14.7: Autoionization of Water

Learning Objectives

• Describe the autoionization of water.
• Calculate the concentrations of H⁺ and OH⁻ in solutions, knowing the other concentration.

We have already seen that H₂O can act as an acid or a base:

NH₃ + H₂O → NH₄⁺ + OH⁻ (H₂O acts as an acid)

HCl + H₂O → H₃O⁺ + Cl⁻ (H₂O acts as a base)

It may not be surprising to learn, then, that within any given sample of water, some H₂O molecules are acting as acids, and other H₂O molecules are acting as bases. The chemical equation is as follows:

H₂O + H₂O → H₃O⁺ + OH⁻

This occurs only to a very small degree: only about 6 in 10⁸ H₂O molecules are participating in this process, which is called the autoionization of water. At this level, the concentration of both H⁺(aq) and OH⁻(aq) in a sample of pure H₂O is about 1.0 × 10⁻⁷ M. If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have

[H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M

for any sample of pure water because H₂O can act as both an acid and a base. The product of these two concentrations is 1.0 × 10⁻¹⁴:

[H⁺] × [OH⁻] = (1.0 × 10⁻⁷)(1.0 × 10⁻⁷) = 1.0 × 10⁻¹⁴

In acids, the concentration of H⁺(aq)—[H⁺]—is greater than 1.0 × 10⁻⁷ M, while for bases the concentration of OH⁻(aq)—[OH⁻]—is greater than 1.0 × 10⁻⁷ M. However, the product of the two concentrations—[H⁺][OH⁻]—is always equal to 1.0 × 10⁻¹⁴, no matter whether the aqueous solution is an acid, a base, or neutral:

[H⁺][OH⁻] = 1.0 × 10⁻¹⁴

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:

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

This means that if you know [H⁺] for a solution, you can calculate what [OH⁻] has to be for the product to equal 1.0 × 10⁻¹⁴, or if you know [OH⁻], you can calculate [H⁺]. 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.

Example $\PageIndex{1}$

What is [OH⁻] of an aqueous solution if [H⁺] is 1.0 × 10⁻⁴ M?

Solution

Using the expression and known value for K_w,

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ = (1.0 × 10⁻⁴)[OH⁻]

We solve by dividing both sides of the equation by 1.0 × 10⁻⁴:

$$\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber$$

It is assumed that the concentration unit is molarity, so [OH⁻] is 1.0 × 10⁻¹⁰ M.

Exercise $\PageIndex{1}$

What is [H⁺] of an aqueous solution if [OH⁻] is 1.0 × 10⁻⁹ M?

Answer

1.0 × 10⁻⁵ 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 H⁺ or OH⁻ ions in the formula unit because [H⁺] or [OH⁻] may not be the same as the concentration of the acid or base itself.

Example $\PageIndex{2}$

What is [H⁺] in a 0.0044 M solution of Ca(OH)₂?

Solution

We begin by determining [OH⁻]. The concentration of the solute is 0.0044 M, but because Ca(OH)₂ is a strong base, there are two OH⁻ ions in solution for every formula unit dissolved, so the actual [OH⁻] is two times this, or 2 × 0.0044 M = 0.0088 M. Now we can use the K_w expression:

[H⁺][OH⁻] = 1.0 × 10⁻¹⁴ = [H⁺](0.0088 M)

Divide both sides by 0.0088:

$$\left [ H^{+} \right ]=\frac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M\nonumber$$

[H⁺] has decreased significantly in this basic solution.

Exercise $\PageIndex{2}$

What is [OH⁻] in a 0.00032 M solution of H₂SO₄? (Hint: assume both H⁺ ions ionize.)

Answer

1.6 × 10⁻¹¹ M

For strong acids and bases, [H⁺] and [OH⁻] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionized by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionization would need to be known before we can determine [H⁺] and [OH⁻].

Example $\PageIndex{3}$

A 0.0788 M solution of HC₂H₃O₂ is 3.0% ionized into H⁺ ions and C₂H₃O₂⁻ ions. What are [H⁺] and [OH⁻] for this solution?

Solution

Because the acid is only 3.0% ionized, we can determine [H⁺] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form:

[H⁺] = 0.030 × 0.0788 = 0.00236 M

With this [H⁺], then [OH⁻] can be calculated as follows:

$$\left [ OH^{-} \right ]=\frac{1.0\times 10^{-14}}{0.00236}=4.2\times 10^{-12}M\nonumber$$

This is about 30 times higher than would be expected for a strong acid of the same concentration.

Exercise $\PageIndex{3}$

A 0.0222 M solution of pyridine (C₅H₅N) is 0.44% ionized into pyridinium ions (C₅H₅NH⁺) and OH⁻ ions. What are [OH⁻] and [H⁺] for this solution?

Answer

[OH⁻] = 9.77 × 10⁻⁵ M; [H⁺] = 1.02 × 10⁻¹⁰ M

Summary

In any aqueous solution, the product of [H+] and [OH−] equals $1.0 \times 10^{−14}$ (at room temperature).