2020 ACJC Prelim Paper 4 Answers

1 © ACJC 2020 9729/04/Prelim/2020 [Turn over ACJC Prelim 2020 Paper 4 Answers 1 Determination of percentage by mass of ethanedioic acid in a mixture of sodium ethanedioate and ethanedioic acid In this experiment, you are to determine the percentage by mass of ethanedioic acid in a mixture of sodium ethanedioate and ethanedioic acid. This experiment will be carried out in two parts. In part one, you will carry out a titration to find out the amount of ethanedioic acid, H2C2O4, present in the mixture. In part two, you will make use of the titration results to find out the total amount of ethanedioate ions, C2O42–, present in the mixture. You will then use the values found in these two parts to calculate the percentage by mass of ethanedioic acid in FA 1. FA 1 is a mixture of aqueous sodium ethanedioate, Na2C2O4, and ethanedioic acid, H2C2O4. FA 2 is 2.00 mol dm–3 NaOH FA 3 is 1.00 mol dm−3 sulfuric acid, H2SO4 FA 4 is 0.0200 mol dm−3 potassium manganate(VII), KMnO4. You are also provided with thymolphthalein solution. Part One: (a) Dilution of FA 2 solution As the NaOH solution provided is too concentrated, you will have to dilute it before conducting the experiment. 1. Fill the burette labelled FA 2 with FA 2. 2. Measure between 17.00 cm3 to 17.50 cm3 of FA 2 into a 250 cm3 volumetric (graduated) flask. 3. Record your burette readings in Table 1.1 below. 4. Fill the flask to the mark with deionised water, using a dropper when nearing the mark. 5. Stopper and mix the contents thoroughly by shaking to obtain a homogeneous solution. Label this solution FA 5. Table 1.1 final burette reading / cm3 17.00 initial burette reading / cm3 0.00 volume of FA 2 added / cm3 17.00 [1]

2 © ACJC 2020 9729/04/Prelim/2020 [Turn over (b) Titration of FA 1 with FA 5 1. Fill the burette labelled FA 5 with FA 5. 2. Pipette 25.0 cm3 of FA 1 into a conical flask. 3. Add 3 to 4 drops of thymolphthalein solution. 4. Run FA 5 from the burette into the flask. The end-point is reached when the solution changes from colourless to a permanent pale blue colour. Record your titration results in Table 1.2 below. 5. Repeat steps 2 to 4 until consistent results are obtained and record your results in the table below. Table 1.2 final burette reading / cm3 25.00 41.00 initial burette reading / cm3 1.00 17.00 volume of FA 5 added / cm3 24.00 24.00 [4] From your titrations, obtain a suitable volume of FA 5, to be used in your calculations. Show clearly how you obtained this volume. Average titre = ½ (24.00 + 24.00) = 24.00 cm3 Suitable volume of FA 5 = ………24.00 cm3……….. [1] Part Two (c) Titration of FA 1 with FA 4 (There is no need to carry out this experiment) 1. Fill the burette labelled FA 4 with FA 4. 2. Pipette 25.0 cm3 of FA 1 into a conical flask 3. Using a measuring cylinder, add 25cm3 of FA 3 to the conical flask. 4. Place the conical flask on a tripod and gauze and heat to about 65 oC. 5. If the neck of the flask is too hot to hold safely, use a folded paper towel to hold the flask. 6. Run 1 cm3 of FA 4 from the burette into the flask and swirl until the colour of the potassium manganate(VII) has disappeared then continue the titration as normal until a permanent pale pink colour is obtained. This is the end-point. Record your titration results in the table below. 7. Repeat steps 2 to 4 until consistent results are obtained and record your results in the Table 1.3 below. Table 1.3 final burette reading / cm3 36.20 36.30 36.40 initial burette reading / cm3 0.90 1.30 1.10 volume of FA 4 added / cm3 35.30 35.00 35.30 chosen values √ √

