Chemistry • Year 12 • Module 6 • Lesson 10

Enthalpy of Neutralisation: Comparing Strong & Weak

Build HSC Band 5–6 extended-response technique on thermodynamic reasoning, source critique, and experimental evaluation.

Master • Extended Response

1. Data + scenario — classify unknown acids and evaluate experimental design (Band 5–6)

8 marks   Band 5–6

Scenario. A food-science research team at UNSW Sydney is analysing three unlabelled acid solutions (A, B, C) each at 1.00 mol/L. They perform two complementary experiments:

Experiment 1 — Calorimetry: Each acid (50.0 mL) is mixed with 50.0 mL of 1.00 mol/L NaOH in a foam cup at Tinitial = 20.0°C. Maximum temperatures recorded: Acid A = 26.6°C; Acid B = 26.8°C; Acid C = 25.3°C.

Experiment 2 — pH: pH of each acid at 1.00 mol/L: Acid A = 1.78; Acid B = 0.00; Acid C = 2.24.

The team suspects one acid is HCl, one is HNO2 (nitrous acid, Ka = 4.5 × 10−4), and one is CH3COOH (Ka = 1.8 × 10−5).

20 23 26 29 30 s 60 s 120 s 180 s Time (s) Temperature (°C) Acid B (Tmax = 26.8°C) Acid A (Tmax = 26.6°C) Acid C (Tmax = 25.3°C)

Figure 1.1 — Temperature–time profiles for three neutralisation experiments (foam cup calorimeter, Tinitial = 20.0°C). Stylised from school calorimetry data.

Q1. Using both experimental datasets, identify which acid (A, B, or C) is HCl, HNO2, and CH3COOH. In your response you must:

  • Calculate ΔHn for each acid from the calorimetry data (show working for at least one).
  • Calculate Ka for each acid identified as weak, using the pH data.
  • Match each acid to its identity, with justification referencing both ΔHn and Ka.
  • Evaluate: the team proposes that calorimetry alone would be sufficient to identify all three acids. Assess this claim, referring to the precision of foam cup calorimetry (±2–3 kJ/mol).
  • Suggest one improvement to the experimental design that would increase the reliability of the classification.
Plan: (a) calculate ΔHn for A, B, C; (b) identify B as strong (close to −57); (c) use pH to get [H+] → Ka for A and C; (d) match Ka values to HNO2 (4.5 × 10−4) and CH3COOH (1.8 × 10−5); (e) evaluate whether ΔHn differences between A and C are resolvable within ±2–3 kJ/mol foam cup precision.

2. Source critique — evaluate a published claim (Band 5–6)

7 marks   Band 5–6

“A commonly observed pattern in school calorimetry experiments is that weaker acids give larger temperature rises during neutralisation with NaOH, because they have stored more potential energy in their un-ionised form that is released when the acid finally reacts. This is why CH3COOH solutions feel warm when you add NaOH to them — the energy released by the forced ionisation of the weak acid adds to the heat of the neutralisation reaction, giving more total heat than you get with HCl.”

— Extract from a student chemistry blog post, published online 2024.

Q2. Critically evaluate this claim. In your response:

  • Identify every factual or conceptual error in the passage above (there are at least three).
  • For each error, explain the correct chemistry at the molecular or thermodynamic level.
  • Explain what measurement a student could perform in a school laboratory to demonstrate that the claim is false, and what the expected result would be.
Errors to find: (1) “larger temperature rises” — wrong direction; weak acid gives smaller ΔT; (2) “forced ionisation adds to the heat” — ionisation is endothermic, not exothermic; (3) “stored potential energy is released” — ionisation is bond-breaking (endothermic), not energy release. The school experiment: measure ΔT for HCl + NaOH and CH3COOH + NaOH under identical conditions — the HCl experiment gives a larger ΔT.

