Weak acids and bases, Ka, for weak acids (A-level only) — AQA A-Level Chemistry
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Weak acids and bases, Ka, for weak acids (A-level only) explained
A weak acid ionises only slightly, so the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ lies well to the left and typically fewer than two molecules in a hundred have donated a proton at any moment.
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That is why 0.100 mol dm⁻³ ethanoic acid has a pH near 2.9 while 0.100 mol dm⁻³ hydrochloric acid, fully dissociated, has a pH of 1.00. A weak base does the same in reverse: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ generates only a small [OH⁻]. Strength and concentration are separate ideas, so a concentrated weak acid can have a lower pH than a dilute strong one. Because the total amount of acid is unchanged, a weak acid still neutralises the same volume of alkali as a strong acid of equal concentration; only the starting pH and the reaction rate differ.
Ka is the dissociation constant for a weak acid.
Ka is the equilibrium constant for the dissociation of a weak acid: for HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA], with units mol dm⁻³. Water does not appear, because it is the solvent. To find the pH of a weak acid on its own, two approximations are made: the H⁺ from water is negligible, so [H⁺] = [A⁻]; and so little acid dissociates that [HA] at equilibrium equals the concentration made up. Then Ka = [H⁺]²/[HA], so [H⁺] = √(Ka × [HA]). For butanoic acid, Ka = 1.51 × 10⁻⁵ mol dm⁻³, so a 0.100 mol dm⁻³ solution gives [H⁺] = 1.23 × 10⁻³ mol dm⁻³ and pH = 2.91. The logarithmic form pKa = −log₁₀Ka ranks acids compactly: a larger Ka, and so a smaller pKa, means a stronger acid.
Your focus
- Write the dissociation equation of propanoic acid with an equilibrium sign and identify its conjugate base.
- Sketch mass-loss curves for equal concentrations of chloroethanoic acid and ethanoic acid reacting with excess sodium hydrogencarbonate.
- Explain why 0.100 mol dm⁻³ ethanoic acid has a pH near 2.9 while hydrochloric acid of the same concentration has pH 1.00.
Show all 6 objectives
- Write the Ka expression for butanoic acid and state its units.
- Calculate the pH of 0.100 mol dm⁻³ butanoic acid from Ka = 1.51 × 10⁻⁵ mol dm⁻³.
- Convert a pKa of 4.82 into a Ka value and use pKa values to rank three acids by strength.
Weak acids and bases, Ka, for weak acids (A-level only) exam tips
Marking Points
- one mark for stating that only a small proportion of the molecules dissociate, so the equilibrium lies to the left
- one mark for writing the dissociation with an equilibrium sign, such as CH₃COOH ⇌ H⁺ + CH₃COO⁻
- one mark for comparing two acids by the initial rate of gas evolution with a carbonate, the stronger acid giving the steeper curve
- one mark for the same final volume or mass loss, because the amount of acid rather than its strength fixes the yield
- one mark for explaining a difference in strength by the effect of a group on the polarity of the O–H bond
- one mark for the expression Ka = [H⁺][A⁻]/[HA]
- one mark for the units mol dm⁻³
- one mark for assuming [H⁺] = [A⁻] and that [HA] at equilibrium equals the initial concentration
- one mark for pKa = −log₁₀Ka and the reverse conversion Ka = 10⁻ᵖᴷᵃ
- one mark for identifying the acid with the larger Ka, or the smaller pKa, as the stronger acid
Examiner Tips
- 💡On a mass-loss or gas-volume graph, strength changes the gradient and amount changes the plateau; check which one the question is testing.
- 💡Always draw the equilibrium arrow for a weak acid; the single arrow is a common and avoidable loss.
- 💡Set out the two approximations explicitly; some mark schemes award a mark for stating them.
- 💡Check the size of your answer: a weak acid at 0.1 mol dm⁻³ should give a pH around 3, not around 1.
Common Mistakes
- confusing weak with dilute, so 5 mol dm⁻³ ethanoic acid is described as a strong acid
- using a single arrow in the dissociation equation of a weak acid
- predicting a smaller final volume of carbon dioxide from the weaker acid, when only the rate differs
- assuming a weak acid needs a different volume of alkali to neutralise it than a strong acid of the same concentration
- including water in the Ka expression
- forgetting to take the square root of Ka × [HA] when finding [H⁺]
- giving pKa a unit, when taking a logarithm removes it
- stating that a larger pKa means a stronger acid
- using the acid concentration directly as [H⁺], as though the weak acid were strong