2025 ASRJC Prelim H2Chem P4 MS

ASRJC JC2 Prelim 2025 9729/04/H2 ANDERSON SERANGOON JUNIOR COLLEGE 2025 JC2 Preliminary Examination Paper 4 Solutions 1 Determination of an enthalpy change of neutralisation, ∆Hneut, by thermometric titration The enthalpy change of neutralisation between an acid and an alkali can be determined using thermometric titration. This involves using a fixed volume of alkali with progressive addition of small volumes of acid and monitoring the temperature of the reaction mixture during the process. FA 1 is an aqueous solution of 1.80 mol dm-3 of a monobasic acid, HA. FA 2 is aqueous sodium hydroxide, NaOH You are to carry out a thermometric titration to determine the enthalpy change of neutralisation for the reaction given below. HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Question continues on Page 4.

2 ASRJC JC2 Prelim 2025 9729/04/H2 (a) Prepare a table in the space provided and record, to appropriate level of precision: • all volumes of FA 1 added, V • the maximum temperature, T, reached after each addition of FA 1. It is important that the volume of FA 1 recorded is the total volume you have added up to the point when the temperature reading was made. Note: If you overshoot on an addition, record the actual total volume of FA 1 added up to that point. Procedure 1. Place a polystyrene cup inside a second polystyrene cup and place both cups in a glass beaker. 2. Fill the burette with FA 1. 3. Use a measuring cylinder to transfer 25 cm3 of FA 2 into the polystyrene cup. 4. Stir the FA 2 solution in the cup gently with the thermometer. Read and record its temperature. 5. Use the burette to add 5.00 cm3 of FA 1 into the cup. Stir the mixture gently with the thermometer. Read and record both the maximum temperature and the actual total volume of FA 1 added. 6. Repeat step 5 until a total of 45.00 cm 3 of FA1 has been added. For each addition of FA 1, read and record both the maximum temperature and the actual total volume of FA 1 added up to that point. Results [1] correct headers + units [1] records V to 2 d.p. and T to 1 d.p. (penalised students if recording last decimal recording is consistently .0 or .5 - competency in reading thermometer) [1] complete set of readings that are of 5.00 cm 3 increment, inclusive of V = 0.00 cm3 and 45.00 cm3 V / cm3 T / oC 0.00 30.8 5.00 33.2 … … 40.00 35.0 45.00 34.4 [3]

3 ASRJC JC2 Prelim 2025 9729/04/H2 [Turn Over (b) Plot a graph of temperature, T, on the y-axis, against total volume of FA 1 added, V, on the x-axis on the grid in Fig.1.1. Your scale on the y-axis should allow for extrapolation above the highest temperature recorded. Draw two lines of best fit, taking into account the points when the temperature of the mixture was rising and the points when the temperature was falling. Each line should have a shape best suited to its plotted points. Extrapolate (extend) the two lines until they intersect. Fig. 1.1 T / oC V / cm3 10.00 20.00 30.00 40.00 50.00 0.00 39.0 37.0 35.0 33.0 31.0 29.0 40.0

4 ASRJC JC2 Prelim 2025 9729/04/H2 • Axes correct way round + correct labels + units + scale (scale must be chosen to allow for the lines to be extrapolated to intersect) • Graph has to occupy more than half the grid for both axes. [1] • Plotting – all points including 0.00 cm3 within ±½ small square. (select two from each side to check) [1] • Both graph lines are best -fit lines and lines are correctly extrapolated to intersect [1] [3] (c) From your graph, read the initial temperature of FA 2, Tinitial, and the maximum temperature of the mixture, Tmax. Use these values to calculate the temperature change in the reaction, ∆T. Read the volume of FA 1 added, Vneut, at the maximum temperature of the mixture. Record all these values below. Tinitial = ………………………………... Tmax = ………………………………… ∆T = ………………………………… Vneut = ………………………………… [4] • Tinital and Tmax correctly read from the graph to ±½ small square. [2] • correct ∆T calculated (Tinitial may come from table or graph plot) [1] • Vneut correctly read from the graph to ±½ small square. [1]

