1. Given thermochemical equations

   We are given:

   $\mathrm{H_2O(l) \rightarrow H^+(aq) + OH^-(aq)} \qquad \Delta H = +57.32\,\text{kJ}$

   and

   $\mathrm{H_2(g) + \tfrac12 O_2(g) \rightarrow H_2O(l)} \qquad \Delta H = -286.20\,\text{kJ}$

   Also,

   $\Delta_f H^\circ\big(\mathrm{H^+(aq)}\big)=0$

2. Target reaction

   We need the enthalpy of formation of $\mathrm{OH^- (aq)}$, i.e. for the reaction:

   $\mathrm{\tfrac12 H_2(g) + \tfrac12 O_2(g) \rightarrow OH^-(aq)}$

   Strictly in aqueous convention, this can be viewed through Hess's law using $\Delta_f H^\circ(\mathrm{H^+})=0$.

3. Apply Hess's law

   Add the two given equations:

   • Formation of water: $\mathrm{H_2(g) + \tfrac12 O_2(g) \rightarrow H_2O(l)} \qquad \Delta H=-286.20$

   • Ionization of water: $\mathrm{H_2O(l) \rightarrow H^+(aq) + OH^-(aq)} \qquad \Delta H=+57.32$

   On adding, $\mathrm{H_2O(l)}$ cancels:

   $\mathrm{H_2(g) + \tfrac12 O_2(g) \rightarrow H^+(aq) + OH^-(aq)}$

   Therefore,

   $\Delta H = -286.20 + 57.32 = -228.88\,\text{kJ}$

4. Use standard enthalpy of formation relation

   For the above reaction,

   $\Delta H = \Delta_f H^\circ(\mathrm{H^+}) + \Delta_f H^\circ(\mathrm{OH^-})$

   since reactants are elements in standard state.

   Given:

   $\Delta_f H^\circ(\mathrm{H^+})=0$

   Hence,

   $\Delta_f H^\circ(\mathrm{OH^-}) = -228.88\,\text{kJ mol}^{-1}$

5. Option check

   • A: $-22.88$ kJ ❌
   • B: $-228.88$ kJ ✅
   • C: $+228.88$ kJ ❌
   • D: $-343.52$ kJ ❌

Final answer: Option B.