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CIE Jun 2016 v1 Paper 4 Q5
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![Dicarboxylic acids dissociate in stages. stage 1 HO2C(CH2)nCO2H HO2C(CH2)nCO2+ H+ stage 2 O2C(CH2)nCO2 PMT+ 2H+ (a) The pKa values for stage 1 and stage 2 for some dicarboxylic acids are listed below. n in HO2C(CH2)nCO2H pKa(1) for stage 1 pKa(2) for stage 2 1 2 3 2.83 4.16 4.31 5.69 5.61 5.41 For comparison, the pKa of ethanoic acid, CH3CO2H, is 4.76. (i) State the mathematical relationship between pKa and the acid dissociation constant Ka.[1] (ii) With reference to the table above, suggest why the pKa(1) values are all smaller than the pKa of ethanoic acid,become larger as n increases.[3] (iii) Suggest why all the pKa(2) values in the table above are larger than the pKa of ethanoic acid.[1] (b) The monosodium salts of edible dicarboxylic acids are added to some foodstuffs as buffers. (i) Explain what is meant by the term buffer solution.[2] (ii) Write two equations to show how monosodium butanedioate, HO2CCH2CH2CO2Na, acts as a buffer.[2]UCLES 2016 9701/41/M/J/16 [Total: 9] [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2016_v1_QP_-_Paper_4_CIE_Chemistry_A-level_5.jpg)
![Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A LevelMay/June 2016 9701 41 Question Answer Marks PMT (a) (i) SCP is the EMF / potential of a cell composed of two electrodes (OR half cells) under standard conditions (OR at 289 K OR 1 mol dm 3 (ii) voltmeter and salt bridge (iii) A is Ag (aq) or AgNO3(aq) B is Ag C is Pt D is Fe 2+ 3+ (aq) and Fe (aq) (combination of A and B can be reversed with combination of C and D) (b) (i) 2+ 3+ Ag + FeAg + Fe (ii) E = E + 0.059log [Ag cell = 0.770.77 = 0.0 V so E ] = 0.800.03 = 0.77 V (a) (i) pKa = log K (ii) diacids are more acidic than CH3CO2H HO2C group is electron-withdrawing, stabilising the monoanion OR HO2C group is electron-withdrawing, weakening the OH bond OR monoanion is stabilised by Hbonding as n increases, the electronwithdrawing group is further away from the ionising CO2H group OR the (intervening) alkyl groups destabilise the anion (iii) removing H from an anion is not electrostatically favourable (b) (i) a solution which resists changes in pH when small amounts of H or OH are addedCambridge International Examinations 2016 [1] [1] [3] [1] [1] [1] [Total: 8] [1] [1] [1] [1] [1] [1] [1]<br />
Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A LevelMay/June 2016 9701 41 Question Answer (ii) HO2CCH2CH2CO2Na + HHO2CCH2CH2CO2H + Na HO2CCH2CH2CO2Na + NaOHNaO2CCH2CH2CO2Na + H2O (a) (i) C6H5NO2 + 6e + 6HC6H5NH2 + 2H2O (ii) 2C6H5NO2 + 14HCl + 3Sn2C6H5NH3Cl + 3SnCl 4 + 4H2O (b) (Mr values: C6H5NO2 = 123 C6H5NH3Cl = 129.5) theoretical yield = 5.0129.5/123 = 5.26 g percentage yield = 1004.2/5.26 = 79.8% (80%) (c) (i) C6H5NH2 = 93 yield of phenylamine = 4.293/129.5 = 3.016 g (ii) mass left in water = 3.0162.68 = 0.336 g part = (2.68/50) / (0.336/25) = 3.99 (d) phenylamine is less basic that ethylamine the lone pair on N is delocalised over the ring making it less available for reaction with a proton / + H step 1: HNO2 OR (NaNO2 + HCl ) at T10 C step 2: boil / heat in water (e) (i) (ii) E is N N (ClCambridge International Examinations 2016 PMT Marks [1] [1] [Total: 9] [1] [2] [1] [1] [1] [1] [1] [2] [1] [1] [1] [Total: 13]<br />](https://learnah.org/wp-content/uploads/2018/08/June_2016_v1_MS_-_Paper_4_CIE_Chemistry_A-level_5-0.jpg)
