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?

Explanation

To find the enthalpy of formation of hydroxide ions (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.

### Step-by-Step Solution:

1. **Identify the given values:**
- Enthalpy of formation of water (H₂O(l)): ΔH_f(H₂O) = -285 kJ/mol
- Enthalpy of neutralization (H⁺ + OH⁻ → H₂O): ΔH_neutralization = -55 kJ/mol

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