Chemistry • Year 12 • Module 6 • Lesson 6
Strong/Weak Mastery — Consolidation
Synthesise strength, concentration, conductivity, and salt hydrolysis across extended scenarios; evaluate flawed student responses at Band 5–6 level.
1. Data and scenario: Which acid is in which beaker?
A technician prepared four unlabelled beakers, each containing a 0.10 mol/L solution of one of the following: HCl, HNO₂, CH₃COOH, or NaOH. Three measurements were made on each beaker: pH, electrical conductivity (mS/cm), and reaction rate with magnesium ribbon (rated qualitatively: fast / moderate / slow / none). Results are shown in Table 1 below.
| Beaker | pH | Conductivity (mS/cm) | Reaction with Mg ribbon |
|---|---|---|---|
| W | 1.0 | 39.1 | Fast (rapid H₂ evolution) |
| X | 2.1 | 3.8 | Moderate (steady H₂ evolution) |
| Y | 2.9 | 0.53 | Slow (slow H₂ evolution initially) |
| Z | 13.0 | 21.8 | None (H₂ not evolved) |
Extended response prompt (7 marks). Using all three data columns from Table 1, identify which substance is in each beaker. In your response you must:
- State which beaker contains each of the four substances, with reasoning that cites specific data values.
- Explain why the conductivity difference between Beakers W and X is consistent with the distinction between a strong acid and a weak acid, even though both are acidic.
- Write the correct ionic equation (with correct arrow type) for the ionisation of the substance in Beaker X.
- Calculate the approximate degree of ionisation of the substance in Beaker X, using the pH data. Show all working.
- Explain why the substance in Beaker Y shows a slow initial reaction with Mg but would eventually consume the same total mass of Mg as a strong acid of the same concentration and volume.
2. Source critique A — Evaluate a student’s exam response
The following is a student’s response to an HSC-style question. The question was: “A 0.10 mol/L solution of HF and a 0.10 mol/L solution of HBr are prepared. Compare the pH, conductivity, and degree of ionisation of the two solutions. Explain what these comparisons reveal about the concept of acid strength. (5 marks)”
Student Response A:
“HF and HBr are both acids. HF has a higher pH than HBr because HF is a weak acid and HBr is a strong acid. This means HBr has completely ionised into H⁺ and Br⁻ ions, while HF has partially ionised into H⁺ and F⁻ ions. The conductivity of HBr will be higher because there are more ions in solution; at 0.10 mol/L HBr gives [H⁺] = 0.10 mol/L, so pH = 1.0. For HF, the pH is higher because it is dangerous — fluoride ion is toxic, so the body treats it differently to chloride, which is why HF has a lower pH than expected. The degree of ionisation of HF at 0.10 mol/L is about 8%, showing it is somewhat strong. Acid strength is defined as how acidic or corrosive an acid is: the more acidic, the stronger.”
Source critique prompt (7 marks). This student response contains multiple scientific errors and imprecisions. Identify and explain at least four distinct errors or imprecisions. For each error, state:
- What the student wrote (quote or paraphrase).
- Why it is incorrect or imprecise.
- The correct scientific statement that should replace it.
Then rewrite a corrected response to the original question (approximately 4–6 sentences).
3. Source critique B — Evaluate a textbook claim
The following passage appears in a Year 12 study guide sold commercially for HSC preparation.
Study Guide Excerpt:
“Acid strength and acid concentration are closely related. When you dilute an acid, you make it weaker, because the ions become more spread out. This is why dilute hydrochloric acid is described as a weaker acid than concentrated hydrochloric acid in many laboratory contexts. Conversely, Ca(OH)₂ is a weak base because, unlike NaOH, it does not fully dissolve in water and so produces far fewer OH⁻ ions per litre than NaOH. Any neutralisation reaction produces a neutral solution, since acids and bases cancel each other out. Students should note that a salt like NaCl is always neutral, which confirms that neutralisation always produces neutral products.”
Source critique prompt (7 marks). Analyse this passage critically. In your response you must:
- Identify and correct the three major scientific errors in the passage.
- Explain, using Ka, why diluting an acid does not change its strength.
- Explain, with reference to degree of dissociation of the dissolved fraction, why Ca(OH)₂ is correctly classified as a strong base.
- Give a named counter-example that refutes the claim that all neutralisation reactions produce neutral solutions, and explain the chemistry using the salt hydrolysis rule.
