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 108 [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 Kw 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 Kw.

Example 6.2.1


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


Using the expression and known value for Kw:

[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?


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


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


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 Kw 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.)



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


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?


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?



[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.


Practice Questions



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