2019 TMJC H2 Chem Prelim P4 ANS

1 1 Determination of the concentration of a base and the enthalpy change of the neutralisation reaction FA 1 is 2.0 mol dm–3 sulfuric acid, H2SO4. FA 2 is aqueous sodium hydroxide, NaOH. The reaction of sulfuric acid and sodium hydroxide is exothermic. In separate experiments, you will add increasing volumes of FA 2 to a fixed volume of FA 1. In each experiment you will measure the maximum temperature rise, ∆T. As the volume of FA 2 is increased, this maximum temperature rise, ∆T, will increase and then decrease. By measuring the maximum temperature rise for differe nt mixtures of the two reagents, you are to determine the following:  the concentration of sodium hydroxide, NaOH, in FA 2  the enthalpy change when 1 mol of H2SO4 is neutralised by NaOH (a) Method  Fill the labelled burette with FA 1.  Support a styrofoam cup in the 250 cm3 beaker.  Run 10.00 cm3 of FA 1 from the burette into the styrofoam cup.  Measure 20.0 cm3 of FA 2 using a measuring cylinder.  Place the thermometer in the FA 2 in the measuring cylinder and record the steady temperature of the solution.  Pour the FA 2 into the styrofoam cup, stir and record the maximum temperature obtained in the reaction.  Empty and rinse the styrofoam cup; shake dry the styrofoam cup. Rinse the thermometer.  Carry out the experiment three more times. Each time use 10.00 cm3 of FA 1.  Use 30.0 cm3, 40.0 cm3 and 50.0 cm3 of FA 2 in these different experiments. Carry out two further experiments. Choose volumes of FA 2 which will allow you to investigate more precisely the volume of FA 2 that produces the highest temperature rise when added to 10.00 cm3 of FA 1. TAMPINES MERIDIAN JUNIOR COLLEGE 2019 JC2 H2 Chemistry Prelim Exam Paper 4 (Suggested Answers)

2 Results Record your results below in an appropriate form showing, for each experiment, the volumes of solutions used, temperature measurements and the temperature rise, T. Expt Volume of FA 1 / cm3 Volume of FA 2 / cm3 Ti / oC Tf / oC ∆T / oC 1 10.00 20.0 31.5 42.5 11.0 2 10.00 30.0 31.5 45.0 13.5 3 10.00 40.0 31.5 44.0 12.5 4 10.00 50.0 31.5 42.0 10.5 5 10.00 25.0 31.5 43.5 12.0 6 10.00 35.0 31.5 45.0 13.5 [4] (b) Plot a graph of temperature rise, ∆T, on the y-axis against the volume of FA 2 added on the x-axis. Draw a line of best fit through the points where the temperature rise is increasing and another line through the points where the temperature rise is decreasing. The intersection of these lines represents the temperature rise for the volume of FA 2 that exactly neutralises the sulfuric acid present in 10.00 cm3 of FA 1. [3] ∆T / oC 15.0 14.0 13.0 12.0 11.0 10.0 Vol of FA 2 / cm3 10.0 20.0 30.0 40.0 50.0 X X X X X X (c) 32.5 (e) 14.0

3 (c) Read from the graph the volume of FA 2 that gives the maximum temperature rise. Volume of FA 2 giving the maximum temperature rise = cm3 [1] (d) (i) Calculate the amount of NaOH required to neutralise the amount of H2SO4 at the maximum temperature rise. Amount of H2SO4 = 2.0  0.010 = 0.0200 mol H2SO4  2 NaOH Amount of NaOH = 2  0.020 = 0.0400 mol Amount of NaOH required = mol [1] (ii) Hence, calculate the concentration of NaOH in FA 2. Concentration of NaOH in FA 2 = 0.0400  32.5 1000 = 1.23 mol dm-3 Concentration of NaOH in FA 2 = mol dm3 [1] (e) Read the maximum temperature rise from the graph and use this to calculate the enthalpy change when 1 mol of H2SO4 is neutralised by NaOH. Give your answer in kJ mol–1. [4.18 J are absorbed or released when the temperature of 1 cm3 of solution changes by 1 °C.] Quantity of heat absorbed by solution = (32.5 10.0) 4.18 14.0   = 2487 J OR 2.487 kJ Enthalpy change = – 2.487 0.0200 = –124 kJ mol-1 Enthalpy change = kJ mol1 [2] (f) The enthalpy change of neutralisation, Hneut, between a strong acid and a strong base is –57 kJ mol1. Explain why the enthalpy change calculated in (e) is significantly more exothermic than Hneut. Enthalpy change in (e) is based on 1 mol of H 2SO4 (or 2 mol of water formed) while the enthalpy change of neutralisation, Hneut (–57 kJ mol-1), is based on 1 mol of water formed. [1] 0.0400 32.5 1.23 –124

