2024 TMJC H2 Chem Prelim P3 (Ans w Markers' Comments)

Section A Answer all questions from this section. 1 Ethanoic acid , CH 3COOH, is a common precursor used in industries to synthesise more complex compounds. (a) Ethanoic acid has a Ka value of 1.74 × 10–5. (i) Calculate the pH of 0.200 mol dm–3 ethanoic acid to 1 decimal place. [1] [H+] = –log10√(0.200 × 1.74 × 10–5) = 2.7 [1] (ii) Write the Ka expression for ethanoic acid. [1] Ka = [H+] × [CH3COO–] / [CH3COOH] [1] (b) An aliquot of 25.0 cm3 of 0.200 mol dm–3 of ethanoic acid is titrated with 25.0 cm3 of 0.200 mol dm–3 NaOH solution. (i) Explain, with the aid of a chemical equation, why the pH at equivalence point is alkaline. [2] CH3COO– + H2O ⇌ CH3COOH + OH– [1] At 25.0 cm3 of NaOH, the ethanoic acid is completely neutralised into ethanoate ions, which then undergoes salt hydrolysis to form an alkaline solution. [1] (ii) A buffer was formed during this titration. Write a chemical equation to show how the buffer solution resists pH change when a small amount of NaOH is added. [1] CH3COOH + OH– → CH3COO– + H2O (single arrow) [1] Marker’s Comments • Almost all candidates could do this calculation, but some candidates ignored the instructions for 1 d.p. Marker’s Comments • Almost all candidates scored this mark. Marker’s Comments • The question specifically asked for a singular chemical equation, but some candidates wrote multiple equations. They need to be aware there is a possibility that actual A-Level markers may only credit the first equation and ignore subsequent answers, so as to not credit candidates who are regurgitating without discernment. • About half of the candidates wrote equation for the neutralisation instead, CH3COOH + OH – → CH3COO– + H2O. Hence they need to read the question carefully to know what it wants. Marker’s Comments • This question only asked for chemical equation and no explanation was required. Nonetheless, many candidates gave lengthy explanations for zero additional credit. • The question asked for change that happened, so the reversible arrow was rejected.

2024 H2 Chem Paper 3 (Suggested Ans) 2 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Chemistry (iii) By using the Ka value from (a), calculate the pH of the solution to 2 decimal places when 15.0 cm3 of NaOH was added. [3] Initial amount of ethanoic acid = 25.0/1000 × 0.200 = 0.00500 mol Amount of NaOH in 15.0 cm3 = 15.0/1000 × 0.200 = 0.00300 mol After neutralisation, amt of ethanoate formed = amt of NaOH added = 0.00300 mol After neutralisation, amt of ethanoic acid remaining = 0.00500–0.00300 = 0.00200 mol New total volume = 25.0 + 15.0 = 40.0 cm3 = 0.0400 dm3 pH = pKa + log10 ( [CH3COO–] [CH3COOH]) = 4.76 + log10 (0.00300 0.0400⁄ 0.00200 0.0400⁄ ) = 4.76 + 0.18 = 4.94 [1] (c) A sample of industrial waste contains only hydrochloric acid and ethanoic acid. 50.0 cm3 of the industrial waste was collected in a volumetric flask and topped up to 250 cm 3. A 25.0 cm3 sample of the resultant solution was then titrated using 0.200 mol dm–3 NaOH solution, giving two equivalence points at 12.40 cm 3 and 23.55 cm 3. The first equivalence point is for the neutralisation reaction with HCl. (i) Both hydrochloric acid and ethanoic acid are Brønsted-Lowry acids . Define the term Brønsted-Lowry acid. [1] A Brønsted-Lowry acid is a H+ donor. [1] Marker’s Comments • Most candidates’ errors arose from misunderstanding the context, or focusing on single components without realising that this is a buffer solution. Marker’s Comments • Almost all candidates scored this mark. [1] [1] Alternative: pH = pKa + log10[(15.0) / (25.0 – 15.0)] = 4.76 + 0.18 = 4.94 [3]

