Terms

• ionizationAny process that leads to the dissociation of a neutral atom or molecule into charged particles (ions).
• autoprotolysisThe autoionization of water (or similar compounds) in which a proton (hydrogen ion) is transferred to form a cation and an anion.
• hydroniumThe hydrated hydrogen ion ( [latex]H_3O^+[/latex] ).

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:

[latex]{H}_{2}O + {H}_{2}O \rightleftharpoons {H}_{3}{O}^{+} + O{H}^{-}[/latex]

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).

The Water Ionization Constant, K_W

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

[latex]{H}_{2}O + {H}_{2}O \rightleftharpoons {H}_{3}{O}^{+} + O{H}^{-}\quad\quad\quad K_W=1.0\times10^{-14}[/latex]

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:

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

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

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

Now, note the definition of pH and pOH:

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

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

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

[latex]\text{pH}=-log(1.0\times 10^{-7})=7.0[/latex]

[latex]\text{pOH}=-log(1.0\times 10^{-7})=7.0[/latex]

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:

[latex]K_W=[H^+][OH^-][/latex]

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

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

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

[latex]\text{pK}_\text{W}=\text{pH}+\text{pOH}[/latex]

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

[latex]\text{pH}+\text{pOH}=14[/latex]

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.