the enthalpy of formation of H₂O(l) is -280.70 kj/mol and enthalpy of neutralisation of strong acid and strong base is -56.70 kj/mol. What is the enthalpy of formation of OH^- ions

1. Formation of H₂O (l)  from H₂ and O₂ gases is as follows :
  
      $H_2\left(g\right) + \frac{1}{2}{O}_2\left(g\right) \to H_{2}O (∆H_f=-280.70kJ/mol)$

2. Neutralisation reaction of strong acid and strong base involves :
     
      $H^{+} +\hspace{0.17em}O{H}^{-} \to H_{2}O \left(l\right) (∆H_n= -56.70kJ/mol)$

3. Formation of OH^- ions :
$$\begin{gathered} H_2\left(g\right) + \frac{1}{2}{O}_2\left(g\right) \to H_{2}O ∆H_f=-280.70kJ/mol \\ - H^{+} \left(l\right) + O{H}^{-}\left(l\right) \to H_{2}O\hspace{0.17em} ∆H_n = -56.70kJ/mol \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ H_2\left(g\right) + \frac{1}{2}{O}_2\left(g\right) \to H^{+} \left(l\right) +\hspace{0.17em}O{H}^{-}\left(l\right) ∆H=-280.70+56.70 =-224kJ/mol \end{gathered}$$

4. The enthalpy of formation of OH^- ions is -224kJ/mol.