Every antacid tablet you've ever swallowed releases heat as it neutralises stomach acid — and the amount of heat released tells you something fundamental about whether the acid and base involved are strong or weak.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A pharmacist is comparing three antacid products for a patient with chronic acid reflux. Product A contains sodium hydroxide solution. Product B contains sodium hydrogen carbonate (baking soda). Product C contains magnesium hydroxide suspension. All three neutralise excess hydrochloric acid in the stomach. The pharmacist measures the temperature change in a simulated stomach acid solution after adding each antacid. Product A produces the largest temperature rise. Products B and C produce noticeably smaller temperature rises — and Product B also produces vigorous bubbling.
Question 1: Why do you think Product A releases the most heat? What is special about NaOH compared to NaHCO₃ and Mg(OH)₂?
Question 2: What is the bubbling from Product B, and why does a neutralisation reaction produce a gas at all?
📚 Core Content
Before any formula is applied, the key question is physical: why does mixing an acid and a base always release heat, and why does the amount of heat depend on whether the acid and base are strong or weak?
Neutralisation is exothermic because forming a covalent O–H bond in liquid water releases more energy than is required to break the bonds that precede it. At the most fundamental level, every neutralisation between a strong acid and a strong base in aqueous solution is the same reaction at the ionic level — regardless of which specific acid or base is used.
HCl, HNO₃, and H₂SO₄ are all fully ionised in solution before the reaction begins, contributing H⁺(aq). NaOH, KOH, and Ca(OH)₂ are fully dissociated, contributing OH⁻(aq). The only reaction that occurs is:
H⁺(aq) + OH⁻(aq) → H₂O(l) ΔHn ≈ −57 kJ/mol
The spectator ions (Na⁺, K⁺, Cl⁻, NO₃⁻) do nothing — they don't react, don't change, and don't contribute to or absorb any energy. Because the reaction is always identical at the ionic level, the enthalpy released is always the same: approximately −57 kJ per mole of water formed, regardless of which strong acid and strong base are combined.
A weak acid or base is not fully ionised in solution before the reaction begins — and the energy needed to complete that ionisation during the neutralisation reaction comes directly from the heat that would otherwise be released, reducing the net enthalpy measured.
When a weak acid such as acetic acid (CH₃COOH) is mixed with a strong base such as NaOH, the neutralisation cannot simply be described as H⁺ + OH⁻ → H₂O. In a 0.1 M CH₃COOH solution, only about 1.3% of the acetic acid molecules have ionised — the other 98.7% remain as intact CH₃COOH molecules.
When NaOH is added, the OH⁻ ions rapidly react with the small available supply of H⁺ ions, driving the acetic acid equilibrium further to the right: CH₃COOH ⇌ H⁺ + CH₃COO⁻. This additional ionisation is endothermic — energy must be absorbed to break the O–H bond in CH₃COOH. This energy comes directly from the heat that the H⁺ + OH⁻ → H₂O reaction releases. The net heat measured is therefore:
(heat from H⁺ + OH⁻ → H₂O) − (energy consumed to ionise CH₃COOH) = |ΔHn| < 57 kJ/mol
The same logic applies to weak bases: NH₃ must accept a proton from the reaction rather than already having it dissociated, which requires energy input. The weaker the acid or base, the less ionised it is before the reaction, and the more energy the ionisation step consumes — and therefore the less heat is released overall.
Energy budget: weak acid ionisation consumes energy that would otherwise be released as heat
The experimental measurement of enthalpy of neutralisation relies on a simple energy transfer principle — all heat released by the reaction is assumed to be absorbed by the surrounding solution, and that absorbed heat is calculated from the temperature change.
In a school calorimetry experiment, a polystyrene (foam) cup is used as an insulated vessel. The procedure is:
A foam cup calorimeter reduces heat loss, but the cup, lid opening, and thermometer still create systematic error.
Every school calorimetry experiment gives a ΔHn value that is slightly less exothermic than the theoretical value — understanding exactly why, and what can be done about it, is a standard HSC extended response target.
