2024 Prelims ACJC H2 Chem P3 (Ans)

This document consists of 28 printed pages and – blank pages. Anglo-Chinese Junior College JC2 Preliminary Examination Higher 2 CANDIDATE NAME Answers FORM CLASS 2 TUTORIAL CLASS 2CH INDEX NUMBER CHEMISTRY Paper 3 Free Response Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/03 27 August 2024 2 hours READ THESE INSTRUCTIONS FIRST Write your index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at the end of this booklet. The question number must be clearly shown. Section A Answer all questions. Section B Answer one question. Circle the number of the question you have attempted. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiners’ use only Section A 1 / 20 2 / 20 3 / 20 Section B 4 / 5 / 20 Presentation Total / 80

2 © ACJC2024 9729/Preliminary Examination/2024 [Turn over Section A Answer all the questions in this section. 1 (a) Describe the variation in the behaviour of Period 3 chlorides NaCl, AlCl3 and PCl5 separately with water. Write equations for any reactions described and state the pH of the resultant solutions. [3] NaCl hydrates (dissolves) to give a neutral solution (pH 7). There is no hydrolysis. NaCl(s) + aq → Na+(aq) + Cl−(aq) AlCl3 undergoes substantial hydrolysis due to the high charge density of Al3+. The resulting solution is acidic. (pH 3-4) Hydration: AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(aq) Hydrolysis: [Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O+(aq) PCl5 undergoes complete hydrolysis to give an acidic solution. pH 1-2 PCl5(s) + 4H2O(l) → H3PO4 (aq) + 5HCl (aq) Comments • Students should take note of using the reversible for partial hydrolysis and the irreversible arrow for complete hydrolysis. • Some students were penalised for using the wrong state symbols such as AlCl3(aq) and PCl5(aq). • Avoid using the equations that involve limited water as we are unable to give pH of the solid or liquid product: AlCl3(s) + 3H2O(l) → Al(OH)3(s) + 3HCl(g) PCl5(s) + 2H2O(l) → POCl3(l) + 2HCl(g) (b) When soil becomes acidic, aluminium leeches out of minerals into the soil. High aluminium content in soil affects root growth and causes the roots to be brittle. These problems are minimised if the soil pH is maintained above 5.5. In a study of a soil condition, a sample of soil water was titrated with EDTA, a hexadentate ligand, to determine its aluminium ion concentration. equation 1 [Al(H2O)6]3+(aq) + [EDTA]4−(aq) [Al(EDTA)]− (aq) + 6H2O(l) (i) State the type of reaction in equation 1. [1] Ligand exchange. (ii) The concentration of aluminium ions in the soil water sample was found to be 2.90 × 10–5 mol dm–3. Calculate the pH of the water sample, given that the aqueous complex of Al3+ has a Ka value of 7.9 × 10–6. You may assume that the complex of Al3+ behaves as a weak monobasic acid, HA. [1] Ka = [𝐻+][𝐴−] [𝐻𝐴] 7.9 × 10–6 = [H+]2 / (2.90 x 10–5) [H+] = 1.514 x 10–5 pH = 4.82

3 © ACJC2024 9729/Preliminary Examination/2024 [Turn over Comments Most students were able to use the Ka expression, Ka value and concentration of the weak acid to determine the pH of the weak acid. (iii) Another 25 cm3 sample of soil water was found to contain 0.250 mol dm–3 of NaH2PO4. 20 cm3 of solution A which contains aqueous Na2HPO4 was added to the water sample to obtain a solution buffered at pH 6.8. Calculate the concentration of HPO42– in solution A, given that the pKa value of H2PO4– is 7.2. [3] Final [H2PO4–] = (0.250 × 25 25+20) = 0.1389 mol dm-3 pH = pKa + lg [HPO4 2−] [H2PO4−] 6.8 = 7.2 + lg [HPO4 2−] [H2PO4−] lg [HPO4 2−] [H2PO4−] = –0.4 [HPO4 2−] [H2PO4−] = 0.3981 [HPO4 2−] 0.1389 = 0.3981 [HPO42–]final = 5.529 × 10–2 mol dm–3 OR Ka = [HPO4 2−][𝐻+] [H2PO4−] 10-7.2 = [HPO4 2−][10−6.8] [H2PO4−] 10-7.2 = [HPO4 2−][10−6.8] 0.1389 [HPO42–]final = 5.529 × 10–2 mol dm–3 Amt of HPO42– in buffer = 5.529 × 10–2 × 25+20 1000 = 0.002488 mol [HPO42–] in solution A = 0.002488 mol 20 x 10−3 = 0.124 mol dm–3 Comments • Many students were able to use the Henderson-Hesselberg equation (i.e. pH = pKa + lg [HPO4 2−] [H2PO4−] ) to relate the pH, pKa and concentrations of the weak acid and salt. They were also able to determine the [H2PO4–] after mixing. • Common mistakes include: o Confusion about the acid and base roles of H2PO4– and HPO42– respectively. o Failing to recognise that the [HPO42–] from the Henderson-Hesselberg equation is the concentration after mixing, not its original concentration in solution A. o Using the Ka expression to calculate [HPO42–] but failing to recognise that for a buffer, [H+] ≠ [salt] (i.e. [H+] ≠ [HPO42–]). (c) The reactions of ethylbenzene to form H, J and K are shown in Fig.1.1.

4 © ACJC2024 9729/Preliminary Examination/2024 [Turn over Fig. 1.1 (i) Suggest the reagents and conditions needed for reactions I and II and suggest the structure of M in Fig. 1.1. [3] I: KMnO4, H2SO4(aq), heat under reflux II: SOCl2 OR PCl5 OR PCl3, heat M: Comments • Reaction I is a side-chain oxidation of alkylbenzene. It results in benzoic acid M. • More students were able to state the correct reagents and conditions to convert M into an acid chloride. However, wrong state symbols such as PCl5(aq) or SOCl2(aq) would lose this mark. (ii) Describe and explain the relative ease of hydrolysis of H, J and K. [3] Relative ease of hydrolysis: K > H > J Due to the presence of the electronegative O and Cl atoms on the acyl chloride in compound K, its carbonyl carbon is more electron deficient than that in H and more susceptible to attack by the nucleophile. Therefore the hydrolysis of K has greater ease than that of H. In aryl chloride J, the p orbital of Cl overlaps with the 𝜋 electron cloud of the benzene ring. As a result the C-Cl bond has partial double bond character which require more energy to break. This makes causes J to have the lowest ease of hydrolysis.

5 © ACJC2024 9729/Preliminary Examination/2024 [Turn over Comments Common misconceptions include: • Confusing the effect of C=O group in –COCl with that for the base strength of – CONH–. • Calling the C=O group electron withdrawing without identifying that the cause is the electronegative O atom. • Thinking that the phenyl group contributes to delocalisation of el