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6.1.4Unit 6. Practical Skills in Chemistry II

Finding Ka Values — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 8 of 14 Unit quiz

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Part 1 of 6Ka and pKa

6.1.4 Finding Ka Values

Measuring a pH during titration gives Ka directly.

Definition
  • For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] ÷ [HA].
pKa
  • pKa = –log₁₀ Ka, so a smaller pKa means a stronger acid.
Buffer region
  • During titration with a strong base, the solution contains both HA and A⁻, so it behaves as a buffer.
Half-equivalence
  • When half the acid is neutralised, [HA] = [A⁻], so Ka = [H⁺] and pH = pKa.

Part 2 of 6Method

1
Pipette a known volume of weak acid into a beaker and place a calibrated pH meter in the solution.
2
Add a strong base such as sodium hydroxide from a burette in 1 cm³ portions, recording pH after each addition.
3
Add smaller portions near the sharp rise in pH so the equivalence volume can be found accurately.
4
Plot pH against volume of base and determine the equivalence point from the steepest part of the curve.
5
Read the pH when the volume equals half the equivalence volume to find pKa.

Part 3 of 6Calculation Routes

Calculation Routes
Half-equivalence method
  • pKa equals the pH at half the equivalence volume, and Ka = 10^(–pKa).
Single pH method
  • For a weak acid of known concentration, Ka ≈ [H⁺]² ÷ [HA].
Assumptions
  • The acid dissociates only slightly and the contribution of water to [H⁺] is negligible.
Calibration
  • The pH meter must be calibrated with buffers before use for reliable readings.

Part 4 of 6Evaluating the Result

Temperature
  • Ka changes with temperature, so record the temperature of the experiment.
Probe rinsing
  • Rinse the electrode with distilled water between readings to avoid contamination.
Curve shape
  • An S-shaped curve with a vertical section above pH 7 is expected for a weak acid and strong base.
Data book
  • Compare the result with a data book value to calculate a percentage difference.

Part 5 of 6Worked Example

Question

  • The pH at half-equivalence is 4.76 for a weak acid.
  • Calculate Ka.

  1. At half-equivalence pH = pKa, so pKa = 4.76.
  2. Ka = 10^(–4.76).
  3. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

Answer Ka = 1.7 × 10⁻⁵ mol dm⁻³

Part 6 of 6Exam Checklist & Tips

6.1.4 – Key Points for the Exam

  • Weak acid An acid that is only partially dissociated in solution.
  • Ka The equilibrium constant for dissociation of an acid.
  • Equivalence point The point at which the acid has been exactly neutralised.
  • Buffer A solution that resists changes in pH when small amounts of acid or base are added.

Common Mistakes

  • Reading pKa at the equivalence point instead of half the equivalence volume.
  • Using the initial acid concentration as [H⁺] in the calculation of Ka.
  • Forgetting to convert pKa into Ka with 10 raised to the power –pKa.
  • Ignoring pH meter calibration.

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