The fundamental assumption is that all heat released by the reaction is absorbed by the solution. In reality, heat is lost to several other pathways. All of these errors produce the same systematic effect: the experimental |ΔHn| is always less than the theoretical value — never more.
| Source of error | Effect on ΔT | Effect on |ΔHn| | Improvement |
|---|---|---|---|
| Heat loss through calorimeter walls and open top | Smaller ΔT (Tmax not fully reached) | Underestimated (less negative) | Add a lid; nest two foam cups; use Dewar flask |
| Heat absorbed by thermometer and cup | Slightly smaller ΔT | Slightly underestimated | Use a thin temperature probe with low heat capacity |
| Density/SHC assumption differs from pure water | Small error in m or c | Small systematic error | Measure actual density; use tabulated SHC values |
| Incomplete mixing | Smaller ΔT (not all reacted) | Underestimated | Swirl thoroughly; use magnetic stirrer |
| Thermometer precision (±0.1°C or ±0.5°C) | Random error in ΔT | Random error in ΔHn | Use a calibrated digital probe (±0.01°C); repeat and average |
Advanced improvement — cooling curve extrapolation: Use a temperature probe and data logger to record temperature continuously before and after mixing. Plot temperature vs time; extrapolate the cooling portion of the curve back to the moment of mixing to find the true Tmax that would have been reached without heat loss. This is more accurate than simply reading the highest temperature from the thermometer.
Reading only the highest measured temperature underestimates the true temperature rise. Extrapolation improves the value of ΔT and therefore ΔH.
Gaviscon (sodium alginate + sodium bicarbonate), Mylanta (aluminium hydroxide + magnesium hydroxide), and Rennie (calcium carbonate + magnesium carbonate) all neutralise excess HCl in the stomach — but they feel different because they release heat at different rates and with different magnitudes. This is direct evidence of ΔHn differences between acid-base combinations.
Rennie contains CaCO₃ — an acid + carbonate reaction (Pattern 2 from L02). This is not a simple H⁺ + OH⁻ → H₂O reaction; it involves CaCO₃ dissolving and CO₃²⁻ accepting protons from HCl, which produces CO₂ gas (the fizzing). The enthalpy is distributed across multiple bond-breaking and bond-forming steps — the net |ΔHn| is lower than for a strong + strong combination. The same experiment you perform in the lab — measuring ΔT with a foam cup — can be used to compare the neutralising capacity and enthalpy of each antacid product, giving quantitative data that a pharmacist or pharmaceutical chemist would use to compare products.
"ΔHn ≈ −57 kJ/mol for all neutralisation reactions." — This value applies only to strong acid + strong base. Any combination involving a weak species gives |ΔHn| < 57 kJ/mol because energy is consumed in the ionisation step during the reaction.
"Use only the mass of the acid in q = mcΔT." — The total mass of solution (acid + base combined) must be used. All heat produced is absorbed by the entire combined solution, not just the acid component.
"ΔT = Tinitial − Tfinal." — Always ΔT = Tfinal − Tinitial. For an exothermic reaction, Tfinal > Tinitial, so ΔT is positive. The negative sign in ΔHn = −q/n is applied separately and deliberately.
"The experimental ΔHn can be more negative than the theoretical value." — All systematic errors in this experiment (heat loss, thermometer heat capacity, etc.) make ΔHn less negative than theoretical, never more negative. If your experimental value is more negative than −57 kJ/mol, check your calculation for errors.
"Human error caused the discrepancy." — This is not an acceptable scientific answer. Name the specific source: "heat loss through the walls of the foam cup calorimeter to the surroundings" → specific improvement: "add a lid and use a better-insulated vessel."
