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This lesson extends the Ka and Kb framework introduced in lesson 2 to the quantitative calculation of pH for weak monoprotic acids and weak monoprotic bases. The central tool is the ICE (Initial-Change-Equilibrium) table applied to the dissociation HA ⇌ H⁺ + A⁻, which yields a quadratic equation in [H⁺]. For most A-Level problems the quadratic simplifies dramatically once a single assumption — that dissociation is small (Ka much smaller than the initial acid concentration c) — is made. Under that assumption the working pH formula is pH = ½(pKa − log c), and the calculation collapses to two arithmetic steps. We will derive that formula from first principles, set out a quantitative rule for when the assumption is safe (dissociation below 5%), and work through the full quadratic for the cases where it is not. The same machinery, applied via Kb, gives the pH of weak bases.
Spec mapping (AQA 7405): This lesson maps to §3.1.12 (acids and bases — weak-acid pH calculations) of the AQA A-Level Chemistry specification. It builds directly on lesson 2 of this course (Ka, Kb, pKa and the Ka·Kb = Kw conjugate-pair relation) and extends the equilibrium ICE methodology developed in §3.1.6 (equilibrium constants Kc) to acid-base systems. The output of this lesson — fluency in computing the pH of weak monoprotic acids and bases — is the prerequisite for lesson 4 (buffers, which extends the calculation to mixtures of a weak acid and its conjugate base) and lesson 5 (pH curves, which tracks pH along the course of a titration). Refer to the official AQA specification document for the exact wording of §3.1.12.
Assessment objectives: Recalling the approximation pH = ½(pKa − log[HA]) and stating the conditions under which it is valid (Ka much smaller than c, dissociation below 5%) is AO1. Computing the pH of a weak acid or weak base solution from Ka or Kb and the initial concentration — including identifying when the approximation breaks down and solving the full quadratic — is AO2. Comparing strong-acid and weak-acid pH at the same nominal concentration, rationalising how pH responds to dilution (a 10-fold dilution changes pH by less than 1 unit for a weak acid), and explaining the limiting behaviour at very low concentration where percent dissociation rises sharply is AO3 reasoning.
For a weak monoprotic acid HA dissolved in water, the equilibrium is:
HA(aq) ⇌ H⁺(aq) + A⁻(aq)
with equilibrium constant:
Ka = [H⁺][A⁻] / [HA]
Let the initial (formal, total) concentration of the acid be c mol dm⁻³ and let x be the equilibrium concentration of H⁺ produced by dissociation. We tabulate the concentrations using the ICE format:
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| Initial (mol dm⁻³) | c | 0 | 0 |
| Change (mol dm⁻³) | −x | +x | +x |
| Equilibrium (mol dm⁻³) | c − x | x | x |
Substituting the equilibrium row into the Ka expression:
Ka = x · x / (c − x) = x² / (c − x)
This is a quadratic in x. Rearranging to standard form:
x² + Ka·x − Ka·c = 0
The physically meaningful root (x > 0) is:
x = (−Ka + √(Ka² + 4 Ka·c)) / 2
That is the exact result. In practice we rarely need it — the assumption developed in the next section reduces the quadratic to a one-line square root for the great majority of A-Level problems.
Key Definition: The two contributions to [H⁺] in a pure weak-acid solution are (i) dissociation of the acid itself (gives x) and (ii) the autoionisation of water (typically negligible: 10⁻⁷ mol dm⁻³). At A-Level we ignore (ii) unless the acid is extremely weak or extremely dilute, in which case [H⁺] approaches 10⁻⁷ from above and water's contribution must be added explicitly.
When Ka is much smaller than the initial concentration c, the amount that dissociates (x) is small compared to c. We can then approximate:
c − x ≈ c
Substituting into Ka = x² / (c − x):
Ka ≈ x² / c
so:
x = √(Ka · c)
Because [H⁺] = x at equilibrium:
[H⁺] = √(Ka · c)
Taking negative logs of both sides:
pH = −log(√(Ka · c)) = ½ · (−log Ka − log c) = ½(pKa − log c)
This is the working formula for the pH of a weak monoprotic acid. It is sometimes written as pH = ½(pKa + p[HA]) where p[HA] = −log c. Either form is acceptable.
