All subjects · Chemistry · WCH14 (Unit 4)

Weak acids, Ka and pKa

A weak acid is only partly dissociated, so its pH depends on its acid dissociation constant, Ka, as well as its concentration.

Ka and pKa (spec 14.8, 14.12)

For HA ⇌ H+ + A-:

Ka = [H+][A-] / [HA], units mol dm-3.

pKa = -log10 Ka and Ka = 10-pKa. A larger Ka (smaller pKa) means a stronger acid. Ethanoic acid has Ka about 1.7 x 10-5 mol dm-3.

pH of a weak acid (spec 14.9)

Two assumptions:

1. [H+] = [A-], because the acid is the only significant source of H+ (the ionisation of water is ignored).

2. [HA] at equilibrium = the starting concentration, because so little dissociates.

Then Ka = [H+]2 / [HA], so [H+] = √(Ka x [HA]). No quadratic equations are needed. Assumption 2 is weakest for more dilute solutions or less weak acids.

Ka from experiment (spec 14.13, 14.14)

Measure the pH of a solution of known concentration (made from a known mass in a known volume). [H+] = 10-pH, then Ka = [H+]2 / [HA].

Comparing equimolar solutions: a strong acid has a lower pH than a weak acid. On dilution by a factor of 10, a strong acid's pH rises by 1, but a weak acid's pH rises by only about 0.5, because more of it dissociates.

Worked example

Calculate the pH of 0.100 mol dm-3 ethanoic acid. Ka = 1.7 x 10-5 mol dm-3.

  1. [H+] = √(1.7 x 10-5 x 0.100) = √(1.7 x 10-6)
  2. [H+] = 1.304 x 10-3 mol dm-3
  3. pH = -log10[H+]

Answer: pH = 2.88

Key idea: Ka = [H+][A-]/[HA]; pKa = -log Ka. For a weak acid [H+] = √(Ka[HA]), assuming [H+] = [A-] and negligible dissociation.

Check yourself

  1. Which is the Ka expression for propanoic acid, CH3CH2COOH?
    • [CH3CH2COOH] / ([CH3CH2COO-][H+])
    • [H+]2 x [CH3CH2COOH]
    • [CH3CH2COO-] / [H+]
    • [CH3CH2COO-][H+] / [CH3CH2COOH]
    Show answer

    [CH3CH2COO-][H+] / [CH3CH2COOH]. Products of dissociation over undissociated acid.

  2. Calculate the pH of 0.0500 mol dm-3 methanoic acid. Ka = 1.6 x 10-4 mol dm-3.
    Show answer

    2.55. [H+] = √(1.6 x 10-4 x 0.0500) = 2.828 x 10-3; pH = 2.55.

  3. Calculate pKa for ethanoic acid, Ka = 1.7 x 10-5 mol dm-3.
    Show answer

    4.77. pKa = -log10(1.7 x 10-5) = 4.77.

  4. A 0.0500 mol dm-3 solution of a weak monoprotic acid has pH 3.00. Calculate Ka.
    Show answer

    2.0 x 10-5 mol dm-3. [H+] = 1.00 x 10-3; Ka = (1.00 x 10-3)2 / 0.0500 = 2.0 x 10-5 mol dm-3.

  5. 1.85 g of propanoic acid (Mr 74.0) is dissolved in water to make 250 cm3 of solution, which has pH 2.94. Calculate Ka.
    Show answer

    1.32 x 10-5 mol dm-3. c = (1.85 / 74.0) / 0.250 = 0.100 mol dm-3; [H+] = 10-2.94 = 1.148 x 10-3; Ka = (1.148 x 10-3)2 / 0.100 = 1.32 x 10-5 mol dm-3.

  6. Which assumption is made when calculating the pH of a weak acid from Ka?
    • The equilibrium concentration of HA equals its starting concentration
    • The acid is fully dissociated
    • [H+] is greater than [A-]
    • Water is fully ionised
    Show answer

    The equilibrium concentration of HA equals its starting concentration. Only a tiny fraction dissociates, so [HA] hardly changes.

  7. A weak acid solution is diluted by a factor of 10. By about how much does its pH rise?
    • 1
    • 2
    • 0.5
    • 0.1
    Show answer

    0.5. [H+] = √(Ka c), so dividing c by 10 divides [H+] by √10, raising pH by 0.5.

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Last updated 7 November 2026. Spotted a mistake? Tell us.