Dissolution Thermochemistry

Systematically measure which salts heat or cool water — and calculate why
Beginner🕐45 min●Low hazardthermochemistry

Summary

Some salts warm water as they dissolve and others chill it; you’ll rank seven of them and use simple calorimetry to estimate each one’s heat of dissolution. You’ll be able to explain the tug-of-war between breaking a crystal lattice (always endothermic1) and hydrating the freed ions (always exothermic2), and why that balance makes hot packs and cold packs.

History

The distinction between exothermic and endothermic processes was cemented in the mid-19th century as thermodynamics matured. Hess’s law (1840) established that energy changes in chemical reactions are additive regardless of path, which made it possible to tabulate dissolution energies systematically. Scientists noticed that some salts heated their containers while others cooled them — a puzzle whose answer turned out to hinge on how strongly each ion grips water molecules.

Commercial instant cold packs were developed in the 1960s using ammonium nitrate (now often replaced with ammonium chloride or urea due to safety concerns). Hot packs using calcium chloride followed. These products brought thermochemistry into everyday life — and into sports medicine bags worldwide.

This experiment takes the commercial pack demonstration further: instead of just observing hot versus cold, you will test seven different solutes, rank them, and use simple calorimetry to estimate their heats of dissolution from your own measurements.

Hazards & preparation

Warning

PPE: safety glasses.

  • Calcium chloride dissolves violently and gets hot (45–55 °C) — add it gradually and expect splashing; don’t touch the cup carelessly.
  • Sodium carbonate is mildly alkaline — rinse splashes off skin.

Disposal: all seven solutions are dilute and harmless — flush down the drain with water. See the Safety page.

Materials

Solute Amount From
Calcium chloride (CaCl₂) 15 g Chemical inventory
Ammonium chloride (NH₄Cl) 15 g Chemical inventory
Sodium chloride (table salt, NaCl) 15 g Kitchen
Sodium carbonate (Na₂CO₃) 15 g Chemical inventory
Epsom salt (MgSO₄·7H₂O) 15 g Chemical inventory
Baking soda (NaHCO₃) 15 g Kitchen
Citric acid 15 g Chemical inventory

Equipment:

  • Thermometer (ideally digital, reading to 0.1°C)
  • 7 × 250 mL cups or beakers
  • 7 × 100 mL portions of water, all at the same starting temperature
  • Kitchen scale or measuring spoons
  • Stirring rod or spoon
  • Notebook for recording temperatures

Procedure

Setting up

  1. Measure 100 mL of water into each of seven cups. Let them all stand together for 5 minutes so they reach the same temperature — small differences in starting temperature will throw off the comparison.
  2. Record the starting temperature of each cup (should be the same, within 0.5°C).
  3. For each solute, weigh out exactly 15 g.

Dissolving and measuring

  1. Add the first solute (calcium chloride) to cup 1, all at once, and stir vigorously for 30 seconds.
  2. Record the temperature at 30 s, 60 s, and 2 min. Note the maximum (or minimum) temperature reached.
  3. Repeat for all seven solutes, working quickly so the water temperature does not drift.

Take care with calcium chloride — it dissolves vigorously and can splash. Add it gradually if needed.

Recording results

Solute Starting T (°C) Final T (°C) ΔT (°C) Hot or cold?
CaCl₂
Na₂CO₃
NaCl
MgSO₄·7H₂O
NH₄Cl
NaHCO₃
Citric acid

What you should see

Calcium chloride is the showstopper — the cup climbs to 45–55 °C, distinctly warm to the touch. Sodium carbonate warms moderately. Ammonium chloride, baking soda, and citric acid each make the cup noticeably cold; Epsom salt cools slightly; and plain table salt barely moves the thermometer at all.

