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When a weak acid HA dissolves in water, only a small fraction of its molecules donate a proton to water:
HA(aq) + H₂O(l) ⇌ A⁻(aq) + H₃O⁺(aq)
Or more simply:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
At equilibrium, the solution contains a mixture of undissociated HA molecules, H⁺ ions, and A⁻ ions. For a typical weak acid like ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³), fewer than 1% of the molecules are dissociated in a 1.0 mol dm⁻³ solution.
Why does this matter for calculations? For a strong acid, [H⁺] simply equals the acid concentration. For a weak acid, [H⁺] is much smaller than the acid concentration. You need Ka to calculate [H⁺].
The equilibrium constant for the dissociation of a weak acid is called the acid dissociation constant, Ka:
Ka = [H⁺][A⁻] / [HA]
The units of Ka are mol dm⁻³ (since there are two concentration terms on top and one on the bottom). Water does not appear in the expression because its concentration is effectively constant in dilute solution (approximately 55.5 mol dm⁻³).
Ka tells you the strength of a weak acid:
| Acid | Formula | Ka / mol dm⁻³ | pKa | Relative Strength |
|---|---|---|---|---|
| Methanoic acid | HCOOH | 1.77 × 10⁻⁴ | 3.75 | Stronger weak acid |
| Benzoic acid | C₆H₅COOH | 6.3 × 10⁻⁵ | 4.20 | Moderate |
| Ethanoic acid | CH₃COOH | 1.74 × 10⁻⁵ | 4.76 | Moderate weak acid |
| Propanoic acid | CH₃CH₂COOH | 1.35 × 10⁻⁵ | 4.87 | Moderate weak acid |
| Carbonic acid | H₂CO₃ | 4.3 × 10⁻⁷ | 6.37 | Weaker weak acid |
| Phenol | C₆H₅OH | 1.3 × 10⁻¹⁰ | 9.89 | Very weak acid |
When calculating the pH of a weak acid, we make an important simplifying assumption. If we start with an initial concentration c of the weak acid HA, and x moles per dm³ dissociate:
Since the acid is weak, x is very small compared to c. Therefore:
c - x ≈ c
This simplification lets us write:
Ka = x² / c
where x = [H⁺]. Rearranging:
[H⁺] = √(Ka × c)
This assumption is valid when Ka is small (typically less than about 10⁻² mol dm⁻³) and the initial concentration is not too dilute. You should always state the assumption in exam answers.
How to state the assumption in an exam: Write: "Assuming that the degree of dissociation is negligible, [HA] at equilibrium ≈ c." Then verify by checking that [H⁺] is less than about 5% of c.
Calculate the pH of a 0.10 mol dm⁻³ solution of ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³).
Step 1: Write the equilibrium.
CH₃COOH ⇌ CH₃COO⁻ + H⁺
Step 2: Apply the assumption that dissociation is small.
[H⁺] = √(Ka × c) [H⁺] = √(1.74 × 10⁻⁵ × 0.10) [H⁺] = √(1.74 × 10⁻⁶) [H⁺] = 1.32 × 10⁻³ mol dm⁻³
Step 3: Calculate pH.
pH = -log₁₀(1.32 × 10⁻³) pH = 2.88
Step 4: Check the assumption. The dissociated concentration (1.32 × 10⁻³) is about 1.3% of the initial concentration (0.10). This is small, so the assumption is valid.
A 0.20 mol dm⁻³ solution of a weak acid HA has a pH of 2.72. Calculate Ka.
Step 1: Find [H⁺].
[H⁺] = 10⁻²·⁷² = 1.91 × 10⁻³ mol dm⁻³
Step 2: Since HA ⇌ H⁺ + A⁻, and [H⁺] = [A⁻]:
Ka = [H⁺][A⁻] / [HA] Ka = [H⁺]² / [HA]
Step 3: Assuming [HA] ≈ 0.20 (since 1.91 × 10⁻³ << 0.20):
Ka = (1.91 × 10⁻³)² / 0.20 Ka = 3.65 × 10⁻⁶ / 0.20 Ka = 1.82 × 10⁻⁵ mol dm⁻³
A 0.50 mol dm⁻³ solution of propanoic acid has Ka = 1.35 × 10⁻⁵ mol dm⁻³. Calculate the percentage dissociation.
