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

Question

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

(A) $-64.55 \mathrm{kcal} / \mathrm{mol}$

(B) $-74.55 \mathrm{kcal} / \mathrm{mol}$

(C) $-84.55 \mathrm{kcal} / \mathrm{mol}$

(D) $-54.55 \mathrm{kcal} / \mathrm{mol}$

Answer 1

The correct option is (C).

SOLUTION —

At $25^{\circ} \mathrm{C},\left[\mathrm{H}^{+}\right]=10^{-7} . \therefore K_{w_1}=10^{-14}$

At $35^{\circ} \mathrm{C},\left[\mathrm{H}^{+}\right]=10^{-6} . \therefore K_{w_2}=10^{-12}$

Now using, $2.303 \log \frac{K_{w_2}}{K_{w_1}}=\frac{\Delta H}{R}\left[\frac{T_2-T_1}{T_1 \times T_2}\right]$

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

$\therefore \Delta H=84551.4 \mathrm{cal} / \mathrm{mol}=84.551 \mathrm{kcal} / \mathrm{mol}$

Thus, $\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} ; \Delta H=84.551 \mathrm{kcal} / \mathrm{mol}$

$\therefore \mathrm{H}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{H}_2 \mathrm{O} ; \Delta H=-84.551 \mathrm{kcal} / \mathrm{mol}$