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pH measurements and finding Ka (Core Practical 11)

Edexcel IAL Chemistry · WCH16 Practical Skills in Chemistry II

Core Practical 11 finds Ka for a weak acid. The theory and titration curves are in the Unit 4 lesson; Unit 6 asks how you actually do it, and what limits the result.

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Using a pH meter

Three ways to find Ka

1. Titration curve: add NaOH from a burette to the acid in small portions, measuring pH after each, with smaller additions near the end point. Find the equivalence volume from the steep section; at half that volume, pH = pKa.

2. Half-neutralisation: divide a volume of acid into two equal portions. Titrate one with NaOH using phenolphthalein to the end point (now all salt). Add the other portion of acid: the mixture now has equal amounts of HA and A-, so its pH = pKa. Neither concentration needs to be known.

3. pH of a known concentration: Ka = [H+]2 / [HA], assuming that [H+] = [A-] and that so little acid dissociates that [HA] is still the starting concentration.

Errors and assumptions

Worked example

A half-neutralised solution of a weak acid has pH 4.76. Find Ka.

  1. At half-neutralisation [HA] = [A-], so Ka = [H+] and pKa = pH.
  2. Ka = 10-4.76

Answer: Ka = 1.74 x 10-5 mol dm-3

Key idea: Calibrate with buffers, rinse with deionised water. pH = pKa at half-neutralisation; make it by adding an equal portion of acid to a neutralised portion. Ka = [H+]2 / [HA] assumes little dissociation.

Check yourself

  1. At half-neutralisation the pH of a weak acid solution is 4.20. Calculate Ka.
    Show answer

    6.3 x 10-5 mol dm-3. Ka = 10-4.20 = 6.3 x 10-5 mol dm-3.

  2. A 0.100 mol dm-3 solution of a weak acid HA has pH 2.90. Calculate Ka.
    Show answer

    1.58 x 10-5 mol dm-3. [H+] = 10-2.90 = 1.26 x 10-3 mol dm-3; Ka = [H+]2 / 0.100 = 1.58 x 10-5 mol dm-3.

  3. Why is a pH meter calibrated with buffer solutions before use?
    • To clean the electrode
    • To make the solution neutral
    • To speed up the reading
    • So that its readings match known pH values, removing systematic error
    Show answer

    So that its readings match known pH values, removing systematic error. Buffers have accurately known pH.

  4. Why are smaller volumes of NaOH added near the end point when plotting a titration curve?
    • The pH changes rapidly there, so small additions locate the equivalence volume precisely
    • To avoid heating the mixture
    • Because the burette is nearly empty
    • To keep the pH below 7
    Show answer

    The pH changes rapidly there, so small additions locate the equivalence volume precisely. Pearson's guide lists large additions near the end point as an error.

  5. In the half-neutralisation method, why is an equal volume of the original acid added to a portion that has been exactly neutralised?
    • It makes the solution neutral
    • It removes excess NaOH
    • It makes the amounts of HA and A- equal, so pH = pKa
    • It calibrates the meter
    Show answer

    It makes the amounts of HA and A- equal, so pH = pKa. The neutralised portion contains A-; the added acid supplies an equal amount of HA.

  6. Why is the pH electrode rinsed with deionised water between measurements?
    • To avoid carrying one solution into the next and changing its pH
    • To recalibrate it
    • To cool it
    • To make the reading steady
    Show answer

    To avoid carrying one solution into the next and changing its pH. Contamination would alter the next reading.

  7. An uncertainty of ±0.01 in a pH reading changes [H+] by a factor of 100.01. Express this as a percentage change in Ka (for Ka = [H+] at half-neutralisation).
    Show answer

    2.3 %. 100.01 = 1.0233, a 2.3% change.

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Last updated 8 November 2026. Spotted a mistake? Tell us.