[H⁺] = √(Ka × c) = √(1.35 × 10⁻⁵ × 0.50) = √(6.75 × 10⁻⁶) = 2.60 × 10⁻³ mol dm⁻³
Percentage dissociation = ([H⁺] / c) × 100 = (2.60 × 10⁻³ / 0.50) × 100 = 0.52%
This confirms the assumption is excellent: only 0.52% dissociates.
Key insight: As you dilute a weak acid, the percentage dissociation increases even though [H⁺] decreases. At 0.0010 mol dm⁻³: [H⁺] = √(1.35 × 10⁻⁵ × 0.0010) = 1.16 × 10⁻⁴ mol dm⁻³, giving 11.6% dissociation. This is why the assumption becomes less valid at very low concentrations.
Just as pH = -log₁₀[H⁺], we define:
pKa = -log₁₀Ka
And the reverse: Ka = 10⁻ᵖᴷᵃ
pKa converts small Ka values into more convenient numbers:
| Acid | Ka / mol dm⁻³ | pKa |
|---|---|---|
| Methanoic acid | 1.77 × 10⁻⁴ | 3.75 |
| Ethanoic acid | 1.74 × 10⁻⁵ | 4.76 |
| Propanoic acid | 1.35 × 10⁻⁵ | 4.87 |
| Carbonic acid | 4.3 × 10⁻⁷ | 6.37 |
| Phenol | 1.3 × 10⁻¹⁰ | 9.89 |
A larger pKa means a weaker acid (the opposite relationship to Ka, because of the negative sign in the log).
Exam tip: pKa values are what you will see at the half-equivalence point of a titration curve. If the titration curve gives pH = 4.76 at the half-equivalence point, you immediately know pKa = 4.76 and Ka = 10⁻⁴·⁷⁶ = 1.74 × 10⁻⁵.
The assumption [HA] ≈ c breaks down when:
In these cases, you must solve the full quadratic equation:
Ka = x² / (c - x)
Ka(c - x) = x²
x² + Ka·x - Ka·c = 0
Then use the quadratic formula:
x = [-Ka + √(Ka² + 4·Ka·c)] / 2
(Take the positive root only.)
Worked example of the quadratic: A weak acid with Ka = 0.050 mol dm⁻³ has concentration 0.10 mol dm⁻³. The assumption gives [H⁺] = √(0.050 × 0.10) = 0.071 mol dm⁻³, which is 71% of 0.10 — clearly not negligible.
Using the quadratic: x² + 0.050x - 0.0050 = 0 x = [-0.050 + √(0.0025 + 0.020)] / 2 = [-0.050 + 0.150] / 2 = 0.050 mol dm⁻³ pH = -log₁₀(0.050) = 1.30
The assumption gives pH = 1.15, which is noticeably different. Edexcel questions will almost always set problems where the assumption is valid, but you should know the quadratic method exists.
| Equation | Use |
|---|---|
| Ka = [H⁺][A⁻] / [HA] | Definition of Ka |
| [H⁺] = √(Ka × c) | pH of weak acid (with assumption) |
| Ka = [H⁺]² / c | Ka from pH (with assumption) |
| pKa = -log₁₀Ka | Converting Ka to pKa |
| Ka = 10⁻ᵖᴷᵃ | Converting pKa to Ka |
| % dissociation = ([H⁺]/c) × 100 | Degree of dissociation |
| Mistake | Correction |
|---|---|
| Using [H⁺] = c for a weak acid | Only valid for strong acids; for weak acids use [H⁺] = √(Ka × c) |
| Forgetting to state the assumption | Always write "Assuming [HA] ≈ c because dissociation is negligible" |
| Confusing Ka with Kw | Ka applies to weak acid dissociation; Kw applies to water autoionisation |
| Thinking a smaller Ka means a stronger acid | Smaller Ka = weaker acid; larger Ka = stronger acid |
| Confusing pKa scale direction | Larger pKa = weaker acid (opposite to Ka) |
| Not checking the assumption is valid | Verify that [H⁺] is less than ~5% of c |
Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, sub-strand 12.3 requires you to write the expression Ka = [A⁻][H⁺]/[HA] for a generic weak monoprotic acid HA; define pKa = −log₁₀ Ka; calculate the pH of a weak acid given Ka and concentration; calculate Ka from pH and concentration; identify the standard simplifying assumptions ([H⁺] = [A⁻] from the same equilibrium and [HA]_eq ≈ [HA]_initial) and the conditions under which they break down (refer to the official specification document for exact wording). Ka is examined alongside pH in Paper 1 Topic 12 problem-solving questions and synoptically in Paper 3 (CP13). The Edexcel data booklet provides selected pKa values; for unknown acids, candidates must derive Ka from experimental data.