✏️ Worked Examples
GIVEN: V(HCl) = 50.0 mL; V(NaOH) = 50.0 mL; c(HCl) = c(NaOH) = 1.00 mol/L; Ti = 21.5°C; Tf = 28.0°C; density = 1.00 g/mL; SHC = 4.18 J g⁻¹ °C⁻¹ | FIND: q (J), n(H₂O) (mol), ΔHn (kJ/mol)
Calculate q: Total volume = 50.0 + 50.0 = 100.0 mL → m = 100.0 g. ΔT = 28.0 − 21.5 = 6.5°C. q = mcΔT = 100.0 × 4.18 × 6.5 = 2717 J
Calculate n(H₂O): n(HCl) = 1.00 × 0.0500 = 0.0500 mol; n(NaOH) = 1.00 × 0.0500 = 0.0500 mol. Equal moles — neither is in excess. n(H₂O) = 0.0500 mol
Calculate ΔHn: ΔHn = −q/n = −2717 J / 0.0500 mol = −54 340 J/mol ÷ 1000 = −54.3 kJ/mol
Comparison: theoretical = −57 kJ/mol. Experimental value is less negative — consistent with heat loss to surroundings from the foam cup calorimeter during the experiment.
Answers: (a) q = 2717 J (b) n(H₂O) = 0.0500 mol (c) ΔHn = −54.3 kJ/mol
Using the same setup, the NaOH is replaced with 50.0 mL of 1.00 mol/L NH₃ (ammonia). The maximum temperature recorded is 26.1°C (Ti = 21.5°C). Calculate ΔHn and explain why it differs from −54.3 kJ/mol.
ΔT = 26.8 − 21.5 = 5.3°C. m = 100.0 g. q = 100.0 × 4.18 × 5.3 = 2215.4 J. n(H₂O) = 0.0500 mol.
ΔHn = −2215.4 / 0.0500 = −44 308 J/mol ÷ 1000 = −44.3 kJ/mol
Explanation (three-step mechanism):
(1) In 1.00 mol/L CH₃COOH solution, only approximately 0.4% of acetic acid molecules are ionised before the reaction (Ka = 1.8 × 10⁻⁵). The other ~99.6% exist as intact CH₃COOH molecules.
(2) When NaOH is added, OH⁻ reacts with the available H⁺, driving the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ further to the right. This further ionisation is endothermic — energy must be absorbed from the surroundings to break the O–H bond in CH₃COOH.
(3) This energy comes directly from the heat released by H⁺ + OH⁻ → H₂O, reducing the net heat available to warm the solution. The result: a smaller ΔT, smaller q, and a less negative ΔHn (−44.3 vs −54.3 kJ/mol).
Answers: (a) ΔHn = −44.3 kJ/mol (b) CH₃COOH is partially ionised before reaction; energy consumed to complete ionisation during the reaction reduces net heat released → less negative ΔHn than HCl + NaOH.
n(HNO₃) = 2.00 × 0.0300 = 0.0600 mol H⁺. n(NaOH) = 1.00 × 0.0500 = 0.0500 mol OH⁻.
OH⁻ is the limiting reagent — 0.0500 mol OH⁻ < 0.0600 mol H⁺. n(H₂O) = 0.0500 mol. Excess H⁺ = 0.0100 mol (does not react further).
Total volume = 30.0 + 50.0 = 80.0 mL → m = 80.0 g. ΔT = 26.4 − 20.0 = 6.4°C.
q = mcΔT = 80.0 × 4.18 × 6.4 = 2140 J
ΔHn = −q/n = −2140 / 0.0500 = −42 800 J/mol ÷ 1000 = −42.8 kJ/mol
Source of error: Heat loss through the walls and open top of the foam cup calorimeter to the surrounding atmosphere. As the solution warms, heat transfers from the warm solution to the cooler surroundings before the thermometer reading reaches the true maximum. This means the recorded Tmax is lower than the true Tmax → smaller ΔT → smaller q → less negative ΔHn.
Improvement: Add a tight-fitting lid to the foam cup to reduce heat loss through the top surface. A more significant improvement: use a temperature probe with a data logger to record temperature every second and extrapolate the cooling curve back to the moment of mixing, recovering the true Tmax without depending on the observed peak temperature.
