HSC Exam Practice

Sample response. A strong acid ionises completely in aqueous solution, donating all available protons to water molecules — [H¹º] = c (adjusted for proton count). A weak acid only partially ionises; equilibrium lies to the left, so [H¹º] << c and must be found from the K_a expression. The six strong acids are: HCl, H₂SO₄ (first ionisation), HNO₃, HClO₄, HBr, HI.

Marking notes. 1 mark for correct definition of strong acid (complete ionisation / [H¹º] = c). 1 mark for correct definition of weak acid (partial ionisation / equilibrium / K_a required). 2 marks for listing all six strong acids correctly (1 mark for any four; 2 marks for all six).

Sample response. Step 1: identify whether the species is an acid or a base. Step 2: identify whether it is strong or weak (for acids: check the six strong acid list; for bases: check the strong base list). Skipping this step is the most common error because students default to the most recently practised formula without first confirming whether it is applicable — most commonly applying [H¹º] = c to a weak acid, giving a [H¹º] that may be 10–100 times too high and invalidating all subsequent working.

Marking notes. 1 mark for Step 1 (acid vs base). 1 mark for Step 2 (strong vs weak, referencing the list). 1 mark for explaining why skipping it causes errors (applying wrong method → cascading errors).

Sample response. NaOH is a strong base. [OH¹¯] = 0.0250 mol/L (1 OH¹¯ per formula unit). pOH = −log(0.0250) = 1.602. pH = 14.00 − 1.60 = 12.40. Sanity check: pH 12.40 > 7 — correct for a base.

Marking notes. 1 mark for [OH¹¯] = 0.0250 with pOH step shown. 1 mark for correct pH = 12.40 (accept 12.4).

Sample response. Propanoic acid is a weak acid (not on the six strong acid list). Assumption check: K_a/c = 1.34 × 10⁻⁵/0.0800 = 1.675 × 10⁻⁴ << 0.0025 — assumption valid. x = √(1.34 × 10⁻⁵ × 0.0800) = √(1.072 × 10⁻⁶) = 1.035 × 10⁻³ mol/L. Verify: 1.035 × 10⁻³/0.0800 × 100% = 1.29% < 5% — valid. pH = −log(1.035 × 10⁻³) = 2.99.

Marking notes. 1 mark for identifying propanoic acid as weak and setting up ICE table / K_a expression. 1 mark for correct assumption check (K_a/c << 0.0025 or x/c < 5%). 1 mark for correct pH = 2.99 (accept 3.00 due to rounding).

Sample response. Using the square root shortcut: x = √(6.8 × 10⁻⁴ × 0.100) = √(6.8 × 10⁻⁵) = 8.25 × 10⁻³ mol/L. Assumption check: x/c × 100% = 8.25 × 10⁻³/0.100 × 100% = 8.25% > 5% — assumption invalid. The quadratic equation must be solved instead: x² + K_ax − K_ac = 0, giving x² + 6.8 × 10⁻⁴x − 6.8 × 10⁻⁵ = 0. Using the quadratic formula: x = [−6.8 × 10⁻⁴ + √((6.8 × 10⁻⁴)² + 4 × 6.8 × 10⁻⁵)]/2 = 7.91 × 10⁻³ mol/L. pH = −log(7.91 × 10⁻³) = 2.10.

Marking notes. 1 mark for correctly calculating x = 8.25 × 10⁻³ mol/L and showing x/c = 8.25% > 5%. 1 mark for stating that the quadratic formula must be used. 1 mark for correct pH = 2.10.

Sample response. An ICE table is used for a pure weak acid or weak base solution where [A¹¯] ≈ 0 initially; it sets up Ka = x²/(c − x) to find [H¹º] = x from equilibrium. The Henderson-Hasselbalch equation (pH = pK_a + log([A¹¯]/[HA])) must be used whenever significant amounts of both HA and A¹¯ are already present in solution — for example, after partial neutralisation or in a buffer. Using an ICE table when [A¹¯] is non-zero violates the initial conditions assumption of the ICE method.

Marking notes. 1 mark for correctly describing the ICE table and its condition ([A¹¯] ≈ 0 initially / pure weak acid). 1 mark for correctly describing Henderson-Hasselbalch and its condition (both HA and A¹¯ present). 1 mark for explicitly stating that using an ICE table when [A¹¯] ≠ 0 is an error.

Sample response (a). n(H¹º) from H₂SO₄ = 2 × 0.120 × 0.0250 = 6.00 × 10⁻³ mol (factor of 2 because H₂SO₄ is diprotic). n(OH¹¯) from NaOH = 0.100 × 0.0600 = 6.00 × 10⁻³ mol.

Sample response (b). n(H¹º) = 6.00 × 10⁻³ mol = n(OH¹¯) = 6.00 × 10⁻³ mol. The moles are exactly equal — this is the equivalence point. Neither species is in excess.

Sample response (c). At the equivalence point of a strong acid + strong base, the products are Na₂SO₄ and H₂O only. No excess H¹º or OH¹¯. The solution is neutral: [H¹º] = [OH¹¯] = 1.00 × 10⁻⁷ mol/L. pH = 7.00.

Marking notes. (a) 2 marks: 1 for n(H¹º) = 6.00 × 10⁻³ mol including the factor of 2 for diprotic H₂SO₄; 1 for n(OH¹¯) = 6.00 × 10⁻³ mol. (b) 2 marks: 1 for identifying equivalence point (moles equal); 1 for stating neither is in excess. (c) 2 marks: 1 for V(total) = 25.0 + 60.0 = 85.0 mL stated (even though no excess makes the volume calculation trivial); 1 for pH = 7.00 with correct reasoning (strong acid/strong base equivalence point).

