Question

The pH of pure water at ${25}^{o}C$ and ${35}^{o}C$ are $7$ and $6$ respectively. The heat of formation of $H^{+}$ and $O{H}^{-}$ ions in $kcalmo{l}^{-1}$ is :

• A. $-84.5$
• B. $+84.5$
• C. $+184.5$
• D. $-184.5$

Answer 1

The correct option is B $+84.5$

At ${25}^{o}C$, pH=$7$, $\left[H^{+}\right]={10}^{-7},K_w={10}^{-14}.$
At ${35}^{o}C$, pH=$6$, $\left[H^{+}\right]={10}^{-6},K_w={10}^{-12}.$
The van't Hoff equation for water is $log\frac{K_2}{K_1}=+\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]=\frac{\Delta H}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$.
Here, $K_1$ and $K_2$ are the values of the equilibrium constants (or ionic products) at temperatures $T_1$ and $T_2$ respectively.
$\Delta H$ is the enthalpy of the reaction and R is the ideal gas constant.
Substituting values in the above expression, we get
$log\frac{{10}^{-12}}{{10}^{-14}}=\frac{\Delta H}{2.303\times 2}\left[\frac{308-298}{298\times 308}\right]$.
Thus, $\Delta H=+84.5kcalmo{l}^{-1}$.