Calorimetry — Neutralisation
In 1840, Germain Hess published measurements showing that mixing aqueous HCl with aqueous NaOH released approximately 57 kJ mol⁻¹ of heat — identical to the result for H₂SO₄ + NaOH and HNO₃ + KOH. Hess's careful neutralisation calorimetry, conducted at the St Petersburg Academy of Sciences, was one of the key observations that led him to formulate the Law of Constant Heat Summation: it didn't matter which strong acid or base you chose — the molar enthalpy of neutralisation was always the same.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Key Facts
- The solution (coffee cup) calorimeter setup for neutralisation
- Why strong acid + strong base gives ΔHn ≈ −57 kJ mol⁻¹
- The net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l)
Concepts
- Why m = total mass of both solutions combined
- Why weak acid + strong base gives a less negative ΔHn
- How neutralisation differs from combustion calorimetry in setup and formula
Skills
- Calculate q and ΔHn from neutralisation data
- Calculate n using n = cV, including for diprotic acids
- Compare experimental ΔHn to the accepted −57 kJ mol⁻¹
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 |
For strong acid + strong base, the net ionic equation is always H⁺(aq) + OH⁻(aq) → H₂O(l), ΔH ≈ −57 kJ mol⁻¹; spectator ions do not affect ΔH. Weak acids give a less negative ΔHn because energy is consumed to fully ionise them during neutralisation.
Pause — copy the highlighted definition into your book before moving on.
Mini-task: H₂SO₄ and HCl are both mixed separately with excess NaOH. State whether the ΔHn per mole of H₂O formed will be the same or different for each experiment, and explain why using the net ionic equation concept.
We just saw that strong acid + strong base always gives ΔHn ≈ −57 kJ mol⁻¹ via the same net ionic equation. That raises a question: how does measuring ΔHn in a polystyrene cup differ from the combustion calorimeter in L02? This card answers it → by comparing what ‘m’ and ‘n’ represent in each experimental context.
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 |
In neutralisation calorimetry: m = total mass of combined solution; n = moles of H₂O formed; polystyrene cup insulates the mixture. In combustion: m = mass of water; n = moles of fuel; copper cup rapidly conducts heat from flame to water.
Add the highlighted point to your notes before the check below.
Explain it: A student is setting up a neutralisation calorimetry experiment using 50 mL HCl and 50 mL NaOH. They use a copper beaker instead of a polystyrene cup, and they measure m as 50 g (the mass of just the acid). Identify both errors and explain the correct approach.
Worked examples · reveal as you go
HCl + NaOH — equimolar. 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.
m = 50.0 + 50.0 = 100.0 mL ≈ 100.0 g
ΔT = 28.0 − 21.4 = 6.6°C = 6.6 K
$q = mc\Delta T = 100.0 \times 4.18 \times 6.6 = 2758.8 \text{ J} = \mathbf{2.759 \text{ kJ}}$
$n = cV = 1.00 \times 0.0500 = \mathbf{0.0500 \text{ mol}}$
$\Delta H_n = \dfrac{-q}{n} = \dfrac{-2.759}{0.0500} = \mathbf{-55.2 \text{ kJ mol}^{-1}}$
Answer: ΔHn = −55.2 kJ mol⁻¹ (experimental). Accepted: ≈ −57 kJ mol⁻¹.
H₂SO₄ + NaOH — diprotic acid. 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)
n(H₂SO₄) = 2.00 × 0.0250 = 0.0500 mol
n(H⁺) = 2 × 0.0500 = 0.100 mol H⁺
n(OH⁻) = 2.00 × 0.0500 = 0.100 mol OH⁻
m = 25.0 + 50.0 = 75.0 g
$q = 75.0 \times 4.18 \times 13.2 = 4138 \text{ J} = \mathbf{4.138 \text{ kJ}}$
$\Delta H_n = \dfrac{-4.138}{0.100} = \mathbf{-41.4 \text{ kJ mol}^{-1}}$
Answer: ΔHn = −41.4 kJ mol⁻¹. Note: diprotic acid → n(H₂O) = 2 × n(H₂SO₄).
Complete this neutralisation calorimetry calculation. Type each missing value then click Check answers.
Mass of solution: m = g
Heat released: q = 100 × 4.18 × 6.8 = J
Moles of water formed: n = mol
ΔH = −q ÷ n = −2842.4 ÷ 0.05 = J mol¹
Formula Reference — This Lesson
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Common errors · the 3 traps that cost marks
"m is just the mass of the acid (or the base)"
Students use only one of the solution volumes as the mass in q = mcΔT, ignoring the other solution.
Fix: m = total mass of both solutions combined. Both the acid and the base are present in the cup and both are heated by the exothermic reaction. Using only one volume will underestimate q and give a ΔHn that is less negative than the true value.
"n = moles of acid added" (missing diprotic factor)
Students calculate n(acid) directly from cV without considering how many H⁺ each mole of acid releases. For diprotic H₂SO₄, this halves the moles used and doubles the calculated ΔHn.
Fix: n = moles of H₂O formed = moles of H⁺ available from the limiting reagent. For H₂SO₄: n(H⁺) = 2 × n(H₂SO₄). Always identify how many protons per mole of acid.
"Using H₂SO₄ gives a different ΔHn per mole than HCl"
Students expect that swapping one strong acid for another changes the molar enthalpy of neutralisation.