Symptom Likely cause Fix
Cups don’t all start at the same temperature Water not equilibrated Let all seven stand together 5 min before starting
Weak or no temperature change Solute not fully dissolved, or lost time Stir hard; read the peak quickly before heat leaks away
CaCl₂ splashes Added too fast Add it in portions with stirring

Simple Calorimetry

Once you have your temperature changes, you can estimate the heat of dissolution for each solute — the energy released or absorbed per mole.

Step 1 — Calculate energy transferred to the water:

\[q = m \times c \times \Delta T\]

Where:

  • \(m\) = mass of water = 100 g
  • \(c\) = specific heat of water = 4.18 J g⁻¹ °C⁻¹
  • \(\Delta T\) = temperature change in °C

For example, if calcium chloride raised the temperature by 30°C:

\[q = 100 \times 4.18 \times 30 = 12{,}540 \text{ J} = 12.5 \text{ kJ}\]

The water gained 12.5 kJ of heat, so the reaction released 12.5 kJ (the reaction’s \(q\) is −12.5 kJ).

Step 2 — Convert to kJ per mole:

\[\Delta H_{\text{measured}} = \frac{-q}{n}\]

For CaCl₂ (molar mass = 111 g/mol), 15 g = 0.135 mol:

\[\Delta H = \frac{-12.5 \text{ kJ}}{0.135 \text{ mol}} = -93 \text{ kJ/mol}\]

Compare this with other solutes using their molar masses:

  • NH₄Cl: 53.5 g/mol
  • NaCl: 58.4 g/mol
  • Na₂CO₃: 106 g/mol
  • MgSO₄·7H₂O: 246 g/mol
  • NaHCO₃: 84 g/mol
  • Citric acid: 192 g/mol

Note on accuracy: Your measurements will differ from literature values because heat also goes into warming the cup, the thermometer, and the surrounding air. Student calorimetry typically captures 50–80% of the true heat. The direction (exo vs. endo) and relative ranking will be accurate even if the absolute numbers are not.

Literature Values

How do your results compare?

Solute Literature ΔH (kJ/mol) Typical ΔT with 15 g in 100 mL
CaCl₂ −81 +30 to +40°C
Na₂CO₃ −28 +8 to +14°C
NaCl +3.9 −1 to +1°C
MgSO₄·7H₂O +13 −3 to −6°C
NH₄Cl +15 −5 to −8°C
NaHCO₃ +27 −6 to −9°C
Citric acid +23 −5 to −8°C

Notice: the ranking from hottest to coldest spans nearly 110 kJ/mol — a huge range for such simple processes.

The Science

Why does dissolving produce heat at all?

Dissolution always involves two competing energy changes:

1. Breaking the crystal lattice — always endothermic

Ions in a solid are held in place by strong electrostatic attraction. Pulling them apart requires energy input. The strength of this attraction is the lattice energy, which scales with ion charge and decreases with ion size.

2. Hydrating the ions — always exothermic

Once free in solution, each ion is surrounded by water molecules oriented with their partial charges facing the ion. This hydration releases energy. Small ions with high charge attract water more strongly, releasing more energy on hydration.

Net effect: If hydration releases more energy than lattice-breaking requires → exothermic (solution heats). If lattice-breaking costs more than hydration releases → endothermic (solution cools).

Why is CaCl₂ so hot?

Calcium(II) is a small, doubly charged ion. Its high charge density pulls water molecules extremely tightly, releasing a large hydration energy that more than compensates for the lattice energy. The result is a strongly exothermic dissolution. Sodium carbonate is moderately exothermic for similar reasons — Na⁺ and CO₃²⁻ both hydrate well.

Why is NaHCO₃ cold?

Bicarbonate (HCO₃⁻) is a large, singly charged ion. It does not pull water as tightly, so hydration energy is modest. But the lattice energy is still significant, leaving a net endothermic balance.

Why does MgSO₄·7H₂O cool water even though anhydrous MgSO₄ would heat it?