3. Extended comparison — two methods of characterising an unknown acid (Band 5–6)

6 marks   Band 5–6

Q3. A chemistry student wants to determine whether an unknown 1.00 mol/L acid solution is strong or weak. They have access to a foam-cup calorimeter, a calibrated pH probe, and 1.00 mol/L NaOH. Compare and evaluate the two methods (calorimetry and pH measurement) for this purpose. Your response must:

  • Describe what each method measures and how the result indicates strong vs weak acid.
  • Identify one specific advantage and one specific limitation of each method.
  • Explain why neither method alone provides complete characterisation of the acid (i.e. both strong/weak classification and Ka value).
  • Recommend the better method for rapid classification only, with justification.
Structure: calorimetry (measures ΔHn, classifies if close to/far from −57; advantage: direct thermodynamic; limitation: can’t determine Ka, precision ±2–3 kJ/mol) vs pH (measures [H+], allows Ka calculation; advantage: precise, gives Ka; limitation: requires known concentration, calibration). Neither alone: calorimetry classifies but not Ka; pH gives Ka but relies on known concentration being accurate. For rapid classification: pH is faster and requires no mixing.
Answers — Do not peek before attempting

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

Calorimetry calculations (m = 100.0 g; n(H2O) = 0.0500 mol):

Acid A: ΔT = 6.6°C; q = 100.0 × 4.18 × 6.6 = 2758.8 J; ΔHn(A) = −2.7588/0.0500 = −55.2 kJ/mol.

Acid B: ΔT = 6.8°C; q = 100.0 × 4.18 × 6.8 = 2842.4 J; ΔHn(B) = −2.8424/0.0500 = −56.8 kJ/mol.

Acid C: ΔT = 5.3°C; q = 100.0 × 4.18 × 5.3 = 2215.4 J; ΔHn(C) = −2.2154/0.0500 = −44.3 kJ/mol.

[1 mark — correct ΔHn calculated for at least two acids with working shown]

Ka calculations from pH:

Acid B: pH = 0.00 → [H+] = 1.00 mol/L — strong acid (complete ionisation). [ΔHn = −56.8 ≈ −57 — confirms strong acid.] Identity: HCl.

Acid A: pH = 1.78 → [H+] = 10−1.78 = 1.66 × 10−2 mol/L. Ka(A) = (1.66 × 10−2)2 / (1.00 − 1.66 × 10−2) = 2.76 × 10−4 / 0.9834 = 2.81 × 10−4 (close to HNO2 Ka = 4.5 × 10−4). Identity: HNO2.

Acid C: pH = 2.24 → [H+] = 10−2.24 = 5.75 × 10−3 mol/L. Ka(C) = (5.75 × 10−3)2 / (1.00 − 5.75 × 10−3) = 3.31 × 10−5 / 0.9943 = 3.33 × 10−5 (close to CH3COOH Ka = 1.8 × 10−5). Identity: CH3COOH.

[2 marks — Ka correctly calculated for each weak acid from pH data]

Matching justification: B = HCl (strong: ΔHn ≈ −57, pH = 0.00); A = HNO2 (weak: ΔHn −55.2, Ka ≈ 2.8 × 10−4); C = CH3COOH (weaker: ΔHn −44.3, Ka ≈ 3.3 × 10−5).

[1 mark — all three acids correctly identified with dual justification]

Evaluation of calorimetry alone: ΔHn(A) = −55.2 and ΔHn(C) = −44.3 differ by 10.9 kJ/mol — easily resolvable within foam cup precision (±2–3 kJ/mol). However, the absolute Ka values cannot be determined from calorimetry alone — calorimetry only classifies strong/weak and measures ΔH(ionisation), not the equilibrium position. So calorimetry alone would correctly distinguish all three in this case, but it cannot confirm identity without a reference library of ΔH(ionisation) values. The claim that calorimetry alone “suffices” is an overstatement for confident identification without prior knowledge of expected values.

[2 marks — quantitative evaluation of precision; recognises calorimetry cannot determine Ka]

Design improvement: Replicate each experiment at least three times and average ΔHn to reduce the effect of random error (e.g. inconsistent swirling, thermometer lag). Using a calibrated digital thermometer rather than a standard mercury/alcohol thermometer would also reduce reading error and allow a more precise Tmax determination.

[1 mark — specific, justified improvement]

Marking criteria: 1 mark — ΔHn for ≥2 acids, working shown. 2 marks — Ka(A) and Ka(C) calculated. 1 mark — all three acids matched with dual justification. 2 marks — quantitative evaluation of calorimetry-alone claim with reference to precision and limitation. 1 mark — specific justified design improvement. Total: 7 marks scored from a possible 8; all criteria above needed for 8/8.

Q2 — Source critique: Sample Band 6 response (7 marks)

Error 1: “weaker acids give larger temperature rises.” This is the opposite of what is observed. When a weak acid is neutralised with NaOH under identical conditions (same volume, same concentration), ΔT is smaller than for HCl + NaOH, not larger. The measured ΔT for CH3COOH + NaOH is approximately 0.2–0.6°C less than HCl + NaOH under equivalent conditions [2 marks: identify error + correct direction with quantitative evidence].

