The enthalpy of formation of `H_(2)O(l)` is -280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong base is -56.70 kJ/mol. What is the enthalpy of formation of `OH^(-)` ions?

Options

A. `-22.9` kJ/mol
B. `-224` kJ/mol
C. `58.7` kJ/mol
D. `214` kJ/mol

Answer

To find the enthalpy of formation of OH⁻ ions, 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.

### Step-by-Step Solution:

1. **Understand the Reactions Involved:**
- The enthalpy of formation of water (H₂O) is given as:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H = -280.70 \, \text{kJ/mol} \]
- The enthalpy of neutralization of a strong acid (H⁺) and a strong base (OH⁻) is given as:
\[ \text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \quad \Delta H = -56.70 \, \text{kJ/mol} \]

2. **Write the Reaction for the Formation of OH⁻:**
- We can rearrange the formation of water to express it in terms of H⁺ and OH⁻:
\[ \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \quad \Delta H = +56.70 \, \text{kJ/mol} \]
- This is the reverse of the neutralization reaction, so we change the sign of ΔH.

3. **Combine the Two Reactions:**
- Now, we can add the two reactions together:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H = -280.70 \, \text{kJ/mol} \]
\[ \text{H}_2\text{O} \rightarrow \text{H}^+ + \text{OH}^- \quad \Delta H = +56.70 \, \text{kJ/mol} \]
- When we add these two reactions, the H₂O cancels out:
\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}^+ + \text{OH}^- \quad \Delta H = -280.70 + 56.70 \]

4. **Calculate the Enthalpy Change:**
- Now we calculate the total ΔH:
\[ \Delta H = -280.70 + 56.70 = -224.00 \, \text{kJ/mol} \]

5. **Conclusion:**
- The enthalpy of formation of OH⁻ ions is:
\[ \Delta H_{\text{f}}(\text{OH}^-) = -224.00 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of OH⁻ ions is **-224.00 kJ/mol**.