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