When you mix a strong acid and a strong base, the cup warms up — no flame, no fuel. Just ions rearranging. Neutralisation calorimetry measures that warmth precisely, and reveals a remarkable fact: it doesn't matter which strong acid or base you use. The answer is always the same.
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A nurse mixes two clear solutions — hydrochloric acid and sodium hydroxide — in a polystyrene cup. Within seconds, the cup feels warm in her hand. No flame, no fuel, just ions rearranging.
What's releasing the energy? And here's the challenge: would it matter if she used sulfuric acid instead of hydrochloric acid — would you get a different amount of heat per mole of water formed?
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📚 Core Content
The reason any strong acid neutralised by any strong base gives approximately the same ΔHn ≈ −57 kJ mol⁻¹ is that the reaction is always, fundamentally, the same reaction.
When a strong acid and strong base are dissolved in water, they fully dissociate into ions. For example:
HCl(aq) → H⁺(aq) + Cl⁻(aq)
NaOH(aq) → Na⁺(aq) + OH⁻(aq)
The net ionic equation for any strong acid + strong base neutralisation is always:
H⁺(aq) + OH⁻(aq) → H₂O(l) ΔH ≈ −57 kJ mol⁻¹
The spectator ions (Na⁺, Cl⁻, K⁺, SO₄²⁻, NO₃⁻, etc.) do not participate in the reaction — so they don't affect ΔH. This is why HCl + NaOH, H₂SO₄ + KOH, and HNO₃ + NaOH all give approximately the same molar enthalpy of neutralisation.
| Acid + Base combination | Spectator ions | Net ionic equation | Expected ΔHn |
|---|---|---|---|
| HCl(aq) + NaOH(aq) | Na⁺, Cl⁻ | H⁺ + OH⁻ → H₂O(l) | ≈ −57 kJ mol⁻¹ |
| HNO₃(aq) + KOH(aq) | K⁺, NO₃⁻ | H⁺ + OH⁻ → H₂O(l) | ≈ −57 kJ mol⁻¹ |
| H₂SO₄(aq) + 2NaOH(aq) | Na⁺, SO₄²⁻ | H⁺ + OH⁻ → H₂O(l) | ≈ −57 kJ mol⁻¹ |
| CH₃COOH(aq) + NaOH(aq) | Na⁺, CH₃COO⁻ | Partial ionisation of weak acid | < −57 kJ mol⁻¹ |
The same formula q = mcΔT applies to both — but what 'm' and 'n' represent, and the apparatus used, are fundamentally different.
| Feature | Combustion (L02) | Neutralisation (L03) |
|---|---|---|
| Apparatus | Copper calorimeter + spirit burner | Polystyrene cup + thermometer |
| 'm' in q = mcΔT | Mass of water in calorimeter | Total mass of combined solution (acid + base) |
| 'n' in ΔH = −q/n | Moles of fuel burned | Moles of H₂O formed |
| Typical ΔT | 5–15°C | 5–10°C |
| Sign of ΔH | Always negative (exothermic) | Negative for strong acid + strong base |
| Main source of error | Heat loss to atmosphere; incomplete combustion | Heat loss through cup walls; incomplete mixing |
| Why apparatus chosen | Copper: high conductivity → fast heat transfer to water | Polystyrene: low conductivity → insulates reaction mixture |
🔧 Worked Examples
50.0 mL of 1.00 mol L⁻¹ HCl is mixed with 50.0 mL of 1.00 mol L⁻¹ NaOH in a polystyrene cup. The temperature rises from 21.4°C to 28.0°C. Calculate the molar enthalpy of neutralisation.
25.0 mL of 2.00 mol L⁻¹ H₂SO₄ is neutralised with 50.0 mL of 2.00 mol L⁻¹ NaOH. The temperature rises by 13.2°C. Calculate ΔHn. (Assume density of solution = 1.00 g mL⁻¹)
🧪 Activities
a Calculate the total mass of solution and ΔT.
b Calculate q in kJ.
c Calculate n(H₂O formed) and then ΔHn. Compare to the expected −57 kJ mol⁻¹.
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Wrong: In calorimetry, m in q = mcΔT is the mass of the fuel burned.
Right: m in q = mcΔT is always the mass of the water (or solution) being heated, not the fuel. The fuel mass is used later to calculate ΔH per mole: ΔH = −q/n where n is moles of fuel. Using the wrong mass is the most common calorimetry error.
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✍️ Short Answer
6. In a neutralisation calorimetry experiment, explain why: (a) the total volume of both solutions is used as the mass in q = mcΔT, and (b) polystyrene is used instead of copper for the calorimeter. 3 MARKS
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7. 40.0 mL of 1.50 mol L⁻¹ HCl is mixed with 60.0 mL of 1.00 mol L⁻¹ NaOH in a polystyrene cup. The temperature rises from 19.0°C to 24.5°C. (a) Identify the limiting reagent. (2 marks) (b) Calculate ΔHn. (3 marks) 5 MARKS
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8. Real-World Application: A nurse measures the heat produced when 50.0 mL of 1.00 mol L⁻¹ HCl is neutralised with 50.0 mL of 1.00 mol L⁻¹ NaOH (ΔHn ≈ −57 kJ mol⁻¹). She then repeats the experiment using 50.0 mL of 1.00 mol L⁻¹ H₂SO₄ with 50.0 mL of 1.00 mol L⁻¹ NaOH.