The standard rule of thumb is that the approximation is acceptable when dissociation is below 5% — that is, when x / c ≤ 0.05. Substituting x = √(Ka · c):
√(Ka · c) / c ≤ 0.05
√(Ka / c) ≤ 0.05
Ka / c ≤ 0.0025
c / Ka ≥ 400
So a quick mental check: if the initial concentration is at least 400 times Ka, the approximation is safe. For ethanoic acid (Ka = 1.74 × 10⁻⁵), 400·Ka = 7.0 × 10⁻³ mol dm⁻³ — so the approximation is fine for 0.01 mol dm⁻³ and above. For chlorous acid HClO₂ (Ka = 1.1 × 10⁻²), 400·Ka = 4.4 mol dm⁻³ — well above any realistic working concentration, so the approximation always fails and the quadratic is mandatory.
Exam Tip: Always state explicitly whether you are using the approximation, and check the validity afterwards. Examiners reward the audit step ("dissociation = x/c = 1.3%, well below 5%, so the approximation is justified") with method marks even if the final number is slightly off.
Calculate the pH of 0.10 mol dm⁻³ ethanoic acid (CH₃COOH), Ka = 1.74 × 10⁻⁵ mol dm⁻³.
Step 1 — Approximation check: c / Ka = 0.10 / 1.74 × 10⁻⁵ = 5750, well above 400. The approximation is safe.
Step 2 — Apply [H⁺] = √(Ka · c):
[H⁺] = √(1.74 × 10⁻⁵ × 0.10) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³
Step 3 — Compute pH:
pH = −log(1.32 × 10⁻³) = 2.88
Step 4 — Validate: dissociation = x / c = 1.32 × 10⁻³ / 0.10 = 1.32%, well below the 5% threshold. ✓
Alternatively, using the log form directly: pKa = −log(1.74 × 10⁻⁵) = 4.76; pH = ½(4.76 − log 0.10) = ½(4.76 + 1.00) = 2.88. ✓
Calculate the pH of 0.001 mol dm⁻³ ethanoic acid. Same Ka.
Step 1 — Approximation check: c / Ka = 0.001 / 1.74 × 10⁻⁵ = 57.5 — well below 400. The approximation is borderline-bad; expect dissociation above 5%.
Step 2 — Approximate answer (for comparison): pH = ½(4.76 + 3.00) = 3.88; [H⁺] = 1.32 × 10⁻⁴; dissociation = 13.2%. Confirmed: approximation invalid.
Step 3 — Solve the quadratic: x² + Ka·x − Ka·c = 0
x² + (1.74 × 10⁻⁵)x − (1.74 × 10⁻⁸) = 0
Discriminant: (1.74 × 10⁻⁵)² + 4 · 1.74 × 10⁻⁸ = 3.03 × 10⁻¹⁰ + 6.96 × 10⁻⁸ = 6.99 × 10⁻⁸
√Discriminant = 2.64 × 10⁻⁴
x = (−1.74 × 10⁻⁵ + 2.64 × 10⁻⁴) / 2 = 2.47 × 10⁻⁴ / 2 = 1.23 × 10⁻⁴ mol dm⁻³
Step 4 — pH: pH = −log(1.23 × 10⁻⁴) = 3.91
The exact answer (3.91) differs from the approximate answer (3.88) by only 0.03 pH units — a small numerical correction, but the quadratic is the methodologically defensible answer in mark-scheme terms.
Calculate the pH of 0.10 mol dm⁻³ hydrocyanic acid (HCN), Ka = 6.17 × 10⁻¹⁰ mol dm⁻³.
Step 1 — Approximation check: c / Ka = 0.10 / 6.17 × 10⁻¹⁰ = 1.62 × 10⁸. Vastly above 400 — approximation extremely safe.
Step 2: [H⁺] = √(6.17 × 10⁻¹⁰ × 0.10) = √(6.17 × 10⁻¹¹) = 7.85 × 10⁻⁶ mol dm⁻³
Step 3: pH = −log(7.85 × 10⁻⁶) = 5.10
Step 4 — Validate: dissociation = 7.85 × 10⁻⁶ / 0.10 = 7.85 × 10⁻³% — utterly negligible. ✓
A weak acid with a small Ka gives a solution whose pH is only modestly below 7 even at 0.10 mol dm⁻³ — HCN is a far weaker acid than ethanoic acid, and the pH reflects that.
Calculate the pH of 0.10 mol dm⁻³ chlorous acid (HClO₂), Ka = 1.1 × 10⁻² mol dm⁻³.
Step 1 — Approximation check: c / Ka = 0.10 / 1.1 × 10⁻² = 9.1 — far below 400. The approximation fails badly; expect dissociation well above 5%.
Step 2 — Approximation (for comparison): [H⁺] ≈ √(1.1 × 10⁻² × 0.10) = √(1.1 × 10⁻³) = 0.0332 — but this implies 33% dissociation, clearly invalid. The quadratic is essential.
