In this investigation, you will measure the enthalpy of neutralisation (ΔHₙₑᵤₜ) for a strong acid and a weak acid reacting with a strong base, and use the difference to explore the relationship between enthalpy change and acid dissociation constants (Kₐ and pKₐ).
This practical is suitable for IB Diploma Chemistry HL and makes an excellent Internal Assessment (IA) topic.
Background Theory
The standard enthalpy of neutralisation for a strong acid reacting with a strong base (e.g. HCl + NaOH) is approximately −57 kJ mol⁻¹. This value corresponds to the formation of one mole of water:
H⁺(aq) + OH⁻(aq) → H₂O(l) ΔH = −57.1 kJ mol⁻¹
Strong acids (e.g. HCl, HNO₃) are fully dissociated in solution, so all the energy released comes from bond formation in water. Weak acids (e.g. ethanoic acid, CH₃COOH) are only partially dissociated. Their dissociation equilibrium is described by the acid dissociation constant:
Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH]
For ethanoic acid, Kₐ = 1.8 × 10⁻⁵ mol dm⁻³, giving pKₐ = 4.74. Because energy is required to fully dissociate the weak acid before neutralisation can occur, the measured ΔHₙₑᵤₜ is less exothermic than for a strong acid — typically around −55 kJ mol⁻¹ for ethanoic acid. The difference represents the enthalpy of dissociation of the weak acid.
This relationship can be extended: acids with lower Kₐ values (higher pKₐ) require more energy to dissociate, and therefore show a larger deviation from the strong acid ΔH value.
Variables
- Independent variable (IV): Type of acid used (strong vs weak; or pKₐ value of the acid)
- Dependent variable (DV): Enthalpy of neutralisation, ΔHₙₑᵤₜ (kJ mol⁻¹)
- Controlled variables (CV): Concentration of acid and base (1.0 mol dm⁻³), volume of solutions (25 cm³ each), same insulated calorimeter, initial temperature measured each time
Suggested Acids to Compare
| Acid | Type | Kₐ (mol dm⁻³) | pKₐ | Expected ΔHₙₑᵤₜ (kJ mol⁻¹) |
|---|---|---|---|---|
| Hydrochloric acid, HCl | Strong | Very large | ~−7 | −57.1 |
| Nitric acid, HNO₃ | Strong | Very large | ~−1.4 | −57.1 |
| Ethanoic acid, CH₃COOH | Weak | 1.8 × 10⁻⁵ | 4.74 | ~−55.2 |
| Methanoic acid, HCOOH | Weak | 1.8 × 10⁻⁴ | 3.75 | ~−56.4 |
| Propanoic acid, CH₃CH₂COOH | Weak | 1.3 × 10⁻⁵ | 4.87 | ~−55.0 |
For an IA, using a series of weak acids with different pKₐ values and plotting ΔHₙₑᵤₜ against pKₐ gives a compelling quantitative relationship.
Equipment
- Hydrochloric acid, HCl (1.0 mol dm⁻³)
- Ethanoic acid, CH₃COOH (1.0 mol dm⁻³) and/or other weak acids
- Sodium hydroxide solution, NaOH (1.0 mol dm⁻³)
- Polystyrene cup calorimeter with lid
- Thermometer or temperature probe (±0.1 °C or better)
- 2 × 25 cm³ pipettes or measuring cylinders
- Stirring rod
Safety
⚠️ All acids and NaOH are corrosive at 1.0 mol dm⁻³ — wear eye protection and gloves. Dispose of all solutions into the appropriate waste disposal bottles provided.
Method
- Measure 25 cm³ of 1.0 mol dm⁻³ HCl into the polystyrene cup. Record the initial temperature (T₁) to the nearest 0.1 °C.
- Measure 25 cm³ of 1.0 mol dm⁻³ NaOH into a separate container. Record its temperature — use the average of the two as T₁ if they differ.
- Add the NaOH to the acid, replace the lid, and stir gently. Record the maximum temperature reached (T₂).
- Calculate ΔT = T₂ − T₁.
- Repeat with ethanoic acid (and any other acids) in place of HCl, keeping all other variables the same.
- Carry out at least three trials for each acid.
Calculations
1. Calculate the heat released: q = mcΔT
Where m = total mass of solution (assume 50 g for 50 cm³), c = 4.18 J g⁻¹ °C⁻¹ (specific heat capacity of water).
2. Calculate moles of water formed: n = 0.025 dm³ × 1.0 mol dm⁻³ = 0.025 mol
3. Calculate enthalpy of neutralisation: ΔHₙₑᵤₜ = −q / n (in kJ mol⁻¹)
Results Table
| Acid | pKₐ | T₁ (°C) | T₂ (°C) | ΔT (°C) | q (J) | ΔHₙₑᵤₜ (kJ mol⁻¹) |
|---|---|---|---|---|---|---|
| HCl | ~−7 | |||||
| CH₃COOH | 4.74 | |||||
| HCOOH | 3.75 |
Analysis
1. Compare your ΔHₙₑᵤₜ values for each acid. Is the strong acid more exothermic than the weak acids?
2. Calculate the enthalpy of dissociation for each weak acid: ΔHᴵᴬ₀₄ = ΔHₙₑᵤₜ(strong) − ΔHₙₑᵤₜ(weak)
3. If you tested multiple weak acids, plot ΔHₙₑᵤₜ (y-axis) against pKₐ (x-axis). What trend do you observe? Does a higher pKₐ (weaker acid) correspond to a less exothermic neutralisation?
4. Use your graph to predict the ΔHₙₑᵤₜ for an acid with pKₐ = 5.5. How confident are you in this prediction?
Discussion Points
- Why is the enthalpy of neutralisation for strong acid + strong base approximately constant at −57 kJ mol⁻¹, regardless of which strong acid is used?
- Why does a weak acid release less energy on neutralisation than a strong acid? Refer to the dissociation equilibrium and Kₐ.
- Why does a higher pKₐ value correlate with a smaller (less negative) ΔHₙₑᵤₜ?
- What are the main sources of heat loss in this experiment and how could they be reduced?
- Why is it important that both solutions start at the same temperature?
IA Guidance
This is an excellent IB Chemistry IA because it combines calorimetry with acid-base theory and generates a clear quantitative relationship. To score highly:
- Research Design: Choose 4–5 acids spanning a range of pKₐ values (e.g. 3.5–5.5). Justify your choice of concentration and volume. Explain why a polystyrene calorimeter is used over a beaker.
- Data Analysis: Plot ΔHₙₑᵤₜ vs pKₐ with error bars. Include a line of best fit and discuss the gradient in terms of dissociation enthalpy per unit pKₐ.
- Conclusion: Compare your ΔH values to literature values. Calculate percentage error and discuss sources of systematic error.
- Evaluation: Consider using a data logger for more precise temperature measurement, or a bomb calorimeter to reduce heat loss.
Practical Science 