Answers: (a) NaOH is limiting (0.0500 mol OH⁻); n(H₂O) = 0.0500 mol (b) q = 2140 J (c) ΔHn = −42.8 kJ/mol (d) Heat loss through foam cup walls → lower Tmax → smaller ΔT → less negative ΔHn. Improvement: add lid + use data logger with cooling curve extrapolation.
🧪 Activities
| Experiment | Acid (50.0 mL, 1.00 mol/L) | Base (50.0 mL, 1.00 mol/L) | Tmax (°C) | q (J) | n(H₂O) (mol) | ΔHn (kJ/mol) |
|---|---|---|---|---|---|---|
| 1 | HCl (strong) | NaOH (strong) | 27.4 | Calculate | Calculate | Calculate |
| 2 | CH₃COOH (weak) | NaOH (strong) | 25.8 | Calculate | Calculate | Calculate |
| 3 | HCl (strong) | NH₃ (weak) | 25.1 | Calculate | Calculate | Calculate |
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
❓ Multiple Choice
1. A student mixes equal volumes of equal concentration HCl and KOH. ΔHn is measured as −55.8 kJ/mol; the theoretical value is −57 kJ/mol. Which explanation best accounts for the difference?
2. Which combination of acid and base would produce the LEAST negative molar enthalpy of neutralisation?
3. In a neutralisation calorimetry experiment, a student uses only the mass of the acid solution (not the total combined mass) in q = mcΔT. What effect does this have on the calculated value of ΔHn?
4. 40.0 mL of 0.500 mol/L H₂SO₄ is mixed with 60.0 mL of 0.500 mol/L NaOH. What is n(H₂O) formed?
5. A student's experimental ΔHn for HCl + NaOH is −54.1 kJ/mol. Which of the following would be the MOST effective single improvement to bring this value closer to the theoretical −57 kJ/mol?
A student's experimental ΔHn for HCl + NaOH is −54.1 kJ/mol. Select the option that would be the MOST effective single improvement to bring this value closer to the theoretical −57 kJ/mol?
✍️ Short Answer
6. Explain why the molar enthalpy of neutralisation is approximately −57 kJ/mol for any strong acid + strong base combination, regardless of which specific acid and base are used. Write the net ionic equation to support your explanation. 3 MARKS
7. 25.0 mL of 2.00 mol/L HCl is mixed with 25.0 mL of 2.00 mol/L NaOH. Ti = 19.5°C; Tmax = 32.7°C. Calculate: (a) q in joules, (b) n(H₂O) formed, (c) ΔHn in kJ/mol. Comment on how your value compares to the theoretical −57 kJ/mol and suggest one reason for the difference. 5 MARKS
8. Real-World Application — Antacid Comparison: A pharmacist tests three antacids against 50.0 mL of 1.00 mol/L HCl (Ti = 20.0°C):
• Product A: NaOH (strong base) — Tmax = 26.8°C
• Product B: Mg(OH)₂ (weak base) — Tmax = 24.1°C
• Product C: NaHCO₃ (acid + carbonate) — Tmax = 23.3°C with vigorous bubbling
(a) Explain why Product A releases the most heat. Write the net ionic equation for this reaction. (2 marks)
(b) Explain why Product B releases less heat than Product A, despite both being bases that neutralise HCl. (2 marks)
(c) What is the gas produced by Product C, and why does this reaction produce less heat? (2 marks) 6 MARKS
Go back to your Think First predictions at the top of this lesson.
Exp 1 (HCl + NaOH): ΔT = 27.4 − 20.0 = 7.4°C. m = 100.0 g. q = 100.0 × 4.18 × 7.4 = 3093.2 J. n(H₂O) = 1.00 × 0.0500 = 0.0500 mol. ΔHn = −3093.2/0.0500 = −61 864 J/mol ÷ 1000 = −61.9 kJ/mol.
