# What is the reaction of an ion with H_2O to produce H^+(aq) and OH^(-)(aq)?

Apr 12, 2017

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

#### Explanation:

OR, MORE COMMONLY..........

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

This is the so-called $\text{autoprotolysis reaction of water}$, which has been meticulously studied, and underlies our conception of acid/base chemistry.

The $\text{hydronium ion}$, ${H}_{3} {O}^{+}$, and $\text{hydroxide ion}$, $H {O}^{-}$, are conceived to be the characteristic cation and characteristic anion of the $\text{WATER SOLVENT}$, i.e. $\text{the acidium ion}$, and $\text{the hydroxide ion}$. I write $\text{conceived}$ because the acidium ion in water solution is probably ${H}_{5} {O}_{2}^{+}$ or ${H}_{7} {O}_{3}^{+}$, a cluster of several water molecules WITH AN EXTRA ${H}^{+}$ associated with the cluster. We write ${H}^{+}$, or ${H}_{3} {O}^{+}$ to represent this species, i.e. as a shorthand.

In water at $298 \cdot K$,

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$ ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$.

Given that this is a bond-breaking reaction, how do you think this equilibrium would evolve AT HIGHER TEMPERATURES? (If you are not an undergrad, don't bother with this question).