Step 3 — Solve the quadratic: x² + (1.1 × 10⁻²)x − (1.1 × 10⁻³) = 0
Discriminant = (1.1 × 10⁻²)² + 4·(1.1 × 10⁻³) = 1.21 × 10⁻⁴ + 4.40 × 10⁻³ = 4.52 × 10⁻³
√Discriminant = 0.0673
x = (−0.011 + 0.0673) / 2 = 0.0563 / 2 = 0.0281 mol dm⁻³
Step 4 — pH: pH = −log(0.0281) = 1.55
The quadratic answer (1.55) differs from the naive approximation (1.48) by 0.07 pH units, but more importantly the dissociation is 28% — neither value comes close to satisfying the small-x assumption. Always solve the quadratic when c / Ka is below 400.
Exam Tip: AQA mark schemes generally award full marks for the approximation answer when it is justified, and for the quadratic answer where the approximation fails. Going straight to the quadratic in every problem is safe but wasteful — the approximation is the expected method for ~90% of weak-acid problems at A-Level.
The same ICE methodology applies to a weak monoprotic base B with equilibrium:
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
with equilibrium constant:
Kb = [BH⁺][OH⁻] / [B]
Let c be the initial concentration of B and let y be the equilibrium concentration of OH⁻. The ICE table is structurally identical to the acid case, and the same approximation gives:
[OH⁻] = √(Kb · c)
from which:
pOH = ½(pKb − log c)
and finally, via pH + pOH = 14.00 (Kw at 298 K):
pH = 14.00 − ½(pKb − log c)
The same 5%-dissociation validity check applies (c / Kb ≥ 400 for the approximation to hold).
When a base is supplied with the Ka of its conjugate acid rather than Kb directly (a very common AQA setup), use the conjugate-pair relation derived in lesson 2:
Ka · Kb = Kw
so:
Kb = Kw / Ka = 10⁻¹⁴ / Ka
Equivalently in log form: pKa + pKb = 14.00. This route is identical in arithmetic — choose whichever value the data sheet supplies.
Calculate the pH of 0.10 mol dm⁻³ aqueous ammonia (NH₃). Data: Ka(NH₄⁺) = 5.6 × 10⁻¹⁰ mol dm⁻³.
Step 1 — Find Kb of NH₃:
Kb = Kw / Ka = 1.00 × 10⁻¹⁴ / 5.6 × 10⁻¹⁰ = 1.79 × 10⁻⁵ mol dm⁻³
Step 2 — Approximation check: c / Kb = 0.10 / 1.79 × 10⁻⁵ = 5590 — well above 400. Safe.
Step 3: [OH⁻] = √(Kb · c) = √(1.79 × 10⁻⁵ × 0.10) = √(1.79 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³
Step 4 — pOH and pH:
pOH = −log(1.34 × 10⁻³) = 2.87
pH = 14.00 − 2.87 = 11.13
Step 5 — Validate: dissociation = 1.34 × 10⁻³ / 0.10 = 1.34%, below the 5% threshold. ✓
The symmetry with the ethanoic acid calculation (Example 1) is striking: both are 0.10 mol dm⁻³, both have conjugate Ka close to 10⁻⁵, both dissociate by about 1.3%, and the two pH values (2.88 and 11.13) sit symmetrically either side of pH 7 — separated by exactly pKa + pKb = 14 worth of arithmetic.
Calculate the pH of 0.10 mol dm⁻³ methylamine (CH₃NH₂), Kb = 4.4 × 10⁻⁴ mol dm⁻³.
Step 1 — Approximation check: c / Kb = 0.10 / 4.4 × 10⁻⁴ = 227 — below 400. Borderline; quadratic recommended.
Step 2 — Approximate answer (for orientation): [OH⁻] ≈ √(4.4 × 10⁻⁴ × 0.10) = √(4.4 × 10⁻⁵) = 6.63 × 10⁻³; dissociation = 6.63% — just above the 5% threshold. The quadratic correction will be small but methodologically required.
Step 3 — Solve the quadratic: y² + Kb·y − Kb·c = 0; that is, y² + (4.4 × 10⁻⁴)y − (4.4 × 10⁻⁵) = 0.