Exp 2 (CH₃COOH + NaOH): ΔT = 25.8 − 20.0 = 5.8°C. q = 100.0 × 4.18 × 5.8 = 2424.4 J. n(H₂O) = 0.0500 mol. ΔHn = −2424.4/0.0500 ÷ 1000 = −48.5 kJ/mol.
Exp 3 (HCl + NH₃): ΔT = 25.1 − 20.0 = 5.1°C. q = 100.0 × 4.18 × 5.1 = 2131.8 J. n(H₂O) = 0.0500 mol. ΔHn = −2131.8/0.0500 ÷ 1000 = −42.6 kJ/mol.
Ranking: Exp 1 (most exothermic) > Exp 2 > Exp 3 (least exothermic). This matches theory: strong+strong > weak acid+strong base > strong acid+weak base.
Exp 1 discrepancy from −57: Note — Exp 1 gives −61.9 kJ/mol, which is MORE negative than −57. This is unusual and suggests the student's data may have experimental variation. In a real experiment, the value should be less than −57 due to heat loss. If the calculated value exceeds −57, check for errors (e.g. ΔT measured incorrectly, or the initial temperature was lower than recorded). For the purpose of this activity, calculate as instructed and note the discrepancy.
Exp 2 less than Exp 1 (3 steps): (1) CH₃COOH is only ~0.4% ionised at 1.00 mol/L before the reaction — most exists as intact molecules. (2) When NaOH is added, the OH⁻ drives the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻ further right; this endothermic ionisation step absorbs energy from the solution. (3) The energy consumed in ionisation reduces the net heat available to raise the temperature → smaller ΔT → smaller q → less negative ΔHn.
1. Glass beaker: Yes, contributes. Glass is a much better heat conductor than foam — heat is lost rapidly through the glass walls to the surroundings → lower Tmax → smaller ΔT → smaller q → less negative ΔHn. This is the most likely cause of the large discrepancy (−38.2 vs −57 kJ/mol).
2. Wrong ΔT order: Yes, contributes — but in a different way. If ΔT = Ti − Tf (negative), then q would be negative, giving ΔHn = −(negative)/n = positive (endothermic), which is wrong in direction. This would not give −38.2 kJ/mol — it would give a positive value. So this error would produce a completely wrong result, not just a different magnitude.
3. Half mass (acid only): Yes, contributes. Using m = 50.0 g instead of m = 100.0 g makes q half as large as it should be. Smaller q → ΔHn = −q/n gives a value that is half the magnitude → less negative. For example, if true q = 2140 J: correct ΔHn = −42.8 kJ/mol; with half mass: q = 1070 J → ΔHn = −21.4 kJ/mol (even less negative).
4. Waited 30s: Yes, contributes. By 30 seconds after mixing, the solution has already begun cooling — the recorded temperature is lower than the true Tmax → smaller ΔT → smaller q → less negative ΔHn. This is the error that a cooling-curve extrapolation technique specifically corrects.
5. Wrong volume unit (L vs mL): Yes, contributes — catastrophically. n(HCl) = 1.00 × 50.0 = 50.0 mol (instead of 0.0500 mol). ΔHn = −q/50.0 = −2140/50.0 = −42.8 J/mol = −0.0428 kJ/mol. This produces a value near zero, far worse than −38.2 kJ/mol. The student would recognise this as wrong immediately.
1. C — HCl is a strong acid and KOH is a strong base — both fully ionised. The theoretical net ionic equation is H⁺ + OH⁻ → H₂O (−57 kJ/mol). The small discrepancy (−55.8 vs −57) is consistent with heat loss to surroundings — a systematic error that always reduces |ΔHn|. Options A and B are wrong — both HCl and KOH are strong and require no ionisation energy.
2. D — The least negative ΔHn occurs when the most energy is consumed in ionisation steps. Options A and B (strong + strong) give ≈ −57 kJ/mol. Option C (weak acid + strong base) gives less — ionisation of CH₃COOH consumes energy. Option D (strong acid + weak base) gives less — ionisation of NH₃ to accept the proton consumes energy. Between C and D: CH₃COOH has Ka ≈ 1.8 × 10⁻⁵; NH₃ has Kb ≈ 1.8 × 10⁻⁵ — similar degree of weakness. In practice, D gives a comparable or slightly less negative ΔHn than C. D is selected as the answer here — if both were weak, it would be even less negative, but that option is not offered.