Question (8 marks):
(a) State the equilibrium expression for the dissociation of ethanoic acid in water and write the expression for Ka. (2)
(b) Calculate the pH of a 0.10 mol dm⁻³ solution of ethanoic acid at 298 K. (pKa = 4.76, so Ka = 1.74 × 10⁻⁵ mol dm⁻³.) State clearly the two simplifying assumptions you make. (4)
(c) A 0.020 mol dm⁻³ solution of an unknown weak acid HA has pH = 3.42 at 298 K. Calculate Ka and pKa for the acid. (2)
Solution with mark scheme:
(a) CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq)
Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]
M1 — correct equilibrium with reversible arrow. A1 — correct Ka expression.
(b) Assumption 1: [H⁺] = [CH₃COO⁻] (every H⁺ comes from the dissociation, ignoring water autoionisation, valid because Ka × c ≫ Kw).
Assumption 2: [CH₃COOH]_eq ≈ [CH₃COOH]_initial = 0.10 mol dm⁻³ (less than 5% dissociates).
Substituting:
Ka = [H⁺]² / [HA]_initial
[H⁺]² = Ka × [HA] = 1.74 × 10⁻⁵ × 0.10 = 1.74 × 10⁻⁶
[H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.32 × 10⁻³) = 2.88
M1 — both assumptions stated. M1 — correct rearrangement to [H⁺]². M1 — correct numerical [H⁺]. A1 — pH = 2.88 (2 d.p.).
(c) [H⁺] = 10⁻³·⁴² = 3.80 × 10⁻⁴ mol dm⁻³.
Using both assumptions: Ka = [H⁺]² / [HA] = (3.80 × 10⁻⁴)² / 0.020 = 1.44 × 10⁻⁷ / 0.020 = 7.22 × 10⁻⁶ mol dm⁻³.
pKa = −log₁₀(7.22 × 10⁻⁶) = 5.14.
M1 — Ka calculation correct (working shown). A1 — pKa = 5.14.
Total: 8 marks (M5 A3, split as shown).
Question (6 marks): Methanoic acid, HCOOH, is a weak monoprotic acid with Ka = 1.78 × 10⁻⁴ mol dm⁻³ at 298 K.
(a) Calculate the pH of a 0.25 mol dm⁻³ solution of methanoic acid. (3)
(b) Calculate the percentage dissociation of the acid in this solution and comment on whether the simplifying assumption [HA]_eq ≈ [HA]_initial is justified. (3)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 1, AO2 = 4, AO3 = 1. Edexcel reserves AO3 marks for explicit critical evaluation of assumptions — bare numerical answers do not earn this.
Connects to:
Topic 11 — Equilibria: Ka is an equilibrium constant. All the Le Chatelier and Kc reasoning from Topic 11 applies — pressure changes don't shift Ka (no gases for typical weak acids), but temperature does (Ka changes because the ionisation reaction has non-zero ΔH).
Topic 17 — Carboxylic acids: the relative pKa values of carboxylic acids reveal substituent effects. CH₃COOH has pKa 4.76; CH₂ClCOOH has pKa 2.86; CCl₃COOH has pKa 0.65 — the trend reflects increasing electron-withdrawing inductive stabilisation of the conjugate base RCOO⁻.
Topic 12.4 — Buffers (Lesson 5): the buffer formula pH = pKa + log₁₀([A⁻]/[HA]) is derived directly from rearranging the Ka expression. Confidence with Ka here is essential.
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