4 (g) A student suggested that the experiments carried out in (a) would be more accurate if volumes of 20.00 cm3 of 1.0 mol dm–3 H2SO4 were used instead. State and explain whether you agree or disagree with the student’s suggestion. Accept any of the following: • Agree – Lower (percentage) error as: acid spray is reduced (since reaction will be slower) OR smaller temperature rise so less heat loss OR larger volume used (accept other reasons) • Disagree – Higher (percentage) error as: smaller temperature change, so higher (percentage) error of reading OR reaction slower so more heat loss (accept other reasons) [1] [Total: 14]

5 2 Determination of the percentage by mass of sodium carbonate in a mixture of sodium hydroxide and sodium carbonate Sodium carbonate is neutralised by hydrochloric acid in two steps: Na2CO3 + HCl  NaCl + NaHCO3 NaHCO3 + HCl  NaCl + H2O + CO2 This step-wise neutralisation can be observed when the acid is added slowly to the sodium carbonate. The percentage by mass of sodium carbonate in a mixture of sodium hydroxide and sodium carbonate can be determined by carrying out titrations using two different indicators. Since both sodium hydroxide and sodium carbonate react with acids, through careful selection of the indicators used for the titration, the volume of acid required to react with only the sodium carbonate can be found. FA 3 is 0.125 mol dm3 hydrochloric acid, HCl. FA 4 is an aqueous solution containing sodium hydroxide, NaOH, and sodium carbonate, Na2CO3. You are also provided with bromophenol blue indicator. In this question, you will carry out titrations to determine the percentage by mass of sodium carbonate in the mixture of sodium hydroxide and sodium carbonate in solution FA 4. (a) (i) Titration of FA 4 against FA 3 using bromophenol blue 1. Fill a burette with FA 3. 2. Pipette 25.0 cm3 of FA 4 into a conical flask. 3. Add four to five drops of bromophenol blue indicator. 4. Titrate the mixture in the flask with FA 3 until the blue-violet colour of the solution changes to yellow. 5. Record your titration results , to an appropriate level of precision, in the space provided. Repeat steps 2 to 5 to obtain consistent results. Titration results [6] 1 2 3 Final burette reading / cm3 20.70 30.55 20.45 Initial burette reading / cm3 0.00 10.00 0.00 Volume of FA 3 / cm3 20.70 20.55 20.45  

6 (ii) From your titration results, obtain a suitable volume of FA 3, VFA 3, to be used in your calculations. Show clearly how you obtained this volume. Volume of FA 3, VFA 3 = 20.55 20.45 2  = 20.50 cm3 (2 d.p.) VFA 3 = [1] (b) When the titrations in (a) were repeated using phenolphthalein as the indicator, 25.0 cm3 of FA 4 required 15.00 cm3 of FA 3. The following explains why different results are obtained using two different indicators. • When phenolphthalein is used as the indicator, the following reactions have taken place at the end-point of the titration. Reaction 1 NaOH + HCl  NaCl + H2O Reaction 2 Na2CO3 + HCl  NaCl + NaHCO3 • When bromophenol blue is used as the indicator in (a), the following reactions have taken place at the end-point of the titration. Reaction 1 NaOH + HCl  NaCl + H2O Reaction 2 Na2CO3 + HCl  NaCl + NaHCO3 Reactio