Error 2: “forced ionisation of the weak acid adds to the heat.” The ionisation of a weak acid (CH3COOH ⇌ H+ + CH3COO−) is an endothermic process — it involves breaking the O–H bond in the carboxyl group, which requires an input of energy (+1.6 kJ/mol for acetic acid). By Hess’s Law, this endothermic ionisation enthalpy is subtracted from, not added to, the exothermic H+ + OH− → H2O step. Net ΔHn = −57 + 1.6 = −55.4 kJ/mol — less exothermic than for HCl [2 marks: error identified + endothermic mechanism explained with sign].

Error 3: “stored potential energy is released” (implying ionisation is exothermic). The un-ionised form of acetic acid does not contain “stored potential energy” that is released during reaction. Ionisation is a bond-breaking process (endothermic). The O–H bond in the carboxyl group is broken when the proton is donated to a base — this requires energy input from the thermal surroundings, reducing the measured temperature rise. Exothermic energy release in neutralisation comes entirely from forming the O–H bonds in liquid water (H+ + OH− → H2O) [1 mark: error identified + bond-breaking endothermic correction].

Experimental test: Mix 50.0 mL of 1.00 mol/L HCl with 50.0 mL of 1.00 mol/L NaOH, and repeat with CH3COOH under identical conditions. Measure Tmax in each case. Expected result: ΔT(HCl + NaOH) ≈ 6.8°C; ΔT(CH3COOH + NaOH) ≈ 6.6°C. The HCl experiment gives a larger ΔT — directly falsifying the claim that weak acids give larger temperature rises [1 mark: specific experiment + expected result].

Marking criteria: 2 marks — Error 1 identified + correct direction established with evidence. 2 marks — Error 2 identified + endothermic ionisation explained with sign and Hess’s Law. 1 mark — Error 3 identified + bond-breaking correction. 1 mark — falsifying experiment described. 1 mark — expected result correctly stated. Total: 7 marks.

Q3 — Compare and evaluate two methods (6 marks)

Calorimetry: Measures heat released during neutralisation (q = mcΔT; ΔHn = −q/n). A value close to −57 kJ/mol indicates a strong acid (no ionisation energy consumed); a significantly more positive value (e.g. −50 to −55 kJ/mol) indicates a weak acid. Advantage: directly measures the thermodynamic consequence of ionisation; does not require knowledge of the acid’s concentration for the classification (only for the per-mole calculation). Limitation: foam cup precision of ±2–3 kJ/mol may not resolve two weak acids with similar Ka values; cannot determine Ka from ΔHn alone [2 marks].

pH measurement: Measures [H+] at equilibrium. For a strong acid at 1.00 mol/L, pH = 0.00; for a weak acid, pH > 0 at the same concentration. Ka can be calculated from pH: Ka = [H+]2/(c − [H+]). Advantage: precise; allows Ka calculation; immediately distinguishes all acids regardless of how close their ΔHn values are. Limitation: requires calibrated pH probe and accurate knowledge of concentration; does not directly measure thermodynamic ionisation enthalpy [2 marks].

Why neither alone gives complete characterisation: Calorimetry classifies strong/weak and measures ΔH(ionisation) but cannot determine Ka. pH measurement gives Ka but does not directly reveal the enthalpy cost of ionisation. Complete characterisation requires both: Ka (equilibrium property, from pH) and ΔH(ionisation) (thermodynamic property, from calorimetry). Knowing only Ka does not tell you the enthalpy; knowing only ΔH(ionisation) does not give Ka [1 mark].

Recommendation for rapid classification only: pH measurement is recommended. It is faster (mix, dip probe, read), requires less preparation (no heating, no calorimeter setup), is more precise (digital readout vs thermometer), and unambiguously distinguishes strong (pH = 0) from any weak acid (pH > 0) without the ±2–3 kJ/mol uncertainty of foam cup calorimetry [1 mark].

Marking criteria: 1 mark each for correctly describing what each method measures and its indicator for strong/weak (2 total). 1 mark each for a specific, accurate advantage and limitation for each method (2 total). 1 mark — explains why neither method alone fully characterises the acid. 1 mark — justified recommendation for rapid classification. Total: 6 marks.

HSCScience • Chemistry • Year 12 • Module 6 • Lesson 10 • Worksheet 3 (Master)