SAJC Prelim P3 ANS

1 2009 SAJC Prelim Paper 3 MARK SCHEME 1. (a) (i) 5C2O4 2- + 2MnO4 - + 16H+  10CO2 + 2Mn2+ + 8H2O No. of moles of C2O4 2- ions 0.025 x 0.1 = 0.0025 Given C2O4 2- : MnO4 - is 5:2, No. of moles of MnO4 - needed to oxidize C2O4 2- = 0.0025 x 2/5 = 0.001 Thus, Volume of MnO4 - needed = (0.001 / 0.1) x 1000 = 10cm3 (ii) Volume of MnO4 - ions needed to oxidize X2+ ions = 15 – 10 = 5cm3 No. of moles of MnO4 - ions = [(15 – 10)/1000] x 0.1 = 0.0005 No of moles of electrons absorbed by MnO4 - = 0.0005 x 5 = 0.0025 Mole ratio of X2+ : e = 1 : 1 Since 1 mole of X2+ loses 1 mole of electrons, Oxidation state of X in the product = +3 (b) Compounds A and B and C all exist as simple covalent molecules. Compound A is held by hydrogen bonding. Compound B is held by weak induced dipole-induced dipole interactions (id-id)/weak VDW. Less energy is needed to overcome the weaker id-id interactions in B. Although Compound C does not have H-bonding, having more electrons causes stronger id-id (or VDW, but not pd-pd) which compensates for the absence of H- bonding. Hence more energy is required to overcome the intermolecular forces of attraction in C. (c) Since the oxide of Y can react with both NaOH and HCl, it is an amphoteric oxide. Among the possible elements, only the oxide of aluminium is amphoteric in nature: Compound Y is Al. Al 2O3 (s) + 6HCl (aq)  2A lCl3 (aq) + 3H2O (l) Al2O3 (s) + 2NaOH (aq) + 3H2O (l)  2NaAl (OH)4 (aq) In the presence of water, AlCl3 undergoes hydrolysis to form an acidic solution, which reacts with the NaOH. A lCl3 (s) + aq  [A l(H2O)6]3+ (aq) + 3Cl (aq) [Al(H2O)6]3+ (aq) + H2O (l) [Al(H2O)5(OH)]2+ (aq ) + H3O+ (aq)

2 Phosphorus chloride and phosphorus oxide will react with water but silicon oxide is insoluble in water. Compound Z is P. PCl 3(l) +3H2O(l) H3PO3 (aq)+3HCl (aq) or PCl 5 (s)+ 4H2O(l)  H3PO4(aq) + 4HCl (aq) P 4O6 (s) + 6H2O (l)  4H3PO3 (aq) or P 4O10 (s) + 6H2O (l)  4H3PO4 (aq) (d) (i) I2, NaOH(aq), heat No yellow ppt for P, yellow ppt for Q (ii) H2SO4(aq) [HCl(aq) not accepted], heat then add acidified KMnO4 or heat with KMnO4 + H2SO4(aq) R will decolourise purple KMnO4, S will not decolourise KMnO4 (e)

3 2. (a) (i) 25 oC (ii)  Complete the circuit by allowing the ions to flow through  maintain electrical neutrality by supplying ions to neutralize any built up of c solution;  prevent mixing of the two solutions in the half-cells. [Any one] (iii) 3Sn(s) + Cr2O7 2-(aq) + 14 H+(aq)  3Sn2+(aq) + 2Cr3+(aq) + 7H2O(l) E = +1.33 + 0.14 = +1.47 V (iv) Q = It = 0.2 x 9000 = 1800 C Since 96500 x 2 C of electricity consumes 1 mole of Sn, 193000 C of electricity consumes 119 g of Sn 1800 C of electricity consumes 1.11 g of Sn (b) (i) Electron flow 2Co 3+ + 2I- 2Co2+ + I2 E0 = + 1.82 – 0.54 = +1.28 V > 0 2Co2+ + S2O8 2- 2Co3+ + 2SO4 2- E0 = + 2.01 – 1.82 = +0.19 V >0 Sn2+ 1 mol dm-3 salt bridge V Sn C Cr2O7 2- , Cr3+ , H+ 1 mol dm-3 From data booklet, Co3+ + e Co2+ E 0 = +1.82 V I2 + 2e 2I- E 0 = + 0.5 V 4 S2O8 2- + 2e 2SO4 2- E0 = + 2.01V

