Chemistry • Year 12 • Module 6 • Lesson 12
Ka, pKa & Comparing Acid Strengths
Synthesise Ka data, evaluate scenarios, and critique scientific claims at Band 5–6 level.
1. Data + scenario — wine acidity, Ka and the AWRI
The Australian Wine Research Institute (AWRI) measures titratable acidity (TA) and pH to ensure wine quality. The dominant acids in wine, with their Ka values, are given in the table below alongside the Ka of H₂CO₃ for comparison. In 2024 the AWRI published research warning that rising temperatures are altering the tartaric-to-malic acid ratio in Australian wine grapes, changing both TA and pH profiles. A winemaker is deciding whether to add tartaric acid or citric acid to a batch to lower the wine's pH into the target range of 3.2–3.5. 8 marks
| Acid | Ka1 | pKa1 (calc.) | Nature | Regulatory status in wine |
|---|---|---|---|---|
| Tartaric acid | 1.0 × 10⁻³ | 3.00 | Diprotic | Permitted additive; standard acidulant |
| Malic acid | 3.5 × 10⁻⁴ | 3.46 | Diprotic | Natural grape acid; permitted |
| Citric acid | 7.4 × 10⁻⁴ | 3.13 | Triprotic | Limited permitted addition (≤1 g/L) |
| Lactic acid | 1.4 × 10⁻⁴ | 3.85 | Monoprotic | Formed during malolactic fermentation |
| H₂CO₃ (reference) | 4.3 × 10⁻⁷ | 6.37 | Diprotic | Not applicable |
Extended response prompt (8 marks). Evaluate which acid — tartaric or citric — is the better choice for the winemaker to use to lower wine pH to 3.2–3.5.
In your response you must:
- Define Ka and explain why it is the correct quantity for comparing how effectively each acid lowers pH at a given concentration.
- Compare tartaric and citric acid on at least three criteria: Ka1 value, pKa1, and degree of ionisation at equal concentrations.
- Use the Ka data to calculate and compare the approximate [H⁺] produced by 0.010 mol/L of each acid (assume the 5% approximation is valid; show working for one acid).
- Refer to the regulatory constraint on citric acid and evaluate how it limits its practical utility despite its Ka.
- Reach an evidence-based judgement recommending one acid, with clear reference to Ka values and the target pH range.
2. Source critique — evaluating a media claim about ocean acidity
Read the following excerpt from an online science-communication article about the Great Barrier Reef and answer the questions below. 7 marks
Source: “Saving the Reef — The Acid Truth” (hypothetical online article, 2024)
“Ocean acidification is making the Great Barrier Reef more acidic every year. As atmospheric CO₂ dissolves into the ocean, it forms carbonic acid (H₂CO₃), which fully ionises in seawater, flooding the reef with hydrogen ions. Scientists measure the increase in acid using pH, which directly tells us how strong the carbonic acid is. The remedy being proposed — adding calcium carbonate (CaCO₃) to the water — would react with the acid and neutralise it by producing CO₂ gas and water, eliminating the acidification problem. Because HCN (Ka = 6.2 × 10⁻¹⁰) is roughly 300,000 times weaker than carbonic acid, it poses no aquatic hazard.”
