SAJC_H2_CHEM_P2_ANS

1 (a) 5Fe2+ + MnO4 ‒ + 8H+  5Fe3+ + Mn2+ + 4H2O [1] (b) If volume of titre is assumed to be 25 cm3: Amount of KMnO4 required = 0.5 x 0.025 = 0.0125 mol Amount of Fe2+ = 0.0125 x 5 = 0.0625 mol Mass of Fe2+ = 0.0625 x 55.8 = 3.4875 g Maximum mass of tablet = (100/80) x 3.4875 = 4.36 g Minimum mass of tablet = (100/90) x 3.4875 = 3.88 g [2] (c)  Weigh a dry and clean weighing bottle.  Add iron supplement tablet into the weighing bottle and weigh the bottle + tablet.  Tip the tablet into a small beaker and reweigh the emptied weighing bottle to determine the actual mass of tablet used.  Add excess dilute sulfuric acid to the small beaker containing the tablet.  Stir with a glass rod to dissolve the tablet.  Transfer the solution with several washings into a clean 250 cm 3 volumetric flask.  Make up to the mark with distilled water.  Stopper the volumetric flask and shake well to obtain a homogeneous solution.  Pipette 25.0 cm 3 of the iron solution prepared into a 250 cm3 conical flask.  Fill the burette the standard solution of KMnO 4.  Titrate the iron solution against KMnO 4 from the burette, with continuous swirling.  Stop when one drop of solution from the burette causes a colour change from colourless to pale pink. [6] (d) Step Expected observation Identity of cation in ppt Pour 2 cm3 of the solution into a test tube. Add NaOH(aq) dropwise until excess. Pale blue ppt formed in colourless solution. Cu 2+ Filter the mixture into a separate test tube. Add excess H 2SO4(aq) to the filtrate. White ppt formed. [½] Ba 2+ * Can identify Ba2+ first [3]

2 a) i) ii) iii) Both compounds are simple covalent. Phosphorus tribromide is polar with permanent dipole-permanent dipole while boron tribromide is non polar with induced dipole-induced dipole. More energy required to break the stronger pd-pd interactions of phosphorus tribromide so PBr3 has a higher boiling point. [5] b) PBr3(l) +3H2O(l)  H3PO3(aq) +3HBr(aq) [1] c) Energy/ kJ mol-1 BE = +250 kJmol -1 [4] d) i) Nucleophilic substitution PBr3(l) PBr3(g)  P(g) + 3/2Br2(l) P(g) + 3Br(g) P(s) + 3/2Br2(l)  0  ‐185  +39  3 x BE  3/2 x (+193) +315

ii)   iii) To form C(CH3)3Br would require the starting alcohol to be a tertiary alcohol which is not feasible for a SN2 reaction due to steric hindrance/large bulky groups/electron donating methyl groups. iv) CH 3COBr [7] e) (i) (ii) [8] Total: 25 marks

3 4 a) a) H 2O b) Fro whe Fro dou The Fro Whe The Rat c) Mec Ste rea   d) Ene                                         i) % Mol Sim n (Co n = 1 O2 +2 I- + 2H m graph 1, en [H+] chan m graph 1, ub