RI_H2_CHEM_P2_ANS Prelim

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 1 RAFFLES INSTITUTION 2011 PRELIMINARY EXAMINATIONS H2 CHEMISTRY – PAPER 2 (Suggested Solutions) 1 (a) Ba 2+ (aq) + SO 4 2– (aq)  → BaSO 4(s) (b) Insoluble barium carbonate and barium sulfite may co -precipitate with barium sulfate . Concentrated hydrochloric acid reacts with any ba rium carbonate and barium sulfite present to form soluble barium chloride , and hence, this ensures that only barium sulfate is pre cipitated during the experiment. (May accept full equations with state symbols involving acid with CO 3 2– and SO 3 2– .) (c) Assumption: MC l2 is soluble in water. 1 Weigh accurately about 0.5 0 g of the impure solid metal sulfate in a clean and dry weighing bottle . (Accept: 0.50 – 1.00 g.) 2 Using a measuring cylinder , add 50 cm 3 of deionised water to dissolve the sulfate completely in a 500 cm 3 beaker . Stir and add more deionised water to dissolve, if necessary. Using a measuring cylinder, add 2 cm 3 of concentrated hydrochloric acid . Stir and warm the mixture (to remove the impurities carbonate and sulfite). 3 Using a 100 cm 3 measuring cylinder , measure 10 cm 3 of the aqueous barium chloride and pour it into the sulfate solution. Stir the resulting mixture with a glass rod. 4 Continue adding aqueous barium chloride dropwise, w hile stirring , until no more white precipitate forms from the further addition of aqueous barium chloride. 5 Filter the mixture using a filter funnel which has been lined with fi lter paper which was pre-weighed . Record the mass of the filter paper. 6 Using a dropper , add aqueous barium chloride to the filtrate . If no precipitate forms, discard the filtrate. If more pr ecipitate forms, repeat steps 4 to 6 . 7 Rinse the white residue with some deionised water and dry the precipitate using an infra-red lamp until a constant mass is obtained. 8 Weigh the dried precipitate with the filter paper and subtract the pre- weighed mass in Step 5 from it to find the mass of the dried BaSO 4.

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 2 (d) MSO 4 + BaC l2  → MC l2 + BaSO 4 amt of MSO 4 in 0.892 g = amt of BaSO 4 obtained = 1.049 137 32.1 4 16.0 + + × = 0.004500 mol molar mass of M = ×0.804 0.892 0.004500 – 32.1 – 4(16.0) = 63.5 g mol –1 The metal is copper. Therefore, the metal sulfate is CuSO 4. 2 (a) (i) To quench the reaction at the specified times by rapidly lowering the temperature of the reaction mixture. (ii) Br 2 may undergo disproportionation in the presence of NaOH(aq) and it would not be possible to determine the [Br 2] accurately . (iii) 2I – + Br 2  → 2Br – + I2 (iv) At the end point, the yellow solution of I2 decolourises . Hence there is no need for any indicator. (b) By m