3 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Chemistry [Turn over (ii) Calculate the concentration of hydrochloric acid present in the 50.0 cm3 sample of industrial waste. [3] amount of NaOH reacted with HCl = 12.40/1000 × 0.200 = 0.00248 mol [1] amount of HCl in 25.0 cm3 = 0.00248 mol amount of HCl in 250 cm3 = 0.00248 × 250/25.0 = 0.0248 mol [1] amount of HCl in 50 cm3 of industrial waste = 0.0248 mol [HCl] in 50 cm3 of industrial waste = 0.0248 ÷ (50.0/1000) = 0.496 mol dm–3 [1] (iii) Hence, calculate an approximate value for the pH of the industrial waste. Give your answer to 1 decimal place. [1] pH = –log100.496 = 0.3 [1] e.c.f. (iv) Sketch the shape of the titration curve and circle the portion that represents a buffer solution. No numerical pH values are needed. [2] [1m – sketch, 1m - buffer] [Total: 15] Marker’s Comments • Some candidates were confused by the context, leading to wrong procedure or simply confusing dilution and extracting aliquots. Otherwise, many candidates scored full marks. Marker’s Comments • Most candidates recognised correctly that the contribution from the weak acid could be ignored. Marker’s Comments • Most candidates correctly derived the shape of the titration curve (1st mark), but the buffer zone (2nd mark) was derived with less success. Some candidates showed clear evidence of analysing the species present at each point on the horizontal axis, and these tend to lead to correct answers. Vol. of NaOH added / cm3 12.40 23.50 pH HCl + NaCl CH3COOH + CH3COO_Na+ + NaCl CH3COO_Na+ + NaCl + NaOH (excess)

2024 H2 Chem Paper 3 (Suggested Ans) 4 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Chemistry 2 Cobalt is a critical component in lithium-ion batteries, which are essential for powering electric vehicles and portable electronics due to their high energy density and reliability. (a) The following sequence of reactions in Fig. 2.1 involves cobalt and its complexes. Fig. 2.1 (i) State the type of reaction that occurred in step 3 and the role of H2O2 in step 8. [2] Step 3: Ligand exchange [1] role of H2O2 in Step 8: oxidising agent [1] (ii) Given that complex A is square planar in shape, suggest the identities of complexes A and B. [2] A – [CoCl4]2– [1] B – Co(OH)2 [1] Marker’s Comments • Candidates should apply their knowledge of copper chemistry to identify compounds A and B. They must recognize that the overall charge of a ppt is zero.

5 Tampines Meridian Junior College 2024 JC2 Preliminary Examination H2 Chemistry [Turn over (iii) When solutions of [Co(H 2O)6]2+ and [Fe(H 2O)6]3+ are each mixed with sodium carbonate solution, different reactions occur. [Fe(H2O)6]3+ produces effervescence of CO2 but not [Co(H2O)6]2+. Explain why [Fe(H 2O)6]3+ results in the evolution of carbon dioxide gas when reacted with carbonate ions. [2] Fe³⁺ has a higher charge density ✓ than Co²⁺, which allows it to polarise water molecules ✓and weaken the O–H bond. ✓This results in the breaking of the O– H bond, ✓ making the solution acidic. ✓ The acidity neutralizes the carbonate, producing effervescence of CO₂. 2 -3 ✓ - [1], 4-5 ✓ - [2], (b) Co3O4, also known as cobalt tetroxide, is a black ionic compound that contains both Co2+ and Co3+ ions. (i) Suggest the ratio of the two different cobalt ions in Co3O4. [1] Co2+ and Co3+ ratio is 1:2 [1] (4O2– = –8, so in order to balance the charge, it must be 2 Co3+ and 1 Co2+) Cobalt tetroxide can be reduced to cobalt oxide at 900 oC. Co3O4(s) 3CoO(s) + ½O2(g) HꝊrxn 1 Some relevant thermochemical data is listed in the Table 2.1 below. enthalpy change value / kJ mol–1 HꝊat Co(s) +426 HꝊf O2–(g) +850 Lattice energy of CoO(s) –3910 HꝊf Co3O4(s) –910 (ii) Using the data given in Table 2.1 and relevant information from the Data Booklet, show that the standard enthalpy change of formation