where \(\ce{H2O}\) acts as an \(\color{blue}{\text{base}}\) (in blue).
It may not surprise you to learn, then, that within any given sample of water, some \(\ce{H2O}\) molecules are acting as acids, and other \(\ce{H2O}\) molecules are acting as bases. The chemical equation is as follows:
This occurs only to a very small degree: only about 6 in 108 \(\ce{H2O}\) molecules are participating in this process, which is called the autoionization of water.
Figure \(\PageIndex{1}\): Autoionization of water, resulting in hydroxide and hydronium ions.
At this level, the concentration of both \(\ce{H3O^{+}(aq)}\) and \(\ce{OH^{−}(aq)}\) in a sample of pure \(\ce{H2O}\) is about \(1.0 \times 10^{−7}\, M\) (at room temperature). If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have
For acids, the concentration of \(\ce{H3O^{+}(aq)}\) (i.e., \(\ce{[H3O^{+}]}\)) is greater than \(1.0 \times 10^{−7}\, M\).
For bases the concentration of \(\ce{OH^{−}(aq)}\) (i.e., \(\ce{[OH^{−}]}\)) is greater than \(1.0 \times 10^{−7}\, M\).
However, the product of the two concentrations—\(\ce{[H3O^{+}][OH^{−}]}\)—is always equal to \(1.0 \times 10^{−14}\), no matter whether the aqueous solution is an acid, a base, or neutral:
This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted \(K_w\):
This means that if you know \(\ce{[H3O^{+}]}\) for a solution, you can calculate what \(\ce{[OH^{−}]}\)) has to be for the product to equal \(1.0 \times 10^{−14}\); or if you know \(\ce{[OH^{−}]}\)), you can calculate \(\ce{[H3O^{+}]}\). This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of \(K_w\).
Warning: Temperature Matters
The degree of autoionization of water (Equation \ref{Auto})—and hence the value of \(K_w\)—changes with temperature, so Equations \ref{eq5} - \ref{eq10} are accurate only at room temperature.
Example \(\PageIndex{1}\): Hydroxide Concentration
What is \(\ce{[OH^{−}]}\)) of an aqueous solution if \(\ce{[H3O^{+}]}\) is \(1.0 \times 10^{−4} M\)?
Solution
Solutions to Example 14.7.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: \(\ce{[H3O^{+}]} =1.0 \times 10^{−4}\, M\)
Find: [OH−] = ? M
List other known quantities.
none
Plan the problem.
Using the expression for \(K_w\), (Equation \ref{eq10}), rearrange the equation algebraically to solve for [OH−].
It is assumed that the concentration unit is molarity, so \(\ce{[OH^{−}]}\) is 1.0 × 10−10 M.
Think about your result.
The concentration of the acid is high (> 1 x 10-7 M), so \(\ce{[OH^{−}]}\) should be low.
Exercise \(\PageIndex{1}\)
What is \(\ce{[OH^{−}]}\) in a 0.00032 M solution of H2SO4?
Hint
Assume both protons ionize from the molecule...although this is not the case.
Answer
\(3.1 \times 10^{−11}\, M\)
When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H3O+ or OH− ions in the formula unit because \(\ce{[H_3O^{+}]}\) or \(\ce{[OH^{−}]}\)) may not be the same as the concentration of the acid or base itself.
Example \(\PageIndex{2}\): Hydronium Concentration
What is \(\ce{[H_3O^{+}]}\) in a 0.0044 M solution of \(\ce{Ca(OH)_2}\)?
Solution
Solutions to Example 14.7.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: \([\ce{Ca(OH)_2}] =0.0044 \,M\)
Find: \(\ce{[H_3O^{+}]}\) = ? M
List other known quantities.
We begin by determining \(\ce{[OH^{−}]}\). The concentration of the solute is 0.0044 M, but because \(\ce{Ca(OH)_2}\) is a strong base, there are two OH− ions in solution for every formula unit dissolved, so the actual \(\ce{[OH^{−}]}\) is two times this: