HSC Exam Practice

Sample response. pH is defined as the negative base-10 logarithm of the hydrogen ion concentration: pH = −log₁₀[H⁺]. The inverse formula to recover [H⁺] from a known pH is [H⁺] = 10^−pH.

Marking notes. 1 mark for pH = −log[H⁺]. 1 mark for [H⁺] = 10^−pH.

Sample response. The relationship pH + pOH = 14 is derived from K₩ = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² L⁻². Taking −log of both sides: −log(K₩) = −log[H⁺] + (−log[OH⁻]); pK₩ = pH + pOH. The numerical value 14 is valid only at 25°C because K₩ is temperature-dependent: K₩ increases with temperature. At other temperatures, use pH + pOH = pK₩ at that temperature.

Marking notes. 1 mark for linking the “14” to K₩ = 1.0 × 10⁻¹⁴ at 25°C specifically. 1 mark for stating K₩ changes with temperature (or pK₩ changes). 1 mark for the correct general expression pH + pOH = pK₩.

(a) HNO₃: Strong monoprotic acid; [H⁺] = 0.040 mol/L. pH = −log(0.040) = −log(4.0 × 10⁻²) = 2 − log(4.0) = 2 − 0.602 = 1.40. Sanity: pH < 7 ✓

(b) H₂SO₄: Dilute diprotic strong acid; [H⁺] = 2 × 0.015 = 0.030 mol/L. pH = −log(0.030) = −log(3.0 × 10⁻²) = 2 − log(3.0) = 2 − 0.477 = 1.52. Sanity: pH < 7 ✓

(c) Ba(OH)₂: Strong diprotic base; [OH⁻] = 2 × 0.030 = 0.060 mol/L. pOH = −log(0.060) = −log(6.0 × 10⁻²) = 2 − log(6.0) = 2 − 0.778 = 1.22. pH = 14.00 − 1.22 = 12.78. Sanity: pH > 7 ✓

Marking notes. 2 marks per part: 1 mark for correct [H⁺] or [OH⁻] (including diprotic factor where needed); 1 mark for correct final pH via correct formula and route.

Sample response. pH = 11.70. pOH = 14.00 − 11.70 = 2.30. [OH⁻] = 10^−2.30 = 5.01 × 10⁻³ mol/L. NaOH is a strong monoprotic base: [NaOH] = [OH⁻] = 5.0 × 10⁻³ mol/L.

Marking notes. 1 mark for pOH = 2.30. 1 mark for [OH⁻] = 10^−2.30 = 5.0 × 10⁻³ mol/L (or equivalent). 1 mark for correctly equating [NaOH] = [OH⁻] for a monoprotic strong base.

Sample response. Step 1: calculate n(KOH) = mass/M = 2.24/56.1 = 0.0399 mol. Step 2: calculate c(KOH) = n/V = 0.0399/0.400 = 0.0998 mol/L. Step 3: KOH is a strong monoprotic base, so [OH⁻] = c(KOH) = 0.0998 mol/L. Step 4: pOH = −log(0.0998) ≈ 1.00. Step 5: pH = 14.00 − pOH = 13.00.

Marking notes. 1 mark for mass → moles → concentration calculation. 1 mark for [OH⁻] = c(KOH) (strong base fully ionised). 1 mark for pH = 14 − pOH route.

Sample response. Before dilution: [HCl] = 0.100 mol/L; [H⁺] = 0.100 mol/L; pH = −log(0.100) = 1.00. Dilution: n(HCl) = c₁ × V₁ = 0.100 × 0.0250 = 2.50 × 10⁻³ mol (moles conserved). New concentration: c₂ = n/V₂ = 2.50 × 10⁻³/0.250 = 0.0100 mol/L. After dilution: [H⁺] = 0.0100 mol/L; pH = −log(0.0100) = 2.00. The pH increases by 1 unit (from 1.00 to 2.00), corresponding to a 10-fold decrease in [H⁺]. Moles of H⁺ are conserved by the dilution equation c₁V₁ = c₂V₂; adding water only increases the volume, spreading the same moles through a larger volume.

Marking notes. 1 mark for correct pH before dilution (1.00). 1 mark for applying c₁V₁ = c₂V₂ correctly. 1 mark for correct pH after dilution (2.00). 1 mark for explaining mole conservation (n(H⁺) unchanged; only volume increases).

(a) Excess column: Row 1 (V = 10.0 mL): excess OH⁻ = 4.50 × 10⁻³ − 2.00 × 10⁻³ = 2.50 × 10⁻³ mol OH⁻. Row 2 (V = 22.5 mL): excess = 0 (equivalence). Row 3 (V = 30.0 mL): excess H⁺ = 6.00 × 10⁻³ − 4.50 × 10⁻³ = 1.50 × 10⁻³ mol H⁺. Row 4 (V = 50.0 mL): excess H⁺ = 1.00 × 10⁻² − 4.50 × 10⁻³ = 5.50 × 10⁻³ mol H⁺. 2 marks: 0.5 per row for correct excess identity and value.

