Acid-Base Properties of Water

Under standard conditions, water will self-ionize to a very small extent. The self-ionization of water refers to the reaction in which a water molecule donates one of its protons to a neighboring water molecule, either in pure water or in aqueous solution. The result is the formation of a hydroxide ion (OH^-) and a hydronium ion (H₃O⁺). The reaction can be written as follows:

${H}_{2}O + {H}_{2}O \rightleftharpoons {H}_{3}{O}^{+} + O{H}^{-}$

This is an example of autoprotolysis (meaning "self-protonating") and it exemplifies the amphoteric nature of water (ability to act as both an acid and a base).

Autoprotolysis of water

The self-ionization of water produces hydronium and hydroxide ions in solution.

The Water Ionization Constant, K_W

Note that the self-ionization of water is an equilibrium reaction:

${H}_{2}O + {H}_{2}O \rightleftharpoons {H}_{3}{O}^{+} + O{H}^{-}\quad\quad\quad K_W=1.0\times10^{-14}$

Like all equilibrium reactions, this reaction has an equilibrium constant. Because this is a special equilibrium constant, specific to the self-ionization of water, it is denoted K_W; it has a value of 1.0 x 10⁻¹⁴. If we write out the actual equilibrium expression for K_W, we get the following:

$K_W=[H^+][OH^-]=1.0\times 10^{-14}$

However, because H⁺ and OH^- are formed in a 1:1 molar ratio, we have:

$[H^+]=[OH^-]=\sqrt{1.0\times 10^{-14}}=1.0\times 10^{-7}\;M$

Now, note the definition of pH and pOH:

$\text{pH}=-log[H^+]$

$\text{pOH}=-log[OH^-]$

If we plug in the above value into our equation for pH, we find that:

$\text{pH}=-log(1.0\times 10^{-7})=7.0$

$\text{pOH}=-log(1.0\times 10^{-7})=7.0$

Here we have the reason why neutral water has a pH of 7.0; it represents the condition at which the concentrations of H⁺ and OH^- are exactly equal in solution.

pH, pOH, and pK_W

We have already established that the equilibrium constant K_W can be expressed as:

$K_W=[H^+][OH^-]$

If we take the negative logarithm of both sides of this equation, we get the following:

$-log(K_W)=-log([H^+][OH^-])$

$-log(K_W)=-log[H^+]+-log[OH^-]$

$\text{pK}_\text{W}=\text{pH}+\text{pOH}$

However, because we know that pK_W = 14, we can establish the following relationship:

$\text{pH}+\text{pOH}=14$

This relationship always holds true for any aqueous solution, regardless of its level of acidity or alkalinity. Utilizing this equation is a convenient way to quickly determine pOH from pH and vice versa, as well as to determine hydroxide concentration given hydrogen concentration, or vice versa.