The ionic product of water, $K_w$ (A-level only)

What you'll learn

Why pure water contains tiny amounts of both hydrogen ions and hydroxide ions.
How $K_w$ comes from the equilibrium for water dissociation.
Why the value of $K_w$ changes with temperature.
How to calculate the pH of a strong base from its concentration.

Before $K_w$ : the key symbols

In acid–base calculations, you will often see square brackets around a formula, such as $[H+][\text{H}^+]$ or $[OH−][\text{OH}^-]$ . These mean concentration.

Definition

Concentration brackets

Square brackets mean the concentration of a species in solution, usually in mol dm⁻³. For example, $[OH−][\text{OH}^-]$ means the concentration of hydroxide ions in mol dm⁻³.

You also need the pH equation from earlier in acids and bases.

Definition

The pH of a solution is a logarithmic measure of its hydrogen ion concentration:

pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+]

In this equation, $[H+][\text{H}^+]$ must be in mol dm⁻³.

Because pH is logarithmic, a change of 1 pH unit means a factor of 10 change in $[H+][\text{H}^+]$ .

Example

Calculating pH from hydrogen ion concentration

A solution has $dm−3[\text{H}^+] = 2.50 \times 10^{-3}\ \text{mol dm}^{-3}$ . Calculate its pH.

Substitute the hydrogen ion concentration into the pH equation:
$pH=−log⁡10(2.50×10−3)\text{pH} = -\log_{10}(2.50 \times 10^{-3})$
Evaluate the logarithm:
$pH=2.602…\text{pH} = 2.602\ldots$
Round appropriately, usually to 2 decimal places in A-level pH calculations:
$pH=2.60\text{pH} = 2.60$

Water is slightly dissociated

Pure water is mostly made of neutral water molecules, but a very small number of water molecules transfer protons between themselves. This is called dissociation, meaning splitting into ions.

A more realistic equation is:

2H2O(l)⇌H3O+(aq)+OH−(aq)2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)

At A-level, the hydronium ion, $H3O+\text{H}_3\text{O}^+$ , is usually simplified and written as $H+\text{H}^+$ :

H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)

Autoionisation of water showing hydronium and hydroxide ions forming reversibly

Key Idea

Water contains both ions

Even pure water contains tiny concentrations of both $H+\text{H}^+$ and $OH−\text{OH}^-$ ions. The equilibrium lies very far to the left, so only a very small fraction of water molecules are ionised.

Deriving the ionic product of water

Because the dissociation of water is reversible, it has an equilibrium constant. For the simplified dissociation,

H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)

we can build a special equilibrium expression called the ionic product of water.

Definition

Ionic product of water

The ionic product of water, $K_w$ , is defined as:

Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]

It links the hydrogen ion concentration and hydroxide ion concentration in aqueous solutions.

Example

Deriving the ionic product expression

Start with the simplified equilibrium for water dissociation:
$H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$
Write an equilibrium expression in the usual way:
$Kc=[H+][OH−][H2O]K_c = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$
The concentration of liquid water is effectively constant because water is the solvent and is present in a huge excess. This constant is combined with $K_c$ to give a new constant:
$Kw=Kc[H2O]=[H+][OH−]K_w = K_c[\text{H}_2\text{O}] = [\text{H}^+][\text{OH}^-]$

The value and units of $K_w$

At 25 °C, or 298 K, the value usually used is:

dm−6K_w = 1.00 \times 10^{-14}\ \text{mol}^2\text{ dm}^{-6}

The units come from multiplying two concentrations:

dm−6(\text{mol dm}^{-3})(\text{mol dm}^{-3}) = \text{mol}^2\text{ dm}^{-6}

In pure water, each dissociation event produces one $H+\text{H}^+$ ion and one $OH−\text{OH}^-$ ion, so:

[H+]=[OH−][\text{H}^+] = [\text{OH}^-]

At 25 °C:

dm−3[\text{H}^+] = [\text{OH}^-] = 1.00 \times 10^{-7}\ \text{mol dm}^{-3}

so pure water has pH 7.00 at 25 °C.

Common Mistake

Neutral does not always mean pH 7

A solution is neutral when $[H+]=[OH−][\text{H}^+] = [\text{OH}^-]$ . Neutral water has pH 7.00 only at 25 °C, because $K_w$ changes with temperature.

