Experiment: enthalpy of solution of copper sulphate

What is this experiment?

In plain words: drop the crystals in water, stir, and watch the thermometer. If it goes down, the dissolving took heat from the water and the beaker. Work out how much heat that was, and scale it to one mole of the salt.

The NCERT textbook gives the theory of enthalpy of solution (Section 5.5, page 156) but not the method; the JEE Main and NEET (UG) syllabuses list this experiment among the practical chemistry experiments of Unit 20. The method below follows the NCERT Laboratory Manual Chemistry for Class 12, Unit 3 (Thermochemical Measurement), Experiment 3.1, pages 20–23.

Theory

When an ionic solid dissolves, its ions leave the crystal (this absorbs energy) and are surrounded by water molecules (this releases energy); Δ_solH is the balance of the two (NCERT page 156). For an endothermic dissolution equilibrium, heating generally favours dissolving while the same solid phase persists. Solubility ultimately depends on Gibbs energy; hydration, entropy, concentration and possible changes of hydrate phase also matter.

Why the blue crystals cool the water but the white powder warms it. In blue CuSO₄·5H₂O each Cu²⁺ ion already holds water molecules, so dissolving it releases little extra energy, while the crystal still has to be broken up: the change is slightly endothermic. White anhydrous CuSO₄ has no water; its ions gain all of their hydration on dissolving, and that releases much more energy than breaking the crystal takes. From standard data (dilute solutions, 298 K):

• CuSO₄·5H₂O(s) → Cu²⁺(aq) + SO₄²⁻(aq) + 5H₂O(l): about +6.0 kJ mol⁻¹;
• CuSO₄(s) → Cu²⁺(aq) + SO₄²⁻(aq): about −73.1 kJ mol⁻¹.

By Hess’s law, the enthalpy change for turning the white powder into blue crystals, CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s), is the difference: −73.1 − 6.0 = −79.1 kJ mol⁻¹. Problem books and JEE/NEET questions call this the “enthalpy of hydration of copper sulphate”; it is not the same as NCERT’s hydration enthalpy of gaseous ions.

The heat is found with the method of mixtures:

where m_soln is the mass of water plus salt in grams (100 g of water, taking 1 g mL⁻¹), 4.184 J g⁻¹ K⁻¹ is taken as the specific heat of the dilute solution, W is the calorimeter constant in J K⁻¹ and Δt = t₂ − t₁. Assuming negligible heat exchange with the room, constant approximate heat capacity and only pressure–volume work, the heat taken in by dissolving is −q, so:

with m_salt the mass dissolved and M its molar mass: 249.7 g mol⁻¹ for CuSO₄·5H₂O (the Laboratory Manual uses 249.5, from rounded atomic masses). When the temperature falls, Δt and q are negative and Δ_solH is positive.

Apparatus

• Three 250 mL beakers and one 500 mL beaker, a small wooden block, cotton wool and a piece of cardboard: the inner 250 mL beaker stands on the block inside the 500 mL beaker, packed round with cotton wool, with the cardboard as a lid.
• A thermometer with 0.1 °C divisions (a digital one reading 0.01 °C is better here), a stirrer and a glass rod.
• About 2 g of well-powdered copper sulphate, weighed accurately.

Procedure

A. Calorimeter constant. Following the method of mixtures, put 50 mL of water in the inner beaker and record its temperature, t_c; add 50 mL of water at 50–60 °C whose exact temperature, t_h, is read just before mixing; stir and record the mixed temperature, t_m. Then W = 4.184 [V_h(t_h − t_m) − V_c(t_m − t_c)]/(t_m − t_c) J K⁻¹, with volumes in mL. This adapts NCERT’s 100 mL + 100 mL calibration to approximately the 100 mL working fill: the manual notes that a glass beaker’s effective heat capacity depends on the wetted region. Use a calibration appropriate to the final fill, thermometer and stirrer configuration.

B. Enthalpy of solution.

1. Put 100 mL of distilled water in the same (dry) beaker inside the insulated set-up; allow the beaker and salt to reach the same initial temperature as the water and record it, t₁.
2. Add the weighed copper sulphate, m_salt g, and stir until it has all dissolved.
3. Record the temperature reached, t₂ (for the crystals, the lowest reading).
4. Work out q with m_soln = 100 + m_salt, then Δ_solH.

Precautions (Laboratory Manual, page 23): use a thermometer with 0.1 °C divisions; read the hot water’s temperature just before mixing; do not use very large amounts of salt; stir until it dissolves but not so much that friction warms the liquid; weigh the copper sulphate carefully; insulate the two beakers with cotton wool.

Slips in the Laboratory Manual. Five points in the manual’s working need correcting:

• page 23 takes the density of water as 1 g L⁻¹; it should be 1 g mL⁻¹, which is what makes 100 mL weigh 100 g;
• page 23 multiplies the calorimeter term W(t₂ − t₁) by 4.184 as well. W is already in J K⁻¹, so only the solution term takes the 4.184;
• page 23 rightly calls the heat gained by the solution and beaker the heat liberated on dissolution, but then sets Δ_solH = +249.5 × (heat liberated per gram). Δ_solH is minus the heat liberated per mole; without the minus sign the expression gives a negative Δ_solH for crystals whose dissolving absorbs heat;
• page 20 drops the minus sign in the formula for W (Eqs. 4 and 5) and calls the heat needed to warm 1 g of a substance by 1 K its heat capacity (that is the specific heat capacity).

