Chemistry • Year 12 • Module 6 • Lesson 10

Enthalpy of Neutralisation: Comparing Strong & Weak

Apply calorimetry calculations, interpret ΔH_n data, and reason about the molecular mechanism behind strong/weak comparisons.

Apply • Data & Reasoning

1. Interpret calorimetry data — four acid–base combinations

A student performs four neutralisation experiments using 50.0 mL of 1.00 mol/L acid mixed with 50.0 mL of 1.00 mol/L NaOH in a polystyrene foam cup. Initial temperature = 20.0°C for all trials. The results are recorded below. 12 marks

Acid	T_max (°C)	ΔT (°C)	q (J)	ΔH_n (kJ/mol)	Strong or Weak?	ΔH(ionisation) (kJ/mol)
HNO₃	26.8
HF (K_a = 6.8 × 10⁻⁴)	26.2
HCOOH (K_a = 1.8 × 10⁻⁴)	25.8
HCN (K_a = 6.2 × 10⁻¹⁰)	24.1

Use m = 100.0 g and c = 4.18 J g⁻¹ °C⁻¹. Calculate n(H₂O) first. Use HNO₃ as the strong acid baseline for ΔH(ionisation) calculations.

1.1 Complete all columns in the table above, showing working for HF below. 7 marks

1.2 The K_a values for HF, HCOOH and HCN differ by factors of roughly 4, 3800 and 10⁶ respectively compared to each other. Describe the trend between K_a value and ΔH(ionisation) that your table reveals. 2 marks

1.3 HCN has a very small K_a (6.2 × 10⁻¹⁰) yet its ΔH_n is still significantly more negative than 0. Explain why the neutralisation is still highly exothermic even for such a weak acid. 3 marks

Stuck? Re-read the energy budget formula in the lesson’s Formula Panel: ΔH_n(obs) = −57 kJ/mol + ΔH(ionisation). Even a large positive ionisation enthalpy does not make the reaction endothermic overall.

2. Interpret graph — ΔH_n bar chart for strong/weak combinations

The bar chart below shows measured ΔH_n values for four acid–base combinations studied in a school calorimetry experiment. All experiments used 1.00 mol/L solutions and 50.0 mL + 50.0 mL volumes. 9 marks

Figure 2.1 — Measured ΔH_n for four acid–base combinations (school calorimetry data, 1.00 mol/L, 50.0 mL + 50.0 mL, foam cup). Dashed red line = theoretical strong+strong baseline (−57 kJ/mol).

2.1 Describe the trend shown in the graph, using the terms “more exothermic” and “strong/weak” appropriately. 2 marks

2.2 Use Hess’s Law to calculate the expected ΔH_n for CH₃COOH + NH₃ from the data shown for the other three combinations. Show your working. 3 marks

2.3 A student claims the HCl + NaOH bar should reach exactly to the −57 dashed line. The measured value is −56.8 kJ/mol. Suggest two reasons for this 0.2 kJ/mol discrepancy, other than incorrect calculation. 2 marks

2.4 Predict the approximate ΔH_n if the experiment were repeated with HCl + NaOH at 2.00 mol/L (50.0 mL + 50.0 mL). Justify your prediction. 2 marks

Stuck? For Q2.2: ΔH_n(weak+weak) = baseline + ΔH(ionisation CH₃COOH) + ΔH(ionisation NH₃). For Q2.4: ΔH_n is per mole of water, so changing concentration should not change the per-mole value — think about what does and does not change.

3. Australian context — acetic acid in Gaviscon and industrial neutralisation

Read the scenario below and answer the questions. 8 marks

Scenario. Gaviscon is an Australian antacid that contains sodium alginate and sodium bicarbonate. Some formulations also include a small amount of acetic acid (vinegar, CH₃COOH) as a mild acidifier to control pH in the stomach. A food scientist is comparing the heat released per gram of acid neutralised when HCl (K_a ≈ 10⁷, strong) and CH₃COOH (K_a = 1.8 × 10⁻⁵) are each neutralised with NaOH. Separately, at Orica’s chemical plant in Kooragang Island (NSW), large quantities of sodium acetate (CH₃COONa) are produced by neutralising acetic acid with sodium hydroxide: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l). The plant engineers must account for the heat released during this industrial-scale reaction when designing reactor cooling systems.

