VJC Prelim P3 Answers

1  VJC 2017 9729/03/PRELIM/17 [Turn over Victoria Junior College 2017 H2 Chemistry Prelim Exam 9729/3 Suggested Answers Section A Answer all the questions in this section. 1 (a) Ethanoic acid and its salt, sodium ethanoate, is often used as an acid buffer to help extend the shelf-life of food products such as meat, fish and dairy. A buffer solution of pH 5.5 was prepared by mixing 0.200 mol dm−3 of ethanoic acid and 0.200 mol dm−3 aqueous sodium ethanoate. (Ka of ethanoic acid = 1.74 x 10−5 mol dm−3) (i) What do you understand by the term buffer solution? [1] A buffer solution can maintain a fairly constant pH when a small amount of acid or base is added to it. (ii) Calculate the ratio of acid][ethanoic ][ethanoate required to prepare the above buffer solution. Hence deduce whether the above buffer is more effective in buffering against added acids or bases. Write an equation to illustrate your answer. [3] Ka = [HA] ]A][[H - = 1.74 x 10 −5 where A − is ethanoate and HA is ethanoic acid. [HA] ]A[ - = 5.5- -5. 10 10741  = 6- -5 . . 10163 10741   = 1 515. As higher [ ethanoate] is present, buffer is better at removing added acids. CH3CO2− + H+  CH3CO2H (iii) Calculate the volume of ethanoic acid and that of aqueous sodium ethanoate required to prepare 60 cm3 of the above buffer. [1] Let x cm 3 be volume of ethanoate and (60 – x) cm 3 to be volume of ethanoic acid. x x 60 = 1 515. x = 50.8 cm3 Hence 50 .8 cm3 of sodium ethanoate and (60- 50.8) = 9.20 cm3 of ethanoic acid will be required.

2  VJC 2017 9729/03/PRELIM/17 [Turn over (iv) The boiling points of the components in the buffer are as shown. Ethanoic acid 118oC Sodium ethanoate 881oC With reference to the structure and bonding present in the compounds, account for the difference in the boiling points. [3] Ethanoic acid has a simple molecular structure with hydrogen bonding between the polar molecules. Sodium ethanoate has a giant ionic structure with very strong electrostatic attractions between the oppositely charged ions. Hence a lot of heat energy is required to overcome the ionic attractions compared to the much weaker hydrogen bonding between acid molecules. (b) The limestone that collects in kettles in hard water areas is mainly calcium carbonate. It can be removed fairly harmlessly by using a warm solution of vinegar, which contains ethanoic acid. The limestone dissolves with fizzing and a solution of calcium ethanoate remains. (i) Write a balanced equation for the reaction between ethanoic acid and calcium carbonate. [1] 2CH3COOH + CaCO3  Ca(CH3COO)2 + CO2 + H2O When the solution in (b) is evaporated and the resulting solid calcium ethanoate is heated strongly in a test-tube, an organic compound Q is formed, which condenses to a colourless liquid. The residue in the tube contains calcium carbonate. When 0.10 g of compound Q was injected into a gas syringe at a temperature of 383 K and a pressure of 101 kPa, 55 cm3 of vapour were produced. (ii) Calculate the relative molecular mass of Q. [2] PV = nRT = M mRT Rearranging gives M = PV mRT = 6-3 .. 105510101 383318100   = 57.3 (iii) Compound Q is neutral and water -soluble. Q does not react with sodium metal nor with Fehling’s solution but it does react with alkaline aqueous iodine. Suggest a structural formula for Q. Justify your answer by reference to these properties of Q. [4] Reaction Deduction Neutral Absence of acidic and basic groups No reaction with Na metal Absence of –OH group so not alcohol ad carboxylic acid No reaction with Fehling’s solution Absence of aliphatic aldehyde

3  VJC 2017 9729/03/PRELIM/17 [Turn over Positive test with alkaline aqueous iodine Presence of CH3CO– group Mass of CH3CO– group = 43.0 hence remaining group has a mass of 57.9 – 43.0 = 14.9 ≈ 15.0  –CH3 group present Q is propanone CH3COCH3 (iv) Construct a balanced equation for the formation of Q by the action of heat on calcium ethanoate. [1] (CH3COO)2  CaCO3 + CH3COCH3 (v) The residue calcium carbonate can also undergo thermal decomposition when heated very strongly. Copper(II) carbonate behaves in a similar manner when heated. State which carbonate, calcium or copper( II) will have a higher decomposition temperature. Explain your answer, with the help of relevant data from the Data Booklet. [3] CaCO3 will have a higher decomposition temperature. Ionic radius of Ca2+ = 0.099 nm; Cu2+ = 0.073 nm Size of Ca2+ is larger so its charge density is lower. Polarising power of Ca2+ is thus lower and less able to distort the electron cloud of CO32- ion. Hence, CaCO3 is more stable to heat and has a higher decomposition temperature. [Total: 19]

4  VJC 2017 9729/03/PRELIM/17 [Turn over 2 (a) In a closed reaction vessel maintained at a high temperature, 1 mol of Al2Cl6 dimers dissociate into AlCl3 according to the following equation. Al2Cl6(g) ⇌ 2AlCl3(g) The average Mr of the equilibrium gas mixture is found to be 178.0. (i) Given that the average Mr of the equilibrium gas mixture is the sum of the mole fractions of each gas, multiplied by the Mr of that gas , calculate the degree of dissociation of Al2Cl6, . [3] Al2Cl6(g) ⇌ 2AlCl3(g) initial amt / mol 1 0 change x +2x eqm amt / mol 1 x 2x Total amt of gases present at eqm = 1 x + 2x = (1 + x) mol Average Mr = [(1 x) / (1+ x)] Mr(Al2Cl6) + [2x / (1 + x)] Mr(AlCl3) = 178.0 178.0 = [(1 x) / (1 + x)] (267.0) + [2x / (1+ x)] (133.5) x = 0.50 Hence,  = x / 1 = 0.50 (ii) Write down the expression of Kp for the above equilibrium system. Hence, show that Kp = KcRT. [2] Kp = (PAlCl3)2 / PAl2Cl6 Using PV = nRT; P = (n/V)RT = CRT Hence, Kp = ([AlCl3]RT)2 / [Al2Cl6]RT = [AlCl3]2RT / [Al2Cl6] = KcRT (iii) An ex periment on the dissociation of A l2Cl6 was conducted in a clos ed container of a fixed volume and the partial pressures of all the components were plotted as shown in the figure below. Copy and complete the diagram to show how the partial pressure of each gas changes when the volume of the container is halved at t1 until t2. [2] P