OCR A Jun 2017 Paper 3 Q4 (with explained solutions) |

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OCR A Jun 2017 Paper 3 Q4

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This question is about weak acids. 10 (a) Compound A is a weak monobasic acid. A student is supplied with a 250.0 cm3 solution prepared from 2.495 g of A. The student titrates 25.0 cm3 samples of this solution with 0.0840 mol dm3 NaOH in the burette. The student carries out a trial, followed by the three further titrations. The diagrams show the initial burette readings and the final burette readings for the students three further titrations. All burette readings are measured to the nearest 0.05 cm3. Titration 1 Titration 2 Titration 3 Initial reading Final reading Initial reading Final reading Initial reading Final reading 23 24 25 23 24 25 45 46 47 10 11 32 33 34 (i) Record the students readings and the titres in an appropriate format. Calculate the mean titre that the student should use for analysing the results. mean titre =cm3 [4]<br />
(ii) The structure of compound A is shown below. 11 HO OH compound A Compound A has four optical isomers. Using this information and the students results, answer the following.Determine the molar mass of A and the formula of the alkyl group R.Draw the structure of compound A and label any chiral carbon atoms with an asterisk*. Show all your working. [6]<br />
(b) The structural formula of compound A is repeated below. 12 HO OH compound A Two reactions of compound A are carried out. Suggest an equation for each reaction and state the type of reaction. In your equations, draw structures for organic compounds. You can use R for the alkyl group. (i) Magnesium ribbon is added to a solution of compound A. Gas bubbles are seen and the magnesium slowly dissolves. Equation Type of reaction[3] (ii) Compound A is heated with a few drops of concentrated sulfuric acid as a catalyst. A cyclic dimer of compound A forms. Equation Type of reaction[3]<br />
(c) Chromium(III) picolinate, shown below, is a neutral complex that can be prepared from the weak acid, picolinic acid. 13 Cr Chromium(III) picolinate is used in tablets as a nutritional supplement for chromium. (i) Draw the structure of the ligand in chromium(III) picolinate. (ii) A typical tablet of chromium(III) picolinate contains 200 g of chromium. Calculate the mass, in g, of chromium(III) picolinate in a typical tablet. 1 g = 106 g. Give your answer to three significant figures. [1] mass =g [2]<br />

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H432/03 Question (a) (i) Burette readings Answer Mark Scheme June 2017 Final (reading)/cm3 23.15 45.95 32.45 Initial (reading)/cm3 0.60 23.15 10.00Correct titration results recorded with initial and final readings, clearly labeled AND all readings recorded to two decimal places with last figure either 0 or 5 Titres Titre/cm3 22.55 22.80 22.45Correct subtractions to obtain final titres to 2 DPUnitsUnits of cm3 for initial, final and titresMean titremean titre = = 22.50 OR 22.5 cm322.55 + 22.45 i.e. using concordant (consistent) titres 15 Marks Guidance Table not required ALLOW initial reading before final reading ALLOW ECF ALLOW units with each value ALLOW brackets for units, i.e. (cm3) ALLOW ECF from incorrect concordant titres<br />
Mark Scheme June 2017 H432/03 Question Answer (a) (ii) ALLOW 3SF or more throughout IGNORE trailing zeroes, e.g. ALLOW 0.084 for 0.0840 ---------------------------------------------------------------------------- n(NaOH) = 0.0840= 1.89103 (mol)22.50 1000 n(A) in 250 cm3 = 101.89103 = 1.89102 (mol)2.495 M(A) = 1.89102 = 132 (g mol1)M(alkyl group) (= 13275) = 57R = C4H9ALLOW alkyl group in drawn structure with straight chain or branch(es) in wrong position, e.g. for R = C4H9, CH3CH2CH2CH2 OR (CH3)3C Structure with chiral carbon atoms identified (see * below) Marks Guidance ALLOW ECF from incorrect mean titre in 4a(i) e.g. From 22.60 cm3 (mean of all 3 titres in (i), n(NaOH) = 1.8984103 (mol) ALLOW ECF from incorrect n(NaOH) ALLOW ECF from incorrect n(A) ALLOW ECF from incorrect M(A)75 ALLOW ECF for alkyl group closest to calculated M(alkyl group), e.g. for M = 45, ALLOW C3H7 (43) ALLOW correct structural OR skeletal OR displayed formula OR mixture of the above as long as non-ambiguous IGNORE poor connectivity to OH groups Given in question ---------------------------------------------------------------- Common error for 4 marks max 25.00 instead of 22.50 and scaling by10 2.101032.10102118.8143.81C3H725.00 instead of 22.50 and scaling by250 22.50 2.101032.33102106.9331.93C2H5No structure with 2 chiral centres possible16<br />
H432/03 Question (b) (i) Answer Marks Mark Scheme June 2017 Guidance ALLOW correct structural OR skeletal OR displayed formula OR mixture of the above as long as non-ambiguous ALLOW 2HOCH(R)COOH + Mg2HOCH(R)COO + Mg2+ + H2 ALLOW multiples IGNORE poor connectivity to OH groups Given in question ALLOW correct structural OR skeletal OR displayed formula OR mixture of the above as long as non-ambiguous ALLOW 1 mark of the 2 equation marks for formation of 3 ring with balanced equation: ALLOW condensation polymerisation ALLOW additionelimination IGNORE elimination IGNORE dehydration 2HOCH(R)COOH + Mg(HOCH(R)COO)2Mg + Equation H2 Type of reaction BalanceRedoxOrganic product(b) (ii) Equation Organic productBalanceType of reaction Condensation OR esterification17<br />
Mark Scheme June 2017 Guidance ALLOW brackets around structure with negative charge outside, i.e. ALLOW ring (Kekul structure) Note: = 3.85106 (at least 3 SF) 200106 52.0 ALLOW ECF from incorrect M OR n(Cr) ALLOW 3 SF up to calculator value correctly rounded H432/03 Question (c) (i) Answer Marks (c) (ii) FIRST CHECK THE ANSWER ON THE ANSWER LINE If answer = 1.61103 award 2 marks M = 418(.0) (g mol1) OR n(Cr) = 3.85106 (mol)Mass = 3.85106418.0 = 1.61103 gTotal 19 18<br />

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