SAJC H2 Chem 2012 Prelim P3 Soln

SAJC Prelim 2012 P3 solutions 1 – SAJC PRELIM 2012 P3 solutions No. Solutions 1(a)(i) Aspirin has a lower pKa hence a higher Ka thus aspirin is a stronger acid Aspirin has a RCOOH group and the conjugate base that is formed is more stable as the negative charge on the carboxylate ion is delocalised over O-C-O bond (charge delocalised over two oxygen atoms) by resonance effect. The conjugate base of phenol is less stable as the negative charge (lone pair of electrons on O) of the phenoxide ion is only delocalised into the benzene ring. (ii) H H CH3N CH3 C N C O O Na(s) -O HO or CH3CH2Br/I/ClNaOH(aq) heat H CH3N CH3CH2 C O O Phenacetin (iii) Add aqueous sodium carbonate / sodium hydrogen carbonate Aspirin gives bubbles that will form white ppt. with aqueous calcium hydroxide. Salicin does not give white ppt or bubbles. OR K2Cr2O7 (aq), aq H2SO4, heat OR Add aqueous KMnO4 (aq), aq H2SO4, heat. Salicin decolourises purple KMnO4 OR turns K2Cr2O7 from orange to green. Aspirin remains purple OR remains orange. (iv) Water is not a suitable solvent as the solute-solvent interaction (pd-pd, id-id, H-bonding) cannot overcome the H-bonding between water and id-id between aspirin. (b)(i) Ka =

2 – SAJC PRELIM 2012 P3 solutions Where HA = O C O CH3 COOH and A- = O C O CH3 COO- (ii) 10-3.5 = Hence, = 3.16 x 10-3 Percentage ionised = (3.16 x 10-3 / 1) x 100 % = 0.316 % Percentage unionised = 100 – 0.316 = 99.7 % (c)(i) Indicators are weak acids hence if added in large amount, the amount of NaOH required will be higher than expected / titration is inaccurate. (ii) [aspirin] = = 0.05 mol dm-3 Using Ka = [H+]2 = 3.98 x 10-3 pH = 2.40 (iii) Using [aspirin] = 0.05 mol dm-3 x = = 15.00 cm3 [paracetamol] = = 0.0267 mol dm-3 y = = 8.00 cm3 (iv) 2(a)(i) Cathode: O2 + 4H+ + 4e à 2H2O Anode: 13H2O + C12H22O11 à 12CO2 + 48H+ + 48e (ii) Using data booklet: Eq O2 + 4H+ + 4e ßà 2H2O + 1.23 V Using Eqcell = Eqcathode - Eqanode + 1.25 = + 1.23 – (Eqanode) Eqcathode = – 0.02 V

3 – SAJC PRELIM 2012 P3 solutions (iii) O2 + 4H+ + 4e ßà 2H2O + 1.23 V When the pH increases, the cathodic reaction shifts to the left to replenish H+. Hence, the Eqcell becomes more negative. (b)(i) A: [Cu(H2O)6]2+ B: NO2 C: Cu(OH)2 D: CuO E: [Cu(NH3)4(H2O)2]2+ F: [CuCl4]2- G/H: HNO2 and HNO3 Equations: · C: [Cu(H2O)6]2+(aq) + 2OH- (aq)à Cu(OH)2 (s) + 6H2O (l) OR Cu(NO3)2 (aq) + 2NaOH (aq) à Cu(OH)2 (s) + 2NaNO3 (aq) · D: Cu(OH)2 (s)à CuO (s) + H2O (l)/ (g) · E: [Cu(H2O)6]2+(aq) + 4NH3 (aq) à [Cu(NH3)4(H2O)2]2+ (aq) + 4H2O (l) · F: [Cu(NH3)4(H2O)2]2+ (aq) + 4HCl (aq) à [CuCl4]2- (aq) + 2H2O(l) + 4NH4+ (aq) OR [Cu(NH3)4(H2O)2]2+ (aq) + 4Cl- (aq) à [CuCl4]2- (aq) + 2H2O(l) + 4NH3 (aq) (ii) 3 bond pairs of electrons repels each other equally OR lp-lp>lp-bp>bp-bp Hence, the shape is trigonal planar. 3(a)(i) ½ N2 (g) + H2 (g) à NH3 (g) (ii) 3N2H4 (l) 4NH3 (g) + N