(b) V(HCl) = 30.0 mL: Excess H⁺ = 1.50 × 10⁻³ mol. V(total) = 60.0 mL = 0.0600 L. c(H⁺) = 1.50 × 10⁻³/0.0600 = 0.0250 mol/L. pH = −log(0.0250) = −log(2.5 × 10⁻²) = 2 − log(2.5) = 2 − 0.398 = 1.60. Sanity: acid excess → pH < 7 ✓

(c) Equivalence point pH = 7.00: At 22.5 mL NaOH added, n(H⁺) = n(OH⁻) = 4.50 × 10⁻³ mol. All H⁺ and OH⁻ are consumed in the neutralisation H⁺ + OH⁻ → H₂O. The solution contains only Na⁺ (from NaOH) and Cl⁻ (from HCl) — spectator ions that are the conjugate of a strong base and a strong acid respectively. Neither hydrolyses water, so [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L and pH = 7.00.

Marking notes (b): 1 mark for correct excess H⁺ = 1.50 × 10⁻³ mol. 1 mark for using V(total) = 0.0600 L. 1 mark for correct pH = 1.60.

Marking notes (c): 1 mark for identifying equivalence (all H⁺ and OH⁻ consumed). 1 mark for spectator ions don’t hydrolyse → [H⁺] = [OH⁻] → pH = 7.00.

(a) The equivalence point is at V(NaOH) = 20.0 mL (equal moles of HCl and NaOH: 0.100 mol/L × 0.0200 L = 2.00 × 10⁻³ mol each). pH = 7.00 because the only ions remaining are Na⁺ and Cl⁻, both spectator ions that do not hydrolyse, so [H⁺] = [OH⁻] from water autoionisation alone.

(b) Before the equivalence point (0–20 mL added), the curve rises slowly because HCl is in large excess and each addition of NaOH barely changes [H⁺]. After the equivalence point (>20 mL added), the curve levels off at high pH because NaOH is in excess and further addition has a smaller relative effect on [OH⁻]. The steep rise near the equivalence point reflects the very small amount of excess acid or base needed to change pH by several units when concentrations are near zero.

Marking notes (a): 1 mark for equivalence at 20.0 mL. 1 mark for spectator ion reasoning.

Marking notes (b): 1 mark for describing the slow rise before and levelling after. 1 mark for accounting using relative excess concentration change.

Sample response. The statement is incorrect: because the pH scale is logarithmic, small numerical differences in pH correspond to large differences in [H⁺], and large numerical differences can sometimes represent less biologically or industrially meaningful changes than they appear. This relationship works in both directions. A small pH difference can be highly significant. In the context of blood physiology, blood pH normally ranges between 7.35 and 7.45 — a difference of only 0.10 pH units. A drop from 7.40 to 7.10 (a difference of only 0.30 units) doubles [H⁺]: [H⁺] at pH 7.10 = 7.94 × 10⁻⁸ mol/L; at pH 7.40 = 3.98 × 10⁻⁸ mol/L; ratio = 2.0. This doubling denatures enzymes, shifts haemoglobin’s oxygen affinity (Bohr effect), and disrupts cardiac function — all life-threatening consequences from a change that appears trivially small on the pH scale. Similarly, in Australian swimming pool regulation, a pH shift from 7.5 to 7.8 (0.3 units) corresponds to a 2-fold decrease in [H⁺], which shifts the active form of chlorine disinfectant from HOCl (effective) to OCl⁻ (much less effective), increasing the risk of bacterial contamination. A large pH difference can also be less meaningful in some contexts. The difference between battery acid (pH 0, [H⁺] = 1.0 mol/L) and stomach acid (pH 2, [H⁺] = 0.01 mol/L) is 2 pH units — a 100-fold change in [H⁺] — but both are highly corrosive to biological tissue; the additional 100-fold difference between them may not represent a proportionally larger practical hazard in many situations. In conclusion, the statement is a misinterpretation of scale: because pH is logarithmic, even small numerical differences can represent dramatic changes in [H⁺] and are frequently the most scientifically and clinically important. The key insight is to always translate pH differences into [H⁺] ratios before judging significance.

Marking notes. 1 mark: states the statement is incorrect and identifies the logarithmic scale as the key issue. 1 mark: quantitative example of a small pH difference being significant (blood pH, pool pH, or ocean pH) with [H⁺] ratio calculated. 1 mark: correctly links small pH change to large proportional [H⁺] change (correct ratio). 1 mark: second real-world Australian context with accurate chemical or biological reasoning. 1 mark: uses NESA language to evaluate (the claim is incorrect because… the flaw is…). 1 mark: reaches an explicit evaluative conclusion that is evidence-based and quantitative.