Q1 — Marking notes (7 marks)
Identifications: W = HCl (pH 1.0 = [H⁺] = 0.10 mol/L = complete ionisation; conductivity 39.1 highest of acids; fastest Mg reaction). X = HNO₂ (pH 2.1 = [H⁺] ≈ 8 × 10⁻³ mol/L = partial ionisation; conductivity 3.8 intermediate). Y = CH₃COOH (pH 2.9; lowest conductivity of acids 0.53; slowest Mg reaction). Z = NaOH (pH 13.0; no H⁺ → no reaction with Mg; high conductivity from Na⁺ + OH⁻). [1 mark for all four correct with data citations]
Conductivity explanation: At the same 0.10 mol/L, HCl (strong) produces 0.10 mol/L H⁺ + 0.10 mol/L Cl⁻ = 0.20 mol/L total ions. HNO₂ (weak, ~8% ionised) produces approximately 0.016 mol/L total ions — roughly 12 times fewer. Conductivity is proportional to ion concentration; hence the ratio 39.1:3.8 ≈ 10:1 is consistent with the strong/weak distinction at this concentration. [1 mark]
Ionic equation for Beaker X (HNO₂): HNO₂(aq) ⇆ H⁺(aq) + NO₂⁻(aq). Equilibrium arrow required (weak acid). [1 mark]
Degree of ionisation calculation: pH = 2.1, so [H⁺] = 10⁻2.1 = 7.94 × 10⁻³ mol/L. Degree = ([H⁺] / c) × 100% = (7.94 × 10⁻³ / 0.10) × 100% ≈ 7.9%. [2 marks: 1 for correct [H⁺] from pH; 1 for applying degree of ionisation formula correctly with working]
Slow initial vs same total Mg: The equilibrium for CH₃COOH lies far to the left at equilibrium — only ~1.3% of molecules are ionised. Initially, [H⁺] is low, so the reaction rate (Mg + 2H⁺ → Mg²⁺ + H₂) is slow. However, as H⁺ is consumed, the equilibrium shifts right (Le Chatelier’s principle), producing more H⁺ to replace what was used. This continues until all 0.10 mol/L of CH₃COOH has been neutralised. The total amount of Mg consumed depends on the total moles of acid available (0.10 mol/L), not on the rate at which it ionises. Given equal concentrations and volumes, both acids provide the same total moles of H⁺ over the full reaction, consuming the same mass of Mg. [2 marks: 1 for Le Chatelier/equilibrium shift explanation; 1 for clearly distinguishing rate from total extent of reaction]
Q2 — Source critique A marking notes (7 marks)
Error 1 — “HF is dangerous, so it has a lower pH than expected”: Danger is not related to Ka or degree of ionisation. HF is dangerous because F⁻ ions penetrate tissue and cause systemic toxicity; this is independent of Ka. The pH of 0.10 mol/L HF is determined solely by Ka(HF) = 6.8 × 10⁻&sup4; and the resulting degree of ionisation (~8%), not by its toxicological profile. [1 mark]
Error 2 — “HF degree of ionisation ~8%, showing it is somewhat strong”: A degree of ionisation of 8% means HF is unambiguously a weak acid; 8% is far below 100%. Strong acids ionise completely (≈100%). 8% places HF firmly in the weak category. [1 mark]
Error 3 — “Acid strength is how acidic or corrosive an acid is”: Acid strength is specifically the degree of ionisation in water (or equivalently, the magnitude of Ka). It is not a measure of corrosiveness or overall acidity in an unqualified sense. A concentrated weak acid can be more corrosive than a dilute strong acid. [1 mark]
Error 4 — Arrow type for HBr: Student does not mention arrow type, but to write the equation correctly: HBr(aq) → H⁺(aq) + Br⁻(aq) requires a single forward arrow (→) because HBr is a strong acid. HF(aq) ⇆ H⁺(aq) + F⁻(aq) requires an equilibrium arrow (↔). [1 mark for noting arrow distinction]
Corrected response (3 marks for quality): At 0.10 mol/L, HBr (strong acid, Ka → ∞) ionises completely: HBr(aq) → H⁺(aq) + Br⁻(aq); [H⁺] = 0.10 mol/L, pH = 1.0. HF (weak acid, Ka = 6.8 × 10⁻&sup4;) partially ionises: HF(aq) ⇆ H⁺(aq) + F⁻(aq); [H⁺] ≈ 8.2 × 10⁻³ mol/L, pH ≈ 2.1, degree of ionisation ≈ 8.2%. HBr has lower pH, higher conductivity, and higher degree of ionisation — all consistent with complete ionisation. HF has higher pH, lower conductivity (≈10× lower), and degree of ionisation of ~8%. Acid strength is defined by Ka (or degree of ionisation), not by danger or corrosiveness. HF is toxic due to F⁻ tissue penetration, not because of its Ka. [3 marks for accurate and complete corrected response]
Q3 — Source critique B marking notes (7 marks)
Error 1 — Diluting makes an acid weaker: False. Ka is an intrinsic property of the molecule determined by its molecular structure and bond energies. It changes only with temperature. Diluting changes concentration (c) and thus [H⁺] and pH, but Ka (and hence the classification as strong or weak) is unchanged. Dilute HCl and concentrated HCl are both strong acids. [1 mark]
Ka explanation: Ka = [H⁺][A⁻] / [HA] is an equilibrium constant that depends only on the Gibbs energy difference between reactants and products — a function of molecular structure and temperature. Adding water changes the volumes and thus concentrations of all species proportionally, but the ratio [H⁺][A⁻] / [HA] remains equal to Ka. Therefore Ka is unchanged by dilution. [1 mark]
Error 2 — Ca(OH)₂ is a weak base: False. Ca(OH)₂ is a sparingly soluble strong base. “Weak” means the dissolved fraction does not fully dissociate. For Ca(OH)₂, every formula unit that dissolves fully dissociates: Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq). The dissolved fraction is 100% dissociated. The low [OH⁻] relative to NaOH at the same “total added” amount is due to low solubility, not weak dissociation. Ca(OH)₂ is correctly described as a sparingly soluble strong base. [2 marks: 1 for correct classification; 1 for explaining solubility vs strength distinction]
Error 3 — All neutralisation reactions produce neutral solutions: False. Neutralisation produces a neutral solution only when both the acid and the base are strong (e.g. HCl + NaOH → NaCl + H₂O, pH = 7). If either is weak, the resulting salt hydrolyses. Counter-example: NH₃(weak base) + HCl(strong acid) → NH₄Cl. NH₄⁺ is the conjugate of the weak base NH₃; it donates H⁺ to water: NH₄⁺(aq) ⇆ H⁺(aq) + NH₃(aq), giving pH < 7. CH₃COONa (from weak acid + strong base) gives pH > 7. Only NaCl (strong acid + strong base) is neutral, so the guide’s generalisation is wrong. [3 marks: 1 for stating the rule; 1 for the counter-example with salt identified; 1 for the hydrolysis equation and pH prediction]