Calorimetry: Weak Acid versus Strong Acid Neutralisation
A student determines the enthalpy change of neutralisation by mixing 50.0 cm3 of 1.00 mol dm-3 hydrochloric acid with 50.0 cm3 of 1.00 mol dm-3 sodium hydroxide in a polystyrene cup. Both solutions are initially at 21.0 °C. The highest temperature reached is 27.8 °C. The student calculates ΔH = -mcΔT, taking the density of the mixture as 1.00 g cm-3 and its specific heat capacity as 4.18 J g-1 K-1. The result is -56.8 kJ mol-1, close to the accepted value of -57.1 kJ mol-1. A second experiment with 1.00 mol dm-3 ethanoic acid instead of hydrochloric acid gives -52.0 kJ mol-1. The thermometer has an uncertainty of ±0.2 °C. The student asks whether the difference between the two values is due to experimental error or to chemistry.
1. Evaluate the assumptions in the calculation and calculate the percentage difference between the hydrochloric acid result and the accepted value.
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The mixture mass is 100.0 g and ΔT = 27.8 - 21.0 = 6.8 °C. Heat released q = 100.0 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.0500 mol, so ΔH = -2842 / 0.0500 = -56.8 kJ mol-1. The percentage difference from -57.1 is (57.1 - 56.8)/57.1 × 100 = 0.5%. Assumptions include: density and specific heat capacity are those of pure water, the cup absorbs negligible heat, and heat loss is negligible. The thermometer uncertainty of ±0.2 °C on each reading gives about ±0.4 °C in ΔT, roughly 6%, so the result is within experimental uncertainty of the accepted value.
2. Explain why ethanoic acid gives a smaller numerical value for the enthalpy of neutralisation.
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Ethanoic acid is a weak acid and is only partially ionised in solution. When it is neutralised by hydroxide ions, some of the energy released is used to ionise the remaining undissociated CH3COOH molecules. The overall enthalpy change is therefore less exothermic than for hydrochloric acid, which is fully ionised. The difference is not simply experimental error: it reflects the enthalpy of ionisation of the weak acid. The measured -52.0 kJ mol-1 is smaller in magnitude because the reaction is less complete in terms of free H+ ions before mixing.
3. Suggest one improvement to the apparatus or procedure and explain its effect on the measured ΔH.
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Use a lid and additional insulation, and stir the mixture continuously. This reduces heat loss to the surroundings. If heat is lost, the measured temperature rise is too small, so the calculated heat release and the magnitude of ΔH are too small. A lid and insulation make ΔT closer to the true adiabatic value. Alternatively, plot a cooling curve and extrapolate back to the time of mixing to correct for heat exchange. Both improvements make the result closer to the accepted value.