$K_w$ changes with temperature

The value of $K_w$ is not fixed. If the temperature changes, the position of the water dissociation equilibrium changes, so the product $[H+][OH−][\text{H}^+][\text{OH}^-]$ changes too.

In exam questions, use the value of $K_w$ given in the question. If no value is given and the temperature is 25 °C, use $1.00 \times 10^{-14}\ \text{mol}^2\text{ dm}^{-6}$ .

Example

Finding the neutral pH at a different temperature

At 50 °C, $dm−6K_w = 5.47 \times 10^{-14}\ \text{mol}^2\text{ dm}^{-6}$ . Calculate the pH of pure water at this temperature.

In pure water, the concentrations of hydrogen ions and hydroxide ions are equal. Let:
$[H+]=[OH−]=x[\text{H}^+] = [\text{OH}^-] = x$
Substitute into the expression for $K_w$ :
$K_w = x^2$
so:
$\sqrt{5.47 \times 10^{-14}} = 2.34 \times 10^{-7}\ \text{mol dm}^{-3}$
Calculate the pH:
$pH=−log⁡10(2.34×10−7)=6.63\text{pH} = -\log_{10}(2.34 \times 10^{-7}) = 6.63$

This water is neutral, even though its pH is below 7, because $[H+]=[OH−][\text{H}^+] = [\text{OH}^-]$ .

Calculating the pH of a strong base

This is the main calculation skill for this sub-topic.

Definition

Strong base

A strong base fully dissociates in water. This means the concentration of hydroxide ions can be found directly from the concentration and formula of the base.

For example:

NaOH(aq)→Na+(aq)+OH−(aq)\text{NaOH}(aq) \to \text{Na}^+(aq) + \text{OH}^-(aq)

One mole of sodium hydroxide produces one mole of hydroxide ions, so for sodium hydroxide:

[OH−]=[NaOH][\text{OH}^-] = [\text{NaOH}]

But for barium hydroxide:

Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)\text{Ba(OH)}_2(aq) \to \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq)

one mole of barium hydroxide produces two moles of hydroxide ions, so:

[OH−]=2[Ba(OH)2][\text{OH}^-] = 2[\text{Ba(OH)}_2]

Key Idea

Strong-base pH method

For a strong base, find $[OH−][\text{OH}^-]$ first, then use $K_w$ to find $[H+][\text{H}^+]$ , then use the pH equation.

The key rearrangement is:

Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]

so:

[H+]=Kw[OH−][\text{H}^+] = \frac{K_w}{[\text{OH}^-]}

Example

Calculating the pH of sodium hydroxide

Calculate the pH of 0.0250 mol dm⁻³ sodium hydroxide at 25 °C.

Sodium hydroxide is a strong base and produces one hydroxide ion per formula unit:
$dm−3[\text{OH}^-] = 0.0250\ \text{mol dm}^{-3}$
Use $Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]$ and rearrange:
$[H+]=1.00×10−140.0250[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0250}$ $dm−3[\text{H}^+] = 4.00 \times 10^{-13}\ \text{mol dm}^{-3}$
Convert hydrogen ion concentration into pH:
$pH=−log⁡10(4.00×10−13)=12.40\text{pH} = -\log_{10}(4.00 \times 10^{-13}) = 12.40$

Example

Using the formula of the base

Calculate the pH of 0.0150 mol dm⁻³ barium hydroxide at 25 °C.

Barium hydroxide produces two hydroxide ions per formula unit, so double the base concentration:
$dm−3[\text{OH}^-] = 2 \times 0.0150 = 0.0300\ \text{mol dm}^{-3}$
Use $K_w$ to find $[H+][\text{H}^+]$ :
$[H+]=1.00×10−140.0300[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0300}$ $dm−3[\text{H}^+] = 3.33 \times 10^{-13}\ \text{mol dm}^{-3}$
Convert to pH:
$pH=−log⁡10(3.33×10−13)=12.48\text{pH} = -\log_{10}(3.33 \times 10^{-13}) = 12.48$

Common Mistake

Using the base concentration as hydrogen ion concentration

In a strong base calculation, the given concentration usually tells you $[OH−][\text{OH}^-]$ , not $[H+][\text{H}^+]$ . If you put the base concentration straight into $pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+]$ , you will get an acidic pH for an alkaline solution.

