# 6.2 Autoionisation of Water

Learning Objectives

- Describe the autoionisation of water.
- Calculate the concentrations of [latex]\ce{H^{+}}[/latex] and [latex]\ce{OH^{-}}[/latex] in solutions, knowing the other concentration.

We have already seen that [latex]\ce{H_{2}O}[/latex] can act as an acid or a base, as shown below:

[latex]\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}[/latex]

It may not surprise you to learn, then, that within any given sample of water, some [latex]\ce{H_{2}O}[/latex] molecules are acting as acids, and other [latex]\ce{H_{2}O}[/latex] molecules are acting as bases. The chemical equation is as follows:

[latex]\ce{H2O}+\ce{H2O}\rightarrow \ce{H3O+}+\ce{OH-}[/latex]

This occurs only to a very small degree: only about 6 in 10^{8} [latex]\ce{H_{2}O}[/latex] molecules are participating in this process, which is called the **autoionisation of water**. At this level, the concentration of both [latex]\ce{H^{+}}{(aq)}[/latex] and [latex]\ce{OH^{-}}{(aq)}[/latex] in a sample of pure [latex]\ce{H_{2}O}[/latex] 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:

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

for *any* sample of pure water because [latex]\ce{H_{2}O}[/latex] can act as both an acid and a base. The product of these two concentrations is [latex]{1.0 × 10^{-14}}[/latex] as shown in the following equation:

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

In acids, the concentration of [latex]\ce{H^{+}}{(aq)}[/latex] — written as [latex]\ce{[H^{+}]}[/latex] — is greater than 1.0 × 10^{−7} M, while for bases the concentration of [latex]\ce{OH^{-}}{(aq)}[/latex] — [latex]\ce{[OH^{-}]}[/latex]— is greater than 1.0 × 10^{−7} M. However, the *product* of the two concentrations — [latex]\ce{[H^{+}]}\ce{[OH^{-}]}[/latex]— is *always* equal to 1.0 × 10^{−14}, no matter whether the aqueous solution is an acid, a base, or neutral, which can be seen from the following:

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

This value of the product of concentrations is so important for aqueous solutions that it is called the **autoionisation constant of water** and is denoted *K*_{w} as shown by the following equation:

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

This means that if you know [latex]\ce{[H^{+}]}[/latex] for a solution, you can calculate what [latex]\ce{[OH^{-}]}[/latex] has to be for the product to equal 1.0 × 10^{−14}, or if you know [latex]\ce{[OH^{-}]}[/latex], you can calculate [latex]\ce{[H^{+}]}[/latex]. 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 6.2.1

# Problem

What is [latex]\ce{[OH^{-}]}[/latex] of an aqueous solution if [latex]\ce{[H^{+}]}[/latex] is 1.0 × 10^{−4} M?

# Solution

Using the expression and known value for *K*_{w}:

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

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

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

It is assumed that the concentration unit is molarity, so[latex]\ce{[OH^{-}]}[/latex]is 1.0 × 10^{−10} M.

# Test Yourself

What is [latex]\ce{[H^{+}]}[/latex] of an aqueous solution if [latex]\ce{[OH^{-}]}[/latex] 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 [latex]\ce{H^{+}}[/latex] or [latex]\ce{OH^{-}}[/latex] ions in the formula unit because [latex]\ce{[H^{+}]}[/latex] or [latex]\ce{[OH^{-}]}[/latex] may not be the same as the concentration of the acid or base itself.

Example 6.2.2

# Problem

What is [latex]\ce{[H^{+}]}[/latex] in a 0.0044 M solution of [latex]\ce{Ca(OH)_{2}}[/latex]?

# Solution

We begin by determining [latex]\ce{[OH^{-}]}[/latex].

The concentration of the solute is 0.0044 M, but because [latex]\ce{Ca(OH)_{2}}[/latex] is a strong base, there are two [latex]\ce{OH^{-}}[/latex] ions in solution for every formula unit dissolved, so the actual [latex]\ce{[OH^{-}]}[/latex] is two times this, or 2 × 0.0044 M = 0.0088 M.

Now we can use the *K*_{w} expression:

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

Dividing both sides by 0.0088:

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

[latex]\ce{[H^{+}]}[/latex] has decreased significantly in this basic solution.

# Test Yourself

What is [latex]\ce{[OH^{-}]}[/latex] in a 0.00032 M solution of [latex]\ce{H_{2}SO_{4}}[/latex]? (Hint: assume both [latex]\ce{H^{+}}[/latex] ions ionise.)

**Answer**

1.6 × 10^{−11} M

For strong acids and bases, [latex]\ce{[H^{+}]}[/latex] and [latex]\ce{[OH^{-}]}[/latex] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionised by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionisation would need to be known before we can determine [latex]\ce{[H^{+}]}[/latex] and [latex]\ce{[OH^{-}]}.[/latex]

Example 6.2.3

# Problem

A 0.0788 M solution of [latex]\ce{HC_{2}H_{3}O_{2}}[/latex] is 3.0% ionised into [latex]\ce{H^{+}}[/latex] ions and [latex]\ce{C_{2}H_{3}O_{2}^{-}}[/latex] ions. What is [latex]\ce{[H^{+}]}[/latex] and [latex]\ce{[OH^{-}]}[/latex] for this solution?

# Solution

Because the acid is only 3.0% ionised, we can determine [latex]\ce{[H^{+}]}[/latex] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form:

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

With this [latex]\ce{[H^{+}]}[/latex], then [latex]\ce{[OH^{-}]}[/latex] can be calculated as follows:

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

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 [latex]\ce{(C_{5}H_{5}N)}[/latex] is 0.44% ionised into pyridinium ions [latex]\ce{(C_{5}H_{5}NH^{+})}[/latex] and [latex]\ce{OH^{-}}[/latex] ions. What are[latex]\ce{[OH^{-}]}[/latex] and [latex]\ce{[H^{+}]}[/latex] for this solution?

**Answer**

[latex]\ce{[OH^{-}]}[/latex] = 9.77 × 10^{−5} M; [H^{+}] = 1.02 × 10^{−10} M

Key Takeaways

- In any aqueous solution, the product of [latex]\ce{[H^{+}]}[/latex] and [latex]\ce{[OH^{-}]}[/latex] equals 1.0 × 10
^{−14}.

Exercises