2.1 The article claims that carbonic acid “fully ionises” in seawater. Identify this error and explain what actually happens to H₂CO₃ in water, using Ka1 = 4.3 × 10⁻⁷ in your explanation. 2 marks
2.2 The article states that “pH directly tells us how strong the carbonic acid is.” Explain the correct chemistry: what pH measures vs what Ka measures, and why only Ka tells you acid strength. 2 marks
2.3 The article compares Ka(HCN) = 6.2 × 10⁻¹⁰ to Ka(H₂CO₃) = 4.3 × 10⁻⁷ and concludes HCN “poses no aquatic hazard.” Calculate the Ka ratio and identify two errors in this reasoning. 3 marks
3. Multi-step calculation — gold cyanidation and HCN equilibria
At the Kalgoorlie Super Pit (WA), leach solutions contain HCN at approximately 5.0 × 10⁻³ mol/L. Ka(HCN) = 6.2 × 10⁻¹⁰. Safety protocols require maintaining solution pH ≥ 10.5 to minimise dissolved HCN (because CN⁻ is the leaching agent and the reaction CN⁻ + H⁺ ⇌ HCN is unwanted). 8 marks
3.1 Calculate the pH of the pure 5.0 × 10⁻³ mol/L HCN solution. Show a full ICE table and verify the 5% assumption. 3 marks
3.2 Calculate Kb for CN⁻ and use it to predict whether a 0.10 mol/L NaCN solution is acidic, basic or neutral. Explain the result using Ka × Kb = Kw. 2 marks
3.3 At pH 10.5, calculate [H⁺] and then use Ka = [H⁺][CN⁻]/[HCN] to determine the [CN⁻]/[HCN] ratio. Interpret what this ratio means for HCN gas evolution hazard at the mine. 3 marks
Q1 — AWRI wine acid evaluation (8 marks)
Criterion 1 — Ka and its role: Ka is the equilibrium constant for HA(aq) ⇌ H⁺(aq) + A⁻(aq); Ka = [H⁺][A⁻]/[HA]. It is an intrinsic molecular property — independent of concentration — so it tells you how effectively each acid lowers [H⁺] per mole added, making it the correct comparison tool. [1 mark]
Criterion 2 — Three-criteria comparison:
- Ka1: Tartaric = 1.0 × 10⁻³; Citric = 7.4 × 10⁻⁴. Ka ratio = 1.35 — tartaric is marginally stronger (~1.4×). [0.5]
- pKa1: Tartaric = 3.00; Citric = 3.13. Tartaric has a smaller pKa — confirming stronger ionisation. [0.5]
- Degree of ionisation at 0.010 mol/L: Tartaric: √(1.0×10⁻³ × 0.010) = √(1.0×10⁻⁵) = 3.16×10⁻³. Check: 3.16×10⁻³/0.010 = 31.6% > 5% — 5% assumption fails. Quadratic needed; but qualitatively tartaric ionises more. [0.5]
Criterion 3 — [H⁺] calculation (showing one fully): For citric at 0.010 mol/L using Ka1 = 7.4 × 10⁻⁴: x = √(7.4 × 10⁻⁴ × 0.010) = √(7.4 × 10⁻⁶) = 2.72 × 10⁻³ mol/L. Check: 2.72 × 10⁻³/0.010 = 27.2% — 5% assumption fails; true [H⁺] is lower, requiring quadratic. pH (approximate) ≈ −log(2.72 × 10⁻³) ≈ 2.57. Tartaric similarly gives higher [H⁺]. Both acids at 0.010 mol/L give pH well below 3.2 — dilution or lower dose needed. [1.5 marks]
Criterion 4 — Regulatory constraint: Citric acid addition is capped at ≤1 g/L (≈ 0.0052 mol/L) under Australian food regulations applied to wine. At this concentration it may not reliably achieve pH 3.2–3.5 without careful blending. Tartaric acid has no such strict upper limit in standard winemaking and is the conventional choice. [1 mark]
Judgement: Tartaric acid is the better choice. Although citric acid has a similar Ka1, the regulatory cap on addition limits how much [H⁺] it can contribute in practice, and tartaric is established practice with a fractionally larger Ka1. The winemaker should use tartaric acid to lower pH reliably to 3.2–3.5, adjusting dose iteratively with pH measurement. [1 mark]
Marking notes: 1 mark — Ka definition and why it is concentration-independent; 1 mark — three-criteria comparison; 1.5 marks — [H⁺] calculation with working shown; 1 mark — regulatory constraint identified and evaluated; 1 mark — explicit evidence-based judgement. Deduct 1 if 5% assumption blindly accepted without checking.
