Edexcel Jan 2015 (IAL) Paper 3 Q3 (with explained solutions) |

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Edexcel Jan 2015 (IAL) Paper 3 Q3

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3 A titration was carried out to find the relative molecular mass of a solid acid. The formula of the acid can be written H2A. (a) 1.05 g of the acid was dissolved in water and the solution made up to 250 cm3. Name the piece of apparatus used for making a solution with volume exactly 250 cm3. (1)(b) 25.0 cm3 of the acid solution was pipetted into a conical flask and titrated with 0.100 mol dm3 sodium hydroxide solution. This titration was repeated three times. The equation for the reaction is shown below. H2A(aq) + 2NaOH(aq) (cid:111) Na2A(aq) + 2H2O(l) (i) The indicator used in the titration was phenolphthalein. What colour change took place at the end point of the titration? H2A and its ions are colourless. Fromto(ii) The following results were recorded. Titration number Burette reading (final) /cm3 Burette reading (initial) /cm3 Volume of NaOH used /cm3 23.60 0.00 23.60 46.90 23.60 23.30 24.35 1.00 23.35 47.65 24.40 23.25 Titration number 1 was a rangefinder, or rough titration. Describe how you would use the rough titration value when carrying out the accurate titrations. (2) (1)*P45043A0816*<br />
(iii) The uncertainty in each burette reading was0.05 cm3. Calculate the percentage uncertainty in titration number 2. (iv) Calculate the mean titre for titration numbers 2, 3 and 4. (1) (1) Mean titre =cm3 (v) Calculate the number of moles of sodium hydroxide in the mean titre and hence calculate the number of moles of H2A in the 25.0 cm3 pipette samples. (vi) Calculate the relative molecular mass of H2A. You must show your working. (2) (2) *P45043A0916* Turn over<br />
(c) The acid, H2A, can be prepared by the oxidation of ethane-1,2-diol, HOCH2CH2OH. (i) State the reagents and conditions needed for this oxidation reaction. (2) ReagentsandConditions(ii) What colour change would occur when the oxidation took place? (1) Fromto(iii) Use the formula of ethane-1,2-diol to deduce the displayed formula of H2A. (1) (Total for Question 3 = 14 marks) 10 *P45043A01016*<br />
BLANK PAGE *P45043A01116* 11 Turn over<br />

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Mark Mark Acceptable Answers Reject (250 cm3) Volumetric flask / graduated flask Flat bottom flask Titration flask Measuring flask Measuring cylinder Conical flask Pipette Burette Beaker Acceptable Answers Reject (From) colourless (1) (to) pink (1) ALLOW (to) red (to) Combination of pink and red/ permanent pink From pink to colourless scores (1) to purple Blue to red loses both marks Question Number 3(a) Question Number 3(b)(i)<br />
Mark Acceptable Answers Reject Just to get an estimate It gives an idea of where the end-point is Use as a control Add slowly when reaches rough value As an indication of when to add drop by drop OR Add slowly when approaching rough value OR Add a significant volume /a stated volume in region 18-23.0 cm3 of alkali/a volume approaching range finder volume (quickly) and then slow down ALLOW It is an indication of when to slow down IGNORE To prevent overshooting Dont use in calculating mean titre The answer should show how the rough titration value is used when carrying out the accurate titration Acceptable Answers Reject (2 x 0.050 x 100 ) 23.30 = ()0.42918/ 0.4292/0.429 / 0.43/ 0.4 (%) Acceptable Answers 23.3(0) (cm3) Reject Mark Mark Question Number 3(b)(ii) Question Number 3(b)(iii) Question Number 3(b)(iv)<br />
Reject Mark Mark Reject 90 with no working ( just deduced from (c)(i)) Acceptable Answers Question Number 3(b)(v) Mol NaOH = (23.3 x 0.1 x 10-3 ) = 2.33 x 10-3 (1) Mol H2A = (23.3 x 0.1 x 10-3) 2 =1.165 x 10-3 / 1.17 x 10-3 (1) Ignore sf except 1 sf TE on 3b(iv) Correct answer with no working scores 2 Question Number 3(b)(vi) mol acid in 250 cm3 = 1.165 x 10-2 Acceptable Answers /0.01165 (1) Mr = 1.05/1.165 x 10-2 = 90.129/ 90.1 / 90 Ignore sf (1) Give both marks for final answer if some working is shown, even if first marking point is not shown separately. TE from 3b(v) Final answer of 901 because mol acid in 25 cm3 is used scores (1) Using 1.17 x 10-3 gives Mr = 89.7 Using 1.2 x 10-2 gives Mr = 87.5 Using 1.15 x 10-2 gives Mr = 91.3<br />
Mark Acceptable Answers Reject Potassium/ sodium dichromate((VI))/ K2Cr2O7 / Na2Cr2O7 and sulfuric acid/ H2SO4 (1) IGNORE concentration of acid alcoholic potassium/ sodium dichromate((VI)) (Heat under) reflux (1) Mark independently. Potassium manganate(VII)/ potassium permanganate hydrochloric acid nitric acid Just heat Reject Mark Acceptable Answers Orange to (dark)green / blue / brown TE if one of the reagents in c(i) is potassium dichromate and the other is not coloured. TE on use of potassium manganate(VII) and sulfuric acid: Purple to colourless No TE on other incorrect reagents Acceptable Answers Reject Mark Allow undisplayed O-H as above or O-H bonds shown. Ignore orientation/ bond angles Total for Question 3 = 14 marks Question Number 3(c)(i) Question Number 3(c)(ii) Question Number 3(c)(iii)<br />

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