IQ2 Mastery: pH Calculations, Mixing & Band 6 Explanations

Q1 — Sample Band 6 response (8 marks), annotated

Student 1 is correct. HNO₃ is on the six strong acid list, so [H¹º] = 0.0500 mol/L and pH = −log(0.0500) = 1.30. No error.

Student 2 has made Error 1 (applying the strong acid shortcut [H¹º] = c to a weak acid). Benzoic acid is NOT on the six strong acid list — it is a weak acid (K_a = 6.3 × 10⁻⁵). Correct calculation for Solution Y: Assumption check first: K_a/c = 6.3 × 10⁻⁵/0.0500 = 1.26 × 10⁻³ < 0.0025 — assumption valid. x = √(6.3 × 10⁻⁵ × 0.0500) = √(3.15 × 10⁻⁶) = 1.775 × 10⁻³ mol/L. Verify: 1.775 × 10⁻³/0.0500 = 3.55% < 5% — valid. Correct pH(Y) = −log(1.775 × 10⁻³) = 2.75. Fix: identify acid type first.

Student 3 has made Error 3 (reporting pOH as pH for a base). The ICE table setup is correct: [OH¹¯] = √(4.2 × 10⁻¹⁰ × 0.0500) = √(2.1 × 10⁻¹¹) = 4.58 × 10⁻⁶ mol/L. pOH = −log(4.58 × 10⁻⁶) = 5.34. But the student wrote pH = 5.29 (slight rounding) and forgot to apply pH = 14 − pOH. Correct pH(Z) = 14.00 − 5.34 = 8.66. Sanity check: pH for a base must be > 7 — pH 5.29 fails immediately. Fix: for any base, always apply pH = 14 − pOH as the final step.

Particle-level explanation (why [H¹º] = c is valid for X but not Y): HNO₃ is a strong acid — every HNO₃ molecule donates its proton completely to water upon dissolving (HNO₃ → H¹º + NO₃¹¯). At the particle level, essentially zero HNO₃ molecules remain intact, so every original molecule contributes one H¹º to the solution: [H¹º] = c exactly. For benzoic acid (a weak acid, C₆H₅COOH ⇌ H¹º + C₆H₅COO¹¯), only a small fraction of molecules — about 3.55% — have donated a proton at equilibrium. The majority remain as intact molecules. Treating it as if it were fully ionised overestimates [H¹º] by a factor of roughly 28, giving a pH that is approximately 1.45 units too low.

Overall judgement: Student 2’s error (Error 1 — wrong method category) is the more costly in HSC because it not only gives a wrong pH but also invalidates any subsequent calculation building on that [H¹º] — for example, a K_a calculation in a follow-up part, or a ΔH_n comparison. Student 3’s error (Error 3 — pOH ≠ pH) produces an answer in the wrong pH region (below 7 for a base), which an examiner recognises immediately as a method mark loss, but the degree of ionisation calculation itself was correct, limiting the cascade damage.

Marking criteria:

• 1 mark — Correctly identifies Student 1 as correct and Student 2 and Student 3 as incorrect.
• 1 mark — Correctly diagnoses Student 2’s error as applying [H¹º] = c to a weak acid (Error 1).
• 1 mark — Correctly calculates pH(Y) = 2.75 with full ICE table working and assumption check.
• 1 mark — Correctly diagnoses Student 3’s error as reporting pOH as pH without applying pH = 14 − pOH (Error 3).
• 1 mark — Correctly calculates pH(Z) = 8.66 with pOH step shown.
• 1 mark — Provides a particle-level explanation of why HNO₃ justifies [H¹º] = c (complete ionisation) but benzoic acid does not (partial ionisation; most molecules remain intact).
• 1 mark — Labels each incorrect student’s error with the correct IQ2 error number and states the fix.
• 1 mark — Reaches an explicit, justified overall judgement about which error costs more marks in an HSC context.

Q2 — Source critique: flaws identified and corrected (7 marks)

Flaw 1: “The K_a of HCl is larger than that of CH₃COOH.” This is incorrect: strong acids like HCl do not have a meaningful K_a in the equilibrium sense because ionisation is essentially complete — there is no equilibrium to define. Saying K_a(HCl) is “larger” implies an equilibrium constant, which is inappropriate for a strong acid [1 mark].

Flaw 2: “Stronger acids have more protons to donate per molecule.” This is incorrect. Both HCl and CH₃COOH are monoprotic acids — each molecule donates exactly one proton. Acid strength (strong vs weak) refers to the degree of ionisation, not the number of protons per molecule [1 mark].

Flaw 3: “The different pH values prove that their K_a values are different, confirming HCl is stronger.” This is circular reasoning and also incorrect in structure. The pH difference does reflect different [H¹º] values, but the conclusion that this “proves” different K_a values is backwards: the K_a (or its absence for a strong acid) determines the degree of ionisation which determines [H¹º] which determines pH — not the reverse [1 mark].

Corrected Band 6 explanation (approximately 5–8 sentences):

HCl is a strong acid: in aqueous solution, every HCl molecule donates its proton completely to water (HCl → H¹º + Cl¹¯). This means every molecule present before dissolution becomes one H¹º ion in solution, so [H¹º] = 0.100 mol/L and pH = 1.00. At the particle level, the frequency of H¹º ions colliding with substrates or indicators is high because ion concentration is maximum for the given starting amount of acid [1 mark]. CH₃COOH is a weak acid (K_a = 1.8 × 10⁻⁵): the equilibrium CH₃COOH ⇌ H¹º + CH₃COO¹¯ lies strongly to the left — only ~1.34% of molecules have donated a proton at equilibrium. At the particle level, most particles in solution remain as intact CH₃COOH molecules; very few H¹º ions are present, giving [H¹º] = 1.34 × 10⁻³ mol/L and pH = 2.87 [1 mark]. The lower [H¹º] in acetic acid means fewer collisions per unit time between H¹º and reacting species, which is why weak acids react more slowly with metals or carbonates than HCl at the same concentration [1 mark]. In summary, both acids are monoprotic (one proton per molecule) but differ in degree of ionisation: HCl ionises completely, so [H¹º] = c; CH₃COOH ionises partially, so [H¹º] << c and must be found from the ICE table. This difference in [H¹º] directly produces the 1.87-unit pH difference [1 mark].

