Edexcel Jan 2013 Paper 6 Q4 (with explained solutions) |

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Edexcel Jan 2013 Paper 6 Q4

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4 The concentration of a solution of sodium dichromate(VI), Na2Cr2O7, can be found by titration with a solution containing Fe2+(aq) ions in acidic conditions. A 20.0 cm3 sample of a solution of sodium dichromate(VI), of unknown concentration, was titrated with a solution of Fe2+(aq) ions, of concentration 0.0500 mol dm3. An indicator, diphenylamine, was used. This turned an intense violet colour at the end point. The titration was repeated several times and some of the results are shown in the table below. Titration number Burette reading (final) / cm3 Burette reading (initial) / cm3 Volume of Fe2+(aq) used / cm3 Titre used to calculate mean (9) 1 (trial) 21.45 1.20 41.35 21.45 21.95 20.00 (a) Explain why a trial titration (titration 1) is carried out. 21.95 19.80 (1)(b) (i) Complete the table and indicate with a tick (9) those titres most suitable for calculating a mean titre. Use the titres you have chosen to calculate the mean titre. (4) Mean titre =cm3 *P41217A01116* 11 Turn over<br />
(ii) Use the equation below, and your mean titre, to calculate the concentration of the sodium dichromate(VI) solution, in mol dm3. 2(aq) + 6Fe2+(aq) + 14H+(aq) o2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) Cr2O7 orange green (c) Assuming the accuracy of the burette is 0.05 cm3 each time the burette is read, calculate the % error of the titre in titration 3. (d) Suggest one reason why the indicator diphenylamine is needed, even though the solution in the titration flask changes colour from orange to green when no indicator is used. (3) (1) (1)12 *P41217A01216*<br />
(e) A student carrying out one titration left an air bubble in the tip of the burette before taking the initial reading. This bubble was no longer present when the student took the final reading. State and explain what effect, if any, this would have on the titre value. What effect would the use of this titre have on the calculated concentration of sodium dichromate(VI)? (3)(Total for Question 4 = 13 marks) TOTAL FOR PAPER = 50 MARKS *P41217A01316* 13<br />

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Question Number 4 (a) Question Number 4 (b)(i) Acceptable answers To find the approximate end point / Rough guide to end point / Estimate of end point / as a range finder Acceptable answers Mark Reject To check the method works First titration is inaccurate Reject Mark 1(Trial) 2 21.45 41.35 21.95 41.75 1.20 21.45 1.95 21.95 20.25 19.90 20.00 19.80 19.9 Titration number Burette reading (final)/ cm3 Burette reading (start)/ cm3 Volume of Fe2+ (aq) used / cm3 (cid:57) (cid:57) (cid:57) Titre used to calculate mean ((cid:57)) All data correct (2) 2 or 3 pieces of data correct (1) 0 or 1 piece of data correct (0) Ticks, consistent with concordant values (1) Mean = 19.9(0) (cm3) (1) First column ticked Allow transferred error from incorrect ticks Allow correct mean titre even if the ticks are incorrect<br />
Reject Mark Reject Mark Reject Mark To find the exact end point Question Number 4 (b)(ii) = 9.95 x 104 (mol) 2 = (9.95 x 104)6 (1) = 1.6583 x 104 (mol in 20 cm3) Acceptable answers Moles of Fe2+ = (19.901000) x 0.05 (1) Moles of Cr2O7 Allow TE from moles of Fe2+ [Cr2O7 Allow TE from moles of Cr2O7 Allow any answer that rounds to 8.3 x 10-3 to 2 sf Ignore sf except 1 2] = 1.6583 x 104 x 1000 / 20 =8.2917 x 103 (mol dm3) (1) 2 Question Number 4 (c) Question Number 4 (d) Acceptable answers (0.05 x 2)20.00 x 100 = 0.5(%) Acceptable answers (indicator makes) the end point more obvious/easier to see OR (indicator makes) the end point sharper<br />
Question Number 4 (e) Reject Mark (1) 2), so calculated Acceptable answers Titre would appear to be greater Because (extra solution needed to fill) the air gap (More Fe2+more Cr2O7 concentration would be greater (than actual value) ALLOW (for third mark only): Increased titre would not be concordant so would not be used to calculate mean. Titre would make no difference to the calculated concentration. ALLOW (for third mark only): TE from smaller titre to smaller concentration (1) (1) (1) Total for Q4 = 13 marks PAPER TOTAL= 50 MARKS<br />

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