VJC H2 Chem 2012 Prelim P3 Soln

 VJC 2012 9647/03/PRELIM/12 [Turn over VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATIONS Higher 2 CHEMISTRY 9647/03 Paper 3 Free Response Candidates answer on separate paper. 17 September 2012 2 hours Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and CT group on all the work you hand in. Write in dark blue or black pen on both sides of the paper. 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 any four questions. A Data Booklet is provided. You are reminded of the need for good English and clear explanation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 11 printed pages.

 VJC 2012 9647/03/PRELIM/12 2 Answer any 4 questions. 1 (a) The elements of Group VII are all oxidising agents , but some are more oxidising than others. Describe and explain, with the aid of equations, t he reactions of the elements with the reducing agent, sodium thiosulfate. [3] Oxidising power of X 2 decreases down the group. For Cl2, Br 2 4X 2 + S 2O3 2– + 5H 2O → 8X – + 2SO 4 2- O.S. of S increases by 4 units from +2 to +6 For I2, I 2 + 2S 2O3 2– → 2I – + S 4O6 2– O.S. of S increases by 0.5 units from +2 to +2½ [3.5, max 3] (b) Potassium chlorate(V), KC lO3, is one of the products formed during the electrolysis of concentrated aqueous KC l solution. (i) State, with the aid of balanced equations, how KC lO3 is produced in the electrolysis of concentrated aqueous KC l solution. Electrolysis of KC l solution gives KOH, H 2 and C l2 KC l + H 2O → KOH + ½H 2 + ½C l2 C l2 undergoes disproportionation and reacts with KOH under hot conditions to give KC lO3. 3C l2 + 6KOH → KC lO3 + 5KC l + 3H 2O (ii) The standard electrode potentials, E/ring2, and standard Gibbs free energy changes, ∆ G/ring2, of different chlorine-containing species are tabu lated below. Half-equation E/ring2/ring2 /ring2/ring2 / V ∆∆ ∆∆ G/ring2/ring2 /ring2/ring2 / kJ mol −− −− 1 2C lO3 − + 12H + + 10e − ⇌ Cl2 + 6H 2O +1.47 − 1420 Cl2 + 2e − ⇌ 2C l− +1.36 − 262 Write a half-equation for the conversion of C lO 3 − to C l− . Using your knowledge of Hess’ Law for ∆ G/ring2 , calculate ∆ G/ring2 for this conversion.

 VJC 2012 9647/03/PRELIM/12 [Turn over 3 ∆ G/ring2 and E/ring2 are related by the following equation: ∆ G/ring2 = − zFE /ring2 where ∆ G/ring2 is the standard Gibbs free energy change in joules per mole , z is the number of moles of electrons transferred du ring the redox reaction and F is the Faraday constant. Show that the standard electrode potential of converting C lO3 − to C l− is not the summation of +1.47 V and +1.36 V. 2C lO3 – + 12H + + 12 e − → 2C l– + 6H 2O OR C lO3 – + 6H+ + 6 e− → Cl– + 3H2O Applying Hess Law, ∆ G /ring2 = –1420 + (–262) = –1682 kJ mol − 1 OR –841 kJ mol − 1 E /ring2 = – [–1682 x 10 3 / (12 x 96500)] = +1.45 V ≠ 1.47 + 1.36 OR E/ring2 = – [–841 x 10 3 / (6 x 96500)] = +1.45 V ≠ 1.47 + 1.36 (iii) One early attempt to make bromic(VII) acid, HBrO 4, was the reaction between bromine water and silver chlorate(VII), AgC lO4. Insoluble silver chloride was formed as a by-product. With the aid o f half-equations, construct a full equation for this process. C lO4 − + 8H + + 8e − → C l− + 4H 2O Br 2 + 8H 2O → 2HBrO 4 + 14H + + 14e − 7AgC lO4 + 4Br 2 + 4H 2O → 7AgC l + 8HBrO 4 [8] 2C lO3 − + 12H + + 12 e −  → 2C l− + 6H 2O Cl2 + 2e − + 6 H2O ∆ G/ring2 ∆ G/ring2 = − 1420 kJ ∆ G/ring2 = − 262 kJ mol − 1

 VJC 2012 9647/03/PRELIM/12 4 (c) The identification of the elements in this question is restricted to proton number 1 to 20. Elements A and B belong to the same group and element C belongs to the same period as B. Elements A and B have high melting points while element C has a low melting point. The various oxides of A and C exist as simple discrete molecules while the oxide of B exists as a giant molecular structure. Both element s A and B can combine to form a compound with a high melting point. Both elements A and B are insoluble in water while C readily reacts with water resulting in an acidic solution and the resultant s olution gives a white precipitate with aqueous silver nitrate. When elements A and C react to form a compound D, it is inert and does not dissolve in water. When elements B and C react to form a compound E, it is reactive towards water leading to the forma tion of an acidic solution and fine white solid. Identify A to E, and explain your reasoning clearly. [9]

 VJC 2012 9647/03/PRELIM/12 [Turn over 5 Elements A and B have high melting points, thus they can be metals or non- metals with giant molecular structures. Element C has a low melting point means it is a non-metal wi th simple molecular structure. The structures of the oxides of A, B and C show that A, B, and C are non- metals. Hence B is Si as SiO 2 has a giant molecular structure. Since A is in the same group as Si, A is C where the oxides of carbon (CO, CO 2) exist as simple discrete molecules. Si and C can form SiC, a giant molecular structure with a high melting point. Carbon and silicon having giant molecular structures are not soluble in water. Element C is Cl OR Cl2. C l2 reacts with water to produce HC lO and HC l where the presence of C l− ions will give a white ppt of AgC l with AgNO 3. D is CC l4 CC l4 is non-polar so it cannot dissolve in water OR No available energetically accessible d orbitals to undergo hydrolysis. E is SiC l4 SiC l4 reacts with water to give acidic solution of HC l and SiO 2, a fine white solid. [Total: 20]

 VJC 2012 9647/03/PRELIM/12 6 2 Phosphoric acid, H 3PO 4, is a triprotic acid and its structure is given below. phosphoric acid The pK a values of the three successive dissociations of ph osphoric acid at 25 ° C are: pK a1 = 2.1; pKa2 = 7.2, pKa3 = 12.3 (a) Write three equations to show the stepwise acid dis sociation of H 3PO 4. Determine the p Kb value of HPO 4 2− . [2] H3PO 4 + H 2O H 2PO 4 − + H 3O+ H2PO 4 − + H 2O HPO 4 2− + H 3O+ HPO 4 2− + H 2O PO 4 3− + H3O+ p Kb of HPO 4 2− = 14 – 7.2 = 6.8 (b) Small quantities of phosphoric acid are extensively used to impart the sour or tart taste to many soft drinks such as cola and roo t beer. A cola having a density of 1.00 g cm − 3 contains 0.05% by weight of phosphoric acid. Determine the pH of the cola (ignoring the second and the third diss ociation steps for phosphoric acid). Assume that the acidity of the cola arises only from phosphoric acid. [3] No. of moles of H 3PO 4 in 1 cm 3 of cola = 98 1x100 05 . 0 = 5.10 x 10 − 6 mol cm − 3 Concentration of H 3PO 4 = 5.10 x 10 − 6 x 1000 = 5.10 x 10 − 3 mol dm − 3 K a = ]PO H [ ]PO H][ H [ 43 - 42 + 10 − 2.1 = 3 - 2+ 10 x 10 .5 ]H [ [H +] = 6.37 x 10 − 3 mol dm − 3 pH of cola = − lg(6.3