The pOH shortcut

You may also see pOH used for alkaline solutions.

Definition

pOH

The pOH of a solution is defined as:

pOH=−log⁡10[OH−]\text{pOH} = -\log_{10}[\text{OH}^-]

At 25 °C:

pH+pOH=14.00\text{pH} + \text{pOH} = 14.00

This shortcut comes from $Kw=1.00×10−14K_w = 1.00 \times 10^{-14}$ at 25 °C. It is often quicker, but the direct $K_w$ method is safer because it works at any temperature if you are given the correct value of $K_w$ .

Tip

Rounding pH answers

Carry extra digits during the calculation and round the final pH at the end. Unless the question says otherwise, pH is usually given to 2 decimal places.

Exam technique

In the exam

Check whether the solution is acidic or alkaline so you know whether the given concentration is likely to be $[H+][\text{H}^+]$ or $[OH−][\text{OH}^-]$ .
For strong bases, use the formula of the base to find $[OH−][\text{OH}^-]$ , then calculate $[H+]=Kw[OH−][\text{H}^+] = \frac{K_w}{[\text{OH}^-]}$ .
Use the value of $K_w$ for the temperature in the question; only use $1.00 \times 10^{-14}$ automatically at 25 °C.

Self review

Check yourself

Why does pure water contain both $H+\text{H}^+$ ions and $OH−\text{OH}^-$ ions?
At 25 °C, how would you calculate the pH of 0.0400 mol dm⁻³ sodium hydroxide?
Why can neutral water at a temperature above 25 °C have a pH below 7?

What you'll learn

Why pure water contains tiny amounts of both hydrogen ions and hydroxide ions.
How $K_w$ comes from the equilibrium for water dissociation.
Why the value of $K_w$ changes with temperature.
How to calculate the pH of a strong base from its concentration.

Before $K_w$ : the key symbols

In acid–base calculations, you will often see square brackets around a formula, such as $[H+][\text{H}^+]$ or $[OH−][\text{OH}^-]$ . These mean concentration.

Definition

Concentration brackets

Square brackets mean the concentration of a species in solution, usually in mol dm⁻³. For example, $[OH−][\text{OH}^-]$ means the concentration of hydroxide ions in mol dm⁻³.

You also need the pH equation from earlier in acids and bases.

Definition

The pH of a solution is a logarithmic measure of its hydrogen ion concentration:

pH=−log⁡10[H+]\text{pH} = -\log_{10}[\text{H}^+]

In this equation, $[H+][\text{H}^+]$ must be in mol dm⁻³.

Because pH is logarithmic, a change of 1 pH unit means a factor of 10 change in $[H+][\text{H}^+]$ .

Example

Calculating pH from hydrogen ion concentration

A solution has $dm−3[\text{H}^+] = 2.50 \times 10^{-3}\ \text{mol dm}^{-3}$ . Calculate its pH.

Substitute the hydrogen ion concentration into the pH equation:
$pH=−log⁡10(2.50×10−3)\text{pH} = -\log_{10}(2.50 \times 10^{-3})$
Evaluate the logarithm:
$pH=2.602…\text{pH} = 2.602\ldots$
Round appropriately, usually to 2 decimal places in A-level pH calculations:
$pH=2.60\text{pH} = 2.60$

Water is slightly dissociated

Pure water is mostly made of neutral water molecules, but a very small number of water molecules transfer protons between themselves. This is called dissociation, meaning splitting into ions.

A more realistic equation is:

2H2O(l)⇌H3O+(aq)+OH−(aq)2\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)

At A-level, the hydronium ion, $H3O+\text{H}_3\text{O}^+$ , is usually simplified and written as $H+\text{H}^+$ :

H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)

Autoionisation of water showing hydronium and hydroxide ions forming reversibly

Key Idea

Water contains both ions

Deriving the ionic product of water

Because the dissociation of water is reversible, it has an equilibrium constant. For the simplified dissociation,

H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)

we can build a special equilibrium expression called the ionic product of water.

Definition

Ionic product of water

The ionic product of water, $K_w$ , is defined as:

Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]

It links the hydrogen ion concentration and hydroxide ion concentration in aqueous solutions.

