You will determine the concentration of sulfuric acid by reaction with a known concentration of sodium hydroxide using a thermometric method. The equation for the reaction is shown. 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) FA 1 is 1.90 moldm-³ sodium hydroxide, NaOH. FA 2 is dilute sulfuric acid, H2SO4.
- (a)Method • Place the cup in the 250 cm³ beaker. • Use the 25 cm³ measuring cylinder to transfer 25.0 cm³ of FA 1 into the cup. • Place the thermometer into the solution in the cup and record its temperature in the table of results. • Fill a burette with FA 2. • Run 5.00 cm³ of FA 2 into the solution in the cup. • Stir the mixture and record the highest temperature reached. • Repeat adding 5.00 cm³ volumes of FA 2 into the solution in the cup until 45.00cm³ has been added. Record the highest temperature reached after each addition. Results volume of FA 2 added/cm³ 0.00 | 5.00 | 10.00 | 15.00 | 20.00 temperature of solution/°C volume of FA 2 added/cm³ 25.00 | 30.00 | 35.00 | 40.00 | 45.00 temperature of solution/°C[3]
- (b)(i)Plot a graph of temperature (y-axis) against volume of acid added (x-axis) on the grid provided. Select a scale on the y-axis to include a temperature 4.0°C above the highest temperature you recorded. Label any points you consider to be anomalous. Draw two lines of best fit, one for the rise in temperature and one for after the maximum temperature has been reached. Extrapolate the two lines so they intersect.[4]
- (b)(ii)Use your graph to determine the volume of sulfuric acid, FA 2, required to neutralise 25.0 cm³ of sodium hydroxide, FA 1. volume of H2SO4 = ...................................... cm³[1]
- (b)(iii)Calculate the concentration of sulfuric acid in FA 2. concentration of H2SO4 = ...................................... moldm-3[1]
- (c)(i)State how the student used their graph to determine the value of ∆T for use in the equation q = mc∆T.[1]
- (c)(ii)The student correctly calculated the value of ∆H for the reaction as ∆H = −55.2 kJ mol⁻¹. The theoretical value for ∆Hneut given in the student's textbook is –57.6 kJ mol⁻¹. Calculate the percentage error in the student's result compared with the theoretical value. percentage error = ...................................... %[1]
- (c)(iii)Suggest why the student's result was less negative than the theoretical value. Explain your answer.[1]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.
- [1] 10 thermometer readings recorded, all to .0 or .5 °C AND at least one reading at .0 and one at .5
- [2] Examiner calculates supervisor's greatest ∆T (= Tmax - Tinitial) and writes it ringed on each candidate script. Examiner calculates candidate's ΔΤ at the same volume then calculates the difference, δ, from the supervisor value. Award 2 marks if δ < 1.5°C for ΔΤ 10.5-15.0 Award 1 mark if 1.5 °C < δ < 2.0°C ΔΤ 10.5-15.0
- [1] Temperature on y-axis and volume of FA 2 on the x-axis AND some numbers for scales AND with unambiguous names or units
- [1] Linear scales chosen so that the graph occupies more than half the available length for both axes
- [1] All points recorded accurately plotted
- [1] Two lines of best fit drawn and extrapolated AND Lines must give a sharp intersection at a temperature equal to or higher than the highest recorded temperature
- [1] Correct volume from intersection to 1 or 2 dp
- [1] Correct expression with answer to 2 – 4 sf Concentration = 25×1.9 / 2x (b)(ii)
- [1] ΔΤ = T at intercept – initial T from table OR ΔΤ = T at intercept – T at intersect on y-axis
- [1] Correctly calculates % error = 4.167, 4.17 or 4.2%
- [1] One of: temperature started dropping between additions (of acid)/ heat loss so ∆T low(er) (than it should be) maximum temperature at intersection is lower (than it should be) as insufficient readings near end point alkali volume less than 25 cm³ as measuring cylinder used.
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