We have seen several examples demonstrating that water is an amphoteric compound as it can act both as an acid and as a base. For example, it reacts with HCl as a base, and with NH₃ as an acid:

 

 

To generalize, we can say that in the presence of an acid, water acts as a base, while in the presence of a base, water acts as an acid.

 

Now, the question is can it react with itself? Can one water molecule be the acid and donate a proton to another one serving as the base?

 

The answer is yes, it can, and this reaction indeed occurs:

 

H₂O(l) + H₂O(l) ⇆ H₃O⁺(aq) + OH^–(aq)

 

This is called the autoionization or dissociation of water which, in a simpler form, can be shown as:

 

H₂O(l) ⇆ H⁺(aq) + OH^–(aq)

 

The equilibrium constant expression for this reaction is:

 

\[K\, = \,{\mkern 1mu} \frac{{[{{\rm{H}}^{\rm{ + }}}][{\rm{O}}{{\rm{H}}^{\rm{ – }}}]}}{{[{{\rm{H}}_{\rm{2}}}{\rm{O}}]}}\; = \;\frac{{[{{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}][{\rm{O}}{{\rm{H}}^{\rm{ – }}}]}}{{[{{\rm{H}}_{\rm{2}}}{\rm{O}}]}}\]

 

It is determined that the concentration of water is constant (55.5 mol/l), so it is omitted from the expression, and the equilibrium constant becomes K_w which is equal to :

This equilibrium constant is called the ion product constant (dissociation constant) of water (K_w).

What is important is that the product of [H⁺] and [OH^–] is constant and equals to 10^-14 at 25 ^oC:

 

 

In pure water, the concentrations of H⁺ and OH^– ions are equal, so by taking the square root of K_w, the concentration of H⁺ and OH^– ions in water are found to be:

                                                                 

\[{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}\; = \;{\rm{[O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}\;{\rm{ = }}\;\sqrt {{K_{\rm{w}}}} \; = \;1.0\, \times \,{10^{ – 7}}\,M\]

 

When [H⁺] = [OH^–], we have a neutral solution, when [H⁺] > [OH^–] the solution becomes acidic, and if [H⁺] < [OH^–], then it is a basic solution.

 

We can change the concentrations of H⁺ and OH^– but because they are correlated and their product is constant, we cannot do it independently. For example, if we increase [H⁺], the [OH^–] will decrease and vice versa.

Given the [H+] or [OH-], we can calculate the other one.

For example, if [H⁺] = 10^-3, we can determine the [OH^–] using the expression for K_w:

 K_w = [H⁺][OH^–], therefore,

                                                                                                                       

\[{\rm{[O}}{{\rm{H}}^{\rm{ – }}}{\rm{]}}\;{\rm{ = }}\;\frac{{\sqrt {{K_{\rm{w}}}} \;}}{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}\;}}\; = \;\frac{{{{10}^{ – 14}}\;}}{{{{10}^{ – 3}}\;}}\, = 1.0\, \times \,{10^{ – 11}}\,M\]

 

So, this is an acidic solution because [H⁺] > [OH^–].

In the next article, we will talk about the pH scale as an indicator of the acidity or basicity of the solution.

 

Check Also

• Definitions of Acids and Bases
• Acid-Base Reactions
• Acid-Base Titrations
• Conjugate Acid and Conjugate Base
• The pH and Acidity
• Acid Strength, K_a, and pK_a
• Base Strength, K_b and pK_b
• K_a, pK_a, K_b, and pK_b Relationship
• The pH of a Strong Acid and Base
• pH + pOH = 14
• The pH of a Weak Acid
• The pH of a Weak Base
• ThepH of Polyprotic Acids
• The acidity of a Salt Solution
• The pH of a Salt Solution
• The pH of Salts With Acidic Cations and Basic Anions
• Acids and Bases Practice Problems