1.4.7Development of Practical Skills in Chemistry
pH – Problem Solving
5 min read 5 parts Lesson 9 of 11
Section 1 of 5Ways to Measure pH

Different tests give different levels of precision, so a solution is usually identified by combining several of them.
Universal indicator
- A mixture of dyes giving a range of colours;
- it is quick but only gives an approximate whole-number pH by comparison with a chart.
pH probe
- A meter gives readings to two decimal places, but it must be calibrated with buffers of known pH before use.
Phenolphthalein
- A single indicator that is colourless in acid and pink in alkaline solution, so it only distinguishes alkalis from acids.
Data logger
- A probe linked to a logger records pH continuously, which is useful during a titration or other changing system.
Section 2 of 5Calculating pH
1
For a strong monoprotic acid, [H⁺] equals the acid concentration, so pH = −log₁₀[H⁺].
2
For a strong base, find [OH⁻] first, then use [H⁺] = Kw ÷ [OH⁻], where Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.
3
For a weak acid, use Ka = [H⁺]² ÷ [HA], assuming negligible dissociation and that [H⁺] equals [A⁻].
4
Rearrange to [H⁺] = √(Ka × [HA]) and then take the negative logarithm to obtain pH.
5
Compare the calculated pH with the measured pH to decide which solution is which in an unknown set.
Section 3 of 5Identifying Unknown Solutions
Identifying Unknown Solutions
Strong acid
- Very low pH such as 1 for 0.1 mol dm⁻³ HCl, and no colour change with phenolphthalein.
Weak acid
- Higher pH than a strong acid of the same concentration, for example about 2.9 for 0.10 mol dm⁻³ ethanoic acid.
Strong base
- Very high pH near 13 for 0.1 mol dm⁻³ NaOH, and phenolphthalein turns pink.
Buffer
- Has a pH close to its pKa, for example near 4.76 for an equimolar ethanoic acid and ethanoate mixture.
| Solution (0.10 mol dm⁻³) | Approximate pH |
|---|---|
| Hydrochloric acid | 1.0 |
| Ethanoic acid | 2.9 |
| Sodium hydroxide | 13.0 |
Section 4 of 5Worked Example
Question Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³.
- Assume [H⁺] = [CH₃COO⁻] and that [CH₃COOH] stays 0.100 mol dm⁻³.
- [H⁺]² = Ka × [HA] = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶, so [H⁺] = 1.32 × 10⁻³ mol dm⁻³.
- pH = −log₁₀(1.32 × 10⁻³) = 2.88.
Answer pH = 2.88.
Section 5 of 5Exam Checklist & Tips
1.4.7 – Key Points for the Exam
- pH The negative logarithm to base 10 of the hydrogen ion concentration, pH = −log₁₀[H⁺].
- Kw The ionic product of water, [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.
- Ka The acid dissociation constant, Ka = [H⁺][A⁻] ÷ [HA].
- Calibration Adjusting a pH probe against buffers of known pH so that its readings are reliable.
Common Mistakes
- Using a probe straight from storage without calibrating it, which gives systematic errors in every reading.
- Treating a weak acid as fully dissociated and equating [H⁺] to the acid concentration.
- Reading a universal indicator colour as an exact pH instead of an approximate value.
- Forgetting to rinse the probe in distilled water between solutions, contaminating the next reading.