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1.4.7Development of Practical Skills in Chemistry

pH – Problem Solving

5 min read 5 parts Lesson 9 of 11

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Section 1 of 5Ways to Measure pH

1.4.7 pH – Problem Solving

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 acid1.0
Ethanoic acid2.9
Sodium hydroxide13.0

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.

  1. Assume [H⁺] = [CH₃COO⁻] and that [CH₃COOH] stays 0.100 mol dm⁻³.
  2. [H⁺]² = Ka × [HA] = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶, so [H⁺] = 1.32 × 10⁻³ mol dm⁻³.
  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.
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