion
The enthalpy of formation of the hydroxide ion (OH⁻) can be concluded as:
\[ \Delta H_{\text{formation of } OH^-} = -230 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of formation of OH⁻ ions is \(-230 \, \text{kJ/mol}\).

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