Question

The pH of pure water at $25^{\circ} C$ and $35^{\circ} C$ are $7$ and $6$ respectively. The heat of formation of water from $H ^{+}$ and $OH ^{-}$ is_____.

Answer 1

At $25^{\circ} C ;\left[ H ^{+}\right]=10^{-7}$

$\therefore K _{ w }=10^{-14}$

At $35^{\circ} C ;\left[ H ^{+}\right]=10^{-6}$

$\therefore K _{ w }=10^{-12}$

Now using

$2.303 \log \frac{ K _{ w _{2}}}{ K _{ w _{1}}}=\frac{\Delta H }{ R }\left[\frac{ T _{2}- T _{1}}{ T _{1} \times T _{2}}\right]$

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

$\therefore \Delta H =84551.4\, cal / mol $

$=84.551\, kcal / mol $

Thus, $H _{2} O \rightleftharpoons H ^{+}+ OH ^{-}$

$\Delta H =-84.551 \,kcal / mol$