Fix: The net ionic equation is always H⁺ + OH⁻ → H₂O regardless of which strong acid or base is used. Spectator ions have no effect on ΔH. ΔHn per mole of H₂O is ≈ −57 kJ mol⁻¹ for all strong acid + strong base combinations.
Quick-fire practice · 5 reps +2 XP per reveal
50.0 mL of 0.500 mol L⁻¹ HNO₃ is mixed with 50.0 mL of 0.500 mol L⁻¹ KOH in a polystyrene cup. The temperature rises from 20.5°C to 23.8°C. Density = 1.00 g mL⁻¹.
(a) Calculate the total mass of solution and ΔT.
(b) Calculate q in kJ.
(c) Calculate n(H₂O formed) and ΔHn. Compare to the expected −57 kJ mol⁻¹.
(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.
40.0 mL of 1.50 mol L⁻¹ HCl is mixed with 60.0 mL of 1.00 mol L⁻¹ NaOH. (a) Identify the limiting reagent. (b) How many moles of H₂O will form?
n(NaOH) = 1.00 × 0.0600 = 0.0600 mol OH⁻
Both are equal — neither is in excess (equimolar). n(H₂O formed) = 0.0600 mol.
A student mixes 25.0 mL of 1.00 mol L⁻¹ H₂SO₄ with 50.0 mL of 1.00 mol L⁻¹ NaOH. The temperature rises by 6.8°C. Calculate ΔHn. (Density = 1.00 g mL⁻¹)
n(OH⁻) = 1.00 × 0.0500 = 0.0500 mol — equimolar; n(H₂O) = 0.0500 mol
m = 25.0 + 50.0 = 75.0 g
q = 75.0 × 4.18 × 6.8 = 2131.8 J = 2.132 kJ
ΔHn = −2.132 ÷ 0.0500 = −42.6 kJ mol⁻¹
An experiment shows that neutralising CH₃COOH (ethanoic acid, a weak acid) with NaOH gives ΔHn = −51.6 kJ mol⁻¹. The expected value for a strong acid is −57 kJ mol⁻¹. (a) Is this result consistent with theory? (b) Explain why the value is less negative.
(b) CH₃COOH only partially ionises in water. During neutralisation, energy is consumed to force complete ionisation of the weak acid. This energy absorption reduces the net heat released, giving a less negative ΔHn. The difference (−57 − (−51.6) = 5.4 kJ mol⁻¹) represents the energy required for the ionisation step.
A nurse neutralises 50.0 mL of 1.00 mol L⁻¹ HCl with 50.0 mL of 1.00 mol L⁻¹ NaOH (ΔHn ≈ −57 kJ mol⁻¹). She then repeats with 50.0 mL of 1.00 mol L⁻¹ H₂SO₄ + 50.0 mL of 1.00 mol L⁻¹ NaOH. Predict whether the temperature rise will be greater, less, or the same in the H₂SO₄ experiment. Justify using the net ionic equation concept and mole calculations.
NaOH is the limiting reagent — only 0.0500 mol H₂O forms (same as the HCl experiment).
q = 0.0500 × 57 = 2.85 kJ; m = 100.0 g; ΔT = 2850 ÷ (100.0 × 4.18) = 6.8°C.
The temperature rise is the same as in the HCl experiment. Even though H₂SO₄ has more H⁺ available, NaOH limits the reaction to the same number of moles of H₂O formed. The net ionic equation (H⁺ + OH⁻ → H₂O) is the same for both.
Go back to your Think First response. Now that you've studied neutralisation calorimetry, revisit Germain Hess's 1840 observation at the St Petersburg Academy of Sciences:
- Hess measured ~57 kJ mol⁻¹ for three different strong acid–base pairs. Using your understanding of net ionic equations, explain in one sentence why the identity of the acid and base doesn't matter. Which ions are the actual "reactants"?
- What is releasing the energy? (Answer: the formation of O–H bonds in water molecules as H⁺ and OH⁻ combine — this bond formation is highly exothermic.) Connect this to the bond energy principle you'll study in L06.
- Were you right about H₂SO₄ vs HCl? Using H₂SO₄ with the same volume gives twice the moles of H⁺, so twice the moles of H₂O (if enough base) — but ΔHn per mole of H₂O is the same ≈ −57 kJ mol⁻¹. Hess observed exactly this — the molar value is unchanged.
Pick your answer, then rate your confidence — that tells the system what to drill next.
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.
Wrong: For neutralisation, n = moles of acid solution added (ignoring stoichiometry).
Right: n = moles of H₂O formed = moles of H⁺ that react with OH⁻. For diprotic acids (H₂SO₄), each mole provides 2 mol H⁺, so n(H₂O) = 2 × n(H₂SO₄). Always trace H⁺ moles, not just acid moles.
Wrong: Using H₂SO₄ instead of HCl will change ΔHn per mole of H₂O formed.
Right: ΔHn per mole of H₂O is ≈ −57 kJ mol⁻¹ for all strong acid + strong base combinations. The net ionic equation H⁺ + OH⁻ → H₂O is always the same; spectator ions don't contribute to ΔH.
Q6. 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
Q7. 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
Q8. 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
Show comprehensive answers ▼
Activity 1 — HNO₃ + KOH Calculation
(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.
Multiple Choice
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.
Short Answer Model Answers
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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