2024 ACJC H2 Chemistry Prelim P4 (Ans)

2 © ACJC 2024 9729/04/Prelim/2024 [Turn over Answer all the questions in the spaces provided. 1 To determine the concentration of NaOH and Na2CO3 in FA 1 Aqueous NaOH can be used to remove CO 2 from exhaled air in spacecrafts. NaOH reacts with CO2 according to the following equation: 2NaOH(aq) + CO2(g) → Na2CO3(aq) + H2O(l) A sample of aqueous NaOH was used to absorb CO2 in the air for some time. The resulting mixture of NaOH and Na2CO3 formed was labelled as FA 1. You are to determine the concentration of NaOH and Na 2CO3 in FA 1 by titration with an aqueous solution of HCl, FA 2. You are provided with the following. FA 1, is a mixture of NaOH and Na2CO3 FA 2, 2.50 mol dm−3 of hydrochloric acid, HCl methyl orange indicator thymol blue indicator distilled water (a) Titration of FA 1 against FA 2 using thymol blue indicator. 1. Fill a burette with FA 2. 2. Using a pipette, transfer 25.0 cm3 of FA 1 into a 250 cm3 conical flask. 3. Add a few drops of thymol blue indicator into the conical flask. 4. Run FA 2 from the burette into the flask. The end point is reached when the solution changes from blue to green. 5. Record your titration results in Table 1.1 6. Repeat points 2 to 5 until consistent titre values are obtained. Table 1.1 1 2 3 final burette reading / cm3 13.80 13.80 initial burette reading / cm3 0.00 0.00 volume of FA 2 added / cm3 13.80 13.80 [3] Students should be mindful to input the burette reading in the correct box. Students should be careful with the subtraction of the burette readings and all entries must be written to 2 dp (to nearest 0.05 cm3). Students should have at least 2 consistent readings within ±0.10 cm3.

3 © ACJC 2024 9729/04/Prelim/2024 [Turn over From your titrations, obtain a suitable volume of FA 2 to be used in your calculations. Show clearly how you obtained this volume. volume of FA 2 = ……………………….. [1] Calculation of average to 2 dp (b) Titration of FA 1 against FA 2 using methyl orange indicator. Repeat points 1 to 6 in (a) but add methyl orange at point 3 in place of thymol blue. Using this indicator, the end point is reached when the solution changes from yellow to orange. Record your titration results in Table 1.2. Table 1.2 1 2 3 final burette reading / cm3 18.80 19.00 initial burette reading / cm3 0.00 0.20 volume of FA 2 added / cm3 18.80 18.80 [3] From your titrations, obtain a suitable volume of FA 2 to be used in your calculations. Show clearly how you obtained this volume. volume of FA 2 = ……………………….. [1] (c) Calculation When thymol blue is used as the indicator in part (a), the following reactions occur. Reaction 1: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Reaction 2: Na2CO3(aq) + HCl(aq) → NaCl(aq) + NaHCO3(aq) When methyl orange is used as the indicator in part (b), the following reactions occur. Reaction 1: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Reaction 2: Na2CO3(aq) + HCl(aq) → NaCl(aq) + NaHCO3(aq) Reaction 3: NaHCO3(aq) + HCl (aq) → NaCl(aq) + H2O(l) + CO2(g) The NaHCO 3 in Reaction 3 is the product from reaction of Na 2CO3 with H Cl in Reaction 2.

