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Ka, pKa and weak acid pHEdexcel A-Level Chemistry: Revision notes

Section 1

The acid dissociation constant, Ka

A weak acid HA is only partially dissociated in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq). Applying the equilibrium law gives the acid dissociation constant:

Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]} (units mol dm⁻³)

Water is in huge excess so its concentration is constant and is not included. A larger KaK_a means more dissociation and a stronger acid. KaK_a changes only with temperature.

Key termsacid dissociation constantweak acid
Exam tip

Write the equation first, then the expression. For HA each dissociation gives one H⁺ and one A⁻.

Section 2

pKa

pKₐ is defined as pKa=−lg⁡KapK_a = -\lg K_a, and so Ka=10−pKaK_a = 10^{-pK_a}. Because of the minus sign, a lower pKₐ means a larger KaK_a and a stronger acid. For ethanoic acid Ka=1.74×10−5K_a = 1.74 \times 10^{-5} mol dm⁻³ gives pKₐ = 4.76. Using pKₐ is convenient because values of KaK_a span many powers of ten.

Key termspKa
Common mistake

Do not mix up the direction: stronger acid means larger KaK_a but smaller pKₐ.

Section 3

Calculating the pH of a weak acid

For a weak acid HA of concentration cc the pH is not −lg cc, because [H+][H^+] is much less than cc. Two assumptions simplify the calculation:

  1. [H+]=[A−][H^+] = [A^-] (water's own ionisation is negligible)
  2. [HA]eq≈c[HA]_{eq} \approx c (very little acid has dissociated)

Then Ka=[H+]2cK_a = \frac{[H^+]^2}{c}, so [H+]=Ka×c[H^+] = \sqrt{K_a \times c} and pH = −lg [H+][H^+].

Worked example: 0.100 mol dm⁻³ ethanoic acid. [H+]=1.74×10−5×0.100=1.32×10−3[H^+] = \sqrt{1.74 \times 10^{-5} \times 0.100} = 1.32 \times 10^{-3} mol dm⁻³, so pH = 2.88.

Key termsassumptionsapproximation
Common mistake

Using pH = −lg 0.100 = 1.00 for a weak acid. That treats it as fully dissociated.

Section 4

Finding Ka from experimental data

If the mass of a weak acid is dissolved in a known volume and the pH is measured, find [HA][HA] from mass ÷ molar mass ÷ volume. Then [H+]=10−pH[H^+] = 10^{-pH} and, with the same two assumptions, Ka=[H+]2/[HA]K_a = [H^+]^2 / [HA].

Example: 0.440 g of an acid (88.0 g mol⁻¹) in 100 cm³ gives [HA][HA] = 0.0500 mol dm⁻³. At pH 3.06, [H+]=8.71×10−4[H^+] = 8.71 \times 10^{-4} mol dm⁻³, so Ka=1.52×10−5K_a = 1.52 \times 10^{-5} mol dm⁻³.

Key termsexperimental Ka

Section 5

Core practical 9: Ka by half-neutralisation

Titrate a known volume of weak acid with strong alkali (e.g. NaOH) using a pH meter, recording pH against volume added. Find the equivalence volume from the steep part of the curve. At half the equivalence volume half the HA has been converted to A⁻, so [HA]=[A−][HA] = [A^-]. These cancel in the expression, so Ka=[H+]K_a = [H^+] and pKₐ = pH at that point.

Take readings at small volume intervals near equivalence, calibrate the meter with buffers, and rinse the electrode between solutions.

Key termshalf-neutralisationequivalence point
Exam tip

State both: [HA]=[A−][HA]=[A^-] at half-neutralisation, so Ka=[H+]K_a=[H^+].

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Ka, pKa and weak acid pH

  1. A food technologist analyses a 0.100 mol dm⁻³ solution of ethanoic acid, CH₃COOH, a weak monobasic acid, at 298 K. For ethanoic acid, KaK_a = 1.74 × 10⁻⁵ mol dm⁻³.
    Calculate the pH of the 0.100 mol dm⁻³ ethanoic acid solution. State the assumptions you make.2 marks
  2. A chemist prepares a 0.0500 mol dm⁻³ solution of methanoic acid, HCOOH, for use as a descaling agent. For methanoic acid, KaK_a = 1.78 × 10⁻⁴ mol dm⁻³ at 298 K. She compares it with propanoic acid, which has pKₐ = 4.87.
    The chemist states that the pH of the methanoic acid solution is greater than that of a 0.0500 mol dm⁻³ solution of a strong monobasic acid. Explain why.2 marks
  3. A student dissolves 0.440 g of a pure solid weak monobasic acid, HA (molar mass 88.0 g mol⁻¹), in water and makes the solution up to 100 cm³. A calibrated pH meter reads 3.06 at 298 K.
    Calculate KaK_a for the acid HA. State the assumptions you make.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).