This is the subtlest result. Anhydrous magnesium sulfate (MgSO₄) has a strongly exothermic dissolution (about −91 kJ/mol) — Mg²⁺ is small and very highly charged, so it hydrates powerfully. But Epsom salt is already hydrated: each formula unit comes with 7 water molecules attached. Before Mg²⁺ can hydrate from bulk water, it must first shed those seven water molecules — an endothermic step that reverses the advantage. The net result tips slightly endothermic.

This is the reason anhydrous CaCl₂ (sometimes sold as a desiccant) heats more dramatically than partially-hydrated forms. The more dehydrated the salt, the more exothermic the dissolution.

Charge density summary

Ion Charge Radius Hydration enthalpy (kJ/mol)
Ca²⁺ 2+ small −1592
Mg²⁺ 2+ very small −1922
Na⁺ 1+ medium −406
NH₄⁺ 1+ large −301
Cl⁻ 1− medium −363
SO₄²⁻ 2− large −1138
HCO₃⁻ 1− large −335

High charge and small size = strong hydration = more exothermic dissolution (all else equal).

Chemicals used in this experiment:

Questions to Explore

  1. Why does dissolving always involve two competing energy changes? Breaking a crystal lattice costs energy; hydrating the freed ions releases it. For any given salt, what determines which process wins and whether dissolution is hot or cold?

    Hint / answer

    It’s a subtraction: hydration energy released minus lattice energy spent. If the ions grip water more strongly than they gripped each other in the crystal, the net is exothermic (hot); if the lattice was harder to break than hydration pays back, it’s endothermic (cold).

  2. Why isn’t Mg²⁺ charge enough to make Epsom salt exothermic? Magnesium is small and doubly charged, so it should hydrate powerfully — yet MgSO₄·7H₂O cools water. What changes when the salt already carries water in its crystal?

    Hint / answer

    Epsom salt’s Mg²⁺ is already hydrated by seven waters, so most of that generous hydration energy has already been “cashed in.” Dissolving it releases little further, but still costs lattice energy to break up — so the net tips endothermic. Anhydrous MgSO₄, with no built-in water, heats strongly.

  3. Why does sodium chloride barely change the temperature? NaCl dissolves with ΔH near zero. Does that mean no energy is involved, or that two big numbers nearly cancel? How could you tell?

    Hint / answer

    Two large energies nearly cancel — the lattice energy and the total hydration energy of Na⁺ and Cl⁻ are close, so their difference is tiny. You could tell by looking up the separate lattice and hydration enthalpies (both are large), not by the small net you measure.

  4. Could you design a better hand warmer? CaCl₂ heats the most here. What would an ideal hot-pack chemistry need — and why isn’t CaCl₂ used in commercial reusable warmers?

    Hint / answer

    You’d want a big, sustained, safe, and ideally reusable heat release. CaCl₂ gives a big but brief, one-shot spike and leaves a salty solution. Reusable warmers instead use supersaturated sodium acetate, which can be melted and re-triggered again and again.

  5. What does your calorimetry error tell you? Measured ΔH is usually 50–80% of the literature value. Where does the missing energy go, and how would you capture more of it?

    Hint / answer

    The “missing” heat leaks into the cup, thermometer, and air instead of the water you’re measuring. Insulate the container (a foam cup with a lid), stir to read the true peak fast, and correct for the cup’s heat capacity to recover more of the real value.

Going further

  • Reverse it. Evaporate your hot CaCl₂ solution in a warm dish; as crystals reform, feel the dish — crystallization releases the same energy dissolution absorbed.
  • Change the solvent. Dissolve a little NaCl in ethanol (specific heat 2.44 J/g·°C) instead of water and recompute ΔH — does it come out the same?
  • Sustained heat: compare this brief CaCl₂ spike with the long, steady warmth of supersaturated sodium acetate in Hot Ice.

Footnotes

  1. Endothermic — Describing a process that absorbs heat from its surroundings, making them feel cold.↩︎

  2. Exothermic — Describing a process that releases heat to its surroundings, making them feel warm.↩︎