MI Prelim P3 Ans

Class Adm No Candidate Name: This question paper consists of 11 printed pages and 1 blank page. 2018 Preliminary Exams Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 18th Sept 2018 2 hours Candidates answer on separate paper. Additional materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number 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. Section A – Answer all questions. Section B – Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for good English and clear presentation in your answers. 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. Question Section A Section B Total 1 2 3 4 5 Marks 23 20 17 20 20 80

2 Section A Answer all questions from this section. 1 Oxalic acid is a weak diprotic Br Ønsted acid with p Ka1 and p Ka2 values of 1.25 and 4.14 respectively. 25 cm 3 of 0.50 mol dm –3 oxalic acid was titrated against a solution of 1 mol dm –3 of sodium hydroxide. Only 1 to 2 drops of phenolphthalein indicator was used in this titration. It was found that the first colour change occurred at 19.40 cm3. A sketch (not drawn to scale) of the pH titration curve is shown below. (a) Suggest why the value of pKa2 is larger than pKa1. [2] Removal of H + for p Ka2 is from a negatively charged species while that for p Ka1 is from a neutral species, hence dissociation for p Ka2 is more difficult / (COOH)COO – is less acidic than (COOH)2 and the value of pKa1 is larger. or The first conjugate base of oxalic acid is stabilised by an intramolecular ion-dipole interaction with the OH group of (COOH)COO –, hence dispersing the negative charge to a greater degree and increasing the acid strength. Thus the first p Ka value is smaller than the second pKa as the second conjugate base formed does not have this extra stability. (b) (i) Calculate the value of pH at point W. [1] [H+] = ඥ(10ିଵ.ଶହ)(0.50) = 0.1677 mol dm-3 pH = 0.776 pKa1 ⇌ volume of NaOH added / cm3 pKa2 ⇌ pH Y X W

3 [Turn over (ii) Given that the pH at point X is 1.25, calculate the volume of NaOH added at point X. [2] Amount of oxalic acid = 0.50 x ଶହ ଵ଴଴଴ = 0.0125 mol pH = pKa1 + log ቀ ሾsaltሿ ሾacidሿቁ Let ݔ