Marking criteria:

• 1 mark — Identifies the K_a flaw (strong acids do not have a meaningful K_a / no equilibrium to define).
• 1 mark — Identifies the “more protons per molecule” flaw (both are monoprotic; strength refers to degree of ionisation, not proton count).
• 1 mark — Identifies the circular reasoning flaw and explains the correct causal direction (K_a → degree of ionisation → [H¹º] → pH).
• 1 mark — Corrected explanation correctly identifies HCl as fully ionised and CH₃COOH as partially ionised.
• 1 mark — Corrected explanation states correct [H¹º] for each (0.100 vs 1.34 × 10⁻³ mol/L) with pH values.
• 1 mark — Links particle level to pH: fewer H¹º ions per unit volume means lower collision frequency and therefore lower reactivity and higher pH.
• 1 mark — Overall explanation is coherent, uses precise terminology (degree of ionisation, equilibrium, ICE table), and reaches a well-structured conclusion.

Q3 — Sample Band 6 response (8 marks), annotated

Why pH(P) + pH(Q) = 14 is a coincidence, not a general rule: For CH₃COOH, [H¹º] = √(K_a × c) = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³, pH = 2.87. For NH₃, [OH¹¯] = √(K_b × c) = √(1.8 × 10⁻⁵ × 0.100) = 1.34 × 10⁻³, pOH = 2.87, pH = 11.13. Sum = 14.00. This occurs because K_a(CH₃COOH) = K_b(NH₃) exactly, causing [H¹º]_P = [OH¹¯]_Q exactly. This is a coincidence of numerically equal equilibrium constants at this concentration, not a universal rule. In general, pH(acid) + pH(base) = 14 only when K_a = K_b and concentrations are equal [2 marks].

Why CH₃COOH and NH₃ are not a conjugate pair: A conjugate acid-base pair consists of two species that differ by exactly one proton. CH₃COOH’s conjugate base is CH₃COO¹¯ (formed by removing one H¹º). NH₃’s conjugate acid is NH₄¹º (formed by adding one H¹º). CH₃COOH and NH₃ differ in both molecular formula and structure — they are unrelated species. The K_a × K_b = K_w relationship applies only within a conjugate pair (e.g. CH₃COOH / CH₃COO¹¯). K_a(CH₃COOH) × K_b(NH₃) = (1.8 × 10⁻⁵)² = 3.24 × 10⁻¹⁰ ≠ K_w = 1.0 × 10⁻¹⁴, confirming they are not conjugate [2 marks].

Verification of the mixing calculation: n(CH₃COOH) = 0.100 × 0.0500 = 5.00 × 10⁻³ mol. n(NaOH) = 0.100 × 0.0500 = 5.00 × 10⁻³ mol. Moles are equal — equivalence point. All CH₃COOH is converted to CH₃COO¹¯. V(total) = 100.0 mL = 0.1000 L. c(CH₃COO¹¯) = 5.00 × 10⁻³/0.1000 = 0.0500 mol/L. CH₃COO¹¯ is a weak base: K_b(CH₃COO¹¯) = K_w/K_a = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. [OH¹¯] = √(5.56 × 10⁻¹⁰ × 0.0500) = √(2.78 × 10⁻¹¹) = 5.27 × 10⁻⁶ mol/L. pOH = 5.28. pH = 14.00 − 5.28 = 8.72. The calculation is correct [2 marks].

Correcting the student’s second conclusion: pH = 8.72 is greater than 7, meaning the solution is basic — not acidic. The student’s conclusion that “the pH dropped below 7” is factually wrong: 8.72 > 7. The basic pH is the expected result because CH₃COO¹¯ (the conjugate base of a weak acid) hydrolyses to produce OH¹¯. The solution at the equivalence point of a weak acid/strong base titration is always basic because the conjugate base of the weak acid is itself a weak base [2 marks].

Marking criteria:

• 1 mark — Identifies that pH(P) + pH(Q) = 14 is a coincidence arising because K_a = K_b.
• 1 mark — Demonstrates numerically that this is not a general rule (e.g. K_a × K_b ≠ K_w, or cites that pH sums to 14 only when K_a = K_b).
• 1 mark — Correctly defines conjugate pair (differs by exactly one proton) and applies this to show CH₃COOH and NH₃ are not conjugate.
• 1 mark — Shows correct K_b(CH₃COO¹¯) = K_w/K_a step.
• 1 mark — Correctly arrives at pH = 8.72 with full working (equivalence point → [CH₃COO¹¯] = 0.0500 mol/L → ICE on conjugate base → pOH → pH).
• 1 mark — Correctly identifies that pH = 8.72 > 7 (basic, not acidic) and names the species responsible (CH₃COO¹¯ hydrolysis producing OH¹¯).
• 1 mark — Overall response uses precise lesson terminology throughout (conjugate pair, equivalence point, degree of ionisation, K_a × K_b = K_w) and is logically structured.
• 1 mark — Corrects the student’s second conclusion explicitly: explains why a weak acid / strong base equivalence point always gives pH > 7.