ACJC H2 CHEM P2 Answers Prelim

2 © ACJC 2014 9647/02/Prelim/14 [Turn over Planning 1 Distribution coefficient, KD, 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 distribution coefficient of acetylsalicylic acid (ASA) in two immiscible mixtures, water and trichloromethane as well as water and methylbenzene. The value s of the dis tribution coefficient can be used to determine Which is a better organic solvent for extracting ASA from water. It is predicted that each of these organic solvents will extract about 80- 90% of ASA dissolved in water. Distribution coefficient, KD (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 ASA (aq). 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 extracti ng 10.0 cm3 of the solution from the aqueous layer and titrating the aliquots against NaOH (aq). 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 cm3 of 0.01 mol dm-3 ASA (aq)  80 cm3 trichloromethane  80 cm3 methylbenzene  100 cm3 separating funnel for mixing and separating the two solutions  0.010 mol dm-3 stock solution of NaOH (aq) For Examiner’s use (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]

3 Independent variable – Organic solvent Dependent variable – Concentration of ASA in water and the organic solvent (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, s uggest why the 0.010 mol dm -3 NaOH stock solution is too concentrated for titrating the aqueous sample. [1] After extracting about 8 0-90% of ASA from the aqueous solution, the concentration of ASA (aq) is around 0.001 mol dm -3. The volume of 0.010 mol dm -3 NaOH used in the titration would be too small resulting in a large percentage error. (c) Procedures 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 ASA (aq). [5] 1. Using a measuring cylinder , measure 40 cm 3 of 0.01 mol dm -3 ASA (aq) and transfer it into the separating funnel. Measure 40 cm3 of trichloromethane and transfer it into the separating funnel. 2. Shake the mixture vigorously. Leave the separatin g funnel to stand for one hour at constant room temperature. 3. Using a 10 cm3 pipette, measure out 10.0 cm3 of the aqueous ASA solution. Transfer this solution into a conical flask. 4. Titrate the solution against the diluted NaOH (aq) using phenolphthalein as the indicator. Titrate until permanent pink colour is seen. 5. Repeat the titration to obtain consistent results. (d) Evaluation and analysis A student used 40 cm3 of 0.01 mol dm -3 ASA (aq) and 40 cm3 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.

4 Outline how you would use her results to determine KD (trichloromethane/water) of ASA. [3] Amount of ASA in aqueous solution = 4n / 1000 x 0.0010 = 4n x 10-6 mol Concentration of ASA in aq solution = (4n x 10-6) / 0.040 = n x 10-4 mol dm-3 Concentration of ASA in trichloromethane = [( x 0.01) – (4n x 10-6)] / = 0.01 – 0.0001n mol dm-3 KD (trichloromethane /water) of ASA = (0.01 – 0.0001n) / (n x 10-4) =(100-n)/n (e) The procedure and calculations above is repeated using methylbenzene and ASA (aq). Suggest how the values of the distribution coefficient for the two solvent mixtures can be used to identify which is the better organic solvent for extracting ASA from water. [1] The larger the value of distribution coefficient the better the organic solvent will be at extracting ASA from water. [Total: 12 marks] 2 Photographic films are coated with fine particles 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 granules will be converted to form th e 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 bromide and washing th e solution away. The fixer solution contains aqueous sodium thiosulfate, Na2S2O3. 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. [Ag(S2O3)2]3- [1]

5 (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. Ag+ (aq) + 2S2O3 2- (aq)  [Ag(S2O3)2]3- (aq) When the film is exposed to thiosulfate ions in the fixer solution, aqueous silver ions form [Ag(S2O3)2]3- (aq) which lowers the concentration of Ag +. This causes the position of equilibrium 1 to shift to the right so that AgBr dissolves. [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, Kc2, so as to justify her suggestion. 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 ΔGrxn against lg Kc3 is given for the reaction of excess 5.0 mol dm -3 NH3 (aq) and Ag + (aq) where Kc3 is the equilibrium constant of Equation 3. The graph shows the relationship ΔGrxn = ΔG° + RT lg Kc3 for Equation 3.

6 (i) State the value of ΔG rxn when the reaction is at dynamic