SAJC H2 Chem 2013 Prelim P2 Soln

1   St Andrew’s Junior College Suggested Answer Scheme Prelim 2013 Paper 2 1 (a) Preparation of saturated solution of lead iodate for titration  Stir and add solid lead iodate(V) to about 100 cm 3 of distilled water in a 250 cm3 beaker until no more solid dissolves.  Filter and collect the filtrate which contains saturated Pb(IO3)2 solution. Titration  Pipette 25.0 / 10.0 cm 3 of this saturated solution into a 250 cm 3 conical flask  Add excess solid K I to the flask. Use a 10 cm3 measuring cylinder to add about 5 cm3 of HCl (aq) to the flask.  Titrate the saturated solution with Na 2S2O3 from the 50 cm 3 burette into the flask  When the brown colour of the iodine fades to a pale yellow colour, add about 5 drops of starch solution to the flask.  Continue adding Na 2S2O3 until the blue-black colour just disappears.  Repeat the titrations to obtain at least two consistent results within ± 0.10 cm3.  Sequence [4] (b) One possible error could be any of the following statements (or more):  Adding the starch indicator too early in the experiment.  Temperature is not constant which might affect the solubility. [1] (c) Let the average titre of Na2S2O3 be v cm3. Amount of Na2S2O3 needed = 2 × 10-5 v mol Amount of iodine liberated = 1 × 10-5 v mol Amount of IO3 - = 3.33× 10-6 v mol [IO3 -] = 1.33 × 10-4 v mol dm-3 [Pb2+] = 6.67 × 10-5 v mol dm-3 Ksp = [Pb2+][IO3 -]2 = (1.33×10-4 v) 2 ( 6.67 × 10-5 v) = 1.179 × 10-12 v3 mol3 dm-9 [3] (d) Step Expected Observations Deduction To 1 cm3 of the mixture in a test tube, add dropwise of AgNO 3(aq) until in excess to the solution. White precipitate and yellow precipitate was formed. AgCl(s) Ag I(s) Add dropwise in excess NH3(aq). White precipitate dissolved in excess NH 3(aq); Yellow precipitate remained insoluble. [Ag(NH 3)2]+Cl-(aq) AgI(s) Filter the mixture and wash the residue with distilled water. Dry the residue by pressing between filter papers. Yellow residue Colourless filtrate obtained. Ag I(s) Cl-(aq) 1174

2   To the filtrate from above, add dropwise of HNO3(aq) in excess. Filter. Wash the residue with distilled water. Dry the residue by pressing between filter papers White residue obtained. AgCl(s) [4] [Total: 12 marks]   2. (a) (i) The VDW interaction / H - bonding between isopentyl acetate and water is not exothermic enough to overcome the hydrogen bonding between water and the id-id between isopentyl acetate which is predominant due to its large electron cloud size. [2] (ii) Isopentyl alcohol is a simple covalent molecule can form hydrogen bonding. Isopentyl acetate is a simple covalent molecule which can form id-id and pd-pd. Isopentyl acetate has more electrons which cause the id-id interactions to be more significant than hydrogen bonding in isopentyl alcohol. Hence more energy is required to overcome the intermolecular forces between isopentyl acetate. [3] (iii) H2O(g) can escape thus shifting the equilibrium to the right to replenish the water vapour lost thus forming more ester as a side product. [2] (iv) To react with the excess ethanoic acid. [1] (v) To remove the CO2 so the cap does not pop out from the pressure of the gas. [1] [9] 2. (b) (i) Step I Excess concentrated H2SO4 170oC Step II NaOH (aq) heat Step III Ethanolic KCN heat Step IV Concentrated H2SO4 heat 1175

3   (ii)   (iii)   [7] (c) (i) It exist in the giant ionic lattice due to its zwitterion form. Energy is required to break the strong electrostatic forces of attraction between oppositely charged ions. [2] 2 (c) (ii) [2] 1176

4   (iii) [1] (iv) Reagents and conditions: NaOH(aq), heat, test any gas with moist red litmus paper / with concentrated HCl Observations: Red litmus turns blue for asparagine / white fumes are formed. Red litmus remains for tryptophan. OR Reagents and conditions: Bromine in CCl4 (in the dark) / Bromine aqueous Observations: Tryptophan: Reddish brown turn colourless / Orange-red turns colourless. Asparagine: Reddish brown or orange-red remains. OR Reagents and conditions: H 2SO4 (aq), KMnO4, heat Observations: Purple turns colourless for tryptophan. Purple remains for asparagine. [2] (v) [2] [9] Total: 25 marks 3 (a) (i) A saturated solution of Ca(OH)2 was to be obtained [1] ii) Additional water will dilute the solution formed. [1] 1177

5   (b) (i) Phenolphthalein.Pink to colourless OR any suitable indicator. [2] (ii) Ksp = [Ca2+][OH-]2 [1] (iii) Total [OH-] = (10.4 / 1000) x 0.05 x 100 = 0.0520 mol dm-3 [OH-] from Ca(OH)2 = 0.0520 – 0.01 = 0.0420 mol dm-3 [Ca2+] = 0.0420/2 = 0.0210 moldm-3 Ksp = 0.0210 x 0.05202 = 5.68 x 10-5 mol3dm-9 [3] (iv) Let x be the solubility of Ca(OH)2. Hence 4x3 = 5.68 x 10-5 x = 0.024215 mol dm-3, 2x = [OH-] = 0.04843 mol dm-3 Moles of HCl in 10 cm3 = Moles of OH- in 10 cm3 = 4.84 x 10-4 Volume of HCl = 4.84 x 10-4/0.05 = 9.68 cm3 [3] [9] Total: 11 marks 4 (a) (i) Al2O3 + 3/2 C  2Al + 3/2 CO2 [1] (ii) Carbon is oxidised to carbon dioxide and is consumed as a result. [1] (iii) ∆G = (568) – (1273)(0.158) = +366.9 kJmol-1 Recycling. ∆G required that is less positive / more negative is favoured. [3] (iv) Copper can delocalise all the 4s and 3d electrons to form the sea of delocalised electrons. Hence, Cu has stronger electrostatic forces of attraction between the sea of delocalised electrons and positively charged cations. More energy required to break the stronger attraction. [3] [8] 4 (b) (i)  Blue solution: [Cr(H 2O)6]2+  Colourless solution: [Al(H 2O)6]3+  Green solution: [Cr(H 2O)6]3+  Yellow solution: CrO 4 2-  Orange solution: Cr 2O7 2-,  Colourless solution: [Al(OH) 4]- or Na[Al(OH)4]  Reagent: HC l (aq) or any acid.  Grey metal: Cr [4] (ii) Transition metal ions contain partially filled 3d orbitals . In the complexes, these 3d orbitals split into two groups with a small energy gap between them. When light shines, the complex absorbs light energy from the visible 1178

6   light spectrum to promote electrons from the lower to the higher energy group. The light not absorbed will be reflected and seen as the colour of the complex. [3] (c) (i) AlCl3 + 6H2O  [Al(H2O)6]3+ + 3Cl- [1] (ii) Entropy increase is due to an increase in the number of ions which is outweighed by the entropy decrease where the water molecules are ordered in the aqua complex / decrease in particle numbers. [2] (iii) Type of reaction: Electrophilic substitution Cl2 + AlCl3  Cl+ + AlCl4 -  Type of reaction  Slow  Intermediate  Generation of electrophile  Arrows  Product (only ortho and para products are accepted) [3] (iv)   Enthalpy change of solution, x = +5924 – 4665 +3(-505) = -256 kJmol-1   [3] [9] Total: 25 marks ~~~~ THE END~~~~ 1179