pH and the ionic product of waterAQA A-Level Chemistry: Revision notes
Section 1
The pH scale
pH is a logarithmic measure of hydrogen ion concentration:
pH = –log₁₀[H⁺] and [H⁺] = antilog(–pH)
A change of one pH unit is a tenfold change in [H⁺]. A lower pH means a higher [H⁺]. Give pH to two decimal places in calculations unless told otherwise.
Check your answer: more concentrated acid must give a lower pH. If your pH is negative or above 7 for an acid, something is wrong.
Section 2
pH of a strong acid
A strong acid is fully dissociated, so for a monoprotic acid [H⁺] equals the acid concentration.
Worked example. 0.0250 mol dm⁻³ HCl: [H⁺] = 0.0250 mol dm⁻³, so pH = –log₁₀(0.0250) = 1.60.
To reverse the calculation: pH 2.40 gives [H⁺] = antilog(–2.40) = 3.98 × 10⁻³ mol dm⁻³.
For a dilution, first find the new concentration (c₁V₁ = c₂V₂), then take the logarithm.
Do not halve the pH when you halve the concentration. Halving [H⁺] raises the pH by only 0.30.
Section 3
The ionic product of water
Water is slightly dissociated: H₂O ⇌ H⁺ + OH⁻. The equilibrium constant is Kc = [H⁺][OH⁻] / [H₂O]. Since [H₂O] is very large and effectively constant, it is included in the constant to give
Kw = [H⁺][OH⁻]
Units are mol² dm⁻⁶. At 298 K, Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. In pure water [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, so pH = 7.00.
Section 4
pH of a strong base
A strong base is fully dissociated, so [OH⁻] comes from the concentration of the base (double it for Ba(OH)₂). Then use Kw:
[H⁺] = Kw / [OH⁻], pH = –log₁₀[H⁺]
Worked example. 0.0400 mol dm⁻³ NaOH at 298 K: [H⁺] = 1.00 × 10⁻¹⁴ / 0.0400 = 2.50 × 10⁻¹³ mol dm⁻³, so pH = 12.60.
Using –log₁₀[OH⁻] as the pH. That gives the pOH, not the pH. Always convert [OH⁻] to [H⁺] with Kw.
Section 5
Kw and temperature
Dissociation of water is endothermic, so raising the temperature shifts the equilibrium right and Kw increases. At 333 K, Kw = 9.61 × 10⁻¹⁴ mol² dm⁻⁶.
Pure water at 333 K: [H⁺] = √Kw = 3.10 × 10⁻⁷ mol dm⁻³, so pH = 6.51. The water is still neutral because [H⁺] = [OH⁻].
So neutral is pH 7 only at 298 K, and the value of Kw must be used for strong bases at other temperatures.
If asked whether warm pure water is acidic, say it is neutral because [H⁺] = [OH⁻], and quote the new pH.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on pH and the ionic product of water
- A student dilutes 10.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid with distilled water to a total volume of 250 cm³. All measurements are at 298 K.Calculate the pH of the diluted solution and the change in pH caused by the dilution.2 marks
- A cleaning product contains sodium hydroxide at a concentration of 0.0250 mol dm⁻³. At 298 K the ionic product of water, Kw, is 1.00 × 10⁻¹⁴ mol² dm⁻⁶.Use Kw to calculate the hydrogen ion concentration in the original sodium hydroxide solution.2 marks
- Pure water is slightly dissociated: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). A data book gives Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K and Kw = 9.61 × 10⁻¹⁴ mol² dm⁻⁶ at 333 K.Show how the expression Kw = [H⁺][OH⁻] is derived from the equilibrium constant for the dissociation of water, and give the units of Kw.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).