Example

Deriving the ionic product expression

Start with the simplified equilibrium for water dissociation:
$H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$
Write an equilibrium expression in the usual way:
$Kc=[H+][OH−][H2O]K_c = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$
The concentration of liquid water is effectively constant because water is the solvent and is present in a huge excess. This constant is combined with $K_c$ to give a new constant:
$Kw=Kc[H2O]=[H+][OH−]K_w = K_c[\text{H}_2\text{O}] = [\text{H}^+][\text{OH}^-]$

The value and units of $K_w$

At 25 °C, or 298 K, the value usually used is:

dm−6K_w = 1.00 \times 10^{-14}\ \text{mol}^2\text{ dm}^{-6}

The units come from multiplying two concentrations:

dm−6(\text{mol dm}^{-3})(\text{mol dm}^{-3}) = \text{mol}^2\text{ dm}^{-6}

In pure water, each dissociation event produces one $H+\text{H}^+$ ion and one $OH−\text{OH}^-$ ion, so:

[H+]=[OH−][\text{H}^+] = [\text{OH}^-]

At 25 °C:

dm−3[\text{H}^+] = [\text{OH}^-] = 1.00 \times 10^{-7}\ \text{mol dm}^{-3}

so pure water has pH 7.00 at 25 °C.

Common Mistake

Neutral does not always mean pH 7

A solution is neutral when $[H+]=[OH−][\text{H}^+] = [\text{OH}^-]$ . Neutral water has pH 7.00 only at 25 °C, because $K_w$ changes with temperature.

$K_w$ changes with temperature

The value of $K_w$ is not fixed. If the temperature changes, the position of the water dissociation equilibrium changes, so the product $[H+][OH−][\text{H}^+][\text{OH}^-]$ changes too.

In exam questions, use the value of $K_w$ given in the question. If no value is given and the temperature is 25 °C, use $1.00 \times 10^{-14}\ \text{mol}^2\text{ dm}^{-6}$ .

Example

Finding the neutral pH at a different temperature

At 50 °C, $dm−6K_w = 5.47 \times 10^{-14}\ \text{mol}^2\text{ dm}^{-6}$ . Calculate the pH of pure water at this temperature.

In pure water, the concentrations of hydrogen ions and hydroxide ions are equal. Let:
$[H+]=[OH−]=x[\text{H}^+] = [\text{OH}^-] = x$
Substitute into the expression for $K_w$ :
$K_w = x^2$
so:
$\sqrt{5.47 \times 10^{-14}} = 2.34 \times 10^{-7}\ \text{mol dm}^{-3}$
Calculate the pH:
$pH=−log⁡10(2.34×10−7)=6.63\text{pH} = -\log_{10}(2.34 \times 10^{-7}) = 6.63$

This water is neutral, even though its pH is below 7, because $[H+]=[OH−][\text{H}^+] = [\text{OH}^-]$ .

Calculating the pH of a strong base

This is the main calculation skill for this sub-topic.

Definition

Strong base

A strong base fully dissociates in water. This means the concentration of hydroxide ions can be found directly from the concentration and formula of the base.

For example:

NaOH(aq)→Na+(aq)+OH−(aq)\text{NaOH}(aq) \to \text{Na}^+(aq) + \text{OH}^-(aq)

One mole of sodium hydroxide produces one mole of hydroxide ions, so for sodium hydroxide:

[OH−]=[NaOH][\text{OH}^-] = [\text{NaOH}]

But for barium hydroxide:

Ba(OH)2(aq)→Ba2+(aq)+2OH−(aq)\text{Ba(OH)}_2(aq) \to \text{Ba}^{2+}(aq) + 2\text{OH}^-(aq)

one mole of barium hydroxide produces two moles of hydroxide ions, so:

[OH−]=2[Ba(OH)2][\text{OH}^-] = 2[\text{Ba(OH)}_2]

Key Idea

Strong-base pH method

For a strong base, find $[OH−][\text{OH}^-]$ first, then use $K_w$ to find $[H+][\text{H}^+]$ , then use the pH equation.

The key rearrangement is:

Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]

so:

[H+]=Kw[OH−][\text{H}^+] = \frac{K_w}{[\text{OH}^-]}

Example

Calculating the pH of sodium hydroxide

Calculate the pH of 0.0250 mol dm⁻³ sodium hydroxide at 25 °C.

