# What is the result of dissociation of water?

May 15, 2017

$\text{Of autoprotolysis..............?}$

#### Explanation:

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

The formation of equilibrium quantities of the characteristic cation, the $\text{acidium}$ ion or so-called $\text{hydronium ion}$, and $\text{hydroxide}$ ion.

At $298 \cdot K$ this equilibrium has been VERY carefully measured, and the ion product, ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$, and taking $- {\log}_{10}$ of both sides we gets............

$p {K}_{w} = p H + p O H = - \left(- 14\right) = 14$.

How do you think $p {K}_{w}$ would evolve at higher temperatures? Would it increase, decrease, stay the same. Remember that this autoprotolysis is manifestly a $\text{bond-breaking reaction}$.