ACJC H2 CHEM P2 Questions Prelim

2 © ACJC 2014 9647/02/Prelim/14 Planning 1 Distribution coefficient, K D, is the ratio of equilibrium concentrations of a compound in a mixture of two immiscible solvents at a fixed temperature. Using two immiscible solvents, the distribution coefficient is a measure of the difference in solubility of a compound in these two solvents. This experiment aims to find out the dist ribution coefficient of acetylsalicylic acid (ASA) in two immiscible mixtures, water and trichloromethane as well as water and methylbenzene. The values of the distribution coefficient can be used to determine which is a better organic solvent for extracting ASA from water. It is predicted that each of t hese organic solvents will extract ASA from water with an efficiency of 80-90%. Distribution coefficient, K D (organic solvent / water) = [ASA] in organic solvent / [ASA] in water COOH O O Acetylsalicylic acid To find KD (organic solvent /water) of ASA, a certain volume of the organic solvent is added to a measured volume of 0.01 mol dm -3 aqueous ASA. Vigorous shaking of the mixture enables ASA to dissolve in the mixture of solvents. The mixture will reach equilibrium when it is left to stand for an hour at a suitable temperature. The concentration of ASA in each solvent is determined by extracting 10.0 cm 3 of the solution from the aqueous layer and titrating the aliquot against aqueous NaOH. By assuming that the remaining amount of ASA is dissolved in the organic solvent, the concentration of ASA in organic solvent can be calculated. You are provided with  80 cm 3 of 0.01 mol dm-3 ASA (aq)  80 cm 3 trichloromethane  80 cm 3 methylbenzene  100 cm 3 separating funnel for mixing and separating the two solutions  0.010 mol dm -3 stock solution of NaOH (aq) For Examiner’s use

3 @ACJC 2014 9647/02/Prelim/14 (a) In an experiment, the independent variable is a parameter that can be manipulated independently, whereas the dependent variable is the data to be collected that varies with the parameters that are manipulated. State the independent variable and dependent variable in this experiment. [2] For Examiner’s use (b) A student who was conducting the experiment discovered that the NaOH stock solution should be diluted to 0.001 mol dm -3 before it is suitable for titrating the aqueous sample. In light of the extraction efficiency of the organic solvents, suggest why the 0.010 mol dm -3 NaOH stock solution is too concentrated for titrating the aqueous sample. [1] (c) Procedure Describe the procedure to determine KD (trichloromethane/water) of ASA. Your plan should include  necessary steps to set up the equilibrium of ASA in water and trichloromethane  appropriate apparatus used  suitable volumes of solutions and solvents  details for titration of aqueous ASA solution, including the indicator and the end point colour. You should note that the same procedure should be suitable for repeating with a mixture of methylbenzene and aqueous ASA. …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… ……………………………………………………………………………

4 @ACJC 2014 9647/02/Prelim/14 …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… …………………………………………………………………………… [5] For Examiner’s use

5 @ACJC 2014 9647/02/Prelim/14 (d) Evaluation and analysis A student used 40 cm 3 of 0.01 mol dm -3 ASA (aq) and 40 cm 3 of trichloromethane in the experiment. From the titration, it was found that 10.0 cm3 of ASA (aq) required n cm3 of 0.0010 mol dm-3 of NaOH (aq) for neutralisation. Outline how you would use her results to determine K D (trichloromethane/water) of ASA. For Examiner’s use [3] (e) The procedure and calculations above is repeated using methylbenzene and aqueous ASA. Suggest how the values of the distribution coefficient for the two solvent mixtures can be used to identify which is a better organic solvent for extracting ASA from water. [1] [Total: 12 marks]

6 @ACJC 2014 9647/02/Prelim/14 2 Photographic films are coated with fine particl es of silver halides, mainly silver bromide. Exposing the film to light promotes reduction of Ag + so that some Ag+ ions are converted into fine granules of silver. During the film development process, the silver granul es will be converted to form the photographic image captured on film. Any excess silver bromide on the film darkens when it is exposed to light and causes the photo to turn black over time. This can be prevented by ‘fixing’ the film to dissolve the excess silver br omide and washing the solution away. The fixer solution contains aqueous sodium thiosulfate, Na 2S2O3. For Examiner’s use (a) (i) Thiosulfate ions act as monodentate ligands and bind to silver ions to form a complex. The silver(I) dithiosulfate ion has a linear shape about the silver ion. Write the formula of silver(I) dithiosulfate ion. [1] (ii) When the film is immersed in the aqueous solution, silver bromide granules dissolve to a small extent, according to Equation 1. AgBr (s) Ag+ (aq) + Br- (aq) ------- Equation 1 With the help of another equation, explain why silver bromide is soluble in aqueous sodium thiosulfate solution. [3] (b) A student suggests that a 5.0 mol dm -3 solution of aqueous ammonia can also be used as a fixer solution for photographic films. She suggests that the following reaction will occur. AgBr (s) + 2NH3 (aq) [Ag(NH3)2]+ (aq) + Br- (aq) ------ Equation 2 The student decides to calculate the equilibrium constant of Equation 2, K c2, so as to justify her suggestion.

7 @ACJC 2014 9647/02/Prelim/14 In order to calculate Kc2, the student considers another equilibrium as shown by Equation 3. Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq) ------- Equation 3 The graph of ∆G rxn against lg K c3 is given below for the reaction of excess 5.0 mol dm -3 NH3 (aq) and Ag + (aq) where K c3 is the equilibrium constant of Equation 3. The graph shows the relationship ∆Grxn = ∆G° + RT lg Kc3 for Equation 3. For Examiner’s use (i) State the value of ∆Grxn when the reaction is at dynamic equilibrium. [1] (ii) Hence calculate the value of K c3 for the reaction between excess 5.0 mol dm-3 NH3 (aq) and Ag + (aq) using ∆Grxn = ∆G° + RT lg K c3, under room condition and R is the molar gas constant. [2]

8 @ACJC 2014 9647/02/Prelim/14 (iii) The Ksp value of AgBr is 5.35 x 10-13. Write an equation to show the relationship between Kc2, Kc3 and Ksp of AgBr. [1] For Examiner’s use (iv) Using your answers in (b)(ii) and (b)(iii), calculate the value of Kc2. [You may use Kc3 = 2.00 x 107 for this calculation if you did not get an answer for (b)(ii).] [1] (v) Use your calculated value fro