3.1 Explain why the Gaviscon acetic acid releases less heat per mole of H₂O formed than HCl when each is neutralised with NaOH. Use the concept of ionisation enthalpy in your answer. 3 marks

3.2 The Orica plant produces sodium acetate at 5000 mol/h. Calculate the heat released per hour (in kJ) using ΔH_n = −55.4 kJ/mol. Is this value larger or smaller than if HCl were the reactant at the same production rate? Account for the sign convention in your answer. 3 marks

3.3 A plant engineer proposes that calorimetry data alone could be used to confirm whether the incoming acid supply is acetic acid (weak) rather than a contaminating strong acid. Describe the expected calorimetry result and the reasoning behind the classification. 2 marks

Stuck? For Q3.1 revisit Card 2 (Molecular Mechanism). For Q3.3 revisit Card 5 (Calorimetry to Classification).

4. Predict and justify — changing experimental conditions

A student replaces the polystyrene foam cup with a metal beaker for the HCl + NaOH experiment (all other conditions identical). She predicts: “The calculated ΔH_n will be more negative than the correct value because the metal absorbs more heat, making ΔT larger.” 4 marks

4.1 Is the student’s prediction about the direction of error correct? Explain fully, using the formula q = mcΔT and the concept of heat loss. 2 marks

4.2 Would switching to a metal beaker affect the comparison between ΔH_n values for HCl + NaOH and CH₃COOH + NaOH, or would both values be shifted by the same amount? Explain. 2 marks

Stuck? Think about whether the metal beaker adds heat to the solution or removes heat from the solution during the experiment. The beaker heats up too — where does that energy come from?

Answers — Do not peek before attempting

Q1.1 — Complete calorimetry table

n(H₂O) = 1.00 × 0.0500 = 0.0500 mol for all trials.

HNO₃: ΔT = 6.8°C; q = 100.0 × 4.18 × 6.8 = 2842 J; ΔH_n = −2.842/0.0500 = −56.8 kJ/mol. Classification: strong acid (within 1 kJ/mol of −57). ΔH(ionisation) = 0 (baseline).

HF: ΔT = 6.2°C; q = 100.0 × 4.18 × 6.2 = 2591.6 J; ΔH_n = −2.5916/0.0500 = −51.8 kJ/mol. Classification: weak acid. ΔH(ionisation) = −51.8 − (−56.8) = +5.0 kJ/mol.

HCOOH: ΔT = 5.8°C; q = 100.0 × 4.18 × 5.8 = 2424.4 J; ΔH_n = −2.4244/0.0500 = −48.5 kJ/mol. Classification: weak acid. ΔH(ionisation) = −48.5 − (−56.8) = +8.3 kJ/mol.

HCN: ΔT = 4.1°C; q = 100.0 × 4.18 × 4.1 = 1713.8 J; ΔH_n = −1.7138/0.0500 = −34.3 kJ/mol. Classification: weak acid. ΔH(ionisation) = −34.3 − (−56.8) = +22.5 kJ/mol.

Q1.2 — Trend: K_a vs ΔH(ionisation)

As K_a decreases (acid becomes weaker), ΔH(ionisation) increases (becomes more positive, more endothermic). HF (largest K_a of the three weak acids, +5.0 kJ/mol) → HCOOH (+8.3 kJ/mol) → HCN (smallest K_a, +22.5 kJ/mol). A weaker acid has a stronger, harder-to-break O–H bond, requiring more energy to ionise. [2 marks: 1 for correct direction, 1 for mechanistic link to bond strength.]

Q1.3 — Why HCN neutralisation is still highly exothermic

Even though HCN has an extremely small K_a, the net ionic equation H⁺ + OH⁻ → H₂O still releases approximately −57 kJ/mol [1]. The ionisation of HCN costs +22.5 kJ/mol, but this is still far less than the 57 kJ/mol released by the water-formation step [1]. The net ΔH_n = −57 + 22.5 = −34.5 kJ/mol — significantly exothermic [1]. Only if ΔH(ionisation) exceeded 57 kJ/mol would the reaction become endothermic overall, which does not occur for any common acid.

