An instant cold pack gets icy the moment you crack it — no freezer, no ice. A bag of white powder and water, and suddenly your ankle is numb. Dissolution calorimetry measures exactly how much energy moves when an ionic lattice breaks apart and its ions disappear into water.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
You've seen the ads: an athlete rolls an ankle, a trainer pulls a cold pack from a bag, cracks it, shakes it — and it goes cold instantly. No freezer. No ice. Just a bag of white powder and water mixing together. The powder is ammonium nitrate.
Why does mixing it with water make the pack cold? And why does NaOH dissolved in water do the exact opposite — it gets so hot the container can warp? They're both ionic solids dissolving in water. What's the difference?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
📚 Core Content
The setup for dissolution calorimetry is very similar to the neutralisation calorimeter from L03 — a polystyrene cup holds a known mass of water, and the ionic solid is added directly into it.
Procedure:
By Lesson 4, you have used q = mcΔT in three different experimental contexts. The formula is the same each time — but what 'm' and 'n' represent, and whether ΔH can be positive or negative, differs for each type.
| Feature | Combustion (L02) | Neutralisation (L03) | Dissolution (L04) |
|---|---|---|---|
| Apparatus | Copper calorimeter, spirit burner | Polystyrene cup, thermometer | Polystyrene cup, thermometer |
| 'm' in q = mcΔT | Mass of water in calorimeter | Total mass of combined solutions (acid + base) | Total mass of solution (water + dissolved solid) |
| 'n' in ΔH = −q/n | Moles of fuel burned (n = m/M) | Moles of H₂O formed (= limiting reagent moles) | Moles of ionic solid dissolved (n = m/M) |
| Typical sign of ΔH | Always negative (exothermic) | Usually negative (exothermic — heat released) | Either sign — exo (NaOH) or endo (NH₄NO₃) |
| Main source of error | Heat loss to atmosphere and copper | Heat loss through cup walls to air | Incomplete dissolution; heat exchange with surroundings |
🔬 Worked Examples
5.35 g of ammonium chloride (NH₄Cl, M = 53.49 g mol⁻¹) is dissolved in 100.0 g of water in a polystyrene cup. The temperature drops from 22.5°C to 18.1°C. Calculate the molar enthalpy of dissolution of NH₄Cl.
2.00 g of NaOH (M = 40.00 g mol⁻¹) is dissolved in 200.0 g of water in a polystyrene cup. The temperature rises from 20.0°C to 22.6°C. Calculate ΔHsoln.
🧪 Activities
a State whether this dissolution is exothermic or endothermic. Justify using the temperature change.
b Calculate ΔT, the total mass of solution, and q in kJ.
c Calculate n(NH₄NO₃) and then ΔHsoln. Compare to the accepted value of +25.7 kJ mol⁻¹.
Show your full working before revealing answers:
Show full working in your workbook.
| Ionic solid | Temperature of solution | Exothermic or endothermic? | Sign of ΔHsoln | Real-world relevance |
|---|---|---|---|---|
| NH₄NO₃ | Falls | Your answer | Your answer | Your answer |
| NaOH | Rises | Your answer | Your answer | Your answer |
| KNO₃ | Falls | Your answer | Your answer | Your answer |
| CaCl₂ | Rises | Your answer | Your answer | Your answer |
Type your table answers and bonus response below:
Complete the table and bonus in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: Ionic compounds conduct electricity in the solid state because they contain charged ions.
Right: Ionic compounds only conduct electricity when molten or dissolved in water. In the solid state, the ions are locked in a fixed lattice and cannot move. Conductivity requires mobile charge carriers, which are only present when the lattice breaks down.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. 8.01 g of NH₄NO₃ (M = 80.05 g mol⁻¹) is dissolved in 100.0 g of water. The temperature falls from 25.0°C to 20.3°C. Calculate the molar enthalpy of dissolution. State whether the dissolution is exothermic or endothermic and explain how the two-step energy model accounts for this. 5 MARKS
Type your answer — show all working:
Show full working in your workbook.
7. 3.00 g of NaOH (M = 40.00 g mol⁻¹) is dissolved in 150.0 g of water. The temperature rises from 21.0°C to 24.1°C.
(a) Calculate the molar enthalpy of dissolution of NaOH. (3 marks)
(b) State whether the dissolution is exothermic or endothermic and explain in terms of lattice energy and hydration energy. (2 marks)
5 MARKS
Type your full answer:
Answer in your workbook.
8. A student performs three calorimetry experiments in the same polystyrene cup:
Experiment A — burns 0.46 g of ethanol (M = 46.07 g mol⁻¹) under a copper calorimeter holding 200 g of water; temperature rises 8.4°C.
Experiment B — mixes 50 mL of 1.00 mol L⁻¹ HCl with 50 mL of 1.00 mol L⁻¹ NaOH; temperature rises 6.6°C.
Experiment C — dissolves 3.74 g of NH₄Cl (M = 53.49 g mol⁻¹) in 150.0 g water; temperature falls 2.2°C.
For each experiment, identify: (i) what 'm' is in q = mcΔT, (ii) what 'n' is in ΔH = −q/n, and (iii) whether the experimental ΔH will be an underestimate or overestimate of the true value, and why. 6 MARKS
Type your answer below:
Answer in your workbook.
Go back to your Think First response. Now that you've studied dissolution calorimetry:
Type your reflection below:
Write your reflection in your book.
