VJC 2015_H1_CHEM_P2_ANS Prelims

©VJC 2015 8872/02/PRELIM/15 1 1 (a) The dissolved oxygen concentration (DOC) in rivers and lakes is important for aquatic life. If the DOC falls below 5 mg dm–3, most species of fish cannot survive. [1 g = 1000 mg] Environmental chemists can determine the DOC in water using the procedure below: Step 1 : A sample of river water is shaken with aqueous Mn 2+ and aqueous alkali. The dissolved oxygen oxidises the Mn 2+ to Mn3+, forming a pale brown precipitate of Mn(OH)3. O2(aq) + 4Mn2+(aq) + 8OH(aq) + 2H2O(l)  4Mn(OH)3(s) Step 2 : The Mn(OH) 3 precipitate is then reacted with an excess of aqueous potassium iodide, which is oxidised to iodine. 2Mn(OH)3(s) + 2I–(aq) → I2(aq) + 2Mn(OH)2(s) + 2OH–(aq) Step 3 : The iodine formed is then determined by titration with aqueous sodium thiosulfate, Na2S2O3(aq). 2S2O3 2–(aq) + I2(aq)  S4O6 2–(aq) + 2I–(aq) 25.0 cm3 of a sample of river water was analysed using the above procedure. The titration requires 24.60 cm3 of 0.00100 mol dm–3 Na2S2O3(aq). (i) Calculate the amount of oxygen present in the 25.0 cm3 sample of river water. Amount of S2O3 2– reacted in titration = 0.00100  24.60 1000 = 2.46  10–5 mol 4S2O3 2–  2I2  4Mn(OH)3  O2 Amount of O2 present = ¼  2.46  10–5 = 6.15  10–6 mol [2] (ii) Hence, calculate the dissolved oxygen concentration (DOC), in mg dm–3, in the river water. DOC = 6.15  10–6  32.0  1000 0.025 = 7.87 mg dm–3 [1] (iii) Comment on whether there is enough oxygen for fish to survive in that river. Since the actual DOC exceeds 5 mg dm –3, there is enough oxygen for the fish to survive. [1] (b) The presence of nitrite ions, NO 2 –, in the river water interferes with this method because NO 2 – ions can also react with iodide ions. During the reaction, NO 2 – is 2015 VJC H1 Chemistry Prelim Exam 8872/2 Suggested Answers Section A Answer all the questions in this section in the spaces provided.

©VJC 2015 8872/02/PRELIM/15 2 reduced to NO while iodide is oxidised to iodine. (i) Construct the half –equation for the reduction of NO 2 – to NO in an alkaline medium. NO2 – + H2O + e–  NO + 2OH– [1] (ii) Hence, give the overall equation for the reaction between NO 2 – and iodide ions. 2I– + 2NO2 – + 2H2O  I2 + 2NO + 4OH– [1] (c) (i) An oxide of manganese contains 72.0% by mass of manganese. Determine the empirical formula of this oxide. Mn O Mass % 72.0 100 – 72.0 = 28.0 Number of moles of atoms 72.0 ÷ 54.9 = 1.311 28.0 ÷ 16.0 = 1.75 ÷ by smaller number 1 1.335 Simplest ratio 1 × 3 = 3 1.335 × 3 = 4 Empirical formula of the oxide is Mn3O4. [2] (ii) The oxide in (i) is actually an equimolar mixture of two oxides. The oxidation state of manganese in one of the oxides is +3. Deduce the formulae of the