4 (ii) [Cu(H2O)6]2++ 4NH3 [Cu(NH3)4(H2O)2]2+ + 4 H2O [Cu(NH3)4(H2O)2]2+ + edta4- [Cuedta]2- + 2 H2O + 4 NH3 Strength of ligands : H2O < NH3 < edta4- (c) (i) CH3 OH H3C CH3 CH3 Cl H3C CH3PCl5 ethanolic KCN heat CH3 CN H3C CH3 aq H2SO4 or aq HCl heat CH3 COOH H3C CH3 (ii) Optical isomerism due to the presence of a chiral carbon Exists as a pair of non superimposable mirror images (iii) A has a higher pKa as A is a weaker acid. Ibuprofen contains a carboxylic acid. Carboxylate anion stabilized by delocalization of the electrons over the carbon atom and both oxygen atoms / distribution of negative charge over the C and 2O atoms

5 3. (a) B is NH3 at 300K while C is CH4 at 500K. NH3 at 300K deviates more than CH4 at 200K. This is because NH3 molecules are held by stronger hydrogen bonding as compared to weaker induced dipole- induced dipole attraction between CH4 molecules. CH4 at 500K deviates less than CH4 at 300K. This is because at higher temperature, particles posses higher kinetic energy and is more able to overcome the forces of attractions between particles such that the collisions are more elastic. (b) (i) 20 4 2 3 H CH H CO P P P P PK  CH 4 H2O CO H2 Initial pressure/atm 3.00 1.00 0 0 Change in pressure/atm -0.66 -0.66 +0.66 +1.98 Partial pressure/atm 2.34 0.34 0.66 1.98 K p = 6.44 atm2 (ii) Equilibrium will shift to the left to lower the pressure of the system / to decrease the number of gaseous particles. No effect on K p as Kp is only dependent on temperature. (iii) From graph, increase in temperature increases fraction of CO indicating forward reaction favoured. Reaction is endothermic to absorb the excess heat /to lower temperature. (c) (i) Electrophilic substitution 2H 2SO4 + HNO3  2HSO4 - + H3O+ + NO2 + (preferred) Or H 2SO4 + HNO3  HSO4 - + H2O + NO2 +

6 (generated) (ii) II: aq. HCl or aq H2SO4 heat followed by careful neutralization using aq. NaOH III: LiAlH4 in dry ether (followed by aq NaOH if neutralization not mentioned in stage II ) (iii) (iv) R has higher pKb. Br atom in R is electronegative/electron withdrawing Lone pair of electrons on N less available for protonation. R is less basic.

7 4. (a) (i) pH of lactic acid = 2.5 [H+] = 10-2. Since lactic acid is a weak monobasic acid, Ka = [H+]2/ [CH3CHOH(COOH)] OR [H+] = ] [HA Ka = (10-2.5)2/ 0.080 = 1.28 x 10-4 mol dm-3 (ii) Maximum buffer capacity occurs when [salt]=[acid] pH = pKa = -log 1.25 x 10-4 =3.90 (iii) When a small amount of H+ is added, CH3CH(OH)COO- (aq) + H+ (aq) CH3CH(OH)COOH The additional acid, H+, is removed by large concentration of CH3CH(OH)COO- from the salt. Thus, H+ changes very slightly and the pH remains almost constant. (iv) At the equivalence point, only basic salt is present. No. of moles of salt formed = 0.08 x 10/1000 = 8 x 10-4 mol [salt] = 8 x 10-4/ 26 x 1000 = 0.031 mol dm-3 [OH-] =√ [( 1 x 10-14/ 1.25 x 10-4) x 0.031] = 1.55 x 10-6 mol dm-3 pOH = 5.8 pH = 8.2 A suitable indicator is phenolphthalein . (b) (i) 2Ba(NO3)2 2BaO + 4NO2 + O2 (ii) The temperature of the thermal decomposition increases down the group. This is because down the group, - the size of the cations increase - hence polarizing power of cations decrease - ability of cations to distort the large anion decrease - nitrates are more stable to heat (b) (iii) L.E is proportionate to ionic charge but inversely proportionate to ionic radius

8 or L.E.       r r q q - Ionic size of nitrate ion bigger than that of chloride - Hence, LE for barium nitrate is SMALLER than barium chloride. (c) Information Deduction Stereoisomer A - contains either an alkene with non- identical groups on the same carbon or chiral carbon A reacts with hot acidified potassium manganate(VII) solution - A undergoes oxidation - A contains C=C B forms a yellow precipitate with hot alkaline iodine solution. - B contains CH3CO- (do not accept CH3CHOH-) Cold alkaline hydrogen cyanide was added to B and the mixture was reduced to form compound C - B is a ketone which undergoes nucleophilic addition with HCN - C contains an amine functional group Compound D formed white precipitat