3. B — Using only the acid mass (e.g. 50 g instead of 100 g) gives a smaller m → smaller q = mcΔT → smaller |q|. Dividing smaller |q| by the same n → smaller |ΔHn| = less negative. The heat produced is absorbed by the entire combined solution, not just the acid, so using only the acid mass underestimates q and gives a ΔHn less negative than the true value.
4. B — n(H₂SO₄) = 0.500 × 0.0400 = 0.0200 mol → n(H⁺) = 2 × 0.0200 = 0.0400 mol. n(NaOH) = 0.500 × 0.0600 = 0.0300 mol OH⁻. Limiting reagent = NaOH (0.0300 mol OH⁻ < 0.0400 mol H⁺). n(H₂O) = n(OH⁻ reacted) = 0.0300 mol. Option A is a distractor — it lists 0.0200 and the label says "NaOH is limiting" but gives the wrong value.
5. C — The main source of error is heat loss, not thermometer precision. The most effective improvement targets the dominant error: adding a lid reduces heat loss through the top surface; data logger + cooling curve extrapolation finds the true Tmax despite heat loss during the measurement period. Together these address the dominant source of discrepancy. Option A improves precision but does not address heat loss. Option B does not help. Option D worsens the calculation.
Q6 (3 marks): Strong acids and strong bases are both fully ionised in solution before the reaction occurs — all H⁺ and OH⁻ ions are already present as free ions [1]. Regardless of which strong acid or base is used, the only species that react are H⁺ and OH⁻: H⁺(aq) + OH⁻(aq) → H₂O(l) [1]. The spectator ions (e.g. Na⁺, Cl⁻, K⁺, NO₃⁻) do not participate — the same bond is formed and the same energy is released each time, giving ΔHn ≈ −57 kJ/mol for any strong + strong combination [1].
Q7 (5 marks): (a) m = 50.0 g; ΔT = 32.7 − 19.5 = 13.2°C; q = 50.0 × 4.18 × 13.2 = 2758.8 J [1]. (b) n(HCl) = 2.00 × 0.0250 = 0.0500 mol; n(NaOH) = 0.0500 mol; equal, so n(H₂O) = 0.0500 mol [1]. (c) ΔHn = −2758.8/0.0500 = −55 176 J/mol ÷ 1000 = −55.2 kJ/mol [1]. The value (−55.2 kJ/mol) is less negative than the theoretical −57 kJ/mol [1]. This is consistent with heat loss from the foam cup calorimeter to the surroundings — the measured Tmax is lower than the true maximum, making q and therefore |ΔHn| smaller [1].
Q8 (6 marks): (a) NaOH is a strong base — fully dissociated in solution. All OH⁻ ions are immediately available to react with H⁺ from HCl [1]. Net ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l) — no ionisation energy is consumed; the full ΔHn ≈ −57 kJ/mol is released [1]. (b) Mg(OH)₂ is a sparingly soluble weak base — it does not fully dissociate in solution before reacting [1]. Energy must be consumed to dissolve and further ionise Mg(OH)₂ during the reaction; this energy comes from the heat released by H⁺ + OH⁻ → H₂O, reducing the net heat available and giving a less negative ΔHn [1]. (c) The gas is CO₂ [1]. NaHCO₃ + HCl → NaCl + H₂O + CO₂ (acid + hydrogen carbonate reaction). CO₂ is produced when the intermediate H₂CO₃ immediately decomposes. This reaction involves multiple bond-breaking and bond-forming steps beyond simple H⁺ + OH⁻ → H₂O, consuming additional energy — the net heat released per mole of acid neutralised is lower than for a simple strong base neutralisation [1].
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