5 ASRJC JC2 Prelim 2025 9729/04/H2 [Turn Over (d) (i) Using your answers from (c), calculate the heat change, q, when FA 1 has completely neutralised 25 cm3 of sodium hydroxide. You should assume that the specific heat capacity of the solution is 4.18 J g−1 K−1, and that the density of the solution is 1.00 g cm−3. Total volume = 25 + Veq (teacher value for Veq is approximate 21 cm3) Heat evolved, q = mcT = (Veq +25) cm3 x 1.00 g cm–3 x 4.18 J g–1 K–1 x ∆T J [1] • correct substitution of total volume • correct substitution of values into mcT (allow ecf) [1] Correct calculation of q q = ………………………………[2] (ii) Calculate the enthalpy change of neutralisation, ∆Hneut, for the reaction. The equation for the reaction is shown. HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) Include the sign of ∆Hneut in your answer. No of moles of water = no of moles of HA = 1.80 x (Veq/ 1000) Hneut = – (q ÷ moles of water) = – (d)(i) ÷ n(water) J mol–1 [1] n(water) formed [1] working to find Hneut (allow ecf) [1] answer (if no negative sign, no marks) [1] 3 s.f. and correct units for (d)(i) & (ii) ∆Hneut = ……………………[4]

6 ASRJC JC2 Prelim 2025 9729/04/H2 (e) Apart from using a thermometer with a greater level of precision, suggest one improvement that could be made to improve the accuracy of results in (d)(ii). Use a burette / pipette to measure volume of FA 2 / cover with lid / use smaller volumes close to Tmax / use windshield. [1] [1] (f) A student decided to perform the same experiment in (a) but used aqueous ammonia instead of aqueous sodium hydroxide. Suggest what effect, if any, would replacing aqueous sodium hydroxide with aqueous ammonia have on the value of enthalpy change of neutralisation calculated in (d)(ii). effect ……………………………………………………………………………… The value of enthalpy change of neutralisation will be smaller in magnitude / less exothermic. [1] explanation ………………………………………………………………………… ……………………………………………………………………………………… Aqueous ammonia is a weak base. Some of the energy released from neutralisation will be absorbed to cause the (complete) dissociation of ammonia, resulting in the overall heat released to be smaller. [1] [2] [Total: 19]

7 ASRJC JC2 Prelim 2025 9729/04/H2 [Turn Over 2 Determination of the amount of water of crystallisation in sodium sulfite crystals, Na2SO3 • xH2O FA 3 is an aqueous solution of 126.0 g dm–3 of hydrated sodium sulfite with the formula Na2SO3 • xH2O. FA 4 is 0.100 mol dm–3 iodine, I2. FA 5 is 0.100 mol dm–3 sodium thiosulfate, Na2S2O3. You are also provided with Solution S. Solid sodium sulfite is often provided as the hydrated salt, Na2SO3 • xH2O, where x is an integer. You will determine the value of x by using a solution of this sodium sulfite salt and reacting it with an excess of aqueous iodine. Na2SO3 + I2 + H2O → Na2SO4 + 2I– + 2H+ The amount of iodine remaining will be determined by titration using a known concentration of sodium thiosulfate, Na2S2O3. I2 + 2S2O32–→ 2I– + S4O62– (a) (i) Dilution of FA 3 FA 3 is too concentrated and needs to be diluted. Use a burette to transfer 25.00 cm3 of FA 3 into a 100 cm3 volumetric flask. Make the solution up to the mark with deionised water. Label this solution FA 6. Titration of FA 6 against FA 5 1. Fill the burette with FA 5. 2. Use a pipette to transfer 10.0 cm3 of FA 6 into a 250 cm3 conical flask. 3. Use another pipette to transfer 25.0 cm3 of FA 4 into the same conical flask. 4. Swirl the flask to mix the contents. 5. Run FA 5 from the burette into the conical flask. Near the end -point, when the brown solution becomes pale,