ACJC H2 CHEM P2 ANS Prelim

2 1 Planning (P) When a solute is added to two solvents, A and B, which do not mix, some of the solute dissolves in each of the solvents and an equilibrium is set up between the two solvents. It has been shown that for dilute solutions, at equilibrium the ratio of the two concentrations is a constant known as the Partition Coefficient, K, and it will remain a constant if the solute remains in the same molecular state in the two solvents. Concentration of solute in solvent A = K Concentration of solute in solvent B An experiment was conducted to verify the above observation. Ethanoic acid (solute) was shaken with two immiscible solvents (water and cyclohexane, (C 6H12)), so that the solute distributed itself between the two solvents. The concentrations of the two solutions were then determined so that the ratio [CH 3COOH]aq/[CH3COOH]S known as the Partition Coefficient, K can be calculated at a given temperature. Using apparatus which are found as standard items in a school laboratory, an experiment was carried out to determine the Partition Coefficient , K for ethanoic acid, CH 3COOH, between water and cyclohexane, (C6H12). The following steps were carried out: 1. 5.0 g of the sample was first dissolved in 50 cm 3 of water in a beaker and the aqueous solution was then transferred into a separating funnel. 2. 50 cm 3 of cyclohexane was then poured into t he separating funnel containing the aqueous solution, stoppered and shaken intermittently. 3. The concentration of ethanoic acid in the two layers was then determined by titration with 0.05 mol dm-3 NaOH. (a) (i) Outline how a fixed volume of the sample of the aqueous layer can be removed from the mixture in the separating funnel for titrimetric analysis. (Note: The organic layer is less dense compared with the aqueous layer.) Either • Place a finger on the open end of a 10.0 cm 3 pipette and insert it into the bottom aqueous layer. • Gently blow lightly into the pipette to dispel any organic layer that may have entered the pipette before withdrawing the required amount of sample Or • Drain the aqueous layer into one dry beaker and the organic layer into another • Pipette 10.0 cm3 of the aqueous layer into a conical flask [2] ACJC 2011 9647/02/Aug/11 [Turn over

3 (ii) Outline stepwise how the removed sample could be analysed by titration to determine the concentration of the solute that had dissolved in that layer. • Place the pipetted sample into a 250 cm 3 conical flask • Add a drop of phenolphthalein • Fill a burette with 0.05 mol dm-3 aqueous NaOH • Titrate the sample until a faint pink colour is obtained • Repeat titration for consistent results [2] (b) When analysing the organic layer, a fixed volume of water (about twice the volume) is added to the pipetted volume of cyclohexane solution in the conical flask and shaken vigorously, prior to the titration. Explain the purpose of this action. • NaOH(aq) will not be able to neutralise the ethanoic acid in the organic layer • The ethanoic acid must be first extracted into the water for complete reaction with the alkali during neutralisation [2] (c) Based on your procedure, state and explain one significant error or limitation that could be encountered in determining an accurate value for the distribution ratio. • Short equilibration time for partitioning • Some organic layer could have been sucked up the pipette when collecting aqueous layer for analysis (if student withdrew sample directly from separating funnel) • Titre reading for organic layer may be lesser than expected due to limited or partial partitioning of acid into water [Any one of the above - 1] [1] (d) Suggest one possible modification that would minimise the error or limitation that you have stated in (c) . Explain how this modification leads to an improvement in the accuracy and reliability of the results. • Allow a longer time for partitioning/distribution of acid between the two solvents • Add/use more water to extract acid quantitatively into aqueous layer before titration or use a fixed volume of standard alkali for extraction and determine solute by back titration with standard acid. [Any one of the above - 1] [1] ACJC 2011 9647/02/Aug/2011 [Turn over

4 (e) A student did the above experiment but carried out onl y one titration using a fixed aliquot of the aqueous layer and then proceeded to use the titre value to calculate a value for K. Assuming the student obtained a titre value of y cm 3 of M mol dm -3 NaOH (aq) when 10.0 cm 3 of the aqueous layer was titrated, outline how the student could have used the result to determine a v a l u e f o r K . No of moles of CH3COOH in 50 cm3 of aqueous layer = (5My / 1000) mol Thus [CH 3COOH]aq = (My / 10) mol dm-3 Total No of moles of CH 3COOH used = 5 / 60 No of moles of CH3COOH in 50 cm3 of organic layer = (5/60 – 5My/1000) Thus, [CH 3COOH]s = (5/60 – 5My/1000) x (1000/50) mol dm-3 = [5/3 - My / 10] mol dm-3 K = (My / 10) / [5/3 - My / 10] or (0.1My) / (1.67 – 0.1My) [3] (f) When a solute is distributed between the two immiscible solvents, it is important to ensure that the resulting solutions are dilute for the distribution to occur and the partition coefficient to remain constant. Suggest why this is critical in this experiment. • Ethanoic acid may dimerise by way of hydrogen bonding in the organic layer when it is concentrated and thus affect the distribution ratio. [1] [Total: 12] 2 (a) A hydrocarbon W contains 92.3 % carbon. On complete combustion, 0.005 mol of W produces 1.76 g of CO2. (i) Determine the molecular formula of W. mole ratio C : H = 1 : 1 ∴ empirical formula of W is CH. Amount of CO2 = 0.0400 mol ∴ mole ratio W : CO2 = 0.005 : 0.0400 = 1 : 8 ∴ Molecular formula of W is C 8H8. (ii) Given that 1 mol of W reacts with 1 mol of bromine in tetrachloromethane, draw the disp