Sodium hydroxide is a strong base and produces one hydroxide ion per formula unit:
$dm−3[\text{OH}^-] = 0.0250\ \text{mol dm}^{-3}$
Use $Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]$ and rearrange:
$[H+]=1.00×10−140.0250[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0250}$ $dm−3[\text{H}^+] = 4.00 \times 10^{-13}\ \text{mol dm}^{-3}$
Convert hydrogen ion concentration into pH:
$pH=−log⁡10(4.00×10−13)=12.40\text{pH} = -\log_{10}(4.00 \times 10^{-13}) = 12.40$

Example

Using the formula of the base

Calculate the pH of 0.0150 mol dm⁻³ barium hydroxide at 25 °C.

Barium hydroxide produces two hydroxide ions per formula unit, so double the base concentration:
$dm−3[\text{OH}^-] = 2 \times 0.0150 = 0.0300\ \text{mol dm}^{-3}$
Use $K_w$ to find $[H+][\text{H}^+]$ :
$[H+]=1.00×10−140.0300[\text{H}^+] = \frac{1.00 \times 10^{-14}}{0.0300}$ $dm−3[\text{H}^+] = 3.33 \times 10^{-13}\ \text{mol dm}^{-3}$
Convert to pH:
$pH=−log⁡10(3.33×10−13)=12.48\text{pH} = -\log_{10}(3.33 \times 10^{-13}) = 12.48$

Common Mistake

Using the base concentration as hydrogen ion concentration

The pOH shortcut

You may also see pOH used for alkaline solutions.

Definition

pOH

The pOH of a solution is defined as:

pOH=−log⁡10[OH−]\text{pOH} = -\log_{10}[\text{OH}^-]

At 25 °C:

pH+pOH=14.00\text{pH} + \text{pOH} = 14.00

Tip

Rounding pH answers

Carry extra digits during the calculation and round the final pH at the end. Unless the question says otherwise, pH is usually given to 2 decimal places.

Exam technique

In the exam

Check whether the solution is acidic or alkaline so you know whether the given concentration is likely to be $[H+][\text{H}^+]$ or $[OH−][\text{OH}^-]$ .
For strong bases, use the formula of the base to find $[OH−][\text{OH}^-]$ , then calculate $[H+]=Kw[OH−][\text{H}^+] = \frac{K_w}{[\text{OH}^-]}$ .
Use the value of $K_w$ for the temperature in the question; only use $1.00 \times 10^{-14}$ automatically at 25 °C.

Self review

Check yourself

Why does pure water contain both $H+\text{H}^+$ ions and $OH−\text{OH}^-$ ions?
At 25 °C, how would you calculate the pH of 0.0400 mol dm⁻³ sodium hydroxide?
Why can neutral water at a temperature above 25 °C have a pH below 7?

What you'll learn

Before $K_w$ : the key symbols

Water is slightly dissociated

Deriving the ionic product of water

The value and units of $K_w$

$K_w$ changes with temperature

Calculating the pH of a strong base

The pOH shortcut

Review The ionic product of water, $K_w$ (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

The ionic product of water, $K_w$ (A-level only)

What you'll learn

Before $K_w$ : the key symbols

Water is slightly dissociated

Deriving the ionic product of water

The value and units of $K_w$

$K_w$ changes with temperature

Calculating the pH of a strong base

The pOH shortcut

Review The ionic product of water, $K_w$ (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

The ionic product of water, $K_w$ (A-level only)

What you'll learn

Before KwK_wKw​: the key symbols

Water is slightly dissociated

Deriving the ionic product of water

The value and units of KwK_wKw​

KwK_wKw​ changes with temperature

Calculating the pH of a strong base

The pOH shortcut

Review The ionic product of water, $K_w$ (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

The ionic product of water, $K_w$ (A-level only) Revision Guide

The ionic product of water, $K_w$ (A-level only)

What you'll learn

Before KwK_wKw​: the key symbols

Water is slightly dissociated

Deriving the ionic product of water

The value and units of KwK_wKw​

KwK_wKw​ changes with temperature

Calculating the pH of a strong base

The pOH shortcut

Review The ionic product of water, $K_w$ (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

The ionic product of water, $K_w$ (A-level only) Revision Guide

Before $K_w$ : the key symbols

The value and units of $K_w$

$K_w$ changes with temperature

Before $K_w$ : the key symbols

The value and units of $K_w$

$K_w$ changes with temperature