(a) Predict whether the temperature rise will be greater, less, or the same for the H₂SO₄ experiment compared to the HCl experiment. Justify using the net ionic equation concept. (2 marks)
(b) Calculate the expected temperature rise for the H₂SO₄ experiment given ΔHn = −57 kJ mol⁻¹. Note: H₂SO₄ is diprotic; check whether NaOH is sufficient to neutralise all H⁺. (3 marks)
5 MARKS
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(a) m = 100.0 g; ΔT = 3.3°C = 3.3 K
(b) q = 100.0 × 4.18 × 3.3 = 1379.4 J = 1.379 kJ
(c) n(HNO₃) = 0.500 × 0.0500 = 0.0250 mol; n(KOH) = 0.0250 mol; n(H₂O) = 0.0250 mol; ΔHn = −1.379 ÷ 0.0250 = −55.2 kJ mol⁻¹. About 3% below the accepted −57 kJ mol⁻¹ — realistic for a polystyrene cup experiment.
Row 1 (methanol combustion): Type = Combustion | m = 150 g (mass of water) | n = moles of methanol burned (= 0.48 ÷ 32.04 = 0.01498 mol)
Row 2 (H₂SO₄ + NaOH): Type = Neutralisation | m = (40 + 80) mL × 1.00 g mL⁻¹ = 120 g | n = moles of H₂O formed
Bonus: n(H⁺) = 2 × 1.5 × 0.040 = 0.120 mol H⁺; n(OH⁻) = 1.5 × 0.080 = 0.120 mol OH⁻. Exactly equimolar — neither is limiting. n(H₂O) = 0.120 mol.
1. C — Both solutions (60.0 mL total) are heated; the combined mass is used.
2. B — Both produce net ionic equation H⁺ + OH⁻ → H₂O(l). Spectator ions don't participate and don't affect ΔH.
3. B — Weak acid partial ionisation consumes energy, reducing net heat released → less negative ΔHn.
4. A — n(H₂SO₄) = 2.00 × 0.0250 = 0.0500 mol; H₂SO₄ diprotic → n(H⁺) = 0.100 mol → n(H₂O) = 0.100 mol.
5. D — The student is partially correct: polystyrene does absorb a small amount of heat (it has low but non-zero heat capacity). However, the dominant source of error is heat loss to the surroundings through the cup walls and open top, not heat stored in the cup material itself. The distinction matters — "absorbed by cup" vs "lost to surroundings" have different implications for calculation correction.
Q6 (3 marks): (a) Both solutions are mixed in the cup and both are heated by the exothermic neutralisation reaction — so both contribute thermal mass that absorbs heat. Using only one volume would underestimate m, making q appear smaller and ΔHn less negative than the true value [2]. (b) Polystyrene is used instead of copper because polystyrene is a poor thermal conductor (low thermal conductivity) — it insulates the reaction mixture, reducing heat loss to the surrounding air. Copper's high conductivity would allow heat to escape rapidly to the bench and atmosphere, making this setup impractical for neutralisation [1].
Q7 (5 marks): (a) n(HCl) = 1.50 × 0.0400 = 0.0600 mol; n(NaOH) = 1.00 × 0.0600 = 0.0600 mol. Both are equal — neither is in excess. Limiting reagent: both equally limiting [1 for each calculation, 1 for correct conclusion = 2 marks]. (b) m = 40.0 + 60.0 = 100.0 g; ΔT = 24.5 − 19.0 = 5.5°C; q = 100.0 × 4.18 × 5.5 = 2299 J = 2.299 kJ [1]; n(H₂O) = 0.0600 mol [1]; ΔHn = −2.299 ÷ 0.0600 = −38.3 kJ mol⁻¹ [1]. (Note: this is much lower than −57 kJ mol⁻¹, suggesting significant heat loss or an error in the experimental setup.)
Q8 (5 marks): (a) Greater temperature rise is expected for H₂SO₄ experiment [1]. Justification: H₂SO₄ is diprotic — 50 mL of 1.00 mol L⁻¹ H₂SO₄ provides 2 × 0.0500 = 0.100 mol H⁺, compared to 0.0500 mol H⁺ from HCl. The net ionic reaction (H⁺ + OH⁻ → H₂O) produces twice as many moles of H₂O, releasing twice the total heat — but ΔHn per mole of H₂O is unchanged at −57 kJ mol⁻¹ [1]. (b) n(H⁺) = 2 × 1.00 × 0.0500 = 0.100 mol; n(OH⁻) = 1.00 × 0.0500 = 0.0500 mol — NaOH is limiting; n(H₂O) = 0.0500 mol [1]. q = 0.0500 × 57 = 2.85 kJ [1]. m = 100.0 g; ΔT = q ÷ (mc) = 2850 ÷ (100.0 × 4.18) = 6.8°C [1]. (Note: with NaOH limiting, only half the H₂SO₄ is neutralised — same temperature rise as the HCl experiment, not double.)
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