VJC H2 CHEM P2 ANS FINAL Prelim

 VJC 2011 9647/02/PRELIM/11 [Turn over 1 Victoria Junior College 2011 H2 Chemistry Prelim Exam 9647/2 Suggested Answers 1 Planning An acid can be represented by the general formula, H xA, where x represents the basicity of the acid. A bottle of a dilute aqueous solution of an acid, either monobasic or dibasic, was found. However, the label on the bottle had been damaged. Only the concentration of the acid, 1.00 mol dm 3 was readable. A student was given a 1.00 mol dm 3 sodium hydroxide solution. She was asked to determine the basicity of the acid in the bottl e by mixing different volumes of the acid and sodium hydroxide. (a) (i) Write an equation to represent the molar enthalpy change of neutralization between sodium hydroxide and HxA. 1 x HxA + NaOH  1 x NaxA + H2O [1] (ii) A student determined the basicity of the acid by performing the following experiments: Experiment 1: 30 cm3 of HxA(aq) was added to 60 cm3 of NaOH(aq) Experiment 2: 60 cm3 of HxA(aq) was added to 30 cm3 of NaOH(aq) The changes in temperature of the mixture were measured for Experiment 1 and Experiment 2 as T1 and T2 respectively. Suggest and explain the basicity of HxA if T1 = 2 x T2 Heat is released when water is formed in the neutralization process. The amount of heat released in Experiment 1 is twice that in Experiment 2. Since total volume of mixture remained unchanged, therefore, the number of moles of water formed in Experiment 1 is twice that in Experiment 2. H xA is a dibasic acid because in Experiment 1, all the acid and alkali are reacted while in Experiment 2, the alkali is the limiting reagent. Alternatively If x = 1 HA + NaOH  NaA + H 2O [2]

 VJC 2011 9647/02/PRELIM/11 [Turn over 2 Experiment H xA NaOH ηH2O 1 30 (limiting) 60 (excess) y 2 60 (excess) 30 (limiting) y Both experiments will produce same amount of water. Hence ∆T1 = ∆T2 (since total volume of mixture is fixed) Experiment H xA NaOH ηH2O 1 30 (limiting) 60 (limiting) 2y 2 60 (excess) 30 (limiting) y If x = 2 H2A + 2 NaOH  Na2A +2 H2O Experiment 1 will produce twice the amount of H2O. Hence ∆T1 = 2∆T2 (since total volume of mixture is fixed) ∴ x = 2 dibasic acid (b) For this part, you may assume that the acid in the bottle is a monobasic acid, HA. A student decided to determine whether the acid, HA, is a strong or weak acid by performing a series of experiments involving mixing of hydrochloric acid with sodium hydroxide and HA with sodium hydroxide. Explain how, from the suggested experiment, the student might be able to determine whether HA is a strong or weak acid. If HA is a weak acid, some of the heat evolve during neutralization is absorbed to dissociate the weak acid, HA, fully. Hence ∆T (HA/NaOH) < ∆T (HCl/NaOH). If HA is a strong acid, full dissociation occurs like HC l, ∆T (HA/NaOH) = ∆T (HCl/NaOH) [2] FA 1 is a solution of sodium hydroxide of unknown concentration. FA 2 is 1.0 mol dm -3 hydrochloric acid. Since neutralization reaction is exothermic, a series of experiments can be performed by mixing different volumes of FA 1 and FA 2 to determine the concentration of sodium hydroxide. (c) (i) Write a procedure to determine the temperature changes for the series of reactions between FA 1 and FA 2. Your answers should include choice of apparatus to measure the volume and temperature of the solutions. Give suitable headings for the columns numbered 1 to 5 as part of the plan of the experiment.

