Calorimetry — Dissolution of Ionic Substances
In 1959, 3M Corporation chemists developing the first commercial instant cold pack measured the dissolution enthalpy of ammonium nitrate (NH₄NO₃) at +25.7 kJ mol⁻¹ — highly endothermic. The product launched in 1961 and could drop the pack temperature to 2°C within 30 seconds using just 35 g of salt and 120 mL of water. Dissolution calorimetry is what made that engineering possible: knowing exactly how much energy is absorbed when the ionic lattice breaks apart.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Key Facts
- Lattice energy (endothermic) and hydration energy (exothermic) as the two steps of dissolution
- Endothermic examples: NH₄NO₃, NH₄Cl, KNO₃ (cold packs)
- Exothermic examples: NaOH, CaCl₂ (warms solution)
Concepts
- Why ΔHsoln sign follows from the temperature change direction
- Why hydration energy can exceed lattice energy, making dissolution exothermic
- The differences in m and n across combustion, neutralisation, and dissolution
Skills
- Calculate q and ΔHsoln from dissolution calorimetry data
- Assign correct sign to ΔHsoln from experimental temperature change
- Identify the correct m and n for all three calorimetry types
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:
- Measure a known mass of distilled water into a polystyrene cup and record the initial temperature
- Weigh the ionic solid accurately on a balance
- Add the solid to the water; stir continuously to ensure complete dissolution
- Record the maximum (or minimum) temperature reached
- Calculate: total mass of solution = mass of water + mass of solid
In dissolution calorimetry: q = mcΔT where m = mass of water; ΔHdiss = −q/n where n = moles of ionic solid dissolved. ΔHdiss can be positive (endothermic, e.g. NH₄NO₃) or negative (exothermic, e.g. NaOH) depending on whether lattice energy or hydration enthalpy dominates.
Pause — copy the highlighted definition into your book before moving on.
Explain it: A student forgets to add the mass of the dissolved solid when calculating m in q = mcΔT. Will their calculated ΔHsoln be too high or too low? Explain in 2–3 sentences.
We just saw the dissolution calorimeter setup and how ΔHdiss is calculated. That raises a question: how does dissolution calorimetry compare with combustion and neutralisation from previous lessons? This card answers it → by systematically comparing ‘m’, ‘n’, and the possible sign of ΔH across all three methods.
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 |
All three calorimetry methods use q = mcΔT, but differ in ‘m’ and ‘n’: combustion (m = water, n = fuel), neutralisation (m = combined solution, n = H₂O formed), dissolution (m = water, n = solute). Only dissolution can give positive or negative ΔH — combustion and neutralisation (strong acid + base) are always negative.
Add the highlighted point to your notes before the check below.
Match it: Match each calorimetry type to its correct description of 'n'.
- Combustion
- Neutralisation
- Dissolution
- mol of H₂O formed
- mol of fuel burned
- mol of ionic solid dissolved
Worked examples · reveal as you go
Endothermic Dissolution: Cold Pack (NH₄Cl). 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.
ΔT = Tfinal − Tinitial = 18.1 − 22.5 = −4.4°C = −4.4 K
m = mass of water + mass of NH₄Cl = 100.0 + 5.35 = 105.35 g
q = mcΔT = 105.35 × 4.18 × (−4.4) = −1934 J = −1.934 kJ
n = m ÷ M = 5.35 ÷ 53.49 = 0.1000 mol
ΔHsoln = −q ÷ n = −(−1.934) ÷ 0.1000 = +19.3 kJ mol⁻¹
Exothermic Dissolution: NaOH Warming. 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.
ΔT = 22.6 − 20.0 = +2.6 K
m = 200.0 + 2.00 = 202.0 g
q = mcΔT = 202.0 × 4.18 × 2.6 = 2195.7 J = 2.196 kJ
n = 2.00 ÷ 40.00 = 0.0500 mol
ΔHsoln = −q ÷ n = −2.196 ÷ 0.0500 = −43.9 kJ mol⁻¹
Ammonium nitrate (NH&sub4;NO&sub3;) dissolves in water — you have seen this in instant cold packs. Predict: is the dissolution exothermic or endothermic? Will ΔH be positive or negative?
How close was your prediction?
Excellent — the cold pack connection is a great memory hook.