4 © ACJC 2024 9729/04/Prelim/2024 [Turn over (i) Calculate the volume of hydrochloric acid that reacted with NaHCO3 in 25.0 cm3 of FA 1 in Reaction 3. NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g) Volume of HCl used = 18.80 (Reaction 1, 2 and 3) – 13.80 (Reaction 1 and 2) = 5.00cm3 volume of hydrochloric acid = ……………………..[1] (ii) By considering Reactions 2 and 3, determine the volume of hydrochloric acid that reacted with sodium carbonate in 25.0 cm3 of FA 1 to form NaHCO3. 1 mol NaHCO3  1 mol Na2CO3 Volume of HCl used to react with Na2CO3 = 5.00 cm3 Students should be mindful of this statement given “The NaHCO3 in Reaction 3 is the product from reaction of Na2CO3 with HCl in Reaction 2.” As such the volume of HCl that reacted with NaHCO3 should be equal to the volume of HCl that react with Na2CO3. volume of hydrochloric acid = ……………………….. [1] (iii) Calculate the volume of hydrochloric acid that reacted with the sodium hydroxide present in 25.0 cm3 of FA 1. Volume of HCl reacted with Na2CO3 = 5.00 cm3 (Reaction 2) Volume of HCl reacted with NaOH = 13.80 – 5.00 = 8.80 cm3 volume of hydrochloric acid = ……………………….. [1] (iv) Using your answer in (c)(ii) and the equation for Reaction 2, calculate the amount of sodium carbonate present in 25.0 cm3 of FA 1. Na2CO3(aq) + HCl(aq) → NaCl(aq) + NaHCO3(aq) Volume of HCl used to react with Na2CO3 = 5.00 cm3 Amount of HCl reacted with Na2CO3 = 2.50 x 5.00 1000 = 0.0125 mol Since Na2CO3 ≡ HCl Amount of Na2CO3 = 0.0125 mol amount of sodium carbonate = ……………………….. [1] (v) Calculate the concentration of sodium carbonate, Na2CO3 in FA 1. [Na2CO3] = 0.0125 x 1000 25 = 0.500 moldm−3 concentration of sodium carbonate = ……………………….. [1]

5 © ACJC 2024 9729/04/Prelim/2024 [Turn over (vi) From your answer to (c)(iii) and the equation for Reaction 1 , calculate the amount of sodium hydroxide present in 25.0 cm3 of FA 1. NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) Amount of HCl used = 2.50 x 8.80 1000 = 0.0220 mol Since NaOH ≡ HCl Amount of NaOH = 0.0220 mol amount of sodium hydroxide = ……………………….. [1] (vii) Calculate the concentration of sodium hydroxide, NaOH in FA 1. [NaOH] = 0.0220 x 1000 25 = 0.880 moldm−3 concentration of sodium hydroxide = ……………………….. [1] (d) Give one reason why after 25.0 cm3 of FA 1 is pipetted into a conical flask, the stock solution FA 1 should be immediately covered with a sealant film. State the effect of not covering FA 1 with a sealant film on the calculated concentration of NaOH. ………..……………………………………………………………….……………….………. ………..……………………………………………………………….……………….………. ………………………………………………………………………………….....…………[2] This will prevent atmospheric carbon dioxide from reacting further with sodium hydroxide present in FA 1. CO2 + 2NaOH → Na2CO3 + H2O If a stopper or sealant is not used, the calculated concentration of NaOH will be lower than actual concentration of NaOH in FA 1 since it reacts with carbon dioxide in the atmosphere. Note: The calculated concentration of NaOH will be lower while the calculated concentration of Na2CO3 will be higher. [Total: 17]

6 © ACJC 2024 9729/04/Prelim/2024 [Turn over 2 Determination of the identity of the acid and the enthalpy change of neutralisation, Hneut You are provided with the following solutions: FA 3 is a solution of 1.60 mol dm-3 sodium hydroxide, NaOH FA 4 is a solution of 2.00 mol dm-3 of either hydrochloric acid, HCl or sulfuric acid, H2SO4. In this question, you are to perform a series of 6 experiments where different volumes of FA 3 and FA 4 are mixed to give a total volume of 50 cm 3. The temperature change, T, of the reaction mixture for each experiment will be determined and a graph of T against the volume of FA 3 will be plotted. You will then use the data from your graph to determine the identity of the acid and the enthalpy change of neutralization for the reaction between aqueous sodium hydroxide and the acid. (a) Procedure Experiment 1 1) Place one polystyrene cup inside a second polystyrene cup. Place these into a