Solved examples

Common mistakes to avoid

Quick revision summary

• Δ_solH: enthalpy change when one mole of a substance dissolves in a specified amount of solvent (NCERT page 156).
• CuSO₄·5H₂O dissolves with a small absorption of heat (about +6 kJ mol⁻¹); anhydrous CuSO₄ gives out heat (about −73 kJ mol⁻¹).
• Method: calorimeter constant first, then dissolve a weighed amount in 100 mL of water and read the temperature change.
• q = (m_soln × 4.184 + W)Δt; Δ_solH = −q × M/m_salt.
• Hess’s law: ΔH(forming the hydrate) = Δ_solH(anhydrous) − Δ_solH(hydrate), about −79 kJ mol⁻¹.

Questions students ask

Is dissolving copper sulphate in water a physical or a chemical change?

Dissolving copper sulphate crystals in water is usually classed as a physical change at school level because the salt can be recovered by suitable crystallisation. Dissolution still changes coordination and interactions as the crystal gives hydrated ions; gentle evaporation under appropriate conditions gives back blue crystals, CuSO₄·5H₂O. The blue colour, in the crystals and in the solution, comes from Cu²⁺ ions bonded to water molecules. White anhydrous CuSO₄ is a less clear case: it binds water, giving out heat, and turns blue, which is why it is used as a test for water, and that hydration step is often described as a chemical change.

How is the enthalpy of hydration of copper sulphate found in the laboratory?

The enthalpy of hydration of copper sulphate, ΔH for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s), is found from two runs in the same calorimeter: measure Δ_solH of the white anhydrous salt and of the blue crystals, then by Hess’s law ΔH = Δ_solH(anhydrous) − Δ_solH(crystals). Use equal amounts, such as 0.0100 mol (1.60 g of CuSO₄ or 2.50 g of crystals), so that both runs end with nearly the same solution. For an exact Hess cycle, match final temperature and composition, accounting for the five waters supplied by each mole of hydrate; otherwise include a dilution correction. Dry the anhydrous salt freshly and weigh it quickly: it takes up moisture from the air and would then give out too little heat.

Can potassium nitrate be used instead of copper sulphate in this experiment?

Yes, potassium nitrate can be used instead: the CBSE 2026–27 Class 12 practical syllabus lists ‘enthalpy of dissolution of copper sulphate or potassium nitrate’, while the JEE Main 2027 and NEET (UG) 2026 syllabuses name only CuSO₄. The method and the calculation are the same, with M = 101.1 g mol^⁻¹ for KNO₃. Potassium nitrate dissolves far more endothermically, about +35 kJ mol^⁻¹, so 2.00 g in 100 mL of water in a calorimeter with W = 34.9 J K^⁻¹ cools it by about 1.5 °C, a fall much easier to read accurately than that of the copper sulphate crystals.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. 4.00 g of CuSO₄·5H₂O is dissolved in 100 mL of water at 26.00 °C in a calorimeter with W = 40.0 J K⁻¹; the lowest temperature is 25.82 °C. Find Δ_solH.

   Show answerHide answer

   Heat capacity = 104 × 4.184 + 40.0, about 475.1 J K⁻¹. q = 475.1 × (−0.18), about −85.5 J. Moles = 4.00/249.7, about 0.01602 mol. Δ_solH = +85.5/0.01602 J mol⁻¹, about +5.3 kJ mol⁻¹.

2. Q2. Some problem books quote Δ_solH = −66.5 kJ mol⁻¹ (anhydrous CuSO₄) and +11.7 kJ mol⁻¹ (CuSO₄·5H₂O), measured in a finite amount of water. What do they give for CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s)?

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   ΔH = −66.5 − (+11.7) = −78.2 kJ mol⁻¹, close to the −79.1 from standard data. The subtraction gives the hydrate-formation enthalpy only if both measurements refer to the same final solution and temperature, including water released by the hydrate. Dilution contributions cancel only under that condition; without full experimental conditions, the 0.9 kJ mol⁻¹ difference cannot be assigned uniquely to data-set differences.

3. Q3. A student applies the Laboratory Manual’s expression to the readings of Example 1. The expression multiplies W(t₂ − t₁) + (100 + m_salt)(t₂ − t₁) by 249.5 × 4.184/m_salt. What does she get, and what two things are wrong?

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   Here W(t₂ − t₁) + (100 + m_salt)(t₂ − t₁) = (34.9 + 105) × (−0.25) = −34.975, and −34.975 × 249.5 × 4.184/5.00 ≈ −7302 J mol⁻¹, about −7.3 kJ mol⁻¹. The sign is wrong (the temperature fell, so Δ_solH is positive), and W, already in J K⁻¹, has been multiplied by 4.184, which makes the size too large. The correct result is about +5.9 kJ mol⁻¹.