Not all dissolution reactions are exothermic. When lattice energy > hydration energy, the dissolution is endothermic.
Formula Reference — This Lesson
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Use the PDF for classwork, homework or revision. Includes key ideas, activities, questions and success-criteria proof.
Common errors · the 3 traps that cost marks
"n is moles of water, not moles of solid"
Students who just finished L03 (neutralisation) reflexively use n = mol H₂O formed for the dissolution calculation.
Fix: In dissolution, n = moles of the ionic solid dissolved (n = m/M for the solid). Moles of water is never n for dissolution. Using the wrong n will give a ΔHsoln that is 50–200 times off.
"m = mass of water only"
Students use only the mass of water (e.g. 100.0 g) and forget to add the mass of the dissolved solid.
Fix: In dissolution calorimetry, m = mass of water + mass of solute. Once dissolved, the solid becomes part of the solution that absorbs or releases heat. Omitting the solute mass underestimates q and therefore underestimates |ΔHsoln|.
"Temperature drop must mean negative ΔHsoln"
Students see the solution cool down and assign ΔHsoln a negative value, confusing the direction of energy flow.
Fix: If the solution cools, heat moved from the solution into the dissolving process — the dissolution is endothermic → ΔHsoln > 0 (positive). The −q/n formula handles this automatically: a negative q gives a positive ΔHsoln. Trust the formula but verify with your experimental observation.
Quick-fire practice · 5 reps +2 XP per reveal
5.00 g of ammonium nitrate (NH₄NO₃, M = 80.05 g mol⁻¹) is dissolved in 95.0 g of water. The temperature falls from 22.5°C to 18.2°C.
(a) State whether this dissolution is exothermic or endothermic. Justify using the temperature change.
Using the same NH₄NO₃ scenario: calculate ΔT, the total mass of solution, and q in kJ.
m = 95.0 + 5.00 = 100.0 g
q = 100.0 × 4.18 × (−4.3) = −1797.4 J = −1.797 kJ
Continue: calculate n(NH₄NO₃) and then ΔHsoln. The accepted value is +25.7 kJ mol⁻¹ — is your answer an overestimate or underestimate? Why?
ΔHsoln = −(−1.797) ÷ 0.06247 = +28.8 kJ mol⁻¹
The experimental value (+28.8) exceeds the accepted value (+25.7 kJ mol⁻¹) — an overestimate. Since this is endothermic, heat from the surroundings (the cup/air) entered the system, making the temperature drop smaller than it should be. The measured ΔT is less negative → q less negative → ΔHsoln appears larger (more positive) than the true value.
Classify each ionic substance as causing an endothermic or exothermic dissolution. What sign is ΔHsoln for each?
(a) NH₄NO₃ — cold packs go cold (b) NaOH — container can warp from heat (c) KNO₃ — solution cools (d) CaCl₂ — used in road de-icing with heat release
(b) NaOH: Exothermic | ΔHsoln negative (hydration energy > lattice energy)
(c) KNO₃: Endothermic | ΔHsoln positive
(d) CaCl₂: Exothermic | ΔHsoln negative
Three-experiment challenge: Identify (i) what 'm' is and (ii) what 'n' is in q = mcΔT and ΔH = −q/n for each:
Exp A — 0.46 g ethanol burned under copper calorimeter with 200 g water
Exp B — 50 mL of 1.00 mol L⁻¹ HCl mixed with 50 mL of 1.00 mol L⁻¹ NaOH
Exp C — 3.74 g NH₄Cl dissolved in 150.0 g water
Exp B (neutralisation): (i) m = 100 g (combined solution, 50+50 mL × 1 g/mL); (ii) n = 1.00 × 0.0500 = 0.0500 mol H₂O formed
Exp C (dissolution): (i) m = 150.0 + 3.74 = 153.74 g (water + solute); (ii) n = 3.74 ÷ 53.49 = 0.06993 mol NH₄Cl dissolved
Go back to your Think First response. Now that you've studied dissolution calorimetry, revisit the 3M engineers' 1959 cold pack design using NH₄NO₃ at +25.7 kJ mol⁻¹:
- The cold pack cools because NH₄NO₃'s lattice energy exceeds its hydration energy — more energy is absorbed to pull the lattice apart than is released when ions are surrounded by water. The 3M engineers measured this as ΔHsoln = +25.7 kJ mol⁻¹: endothermic, meaning the solution absorbs energy from your ankle.
