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Water ionisation, pH and neutrality

Key idea: Neutrality means equal hydrogen and hydroxide ion concentrations, so neutral pH is pKw/2. Water ionisation increases with temperature, and neutral pH falls while the water remains neutral.

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  • GCE A-Level H2 Chemistry 2027
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Even pure water contains a few hydrogen ions and hydroxide ions, because water molecules pass protons to one another. The product of their concentrations is fixed at a given temperature, so knowing one tells you the other. This lesson defines that product, the pH and pOH scales, and what “neutral” really means.

The ionic product of water

Water ionises very slightly:

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

The ionic product of water, Kw, is

Kw = [H⁺][OH⁻]

At 25 °C, Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. The dilute aqueous solutions considered here contain both ions, with their concentration product equal to Kw. Adding acid raises [H⁺], so [OH⁻] falls, but it never becomes zero.

Check your understanding 1

At a fixed temperature, acid is added to water. Predict the change in [OH⁻] before using the logarithmic relations.

Hydroxide concentration
Show answer

[OH⁻] falls so that [H⁺][OH⁻] = Kw remains fixed. Both ions remain present.

pH and pOH

Hydrogen-ion concentrations span many powers of ten, so they are expressed on a logarithmic scale:

pH = - log ₁₀[H⁺] [H⁺] = 10^(-pH)
pOH = - log ₁₀[OH⁻] [OH⁻] = 10^(-pOH)

Taking - log ₁₀ of both sides of Kw = [H⁺][OH⁻] gives

pH + pOH = pKw

which is 14.00 at 25 °C. Because the scale is logarithmic, a tenfold fall in [H⁺] raises the pH by exactly one unit. pH and pOH have no unit.

Worked example 1

A tenfold change in concentration

Problem

Two solutions have [H⁺] = 2.0 × 10⁻³ and 2.0 × 10⁻⁴ mol dm⁻³. Find their pH values and explain the difference.
Show full solution

  1. Calculate each pH

    Method

    Take - log ₁₀ of each concentration.

    Reason

    That is the definition of pH.

    Working

    Answer

    - log ₁₀(2.0 × 10⁻³) = 2.70; - log ₁₀(2.0 × 10⁻⁴) = 3.70

  2. Explain the difference

    Method

    The second solution has one tenth of the hydrogen-ion concentration, and its pH is one unit higher.

    Reason

    Dividing by 10 adds 1 to - log ₁₀.

    Working

    Answer

    - log ₁₀(h/10) = - log ₁₀h + 1

Exam-style question 1

Hydroxide concentration from Kw

3 marks

Problem

At 25 °C, Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶. Calculate [OH⁻] in a solution in which [H⁺] = 2.0 × 10⁻⁵ mol dm⁻³. [3 marks]

Try this before viewing the solution

Unit: mol dm⁻³
Show solution step by step

  1. Write the expression

    1 mark

    Method

    Write Kw in terms of both ions.

    Reason

    Their product is fixed at this temperature.

    Working

    Answer

    Kw = [H⁺][OH⁻]

  2. Rearrange

    1 mark

    Reason

    Divide by [H⁺].

    Working

    Answer

    [OH⁻] = Kw/[H⁺]

  3. Substitute

    1 mark

    Method

    Divide and give the unit.

    Reason

    mol² dm⁻⁶÷mol dm⁻³ = mol dm⁻³.

    Working

    Answer

    [OH⁻] = (1.0 × 10⁻¹⁴)/(2.0 × 10⁻⁵) = 5.0 × 10⁻¹⁰ mol dm⁻³

Kw changes with temperature

The ionisation of water is endothermic. Raising the temperature shifts the equilibrium to the right, so Kw increases and pKw falls. At 50 °C, for example, pKw = 13.26. Away from 25 °C, use the value given for that temperature, not 14.00.

Try it yourself 2

Use pKw away from 25 °C

Minimal support

Problem

At 50 °C, pKw = 13.26. A solution has pH 4.00. Calculate its pOH to two decimal places and [OH⁻] at this temperature.

Try this before viewing the solution

Show the steps Optional working
Unit: mol dm⁻³

Hints

Hint 1

the right total

Which value of pKw applies at this temperature?
Hint 2

back to concentration

Then use [OH⁻] = 10^(-pOH).
Show solution step by step

  1. Find pOH

    Method

    Subtract the pH from pKw at 50 °C.

    Reason

    pH and pOH add up to pKw at the same temperature.

    Working

    Answer

    pOH = 13.26-4.00 = 9.26

  2. Convert to concentration

    Method

    Take 10^(-9.26).

    Reason

    That reverses pOH = - log ₁₀[OH⁻].

    Working

    Answer

    [OH⁻] = 5.5 × 10⁻¹⁰ mol dm⁻³

What neutral means

A solution is neutral when [H⁺] = [OH⁻]. Then pH = pOH, so

pHₙₑᵤₜᵣₐₗ = pKw/2

At 25 °C this is 7.00. At higher temperatures pKw is smaller, so the neutral pH is below 7, even though pure water is still exactly neutral. At a given temperature, a solution is acidic if its pH is below pKw/2 and alkaline if it is above.

Check your understanding 2

At 50 °C, pKw = 13.26. Find the neutral pH to two decimal places, then classify a solution of pH 6.80 at this temperature. Why can you not decide by comparing with 7.00?

Solution of pH 6.80 at 50 °C
Show answer

The neutral pH is 13.26/2 = 6.63. A pH of 6.80 is above that, so the solution is slightly alkaline at 50 °C. Comparing with 7.00 would use the neutral point for 25 °C, which does not apply at 50 °C.

Check your understanding 3

Explain your classification of pH 6.80 at 50 °C using both ion concentrations. Why does comparing with pH 7.00 give the wrong conclusion?

Show answer

pOH = 13.26 − 6.80 = 6.46. Since 10^(-6.46) > 10^(-6.80), hydroxide concentration exceeds hydrogen-ion concentration, so the solution is alkaline. Neutrality means equal concentrations at any temperature. Its neutral pH here is 6.63; 7.00 belongs to 25 °C.

Common mistakes

  • Thinking an acidic solution contains no hydroxide ions. Both ions are always present.
  • Using pH + pOH = 14.00 at a temperature other than 25 °C.
  • Saying pure water at 50 °C is acidic because its pH is below 7. It is neutral, because [H⁺] = [OH⁻].
  • Using ln instead of log ₁₀.
  • Giving Kw the unit mol dm⁻³. It is mol² dm⁻⁶.

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