Discriminant = (4.4 × 10⁻⁴)² + 4·(4.4 × 10⁻⁵) = 1.94 × 10⁻⁷ + 1.76 × 10⁻⁴ = 1.76 × 10⁻⁴
√Discriminant = 1.327 × 10⁻²
y = (−4.4 × 10⁻⁴ + 1.327 × 10⁻²) / 2 = 1.283 × 10⁻² / 2 = 6.41 × 10⁻³ mol dm⁻³
Step 4 — pOH and pH:
pOH = −log(6.41 × 10⁻³) = 2.19
pH = 14.00 − 2.19 = 11.81
Methylamine is a stronger base than ammonia (pKb = 3.36 vs pKb = 4.75) — the methyl group is electron-donating, raising electron density on N and stabilising the protonated form. The pH at the same concentration is correspondingly higher.
Compare 0.10 mol dm⁻³ HCl and 0.10 mol dm⁻³ CH₃COOH:
| Acid | Type | [H⁺] / mol dm⁻³ | pH | % dissociation |
|---|---|---|---|---|
| HCl | strong | 0.10 (fully dissociated) | 1.00 | 100% |
| CH₃COOH | weak (Ka = 1.74 × 10⁻⁵) | 1.32 × 10⁻³ | 2.88 | 1.3% |
At the same formal concentration, the weak acid has ~76× lower [H⁺] than the strong acid — and its pH is therefore 1.88 units higher. The two acids contain identical moles of acid per dm³; the difference is purely a matter of how much of that acid is ionised at equilibrium.
This explains an important practical observation: a weak acid will deliver less of its acid character (lower [H⁺], slower reactions with metals or carbonates) at the same concentration, but if more of it is consumed by a reaction the equilibrium will shift to release more H⁺ — so the total available acid (the concentration c) ultimately limits a neutralisation titration. Strong-acid and weak-acid titrations of the same concentration and volume require the same volume of base to reach the equivalence point; they differ only in the shape of the pH curve (lesson 5).
For a strong acid, halving the concentration doubles the pH-decrease — pH rises by log 2 ≈ 0.30 per halving. For a weak acid, the dependence is weaker because [H⁺] = √(Ka · c) — diluting 10-fold reduces [H⁺] by only a factor of √10 ≈ 3.16, so the pH rises by only ½ unit.
Calculate the pH of ethanoic acid (Ka = 1.74 × 10⁻⁵) at 0.10, 0.010, and 0.001 mol dm⁻³.
| c / mol dm⁻³ | √(Ka·c) | [H⁺] (approx) / mol dm⁻³ | pH (approx) | dissociation | quadratic pH |
|---|---|---|---|---|---|
| 0.10 | √(1.74 × 10⁻⁶) | 1.32 × 10⁻³ | 2.88 | 1.32% | 2.88 |
| 0.010 | √(1.74 × 10⁻⁷) | 4.17 × 10⁻⁴ | 3.38 | 4.17% | 3.39 |
| 0.001 | √(1.74 × 10⁻⁸) | 1.32 × 10⁻⁴ | 3.88 | 13.2% | 3.91 |
Each 10-fold dilution raises pH by approximately 0.50 — exactly as ½·log 10 predicts. Note also that percent dissociation rises sharply as the solution is diluted: from 1.3% at 0.10 mol dm⁻³ to 13% at 0.001 mol dm⁻³. This is the celebrated dilution law of Ostwald: at infinite dilution, α → 1 (the acid becomes effectively fully dissociated). The qualitative reasoning: Ka is fixed, so as c falls the equilibrium shifts to maintain Ka constant by increasing the fractional dissociation. In the limit c → 0, the acid behaves as if it were strong, although [H⁺] itself tends towards 10⁻⁷ (the autoionisation of water dominates).
Common Misconception: "Dilution always raises pH." False in the limit. A pure weak acid solution can never have pH > 7 no matter how dilute; pH asymptotes to 7 from below as c → 0. Once c is so small that the approximation gives [H⁺] < 10⁻⁷, water's autoionisation contribution must be added explicitly (this is rarely tested at A-Level but is essential at undergraduate level).
Practical-Skills Box: Errors that recur in marked exam scripts (and the fixes):
- Forgetting the square root. [H⁺] = √(Ka · c), not Ka · c. The unit check ((mol dm⁻³)² vs mol dm⁻³) catches this immediately.
- Not checking validity of the approximation. Skipping the c / Ka ≥ 400 audit means missing the cases where the quadratic is mandatory. Examiners reward the audit step explicitly.
- Using [H⁺] = √(Ka · c) for a weak base. For a base, the formula is [OH⁻] = √(Kb · c) — and pH = 14 − pOH at 298 K.
- Confusing Ka with [H⁺]. Ka is a property of the acid, not of the solution. [H⁺] depends on both Ka and c.
- Forgetting Kw varies with temperature. Kw = 1.00 × 10⁻¹⁴ only at 298 K; AQA assumes 298 K unless stated otherwise.
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