2015 HCI H1 Chemistry Prelims P2 Answers

1 HWA CHONG INSTITUTION Preliminary Examination Higher 1 CANDIDATE NAME CT GROUP 14S CHEMISTRY Paper 2 Candidates answer Section A on the Question Paper. Additional Materials: Data Booklet Writing paper 8872/02 2 Sept 2015 2 hours READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue, correction fluid or tapes. Section A Answer all questions. Section B Answer two questions on separate answer paper. 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 Paper 1 Paper 2 TOTAL Multiple Choice Section A (Structured) Section B (Free Response) Q1 /18 Q4 / 20 Q2 /10 Q5 / 20 Q3 /12 Q6 / 20 / 30 Subtotal / 40 Subtotal / 40 This question booklet consists of 14 printed pages. 110

2 Section A Answer all the questions in this section in the spaces provided. 1 There is concern over the way the oceans are becoming more acidic as more carbon dioxide dissolves in them (a) (i) Draw a “dot-and –cross” diagram for carbon dioxide [1] (ii) Use your diagram to state and explain the shape and bond angle of a carbon dioxide molecule. Linear and 180 o  Two areas of electron density ( bonding pairs) around central C atom  Bonding pairs / electron pairs repel each other; get as far away from each other as possible to minimize repulsion.  ………………………………………………………………………………… [3] (iii) Carbon dioxide forms hydrogen bonds with water. Draw a diagram to illustrate this. Partial charges ; lone pairs ; [2] (iv) Carbon dioxide is not very soluble in water. Suggest an explanation for this in terms of hydrogen bonding. Hydrogen bonds in water Fewer hydrogen bonds between CO 2 and water.  ……………………………………………………………………………… [2] Hydrogen bond

3 (b) When carbon dioxide dissolves in water of the ocean, the following reactions occur. CO2(g)  CO2(aq) equation 1 CO2(aq) + H2O(l)  HCO3 r (aq) + H+(aq) equation 2 HCO3 r (aq)  H+(aq) + CO3 2r (aq) equation 3 (i) The reaction in equation 3 can reach a state of dynamic equilibrium. Explain what is meant by the term dynamic equilibrium. Rate of forward = rate of backward reaction Concentrations of reactants and products remain constant ( formed at same rate) [2] (ii) Suggest why the balance of CO 2(g) in the atmosphere and CO2(aq) in the oceans cannot be regarded as a dynamic equilibrium. System is not close / CO2 moves away from the surface. ………………………………………………………………………………………. [1] (iii) Explain why an increase in the concentration of dissolved carbon dioxide leads to an increase in the acidity of the water. Equilibrium position in equation 2 moves to right, increased H+ concentration ……………………………………………………………………………………. ……………………………………………………………………………………. [1] (iv) The pH of the oceans is buffered by the reaction in equation 2. Explain the meaning of buffered. pH remains almost constant buffered solution resists pH change when small amounts of acid or alkali added. [1] (v) Give the important condition necessary for this equilibrium to result in buffering, in terms of concentration of species present. Large HCO 3 - concentration.( reservoir) Or HCO3 - concentration similar to the large CO2 concentration …………………………………………………………………………………….. [1]

4 (vi) Reference books states that the pH of the oceans has changed from 8.179 in pre-industrial times to 8.069 today. Calculate the percentage increase in [H +] 10-8.179 = 6.62 x 10-9 10-8.069 = 8.53 x 10-9 1.91/6.62 x 100 = 29% % increase in [H +] = ………………………………………….[1] (c) The shells of some sea creatures are made of calcium carbonate. Use the equations below to explain a possible effect of increased acidity on the shells of these sea creatures. CaCO 3(s)  Ca2+(aq) + CO3 2r (aq) equation 4 CO3 2r (aq) + H+(aq)  HCO3 r (aq) equation 5 Equilibrium (position) in equation 5 moves to right with increase H+ Equilibrium in Equation 4 moves to the right CaCO3 dissolves [2] (d) The concentration of a saturated solution of carbon dioxide in water is 3.3 x 10-3 mol per 100 g at room temperature and pressure. 1.0 kg of this saturated solution is boiled, releasing all the CO2. Calculate the volume that this CO 2 would occupy at room temperature and pressure. One mole of gas at room temperature and pressure occupies 24 dm3. 3.3 x 10-3 x 10 x 24000 = 790 or 792 cm3 Volume of CO 2 : …………………………..………………... [1] Total [18]

5 2 (a) E300 is an oxidant used in white wines. The maximum allowed concentration of E300 in drinks is 150 mg dm r3. A student performed the following redox titration procedure to find out if a 250.0 cm3 sample of a drink containing E300 was within this limit. The sample was acidified followed by the addition of 25.0 cm 3 of 0.00500 mol dm-3 KIO3(aq). Excess KI(aq) was then added to form I2 in solution. IO3 r (aq) + 5I r (aq) + 6H+ (aq)  3I2(aq) + 3H2O(l) (i) Calculate the amount, in moles, of iodine, I2, formed in this reaction. n KIO3 = 0.00500 x 25.0 x 10-3 =0.000125 mol n I 2 = 3 x 0.000125 = 3.75 x 10-4mol [1] (ii) Some of the I 2 formed reacted with the E300 in the 250.0 cm3 sample of the drink. C6H8O6 + I2  C6H6O6 + 2H+ + 2I r E300 The amount of unreacted I2 was found by titrating with sodium thiosulfate, Na2S2O3(aq), using starch indicator. At the end point, 20.4 cm3 of 0.00500 mol dm-3 Na2S2O3(aq), had been added. The following reaction occurred: I2(aq) + 2S2O3 2r (aq)  2I r (aq) + S4O6 2r (aq) Calculate the amount, in moles, of iodine, I2, remaining after the E300 had reacted. n S2O3 2- = 0.00500 x 20.4 x 10-3 =0.000102 mol n I 2 = 0.5 x 0.000102 = 0.000051 mol = 5.10 x 10-5 mol [1] (iii) Determine the concentration of the E300 in the 250.0 cm3 sample of the drink and hence whether the drink is within the limit allowed. M r ( E300) = 176 n E300 = nI 2 – I2n = 0.000375 - 0.000051 = 0.000324 mol a(i) a(ii) conc E300 = 0.000324 x 1000/250.0 =0.001296 mol dm-3 0.001296 x 176 = 0.228 g dm -3 = 228 mg dm-3 > 150 mg dm-3 Concentration = 228 mg dm-3 units: mg dm-3 Is the drink within the allowed limit for E300? No [3]

6 (b) E300 has a C=C bond with two different groups on each carbon. It does not, however, show geometrical isomerism whereas 1,2-dichloroethene does. Explain why 1,2-dichloroethene shows geometrical isomerism and suggest a reason why E300 does not. Restricted rotation around C=C bond each carbon atom has two different groups/atoms attached to it The two –OH groups in E300 can only be on the same side of C=C; ring structure will not allow them to be on opposite side/ ring cannot rotate. [3] (c) The primary alcohol group in E300 reacts with C 17H35COOH to form another antioxidant. (i) Draw the structural formula of the compound formed in this reaction. [1] (ii) What else must be added to a mixture of E300 and C 17H35COOH, to make the new antioxidant? Conc H2SO4 [1] Total [10]

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