 VJC 2011 9647/02/PRELIM/11 [Turn over 3 Volume of FA 1 /cm3 Volume of FA 2 /cm3 1 2 3 4 5 30.00 40.00 33.00 37.00 36.00 34.00 40.00 30.00 44.00 26.00 48.00 22.00 50.00 20.00 Headings for 1. Temperature of FA 1 /oC 2. Temperature of FA 2 /oC 3. Average temperature of solutions before mixing /oC 4. Final temperature of solution after mixing /oC 5. Temperature change /oC Procedure 1. Use a burette to transfer 30.00 cm 3 of FA 1 into the Styrofoam cup labeled FA 1. Place the cup in a 250 cm 3 beaker to prevent it from tipping over. 2. Use another burette to transfer 40.00 cm 3 of FA 2 into the styrofoam cup labeled FA 2. 3. Use two thermometers to stir and measure the temperature of the FA 1 and FA 2 solution. 4. Calculate the average temperature of the solutions. 5. Add the contents of FA 2 cup to the FA 1 cup. Use the thermometer to stir the mixture and measure the maximum temperature of the mixture. 6. Wash and carefully dry both the FA 1 and FA 2 styrofoam cups. 7. Repeat steps 1 to 5 using 40.00 cm 3 and 50.00 cm 3 of FA 1, each time using the appropriate volume of FA 2 so that the total volume of reacting mixture is 70.00 cm3. [4]

 VJC 2011 9647/02/PRELIM/11 [Turn over 4 (ii) The following points were plotted on a grid after an experiment. Draw a suitable graph through the plotted points. [1] (iii) By using the graph in (ii), calculate the concentration of sodium hydroxide in the experiment. Volume of FA 1 used when temperature change is max = 36.5 cm 3 No. of moles of FA 2 (hydrochloric acid) used = (70–36.5) / 1000 × 1.0 = 0.0335 mol Concentration of FA 1 = 0.0335 / (36.5 / 1000) = 0.918 mol dm -3 [2] [Total: 12] 15.0 14.0 13.0 12.0 11.0 10.0 9.0 30 50 454035 Volume of FA 1 / 3

 VJC 2011 9647/02/PRELIM/11 [Turn over 5 2 Oxoanions of Group VII elements have the general formula XOm -, where m = 1, 2, 3 or 4. These oxoanions are strong oxidizing agents. (a) Explain why fluorine does not form oxoanions. Fluorine is the most electronegative element, and is unable to form a compound in which it has a positive oxidation number. In addition, fluorine is in Period 2 of the Periodic Table, thus unable to expand its octet structure to form more than 1 covalent bond. [2] (b) 1.25 x 10 -3 mol of an aqueous bromate salt containing the BrO m - anion was added to excess potassium iodide. The resulting mixture was washed with chloroform to dissolve the iodine, and the aqueous and organic layers were separated. Silver nitrate solution was added to the aqueous layer, and a mixture of two precipitates was obtained. (i) Identify the two precipitates formed. AgBr (redox product) and AgI (excess KI) [1] (ii) The iodine collected in the organic layer was titrated against 0.500 mol dm -3 sodium thiosulfate. 14.90 cm 3 of titrant was required to discharge the blue- black colour of the starch indicator. Calculate the value of m. I2 + 2S2O3 2-  2I- + S4O6 2- nthiosulfate = 14.90 0.5001000  = 7.45 x 10-3 mol niodine = ½ nthiosulfate = 3.725 x 10-3 mol 31025.1 10725.3 3 3     bromate iodine n n (nearest whole number) I2 + 2e-  2I- 6 mol of electrons were transferred Original oxidation state of Br in BrOm - = -1 + 6 = +5 5 + m(-2) = -1 m = 3 [3] (iii) The two precipitates can be separated by addition of concentrated aqueous ammonia, followed by filtration. Describe what you expect to observe and explain the chemical principles behind this method. The cream ppt dissolves, while yellow ppt is obtained as the residue.

 VJC 2011 9647/02/PRELIM/11 [Turn over 6 High [NH3] causes formation of soluble [Ag(NH 3)2]+ complex, lowering [Ag+]. As a result, the equilibrium Ag+(aq) + Br-(aq) AgBr(s) shifts to the left. The ionic product [Ag +][ Br -] falls below Ksp and hence the salt dissolves. The Ksp of Ag I is so low that even at high [NH 3], the ionic product [Ag+][ I-] is still higher than Ksp. [4, max 3] (c) 0.500 g of a Group II iodate(V) salt, M( IO3)2, was heated and decomposed to give a white solid, a purple gas, and a colourless gas that rekindles a glowing splint. (i) Identify the 3 decomposit