(a) Temperature fell → endothermic dissolution. ΔHsoln will be positive.
(b) ΔT = 18.2 − 22.5 = −4.3 K; m = 95.0 + 5.00 = 100.0 g; q = 100.0 × 4.18 × (−4.3) = −1797.4 J = −1.797 kJ
(c) n = 5.00 ÷ 80.05 = 0.06247 mol; ΔHsoln = −(−1.797) ÷ 0.06247 = +28.8 kJ mol⁻¹. Accepted ≈ +25.7 kJ mol⁻¹; the higher experimental value suggests some heat was absorbed from the surroundings (cup or atmosphere) during the slow dissolution, making the measured temperature drop smaller and the apparent ΔHsoln larger.
NH₄NO₃: Endothermic | ΔHsoln positive | cold packs / sports injury packs
NaOH: Exothermic | ΔHsoln negative | lab safety hazard (solution heats up dangerously); some industrial heating applications
KNO₃: Endothermic | ΔHsoln positive | occasionally used in improvised cooling; also found in fertilisers where the cooling effect is observed when dissolved in soil water
CaCl₂: Exothermic | ΔHsoln negative | road de-icing (generates heat to help melt ice in addition to lowering freezing point)
Bonus: For NH₄NO₃ and KNO₃, lattice energy > hydration energy. This suggests both compounds have relatively strong ionic lattices compared to the strength of ion-water interactions — the electrostatic forces holding the lattice together are not fully compensated by the energy released when the separated ions are hydrated.
1. C — Temperature drop → endothermic → lattice energy exceeds hydration energy. The dissolving process absorbed more from the solution than hydration returned.
2. A — m = 150.0 + 4.00 = 154.0 g; q = 154.0 × 4.18 × 3.5 = 2253.0 J = 2.253 kJ; n = 4.00 ÷ 111.1 = 0.03600 mol; ΔHsoln = −2.253 ÷ 0.03600 = −62.6 kJ mol⁻¹. Option C is the common error of using only 150 g (forgetting to add the solute mass).
3. B — Exothermic dissolution occurs when hydration energy exceeds lattice energy. For NaOH, the strong hydration of both Na⁺ and OH⁻ ions releases more energy than is required to break the NaOH lattice.
4. C — 'm' is the total mass of solution: water + dissolved solid. Using only the water mass underestimates m and gives an inflated ΔHsoln.
5. B — Temperature rise → the solution gained energy → the dissolution process released energy → exothermic → ΔHsoln is negative. The −q/n formula: q is positive (ΔT > 0), so −q/n is negative.
Q6 (5 marks): ΔT = 20.3 − 25.0 = −4.7 K [1]; m = 100.0 + 8.01 = 108.01 g; q = 108.01 × 4.18 × (−4.7) = −2122 J = −2.122 kJ [1]; n = 8.01 ÷ 80.05 = 0.1001 mol [1]; ΔHsoln = −(−2.122) ÷ 0.1001 = +21.2 kJ mol⁻¹ [1]. Endothermic — the temperature fell, meaning the solution lost energy to the dissolution process. Step 1 (lattice dissociation) requires more energy than Step 2 (hydration) releases — the lattice energy of NH₄NO₃ exceeds its hydration energy, giving a net energy absorption [1].
Q7 (5 marks): (a) ΔT = 24.1 − 21.0 = 3.1 K; m = 150.0 + 3.00 = 153.0 g; q = 153.0 × 4.18 × 3.1 = 1981.3 J = 1.981 kJ [1]; n = 3.00 ÷ 40.00 = 0.0750 mol [1]; ΔHsoln = −1.981 ÷ 0.0750 = −26.4 kJ mol⁻¹ [1]. (b) Exothermic — ΔT is positive, meaning the solution gained energy from the dissolution [1]. This indicates that the hydration energy released when Na⁺ and OH⁻ ions are surrounded by polar water molecules (ion-dipole interactions) is greater than the lattice energy required to separate the ions in the NaOH crystal — net energy is released to the solution [1].
Q8 (6 marks — 2 per experiment):
Experiment A (combustion): (i) m = 200 g (mass of water in copper calorimeter — not the ethanol, which is outside) [½]; (ii) n = 0.46 ÷ 46.07 = 0.00998 mol of ethanol burned [½]; (iii) Underestimate of |ΔHc| — the copper calorimeter conducts heat to the surroundings; q measured is less than q actually released, so ΔHc (negative) appears less negative than the true value [1].
Experiment B (neutralisation): (i) m = (50 + 50) × 1.00 = 100 g (total solution) [½]; (ii) n = 1.00 × 0.0500 = 0.0500 mol H₂O formed [½]; (iii) Underestimate of |ΔHn| — the polystyrene cup loses some heat to the air; the measured temperature rise is less than if no heat escaped, giving a less negative ΔHn [1].
Experiment C (dissolution): (i) m = 150.0 + 3.74 = 153.74 g [½]; (ii) n = 3.74 ÷ 53.49 = 0.06993 mol NH₄Cl [½]; (iii) Underestimate of ΔHsoln (positive value appears smaller than true value) — heat from the surroundings enters the cup slightly, partially offsetting the temperature drop. The measured ΔT is less negative than the true value, making q less negative and ΔHsoln less positive than the true value [1].
Calorimetry — Dissolution of Ionic Substances
Tick when you've finished all activities and checked your answers.