Thermometric titrations can be used to determine the standard enthalpy change of neutralisation. The maximum temperature reached in a thermometric titration occurs at the point of neutralisation between an acid and an alkali. A diagram of the apparatus used is shown in [Figure 1.1]. A student uses the following method. Step 1 Transfer 25.00 cm³ of 1.00 moldm⁻³ dilute hydrochloric acid, HCl(aq), to a polystyrene cup. Step 2 Place a thermometer with 0.2°C divisions into the HCl(aq) in the polystyrene cup and leave it for 3 minutes. Record the temperature. Step 3 Add 5.00cm³ aqueous sodium hydroxide, NaOH(aq), from a burette. Stir and record the temperature of the solution in the polystyrene cup. Step 4 Immediately add another 5.00cm³ of NaOH(aq). Stir and record the temperature of the solution in the polystyrene cup. Step 5 Repeat Step 4 until there is no further increase in temperature. Once the temperature starts to decrease, repeat Step 4 three more times. The student obtains the results shown in Table 1.1.
- (a)(i)Plot a graph of temperature (y-axis) against volume of NaOH(aq) added (x-axis) on the grid. Use a cross (x) to plot each data point. Draw two straight lines of best fit. One for the rise in temperature and one for the fall in temperature. Extrapolate the two lines so they intersect.[2]
- (a)(ii)Use your graph to determine the maximum temperature change of the mixture. Assume the initial temperature of NaOH(aq) is 18.8°C. maximum temperature change of the mixture = ................................... °C[1]
- (a)(iii)Use your graph to determine the volume of NaOH(aq) needed to neutralise 25.00 cm³ of 1.00 moldm⁻³ HCl(aq). volume of NaOH(aq) = ................................... cm³[1]
- (a)(iv)Use your answer to (iii) to calculate the concentration of NaOH(aq) in moldm⁻³. concentration of NaOH(aq) = ................................... moldm⁻³[2]
- (a)(v)Suggest why a titration using an indicator is more accurate than a thermometric titration.[1]
- (b)Suggest a suitable piece of apparatus for the transfer of 25.00 cm³ of 1.00 moldm⁻³ HCl(aq) in Step 1.[1]
- (c)Determine the percentage error of the measured temperature increase when the first 5.00cm³ of NaOH(aq) is added. Show your working. percentage error =[1]
- (d)The standard enthalpy change of neutralisation, ΔH°neut, is defined as the enthalpy change when one mole of H₂O(l) forms from H⁺(aq) and OH⁻(aq). In another experiment a student finds that 22.10cm³ of 1.00 moldm⁻³ of NaOH(aq) increases the temperature by 6.0°C when added to 25.00 cm³ of 1.00moldm⁻³ of HCl(aq). The equation for the reaction between HCl and NaOH is shown. HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Use the formula ΔH = −mcΔT to determine the standard enthalpy change of neutralisation, ΔH°neut, in kJ mol⁻¹. Assume the mass of 1.00cm³ of solution is 1.00g. ΔH°neut = ................................... kJ mol⁻¹[2]
- (e)The theoretical value for the standard enthalpy change of neutralisation in the reaction between HCl(aq) and NaOH(aq) is –57.6 kJ mol⁻¹. Give one reason why the value you obtained in (d) differs from the theoretical value. If you were unable to obtain an answer to (d), use –46.4kJmol⁻¹. This is not the correct answer.[1]
- (f)Suggest why the standard enthalpy change of neutralisation determined using ethanoic acid is less exothermic than the standard enthalpy change using hydrochloric acid.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [2] 9 points plotted correctly Two straight lines of best fit which correctly intersect
- [1] intersection point temperature (read from graph) – 18.8
- [1] intersection point volume (read from graph)
- [1] n(HCl) = 1.00 × (25.00/1000) = 0.0250
- [1] [NaOH] = 0.0250 × (1000/1(a)(iii))
- [1] The indicator: changes colour rapidly / instantly OR detects end-point / neutralisation rapidly / instantly
- [1] (25 cm³) Volumetric pipette
- [1] ((2 x 0.1)/2.5) × 100 = 8(.00) (%)
- [1] mcΔT = (25.00 + 22.10) × 4.18 x 6.0 = 1181.268 (J)
- [1] = q/n(NaOH) (+1000) = 1181.268/(1.00 × 22.10/1000) ÷ 1000 = –53.5 (kJ mol⁻¹)
- [1] Heat loss (to the surroundings)
- [1] Ethanoic acid is a weak acid
- [1] Energy is required for dissociation of ethanoic acid
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