# On the basis of the following thermo chemical data $[ \Delta G_f^{\circ} (H^+ )(aq) = 0]$  $H_2O(l) \rightarrow H^+ (aq) + OH^-(aq);\: \: \: \: \: \Delta H = 57.32\: kJ$  $H_2(g) + \large\frac{1}{2} (g) \rightarrow H_2O(l); \: \: \: \: \: \: \Delta H = -286\: kJ$  The value of enthalpy of formation of $OH^-\: ion$ at $25^{\circ}C$ is

$\begin {array} {1 1} (A)\;-22.88\; kJ & \quad (B)\;-228.88\: kJ \\ (C)\;228.88 \: kJ & \quad (D)\;-343.52\: kJ \end {array}$

For heat of formation of $H_2O,$
$\Delta H_r = \Delta H_f (H_2O,l) - \Delta H_f\: (H_2,g) – \large\frac{1}{2}$$\Delta H_f(O_2,g) – 286.2 = -286.2\: kJ$
For ionization of $H_2O,$
$\Delta H_r = \Delta H_f(H^+, aq) + \Delta H_f ( OH^-, aq) - \Delta H_f (H_2, O, l)$
$\Rightarrow 57.32 = 0 + \Delta H_f (OH^- , aq) – (-286.2)$
Thus, $\Delta H_f (OH^- , aq) = 57.32 – 286.2 = -228.88\: kJ$
Ans : (B)