Edexcel Jun 2012 Paper 6 Q3 (with explained solutions) |

Skip to content

Edexcel Jun 2012 Paper 6 Q3

Answers available below

3 Some old coins with a high copper content were analysed as follows. Procedure 1. The coins were weighed and dissolved in concentrated nitric acid, producing a solution which contained copper(II) nitrate. 2. The solution containing copper(II) nitrate was neutralized by adding sodium carbonate solution until a precipitate of copper(II) carbonate just appeared. Dilute ethanoic acid was then added, drop by drop, until the copper(II) carbonate precipitate just dissolved. 3. The solution containing copper(II) nitrate was transferred to a volumetric flask and made up to 250 cm3 with distilled water. 4. 25 cm3 portions of this solution were transferred to separate conical flasks. Then 10 cm3 of 1.0 mol dm3 potassium iodide (an excess) was added to each flask. 5. The liberated iodine was titrated with 0.125 mol dm3 sodium thiosulfate solution. (a) One reason why the solution for titration must be neutralized is because sodium thiosulfate reacts with acid as shown below. S2O3 2 + 2H+ o S + SO2 + H2O State one observation you would make when an acid reacts with sodium thiosulfate solution. (1)(b) (i) What colour is the diluted solution containing copper(II) nitrate? (1)(ii) What would you observe in Step 2, before the formation of the copper(II) carbonate precipitate, when the sodium carbonate was added? (1)*P39311A0716* Turn over<br />
(c) The equation for the reaction producing iodine in Step 4 is shown below. 2Cu2+(aq) + 4I(aq) o 2CuI(s) + I2(aq) (i) Give the name of the precipitate formed in this reaction. (1)(ii) Suggest, by considering the electronic configuration of the relevant ion, why the precipitate is white. (1)(d) The equation for the reaction of thiosulfate ions in the titration is 2S2O3 2(aq) + I2(aq) o S4O6 2(aq) + 2I(aq) Results: Mass of coins Mean (average) volume of 0.125 mol dm3 sodium thiosulfate used in titration 2.10 g 24.40 cm3 (i) Calculate the number of moles of sodium thiosulfate used in the titration. (ii) Calculate the number of moles of Cu2+ in the 25 cm3 samples used for the titration. (1) (2) *P39311A0816*<br />
(iii) Hence calculate the mass of copper present in the original mass of coins. (iv) What is the percentage of copper in the coins? (e) (i) The balance used to weigh the coins produced a total error in the weighing of0.01 g. Calculate the percentage error in the weighing. (ii) The error in the mean titre of 24.40 cm3 was0.10 cm3. Show, by calculation, that the percentage error in the titration reading is less than the percentage error in the balance reading. (f) Starch solution can be used to show the end point for this titration, or the titration can be self-indicating. What colour change would be observed at the end point if starch was not used? (2) (1) (1) (1) (1)(Total for Question 3 = 14 marks) *P39311A0916* Turn over<br />

Show answer

Correct Answer Reject (white/yellow) precipitate (of sulfur) /goes cloudy/solid forms ALLOW choking/pungent smell (of sulfur dioxide) IGNORE effervescence/bubbles/gas forms/Gas turns blue litmus red /Heat evolved Correct Answer Blue ACCEPT any blue eg blue-green Correct Answer Yellow solution Colours other than white or yellow e.g. black solid Smell of bad eggs/ bad smell Reject Reject Effervescence /fizzing /bubbling IGNORE identity of gas Just Gas given off Effervescence with brown fumes Correct Answer Reject Copper(I) iodide Oxidation number is essential Just Copper iodide Oxidation number after iodide Correct Answer Reject (In CuI) the copper (ion) has a full d (sub) shell/does not have an incomplete d (sub) shell/has configuration (3)d10 ALLOW Cu+ has full d orbital(s) Configuration of element Just d-d transitions cannot occur Just all the shells are full Mark Mark Mark Mark Mark Question Number 3(a) Question Number 3(b)(i) Question Number 3(b)(ii) Question Number 3(c)(i) Question Number 3(c)(ii)<br />
Reject 3.00 x 103 3.10 x 103 Reject Mark Mark Question Number 3(d)(ii) Mol I2 = half answer to (i) (1) Correct Answer =1.525 x 103 Mol Cu2+ = 2 x mol I2 (1) =3.05 x 103 IGNORE sf unless 1 sf Correct final answer without working (2) Question Number 3(d)(i) Correct Answer ((24.40 x 0.125) / 1000) = 3.05 x 103 / 0.00305 (mol) ALLOW 3.1 x 103 (mol)<br />
Reject Mark Reject Mark Reject Reject 0.40/ 0.409 Mark Mark Question Number 3(d)(iii) mass Cu in 25 cm3 = (63.5 x 3.05 x Correct Answer 103 = 1.93675 x 10-1 ) = 1.94 x 10-1 / 0.194 (g) (1) Mass in original = (1.93675 ) = 1.94 (g) (1) TE for 10x mass in 25 cm3 Ignore sf except 1 sf ALLOW use of Cu = 64 which gives 1.95 (g) Correct final answer without working (2) Question Number 3(d)(iv) % Cu = (1.93675 x 100 / 2.10 = Correct Answer 92.2261) = 92.2 % OR (1.94 x 100) / 2.10 = 92.38095) =92.4% ALLOW TE from use of Cu = 64 which gives 92.9% ALLOW TE from mass of Cu provided less than 100% IGNORE sf except 1 sf Correct Answer (0.01 / 2.10 x 100 =0.4761904) = () 0.48 (%) / ( )0.5 (%) IGNORE sf Correct Answer (0.10 / 24.40 x 100 = 0.4098) = () 0.41 (%) / ( 0).4 (%) IGNORE sf If errors in (i) and (ii) are both doubled allow 1 mark in e(ii) Question Number 3(e)(i) Question Number 3(e)(ii)<br />
Correct Answer Reject Brown /straw-coloured/yellow to colourless solution/white solid Both colours in change needed Blue to colourless Red-brown to colourless Clear for colourless Mark Question Number 3(f)<br />

Question 1 2 3 4

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.