# Ionization of Water Water tends to dissociate into hydrogen ions and hydroxide ions, and then re-form again.

A water molecule comprises an oxygen atom and two hydrogen atoms covalently bonded together. Water tends to separate into its component parts, creating H+ (hydrogen) ions and OH (hydroxide) ions. When this happens, water is said to be self-ionized. While reversible, this reaction is very important to living systems because H+ and OH ions are very reactive. Too much of either ion relative to the other within the cell can cause proteins to denature. The free hydrogen ions (which are essentially bare protons) bond with a nearby water molecule, forming hydronium ions (H3O+).

The equilibrium for water self-ionization is described by this equation:
$2{\rm {H}_2 \rm O(l)}\leftrightarrows{\rm H}_3{\rm O}^+(aq)+{\rm{OH}}^-(aq)$
Step-By-Step Example
Calculating the Equilibrium for Water Self-Ionization
Kw, the equilibrium constant for the reaction, known as the ionization product constant of water, can be defined as the product of the hydronium ion concentration, [H3O+], and the hydroxide ion concentration [OH].
$K_{\rm w}=\lbrack{\rm H}_3\text O^+\rbrack\lbrack\rm{OH}^-\rbrack$
Step 1
At 25°C, these concentrations in pure water are equal. Therefore, knowing the value of Kw allows the calculation of [H3O+] and [OH].
$K_{\rm w}=\lbrack{\rm H}_3\text O^+\rbrack\lbrack\rm{OH}^-\rbrack=1.0\times10^{-14}$
Step 2
Since the two concentrations are the same, the square of one of either concentration can be used to find the pH:
$1.0\times10^{-14}=\lbrack{\rm H}_3\rm O^+\rbrack^2$
Solution
Take the square root of both sides of the equation to find the concentration.
$\lbrack{\rm H}_3\rm O^+\rbrack=1.0\times10^{-7}$
The exponent indicates that the pH is 7. When there is an equal concentration of H+ and OH ions in a solution, the solution is neutral.