- NaOH warms its solution because the reverse is true — the hydration energy of Na⁺ and OH⁻ exceeds the NaOH lattice energy. The energy difference is released to the solution. NaOH spills are dangerous because the hot solution can burn before you notice.
- Both are ionic solids dissolving. The difference is entirely in the relative magnitudes of lattice energy vs hydration energy — the same Born–Haber cycle you can construct using Hess's Law (L08).
Pick your answer, then rate your confidence — that tells the system what to drill next.
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.
Wrong: Temperature drop → ΔHsoln is negative.
Right: Temperature drop means heat flowed from the solution into the dissolution process — endothermic → ΔHsoln is positive. The formula handles this: negative ΔT → negative q → −q/n is positive.
Wrong: m = mass of water only (forget to add solute).
Right: m = mass of water + mass of dissolved solid. The entire solution absorbs or releases heat, including the dissolved ions.
Q6. 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
Q7. 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
Q8. A student performs three calorimetry experiments:
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
Show comprehensive answers ▼
Activity 1 — NH₄NO₃ Calculation
(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 heat was absorbed from the surroundings (cup or atmosphere) during the slow dissolution, making the measured temperature drop smaller and the apparent ΔHsoln larger.
Classification Table
NH₄NO₃: Endothermic | ΔHsoln positive | cold packs / sports injury packs
NaOH: Exothermic | ΔHsoln negative | lab safety hazard (solution heats up dangerously)
KNO₃: Endothermic | ΔHsoln positive | cooling applications
CaCl₂: Exothermic | ΔHsoln negative | road de-icing (generates heat)
Bonus: For NH₄NO₃ and KNO₃, lattice energy > hydration energy — the ionic lattices are relatively strong compared to the ion-water interaction strength.
Multiple Choice
1. C — Temperature drop → endothermic → lattice energy exceeds hydration energy.
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.
3. B — Exothermic dissolution occurs when hydration energy exceeds lattice energy.
4. C — 'm' is the total mass of solution: water + dissolved solid.
5. B — Temperature rise → exothermic → ΔHsoln is negative. q is positive (ΔT > 0), so −q/n is negative.
Short Answer Model Answers
Q6 (5 marks): ΔT = 20.3 − 25.0 = −4.7 K [½]; m = 100.0 + 8.01 = 108.01 g [½]; q = 108.01 × 4.18 × (−4.7) = −2122.5 J = −2.123 kJ [1]; n = 8.01 ÷ 80.05 = 0.1001 mol [½]; ΔHsoln = −(−2.123) ÷ 0.1001 = +21.2 kJ mol⁻¹ [½]. Endothermic — ΔT < 0, ΔHsoln positive [1]. Two-step model: Step 1 (lattice dissociation): NH₄NO₃(s) → NH₄⁺(g) + NO₃⁻(g) — endothermic, requires input of lattice energy [½]; Step 2 (hydration): ions surrounded by water — exothermic, hydration energy released [½]. Net result: for NH₄NO₃, lattice energy exceeds hydration energy; overall ΔHsoln positive [1].
Q7 (5 marks): (a) ΔT = 24.1 − 21.0 = +3.1 K [½]; m = 150.0 + 3.00 = 153.0 g [½]; q = 153.0 × 4.18 × 3.1 = 1981.7 J = 1.982 kJ [1]; n = 3.00 ÷ 40.00 = 0.0750 mol [½]; ΔHsoln = −1.982 ÷ 0.0750 = −26.4 kJ mol⁻¹ [½]. (b) Exothermic — temperature rose and ΔHsoln negative [1]. NaOH dissolves because the hydration energy released when Na⁺ and OH⁻ ions are surrounded by water molecules exceeds the lattice energy required to separate the NaOH lattice. The net energy is released to the solution [1].
Q8 (6 marks — 2 per experiment):
Exp A (combustion): (i) m = 200 g (water in copper calorimeter — not the ethanol) [½]; (ii) n = 0.46 ÷ 46.07 = 0.00998 mol 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 appears less negative than the true value [1].
Exp 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].
Exp 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].
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