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:

$\begin{array}{lcll} \ce{NH3}+\ce{H2O}&\rightarrow&\ce{NH4^+}+\ce{OH^-}&\hspace{5mm}\ce{H2O}\text{ acts as an acid} \\ \ce{HCl}+\ce{H2O}&\rightarrow&\ce{H3O^+}+\ce{Cl^-}&\hspace{5mm}\ce{H2O}\text{ acts as a base} \end{array}$

It may not surprise you 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:

$\ce{H2O}+\ce{H2O}\rightarrow \ce{H3O+}+\ce{OH-}$

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:

$[\ce{H+}]=[\ce{OH-}]=1.0\times 10^{-7}\text{ M}$

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:

$[\ce{H+}]\times [\ce{OH-}]=(1.0\times 10^{-7})(1.0\times 10^{-7})=1.0\times 10^{-14}$

In acids, the concentration of H⁺(aq) — written as [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:

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

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:

$K_{\text{w}}=[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}$

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 12.9

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:

$K_{\text{w}}=[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}=(1.0\times 10^{-4})[\ce{OH-}]$

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

$[\ce{OH-}]=\dfrac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}\text{ M}$

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

Test Yourself
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 12.10

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:

$[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}=[\ce{H+}](0.0088\text{ M})$

Dividing both sides by 0.0088:

$[\ce{H+}]=\dfrac{1.0\times 10^{-14}}{0.0088}=1.1\times 10^{-12}\text{ M}$

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

Test Yourself
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 12.11

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:

$[\ce{H+}]=0.030\times 0.0788=0.00236\text{ M}$

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

$[\ce{OH-}]=\dfrac{1.0\times 10^{-14}}{0.00236}=4.2\times 10^{-12}\text{ M}$

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

Test Yourself
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

Key Takeaways

In any aqueous solution, the product of [H⁺] and [OH⁻] equals 1.0 × 10⁻¹⁴.

Exercises

Questions

Does [H⁺] remain constant in all aqueous solutions? Why or why not?
Does [OH⁻] remain constant in all aqueous solutions? Why or why not?
What is the relationship between [H⁺] and K_w? Write a mathematical expression that relates them.
What is the relationship between [OH⁻] and K_w? Write a mathematical expression that relates them.
Write the chemical equation for the autoionization of water and label the conjugate acid-base pairs.
Write the reverse of the reaction for the autoionization of water. It is still an acid-base reaction? If so, label the acid and base.
For a given aqueous solution, if [H⁺] = 1.0 × 10⁻³ M, what is [OH⁻]?
For a given aqueous solution, if [H⁺] = 1.0 × 10⁻⁹ M, what is [OH⁻]?
For a given aqueous solution, if [H⁺] = 7.92 × 10⁻⁵ M, what is [OH⁻]?
For a given aqueous solution, if [H⁺] = 2.07 × 10⁻¹¹ M, what is [H⁺]?
For a given aqueous solution, if [OH⁻] = 1.0 × 10⁻⁵ M, what is [H⁺]?
For a given aqueous solution, if [OH⁻] = 1.0 × 10⁻¹² M, what is [H⁺]?
For a given aqueous solution, if [OH⁻] = 3.77 × 10⁻⁴ M, what is [H⁺]?
For a given aqueous solution, if [OH⁻] = 7.11 × 10⁻¹⁰ M, what is [H⁺]?
What are [H⁺] and [OH⁻] in a 0.344 M solution of HNO₃?
What are [H⁺] and [OH⁻] in a 2.86 M solution of HBr?
What are [H⁺] and [OH⁻] in a 0.00338 M solution of KOH?
What are [H⁺] and [OH⁻] in a 6.02 × 10⁻⁴ M solution of Ca(OH)₂?
If HNO₂ is dissociated only to an extent of 0.445%, what are [H⁺] and [OH⁻] in a 0.307 M solution of HNO₂?
If (C₂H₅)₂NH is dissociated only to an extent of 0.077%, what are [H⁺] and [OH⁻] in a 0.0955 M solution of (C₂H₅)₂NH?

Answers

[H⁺] varies with the amount of acid or base in a solution.

[H⁺] = K_w[OH⁻]

H₂O + H₂O → H₃O⁺ + OH⁻; H₂O/H₃O⁺ and H₂O/OH⁻

1.0 × 10⁻¹¹ M

1.26 × 10⁻¹⁰ M

1.0 × 10⁻⁹ M

2.65 × 10⁻¹¹ M

[H⁺] = 0.344 M; [OH⁻] = 2.91 × 10⁻¹⁴ M

[OH⁻] = 0.00338 M; [H⁺] = 2.96 × 10⁻¹² M

[H⁺] = 0.00137 M; [OH⁻] = 7.32 × 10⁻¹² M

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Autoionization of Water

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