3 © ACJC 2020 9729/04/Prelim/2020 [Turn over From the titrations above, obtain a suitable volume of FA 4, to be used in your calculations and place ticks below the chosen volumes of FA 4. Show clearly how you obtained this volume. Average titre = ½ (35.30 + 35.30) = 35.30 cm3 Suitable volume of FA 4 = ……35.30 cm3…….. [2] Calculations Show your working and appropriate significant figures in all of your calculations. (d) (i) Using your values in (a), calculate the concentration of NaOH in FA 5 that you prepared. Volume of FA2 = 17.00 cm3 Amount of NaOH in 250 cm3 = 17.00 x 10-3 x 2.00 = 0.0340 mol [NaOH] in FA5 = 0.0340 1000250  = 0.136 mol dm-3 concentration of NaOH in FA 5 = ……0.136 mol dm-3……..[1] (ii) Using your answer from (b) and d(i), calculate the amount of NaOH required to react with 25.0 cm3 of FA 1 in Part One. Vol in (b) x concentration from d(i) = amount of NaOH [1] Amount of NaOH = 0.136 x 24.00 x 10-3 = 0.00326 mol amount of NaOH required = ……0.00326 mol ……….. [1] (iii) The equation for the reaction between sodium hydroxide and ethanedioic acid is shown below 2NaOH(aq) + H2C2O4(aq)  Na2C2O4(aq) + 2H2O(l) Using your answer in d(ii), Calculate the amount of ethanedioic acid present in 25.0 cm3 of FA 1. amount of NaOH in d(ii)/2 = amount of H2C2O4 [1] Amount of H2C2O4 in 25.0 cm3 = ½ x 0.003264 = 0.00163 mol amount of H2C2O4 in 25.0 cm3 of FA 1 = ……0.00163 mol….. [1]

4 © ACJC 2020 9729/04/Prelim/2020 [Turn over (e) (i) Using your answer from (c), calculate the amount of potassium manganate(VII) required to react with 25.0 cm3 of FA 1 in Part Two. Volume in (c) x 0.0200 mol dm–3 = amount of KMnO4 required [1] Amount of KMnO4 = 35.30 x 10-3 x 0.0200 = 7.06 x 10-4 mol amount of KMnO4 required = ……7.06 x 10-4 mol.. [1] (ii) The balanced equation for the reaction between acidified manganate(VII) ions and ethanedioate ions is given below. 2MnO4(aq) + 5C2O42(aq) + 16H+(aq)  2Mn2+(aq) + 10CO2(g) + 8H2O(l) Using your answer in e(i), calculate the amount of ethanedioate ions present in 25.0 cm3 of FA 1. amount of KMnO4 from e(i) x 5/2 = amount of C2O42– [1] Amount of C2O42- in 25.0 cm3 = 5 2 x 7.06 x 10-4 mol = 0.001765 mol amount of C2O42– in 25.0 cm3 of FA 1 = ………0.001765 mol …….. [1] (iii) Hence, calculate the amount of ethandioate ions which came from the sodium ethanedioate dissolved in 25.0 cm3 of FA 1. amount in e(ii) – amount in d(iii) = amount of C2O42– from Na2C2O4 Amt of C2O42- from Na2C2O4 = 0.001765 – 0.00163 = 1.35 x 10-4 mol amount of C2O42– from Na2C2O4 in 25.0 cm3 of FA 1 = ………1.35 x 10-4 mol …….. [1] (f) (i) Using your answer to (d)(iii), calculate the mass of ethanedioic acid in 25.0 cm3 of FA 1. [Ar: H, 1.0; C, 12.0; O, 16.0] amount of ethanedioic acid from d(iii) x 90.0 = mass Mass of ethanedioic acid = 0.00163 x 90.0 = 0.147 g mass of ethanedioic acid in 25.0 cm3 of FA 1 = ………0.147 g……….. [1]

5 © ACJC 2020 9729/04/Prelim/2020 [Turn over (ii) Using your answer to (e)(iii), calculate the mass of sodium ethanedioate in 25.0 cm 3 of FA 1. [Ar: C, 12.0; O, 16.0; Na, 23.0] amount of sodium ethanedioate from e(iii) x 134.0 = mass Mass of sodium ethanedioate = 1.35 x 10-4 x 134.0