Sample response (a). The equivalence point is at 50 mL of NaOH added, where the graph shows its steepest inflection and pH ≈ 8.72. The pH is not 7 because the neutralisation product is CH₃COO¹¯ — the conjugate base of the weak acid CH₃COOH. CH₃COO¹¯ is a weak base that undergoes hydrolysis (CH₃COO¹¯ + H₂O ⇌ CH₃COOH + OH¹¯), producing excess OH¹¯ and making the solution basic. [3 marks: 1 for 50 mL/pH≈8.72; 1 for identifying CH₃COO¹¯ as the product; 1 for hydrolysis explanation]

Sample response (b). At the half-equivalence point, [CH₃COO¹¯] = [CH₃COOH], so log([A¹¯]/[HA]) = log(1) = 0. Therefore pH = pK_a. From the graph, pH at 25 mL ≈ 4.74, so pK_a ≈ 4.74. K_a = 10^−4.74 = 1.8 × 10⁻⁵. [2 marks: 1 for pH = pK_a reasoning; 1 for K_a = 1.8 × 10⁻⁵]

Sample response (c). Near the equivalence point (45–55 mL), n(H¹º) remaining from CH₃COOH approaches zero as the last few millimoles of weak acid are consumed. Because [H¹º] is already extremely small, adding even a tiny amount of NaOH changes the ratio [A¹¯]/[HA] dramatically, causing a large change in pH (logarithmic relationship: pH = pK_a + log([A¹¯]/[HA])). After the equivalence point, excess NaOH dominates, keeping pH high. [2 marks: 1 for linking small n(H¹º) remaining to large fractional change; 1 for logarithmic / H-H relationship]

Sample response. At the same molar concentration, a strong acid (e.g. HCl) ionises completely in water (HCl → H¹º + Cl¹¯), so [H¹º] = c = 0.100 mol/L and pH = 1.00. A weak acid (e.g. CH₃COOH, K_a = 1.8 × 10⁻⁵) partially ionises (CH₃COOH ⇌ H¹º + CH₃COO¹¯); the equilibrium lies far to the left because K_a is small. Only ~1.34% of molecules donate a proton, so [H¹º] = 1.34 × 10⁻³ mol/L and pH = 2.87 — approximately 2 units higher than HCl. [2 marks: 1 for degree of ionisation link to [H¹º]; 1 for correct pH values with K_a connection]

At the particle level, the frequency of H¹º ions in solution is directly proportional to [H¹º]. In HCl solution, approximately 100 H¹º ions exist for every 1.34 in the acetic acid solution of the same concentration. Electrical conductivity is proportional to the total concentration of charge carriers (ions). HCl solution has [H¹º] + [Cl¹¯] ≈ 0.200 mol/L total ion concentration; acetic acid has [H¹º] + [CH₃COO¹¯] ≈ 2.68 × 10⁻³ mol/L — about 75 times fewer ions, so conductivity is correspondingly lower. [2 marks: 1 for linking ion concentration to conductivity; 1 for quantitative comparison]

The rate of reaction with a metal carbonate (e.g. CaCO₃) depends on the collision frequency of H¹º ions with the carbonate surface. Because [H¹º] is ~75 times higher in HCl than in acetic acid at equal concentration, H¹º ions collide with the solid surface much more frequently per unit time, giving a faster initial reaction rate. As the weak acid reacts, Le Chatelier’s principle drives the equilibrium CH₃COOH ⇌ H¹º + CH₃COO¹¯ to the right to partially replace the consumed H¹º, so the weak acid reaction continues (but more slowly) even as [H¹º] is depleted. [2 marks: 1 for collision frequency explanation; 1 for Le Chatelier’s principle restoring H¹º in weak acid]

Overall, the degree of ionisation is the single quantity that unifies all three differences: higher degree of ionisation → higher [H¹º] → lower pH, higher conductivity, and faster reaction rate. [1 mark for unified conclusion]

Marking criteria. 1 mark — Correctly identifies strong acid as fully ionised and weak acid as partially ionised; links this to [H¹º]. 1 mark — Gives correct pH values (or at least states pH(strong) < pH(weak) with correct K_a reasoning). 1 mark — Links ion concentration (from degree of ionisation) to electrical conductivity. 1 mark — Makes a quantitative comparison showing HCl has many more ions than acetic acid at equal concentration. 1 mark — Explains reaction rate difference using collision frequency of H¹º. 1 mark — Invokes Le Chatelier’s principle for the weak acid (equilibrium shifts to replenish H¹º as it is consumed). 1 mark — Provides a unified conclusion linking degree of ionisation to all three properties.

Sample response. The student’s claim is incorrect: the equivalence point of a weak acid / strong base titration produces a basic solution, not a neutral one. At the equivalence point, all CH₃COOH has been converted to CH₃COO¹¯ (acetate ion). K_b(CH₃COO¹¯) = K_w/K_a = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. Assuming equal volumes (total volume halved concentration): c(CH₃COO¹¯) = 0.0500 mol/L. [OH¹¯] = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ mol/L. pOH = 5.28. pH = 14.00 − 5.28 = 8.72. The pH is 8.72 > 7, confirming the solution is basic. The student’s error is treating “all acid neutralised” as equivalent to “neutral solution”: neutralisation means no excess strong acid or base, but the conjugate base of a weak acid is itself a weak base that hydrolyses water, producing OH¹¯ and making the solution basic.

Marking criteria. 1 mark — Identifies the product as CH₃COO¹¯ and states it is a weak base. 1 mark — Calculates K_b(CH₃COO¹¯) = K_w/K_a = 5.56 × 10⁻¹⁰. 1 mark — Correctly arrives at pH = 8.72. 1 mark — Explicitly corrects the student’s conceptual error: “neutralised” ≠ “neutral”; the conjugate base hydrolyses to produce OH¹¯, making the solution basic.