# How does water react to form ions?

Dec 29, 2015

Water undergoes autoprotolysis. The reaction is given in the next section.

#### Explanation:

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

Alternatively:

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

Reaction (ii) is a bit more common these days. What we conceive to be the acidium ion, ${H}^{+} \mathmr{and} {H}_{3} {O}^{+}$, is actually a cluster of water atoms (4-5) with an extra proton, i.e. ${H}_{9} {O}_{4}^{+}$.

In any case the equilibrium is known for standard conditions ($1$ $a t m$, $298 K$), and the ion product, ${K}_{w}$ $=$ $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ $=$ ${10}^{- 14}$.

Under non-standard conditions (i.e. > $298 K$), would you expect ${K}_{w}$ to increase or decrease? Why?