CJC H2 CHEM P2 SOL Prelim

CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 2 CANDIDATE NAME CLASS 2T CHEMISTRY 9647/02 Paper 2 Structured Questions Monday 24th August 2015 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class in the spaces provided above. Write in dark blue or black pen in the spaces provided, on the Question Paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. You are remind ed of the need for good English and clear presentation in your answers. The number o f marks is given in brackets [ ] at the end of each question or part of the question. For Examiner’s Use Paper 1 40 Paper 2 Q 1 12 Q 2 15 Q 3 15 Q 4 15 Q 5 15 72 Paper 3 Q 1 20 Q 2 20 Q 3 20 Q 4 20 Q 5 20 80 Total 192 ANSWER SCHEME

2 1 Planning (P) Cream of tartar (potassium hydrogen tartrate, KHC 4H4O6) is one of the ingredients in baking powder which is used to make cakes rise. Potassium hydrogen tartrate is a weak acid that is not very soluble in water. KHC4H4O6(s) ⇌ K+(aq) + HC4H4O6 –(aq) The HC 4H4O6 –(aq) ion contains one acidic hydrogen, so the quantity of potassium hydrogen tartrate in solution can be determined by titration with a base. HC4H4O6 –(aq) + OH–(aq)  H2O(l) + C4H4O6 2–(aq) You are to design an experiment to determine the solubility of potassium hydrogen tartrate in two solvent systems: pure water and 0.100 mol dm–3 potassium chloride. In addition to the standard apparatus available in a school laboratory, you are provided with the following materials:  potassium hydrogen tartrate  distilled water  methyl orange indicator  phenolphthalein indicator  standard sodium hydroxide solution (a) Write an expression for the solubility product, Ksp, of potassium hydrogen tartrate, stating its units. …………………………………………………………………………………………………. [1] (b) When a solution is saturated, the undissolved solid is in equilibrium with its aqueous solution. X(s) + aq ⇌ X(aq) Describe how you would prepare a saturated solution of potassium hydrogen tartrate at room temperature. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. [3]  Dissolve solid KHC4H4O6 in water until some solid remains undissolved.  Allow solution to stand for few hours (to establish equilibrium).  Filter to remove undissolved solid. The saturated solution of KHC4H4O6 is collected as the filtrate. OR  Dissolve solid KHC4H4O6 in hot water until some solid remains undissolved.  Cool the solution to room temperature to saturate. Excess solid would crystallise out.  Filter to remove excess solid/crystals. The saturated solution of KHC4H4O6 is collected as the filtrate. Ksp = [K+ ] [HC4H4O6 – ] mol2 dm–6

3 9647/02/CJC JC2 Preliminary Exam 2015 (c) The concentration of a saturated solution of potassium hydrogen tartrate may be determined by titration with standard sodium hydroxide. (i) Given that, at room temperature, Ksp of potassium hydrogen tartrate has a numerical value of 6.27 × 10 –4, suggest an appropriate concentration of the standard sodium hydroxide solution to be prepared. Show your working. (ii) Name the indicator used in this titration. ……………………………………………………….………………………………….. [3] (d) Describe what further experiments you would carry out to determine the solubility of potassium hydrogen tartrate in 0.100 mol dm–3 potassium chloride. State clearly the volume of solution used and the expected observation. …...…………………………………………………………………………………………….. …...…………………………………………………………………………………………….. ……………...………………………………………………………………………………….. ………………………...……………………………………………………………………….. …………………………………...…………………………………………………………….. ……………………………………………...………………………………………………….. ……………………………………………………………………………...………………….. State and explain how the titration result of this further experiment would compare with that of the solution prepared in (b). …...…………………………………………………………………………………………….. ………………………………………………………………………………………...……….. ………………………………………………………………………...……………………….. [5] [Total: 12]  for a saturated solution of KHC4H4O6, [K+ ] [HC4H4O6 – ] = Ksp   [HC4H4O6 – ] = spK (since [K+ ] = [HC4H4O6 – ]) = 41027.6  = 0.0250 mol dm–3  To avoid too small a titre value which has high percentage error, or too large a titre value which requires more than one burette full of titrant, [titrant] is chosen such that volume of titrant at end-point is about equal to volume of analyte. Since, HC4H4O6 – + OH–  C4H4O6 2– + H2O  [NaOH ] = [HC4H4O6 – ] = 0.0250 mol dm–3  a suitable [NaOH ] = 0.0250 mol dm–3 [Accept answers between 0.0200 and 0.0300 mol dm–3 ] phenolphthalein indicator  Prepare a saturated solution of KHC4H4O6 in 0.100 mol dm–3 KCl; i.e. repeat procedure stated in (b) but use 0.100 mol dm–3 KCl instead of water.  Pipette 25.0 cm3 of the saturated solution of KHC4H4O6 prepared above (into a 250 cm3 conical flask). Add (2-3 drops of) phenolphthalein indicator. Titrate with standard / 0.0250 mol dm–3 NaOH (placed in a 50 cm3 burette) until the solution in the conical flask changed from colourless to pink. [1] correct indicator + end-point colour change  vol of NaOH less than that for saturated solution of KHC4H4O6 in water.  Presence of common ion, K+, lowers/depresses the solubility of KHC4H4O6.

4 9647/02/CJC JC2 Preliminary Exam 2015 2 Carbon is the fourth most abundant element in the universe by mass after hydrogen, helium and oxygen. It is present in all forms of carbon -based life, and in the human body. This abundance, together with the unique diversity of or ganic compounds , makes this element the chemical basis of all known life. (a) (i) Organic compound P contains only the elements carbon, hydrogen and oxygen in the following composition by mass: C, 40 %; H, 6.7 %. Calculate the empirical formula of P. (ii) When a 0.102 g sample of compound P was vapourised in a suitable apparatus, the vapour occupied 59 cm3 at 150 °C and 101 kPa. Calculate the relative molecular mass of compound P, and hence determine its molecular formula. (iii) Compound P gives a silver mirror with Tollens’ reagent. Draw the displayed formula of compound P. [5] % by mass of O = 100 – 40 – 6.7 = 53.3% Let mass of compound P be 100 g. C H O Mass / g 40 6.7 53.3 Amount / mol 40 ÷ 12.0 = 3.33 6.7 ÷ 1.0 = 6.7 53.3 ÷ 16 = 3.33 Simplest ratio 1 2 1 Hence, the empirical formula of P is CH2O. with working Using ideal gas equation, pV = nRT = rM m RT  Mr = pV mRT = )1059)(10101( )273150(31.8)102.0( 63   correct conversions = 60.2 correct value to 1 dp; NO units Let the molecular formula of P be CnH2nOn. Since Mr of CnH2nOn = 60.2 n(12.0) + 2n(1.0) + n(16.0) = 60.2  n = 2 Hence, molecular formula of P is C2H4O2. structure & all bonds shown

5 9647/02/CJC JC2 Preliminary Exam 2015 (b) Starch is another organic compound consisting of carbon, hy drogen and oxygen. The hydrolysis of starch into simpler sugars can be catalysed by the enzyme amylase. Amylase is present in the saliva of humans and acts as a biological catalyst in the digestion of starch. (i) A few experiments are carried out wi