RI 2021 Year 6 March TP Suggested Solutions

© Raffles Institution 2020 9729/M/20 2021 Y6 H2 Chemistry Timed Practice – Suggested Solutions Section B B1 (a) 3 3 [NH (aq)] [NH (org)] ≈ 25 ⇒ [NH3(aq)] = 25[NH3(org)] [NH3(aq)] + [NH3(org)] = 1.04 25[NH3(org)] + [NH3(org)] = 1.04 [NH3(org)] = 0.04 mol dm–3 [NH3(aq)] = 1.04 – 0.04 = 1.00 mol dm–3 Amount of NH3 in 25.0 cm3 sample of NH3(aq) = 25.0 1.00 0.0250 mol1000= Since NH3 and HCl react in a 1:1 ratio, Amount of HCl required for titration = 0.0250 mol Volume of HCl = 330.0250 0.250 dm 250 cm0.100 == Volume of HCl is unsuitable because it exceeds the burette capacity of 50.00 cm3, therefore it is tedious to refill the burette multiple times . This is a result of the concentration of ammonia being too high. Examiners’ Comments • Some students did not read the question and failed to realise that there are two parts to the question – 1. To calculate the volume of HCl required; 2. To explain why [NH 3] is unsuitable. • Careful reading of the information provided will lead students to realise that the NH3 is dissolved in both layers. Some students assumed that 1.04 mol dm–3 was the [NH3] in the aqueous layer. • To calculate the volume of HCl required to titrate the aqueous layer, the concentration of NH3 in the aqueous layer should be calculated first. • Students need to elaborate beyond the fact that the volume of HCl required exceeds the burette capacity by explaining, for example, that there is thus a need to refill the burette, causing the experiment to become tedious. (b) Ideally, 25.0 cm3 of diluted ammonia would require 25.0 cm3 of 0.100 mol dm–3 hydrochloric acid. Since 250 cm 3 of hydrochloric acid was required, a sample of the aqueous layer needs to be diluted 10 times i.e. obtain 25.0 cm 3 of aqueous ammonia and dilute to 250 cm3. 1. Drain the bottom organic layer into a 100 cm 3 conical flask / beaker to discard. Pour/drain the remaining aqueous layer into another 100 cm3 conical flask / beaker. 2. Use a 25.0 cm 3 pipette to t ransfer 25.0 cm 3 of the aqueous layer solution into a 250 cm3 volumetric flask. This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2020 9729/M/20 3. Fill the graduated flask to the 250 cm3 mark with more deionised water. Use a dropping pipette (or teat pipette or dropper) to add the deionised water drop by drop when nearing the mark. 4. Stopper the graduated flask and shake the solution thoroughly to ensure that it is homogeneous. Label the solution FA 1. 5. Pipette 25.0 cm 3 of FA 1 into a 250 cm 3 conical flask. Add 2 drops of methyl orange indicator and swirl the conical flask. 6. Titrate the solution in the conical flask with the standard 0.100 mol dm –3 hydrochloric placed in the burette. 7. Stop the titration when the end -point of the titration is reached i.e. when one drop of the hydrochloric acid added changes the colour of the solution in the conical flask from yellow to orange. 8. Record the titration readings using an appropriate table. 9. Repeat the titration until at least two consistent results are obtained (i.e. the two titre volumes are within 0.10 cm3 of each other). Examiners’ Comments Students should pay attention to the points to be included in their plan: (1) Calculate a suitable volume of aq. layer to use for dilution - Since the diluted aq. NH 3 will be titrated against HCl, an ideal titre volume of HCl would be 25.00 cm 3 (with 25.0 cm 3 NH3 pipetted). However the undiluted aq. NH 3 requires a titre volume of 250 cm3 HCl – this means that the undiluted aq. NH3 is 10 times too concentrated, or the dilution factor is 10. - If a 250 cm3 graduated flask is used to prepare the diluted solution, the volume of aq. layer required = 250 10 = 25 cm3. This means 25 cm 3 of aq. layer is transferred to the graduated flask, then the flask is topped up with water