Q2.1 — Graph trend description

As the number of weak species in the combination increases, ΔH_n becomes less negative (less exothermic). Strong + strong gives the most negative value (−56.8 kJ/mol); weak acid + weak base gives the least negative value (−49.5 kJ/mol). Each weak species contributes a positive ionisation enthalpy that is subtracted from the baseline −57 kJ/mol. [1 mark for correct direction; 1 mark for linking to number of weak species / ionisation enthalpy.]

Q2.2 — Hess’s Law prediction for CH₃COOH + NH₃

ΔH(ionisation CH₃COOH) = −55.4 − (−56.8) = +1.4 kJ/mol [1]
ΔH(ionisation NH₃) = −52.2 − (−56.8) = +4.6 kJ/mol [1]
Expected ΔH_n(CH₃COOH + NH₃) = −56.8 + 1.4 + 4.6 = −50.8 kJ/mol [1] (close to the observed −49.5; small discrepancy is within experimental error).

Q2.3 — Reasons for discrepancy from −57

Accept any two from: (1) heat loss to the surroundings and to the foam cup itself (calorimeter absorbs some heat, reducing ΔT measured); (2) the foam cup has non-zero heat capacity, so not all heat goes into the solution; (3) evaporation of the solution removes energy; (4) thermometer lag — T_max may have been reached before the reading was taken.

Q2.4 — Prediction for 2.00 mol/L experiment

ΔH_n is defined per mole of water formed and should remain approximately −56.8 kJ/mol [1]. Although the temperature rise (ΔT) would roughly double (more moles reacting in the same mass of solution), dividing by twice the moles of H₂O gives the same per-mole value — the intensive property ΔH_n does not change with concentration for a strong acid + strong base [1].

Q3.1 — Gaviscon acetic acid: less heat per mole

CH₃COOH is a weak acid (K_a = 1.8 × 10⁻⁵); before mixing with NaOH, most molecules are intact and not ionised [1]. When OHˉ is added, it drives the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ to the right, forcing ionisation of the O–H bond [1]. This ionisation is endothermic (ionisation enthalpy ≈ +1.6 kJ/mol), consuming energy from the solution that would otherwise contribute to the measured ΔT. Net heat = 57 − 1.6 = ≈ 55.4 kJ/mol — less than for HCl [1].

Q3.2 — Orica: heat released per hour

Heat released = 5000 mol/h × 55.4 kJ/mol = 277 000 kJ/h (or 277 MJ/h) [1]. If HCl were used: 5000 × 57.0 = 285 000 kJ/h. The acetic acid process releases less heat per hour (277 000 vs 285 000 kJ/h) [1]. Sign convention: ΔH_n is negative (exothermic), so the system releases heat — the magnitude per hour is 277 MJ (heat released = +277 MJ/h to the surroundings) [1].

Q3.3 — Calorimetry to classify acid supply

If the incoming acid is acetic acid (weak), neutralisation with NaOH will give ΔH_n significantly more positive than −57 kJ/mol (e.g. −55 to −56 kJ/mol) [1]. If a strong acid contaminant is present, ΔH_n will be close to −57 kJ/mol. The classification rests on whether the deviation from the strong-acid baseline is within experimental error (±2–3 kJ/mol) or exceeds it [1].

Q4.1 — Direction of error with metal beaker

The student’s prediction about the direction is incorrect. A metal beaker has a much higher heat capacity than foam and absorbs significant heat from the solution during the reaction [1]. This means ΔT of the solution is smaller than it would be in an insulating foam cup. Substituting a smaller ΔT into q = mcΔT gives a smaller q, and therefore a less negative (more positive, smaller magnitude) calculated ΔH_n — not more negative [1].

Q4.2 — Effect on strong vs weak comparison

Both ΔH_n values would be shifted toward less negative values by approximately the same systematic amount, because the metal beaker absorbs heat regardless of which acid is used [1]. However, the difference between the two (ΔH(ionisation of CH₃COOH) ≈ +1.6 kJ/mol) is likely to still be detectable, since the systematic error affects both equally. The comparison remains valid in principle, though the absolute magnitudes would both be underestimated [1].