Beatty 2. VA2 6092 Chemistry Paper 3 2018 Q2_Answers

1 Beatty Secondary School Science Department (Chemistry Unit) Chemistry 6092 Name: ___________________________________ ( ) Date: ________________ Class: 4E___ Practical 2: VA2 – O level 6092 Chemistry Paper 3 Nov 2018 Q2 & Q3 1 Tablets taken for the prevention and treatment of iron deficiency contain an iron(II) compound. You are going to determine the mass of iron present in a tablet by performing a titration using aqueous potassium manganate(VII). No indicator is needed for this titration as the products of the reaction are almost colourless and one drop of aqueous potassium manganate(VII) in excess produces a pale pink colour. Read all the instructions below carefully before starting the experiment in Question 2. Instructions Q is 0.0105 mol/dm3 potassium manganate(VII). R is an aqueous solution of iron(II) ions. The solution was made by dissolving nine tablets containing the iron( II) compound completely in sulfuric acid and making the final volume up to 250 cm3 by adding water. (a) (i) Put Q into the burette. The colour of Q makes it difficult to see the bottom of the meniscus so you should take all your readings using the top of the meniscus. Pipette 25.0 cm3 of R into a flask. Add Q from the burette. At first the purple colour disappears quickly but as more Q is added the colour disappears less quickly. At the end -point, one drop of Q produces a pale pink colour that does not disappear on swirling. Record your titration results in the space provided, repeating the titration as many times as you consider necessary to achieve consistent results. Results titration number 1 2 3 final burette reading / cm3 20.50 20.00 20.00 initial burette reading / cm3 0.00 0.00 0.00 volume of solution Q added / cm3 20.50 20.00 20.00 best titration results   Range 19.80 to 20.20cm3 [5] 21

2 (ii) From your titration results, obtain an average volume of Q to be used in your calculations. Show clearly how you obtained this volume. Average volume of Q used = 20.00 20.00 2  = 20.00 cm3 (2 dp) average volume of Q 20.00 cm3 [1] (b) (i) Q is 0.0105 mol/dm3 aqueous potassium manganate(VII). Calculate the number of moles of potassium manganate(VII) present in the average volume of Q. No of mol of KMnO4 = 20.000.0105 1000 = 0.000210 mol (3 sf) number of moles of potassium manganate(VII) 0.000210 mol [1] (ii) In the reaction five moles of iron(II) ions react with one mole of potassium manganate(VII). Using your answer from (i), calculate the number of moles of iron(II) ions present in 25.0 cm3 of R. Comparing mole ratio: KMnO4 : Fe2+ 1 : 5 0.000210 : 0.00105 No of mol of Fe2+ = 0.00105 mol (3 sf) number of moles of iron(II) ions 0.00105 mol [1] (iii) Using your answer from (ii), c