The pH of pure water at $25°Cand35°C$ are 7 and 6 respectively. Calculate the heat of formation $(inK.calmo{l}^{-1})$ of water from $H^{+}$ and $O{H}^{-}.$ $(GivenR=2cal{K}^{-1}mo{l}^{-1})$

At $25°C,pH=7,\left[H^{+}\right]={10}^{-7},K_W={10}^{-14}.$
At $35°C,pH=6,\left[H^{+}\right]={10}^{-6},K_W={10}^{-12}.$
The van’t Hoff equation for water is $\log \frac{K_2}{K_1}=+\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]=\frac{\Delta H}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right].$

$\log \frac{{10}^{-12}}{{10}^{-14}}=\frac{\Delta H}{2.303\times 2}\left[\frac{308-298}{298\times 308}\right].$