until a 250 cm 3 diluted NH3 solution is obtained. - Most students were able to deduce the dilution factor, but some were unable to calculate the correct volume of aq. layer to use for dilution. - A graduated flask must be used for the accurate dilution of solutions for titration. Many students were unable to recal l the correct name of the flask and some used imprecise apparatus such as measuring cylinders. - Graduated flasks of 100 cm 3 or 250 cm 3 capacity can be used but the correct corresponding volume of the aqueous layer needs to be used to achieve the 10-time dilution. (2) Procedure to obtain the vol. of aqueous layer WITHOUT the organic layer - An important clue is given in the diagram in the question which shows that the aqueous layer lies above the organic layer. - One way to obtain the aqueous layer is to drain the organic layer first, then drain the aq. layer into a separate flask. - Alternatively, a pipette can be used to withdraw the aqueous sample from the top of the funnel, but care must be taken not to disturb or stir the organic layer. (3) Procedure to dilute the aq. layer to an appropriate volume - The detailed procedure for using a graduated flask to dilute a solution should be given, including transferring th e aqueous layer to the flask by suitable means, topping up to the mark with water, stopper and shaking the flask. This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2020 9729/M/20 (4) Procedure for titration of diluted aq. NH3 against HCl - This is a strong acid-weak base titration, thus suitable indicators are methyl orange and screened methyl orange. Some students chose inappropriate indicators. - The colour change at end -point must be stated correctly – for methyl orange, the colour change is from yellow to orange, since the indicator was added to basic NH 3 in the conical flask. - For experiments involving titrations, the titration must be carried out until at least two consistent results are obtained. • In writing procedures for planning questions, all apparatus used together with their capacities must be clearly stated, e.g. 25.0 cm3 pipette, 250 cm3 conical flask, 250 cm3 graduated flask. (c)(i) When a small amount of acid is added, NH3(aq) + H+(aq) → NH4+(aq) When a small amount of base is added, NH4+(aq) + OH–(aq) → NH3(aq) + H2O(l) Examiners’ Comments • The “→” should be used in this question to depict the response of the buffer to H + or OH– being added. • Common incorrect answers show the ionisation of NH3 and hydrolysis of NH4+. These do not show how the buffer responds when its pH is being affected by the addition of acid or base i.e. how a buffer controls pH. (c)(ii) pOH = 14 – 9.0 = 5.0 + 4 3 + 4 +3 4 [NH ]pOH = pK + lg [NH ] [NH ]5.0 4.75 + lg 0.80 [NH ] 1.423 mol dm b − = = Amount of ammonium chloride in 1 dm3 = 1.423 x 1 = 1.423 mol Mass of ammonium chloride = 1.423 [14.0 + 4(1.0) + 35.5] = 76.1 g Examiners’ Comments • Generally well answered. • Common mistakes made involved: o Incorrectly remembering the formula to calculate the pH of buffer solution. o Incorrectly using 9 instead of 5 as pOH in the above equation. o Incorrectly using the Mr of NH4+ instead of NH4Cl as required by the question. This document is copyrighted, please do not reproduce it without permission

© Raffles Institution 2020 9729/M/20 B2 (a)(i) As phenol is a weak acid and dissociates to a very small extent ( deduced from the very small Ka value), it can be assumed that equilibrium [C6H5OH] ≈ initial [C6H5OH]. [𝐻+] ≈ √𝐾𝑎[C6H5OH] = √10−10 × 0.750] = 8.660 × 10−6 mol dm–3 pH = –lg (8.660 x 10–6) = 5.06 Examiners’ Comments • Most students were able to first convert the pKa value given to Ka, and then correctly determined the [H+] and pH value. However, some candidates substituted 10.0 as Ka when it is the pKa value for phenol. • So