Q2 — Source critique (7 marks)
2.1 Error: H₂CO₃ does not fully ionise in seawater — it is a weak acid. Ka1 = 4.3 × 10⁻⁷ « 1, meaning the equilibrium lies far to the left. At ocean pH ~8.1, only a tiny fraction of H₂CO₃ molecules ionise at any given moment. Full ionisation would require Ka ≫ 1 (as for HCl). The acidification occurs because increasing dissolved CO₂ increases the total concentration of the carbonic acid system, not because it fully ionises. [2 marks: 1 for identifying "fully ionises" as wrong + 1 for Ka1 ≦ 1 explanation]
2.2 pH = −log[H⁺] measures the concentration of H⁺ ions in a specific solution — it depends on both Ka and the concentration of the acid. Ka, by contrast, is an intrinsic molecular property that tells you how much the acid ionises regardless of concentration. Two solutions can have identical pH but different Ka values (different concentrations of different acids). Only Ka enables a valid comparison of inherent acid strength. [2 marks]
2.3 Ka ratio = 4.3 × 10⁻⁷ / 6.2 × 10⁻¹⁰ ≈ 694 — H₂CO₃ is ≈694 times stronger than HCN (not 300,000 — that figure is numerically wrong in the article). [1 mark for correct ratio]
Error 1: The conclusion that HCN “poses no aquatic hazard” conflates acid strength (Ka) with toxicity. Ka measures ionisation extent, not biological or ecological danger. CN⁻ at even nanomolar concentrations is acutely toxic to fish by blocking cellular respiration. [1 mark]
Error 2: Even if HCN has a small Ka, the un-ionised HCN molecule itself is volatile and toxic by inhalation. A weak Ka means less ionisation, which means more un-ionised HCN present — potentially increasing volatilisation hazard rather than reducing it. [1 mark]
Q3 — Kalgoorlie HCN calculation (8 marks)
3.1 ICE table for HCN(aq) ⇌ H⁺(aq) + CN⁻(aq):
I: [HCN] = 5.0 × 10⁻³, [H⁺] = 0, [CN⁻] = 0
C: −x, +x, +x
E: 5.0 × 10⁻³ − x, x, x
Ka = x² / (5.0 × 10⁻³ − x) = 6.2 × 10⁻¹⁰
Assumption: x « 5.0 × 10⁻³ (valid if x/c < 5%)
x = √(6.2 × 10⁻¹⁰ × 5.0 × 10⁻³) = √(3.1 × 10⁻¹²) = 1.76 × 10⁻⁶ mol/L
Check: 1.76 × 10⁻⁶ / 5.0 × 10⁻³ = 0.035% « 5% — assumption valid.
pH = −log(1.76 × 10⁻⁶) = 6 − log(1.76) = 6 − 0.246 = 5.75 [3 marks: 1 ICE + 1 calculation + 1 assumption check]
3.2 Kb(CN⁻) = Kw/Ka(HCN) = (1.0 × 10⁻¹⁴)/(6.2 × 10⁻¹⁰) = 1.6 × 10⁻⁵.
NaCN dissolves to give Na⁺ (spectator) and CN⁻. CN⁻ has Kb = 1.6 × 10⁻⁵ — a moderately strong weak base. It hydrolyses water: CN⁻ + H₂O ⇌ HCN + OH⁻, producing excess OH⁻. Therefore 0.10 mol/L NaCN is basic (pH > 7). This is consistent with Ka × Kb = Kw: HCN is a very weak acid (tiny Ka), so its conjugate base CN⁻ is relatively strong (moderate Kb), making sodium cyanide solutions noticeably alkaline. [2 marks]
3.3 [H⁺] at pH 10.5: [H⁺] = 10⁻¹⁰⋅⁵ = 3.16 × 10⁻¹¹ mol/L.
Rearranging Ka: [CN⁻]/[HCN] = Ka/[H⁺] = (6.2 × 10⁻¹⁰)/(3.16 × 10⁻¹¹) = 19.6 ≈ 20.
Interpretation: At pH 10.5, the ratio [CN⁻]/[HCN] ≈ 20:1, meaning about 95% of the cyanide exists as CN⁻ and only ~5% as HCN(aq). This dramatically reduces HCN gas evolution because it is the un-ionised HCN that evaporates. Maintaining pH ≥ 10.5 is the key safety protocol — if pH falls (e.g. due to acid contamination), the equilibrium shifts toward HCN, rapidly increasing volatile HCN concentration and inhalation risk. [3 marks: 1 [H⁺] calculation + 1 ratio + 1 correct interpretation of hazard implication]