HCI H2 CHEM P2 ANSWERS Prelim

2014 HCI C2 Chemistry Prelim Paper 2 – Answers 1 (a) Cathode: 2H+(aq) + 2e-  H2(g) [1] (b) Q = It = (0.50)(10)(60) = 300 C No. of moles of electrons = 300 / 96500 = 0.003109 No. of moles of hydrogen gas = 1.554 x 10-3 Volume of hydrogen gas = 1.554 x 10-3 x 24000 = 37.3 cm3 [1] (c) Diagram – 3 marks:  Battery (correct polarity)  complete circuit with ammeter  appropriate choice of electrodes – Pt or graphite  appropriate choice of apparatus for collecting H2 and holding electrolyte – E.g. inverted measuring cylinder/burette, beaker/ water trough Electrolysis – 2 marks:  Record initial and final current to find average current  Stop experiment after 10 min  Record initial and final burette / measuring cylinder Titration – 3 marks:  Stir/swirl the electrolyte after disconnecting the set-up  Pipette out 25.0 cm3 of electrolyte into a conical flask, fill up burette  Record initial and final burette readings  Add starch indicator at correct juncture  State colour change – blue-black to colourless Calculations – 2 marks:  Determine number of moles of electrons at anode  Determine L using the equation L = (average current x time) / (number of moles of electrons x charge of an electron) Sample answer: 1) Using the given 500 cm3 of acidified KI, fill a 50 cm3 measuring cylinder with some acidified KI and transfer the remaining solution into the water trough. Invert the measuring cylinder over the Pt electrode and set up the apparatus as shown above. Do not connect the battery yet. 2) Record the initial reading on the measuring cylinder.

3) Connect the battery and immediately record the initial current shown on the ammeter. Start the stopwatch at the same time. 4) At 10 minutes, record the final current and final reading on the measuring cylinder. Disconnect the battery. Dismantle the setup and empty the solution in the measuring cylinder into the water trough. 5) Stir the electrolyte using a glass rod. 6) Pipette 25.0 cm3 of the electrolyte into a 250 cm3 conical flask. 7) Fill up a burette with aqueous Na2S2O3. Record initial burette reading. 8) Titrate the solution in the conical flask with aqueous Na2S2O3 until the solution turns pale yellow. Add about 1 cm3 of starch. 9) Continue titration until the solution turns from blue-black to colourless. Record final burette reading. Processing of Data Let the titre value be V cm3. No. of moles of Na2S2O3 used = = 1 x 10-5 V No. of moles of iodine used for titration = ½ x 1 x 10-5 V No. of moles of iodine produced in electrolysis = x ½ x 1 x 10-5 V No. of moles of electrons = 2 x x ½ x 1 x 10-5 V = 2 x 10-4 V Let average current be I Total charge passed through = It = I x (10 x 60) = 600I L = = - - Total: [12] 2 (a) (i) Mg2+(aq) + 2F–(aq)  MgF2(s) Ksp = [Mg2+][F-]2 mol3 dm-9 [1] (ii) [F−] = )19.0 101( 3  = 5.26 x 10-5 mol dm-3 IP of MgF2 = [Mg2+][F−]2 = [Mg2+](5.26 x 10-5)2 Precipitation first occurs when [Mg2+][F–]2 = Ksp  [Mg2+](5.26 x 10-5)2 = 5.16  10–11 [Mg2+] = 0.0186 mol dm–3 The minimum concentration of Mg2+ is 0.0186 mol dm-3. [2] (b) (i) Hydrogen gas (H2), chlorine gas (Cl2) and sodium hydroxide [1] (ii) 3Cl2(g)/(aq) + 6NaOH(aq)  5NaCl(aq) + NaClO3(aq) + 3H2O(l) OR 3Cl2(g)/(aq) + 6OH(aq)  5Cl(aq) + ClO3 (aq) + 3H2O(l) [1]

(c) Hydrogen iodide formed is oxidised by conc, H2SO4 to iodine since HI is easily oxidised any of these equations (state symbols not required) 2HI(g) + H2SO4(l)  I2(g) + SO2(g) + 2H2O(l) 6HI(g) + H2SO4(l)  3I2(g) + S(s) + 4H2O(l) 8HI(g) + H2SO4(l)  4I2(g) + H2S(g) + 4H2O(l) [2] (d) (i)  molecules/particles of an ideal gas have zero/negligible volume. (reject: Ideal gases have zero volume)  there are negligible (or NO) intermolecular forces of attraction between ideal gas molecules/particles  collisions between ideal gas molecules are p erfectly elastic (i.e. no loss of energy during collision) [3] (ii) HF deviates more from ideal behaviour because it experiences stronger Intermolecular hydrogen bonding between the molecules compared to F 2 which has only (instantaneous dipoleinduced dipole interaction or dispersion forces) between its molecules. [2] Total: [12] 3 (a) (i) Type of bonding Position in the triangle NaCl ionic any point below CsF AlCl3 covalent any point below and to the right of NaCl Ti-Al alloy metallic any point close to Cs pV/RT p 1.0 F2 Ideal gas HF

(ii) [3] (b) (i) The discrepancies in lattice energy values are due to the presence of covalent character in the silver halides. A greater difference between lattice energy values is observed for AgI due to a more polarisable electron cloud of the anion / smaller difference in electronegativity between Ag and I. [1] (ii)   [2]

(iii) By H ss’s L w, ∆Hf ½ x 44 5 4 4 kJ mol 1 [4] (iv) 2CaCl(s) Ca(s) + CaCl2(s) ∆Hr 5 5 ) = 55 kJ mol 1 [3] (v) Provides support : (b)(iv) Explanation : 2Ca+(g) + 2Cl (g) Ca+(g) + e + Ca2+(g) + 2Cl (g) Ca (g) + Ca2+(g) + 2Cl (g) Ca (s) + Ca2+(g) + 2Cl (g) 5 +1150 2(+687)

Although a negative ∆Hf (= 146 kJ mol1) means that Ca Cl is stable with respect to its elements, CaCl is unstable with respect to its disproportionation products. So the reason for the non-existence of Ca Cl is because th ∆H r of the disproportionation reaction is strongly exothermic (= 556 kJ mol1). [1] (c) (i) 2Ba(NO3)2(s)  2BaO(s) + 4NO2(g) + O2(g) [1] (ii) Cs+ has a lower positive charge and larger ionic radius, hence a lower charge density. It is unable to polarize the electron cloud of the nitrate ion sufficiently / completely to cause it to break up into O2. [2] (d) (i) Ka = 100.0 ][ 2H (assume that [H+] << 0.100) [H+] = 0.10010 5.0 = 103 pH = 3 [1] (ii) Cation: Al3+ or Fe3+ Explanation: The aqua complex of Al3+ (or Fe3+) has a lower pK a value than that of H 2CO3. Hence, it is a stronger acid than H2CO3 / is able to protonate HCO3 - to form H2CO3, which decomposes to give CO2 and H2O. Observations and products:: Al3+: effervescence of CO2 and white ppt of Al(OH)3 Fe3+: efferverscence of CO2 and reddish brown ppt of Fe(OH)3 [3] Total: [21] 4 (a) reagent compound A, B, C or D structural formula of the organic product dilute HNO3 D Na D

dilute H2SO4, heat A F h i g’s reagent B [4] (b) Step I: Electrophilic substitution of benzene ring with Br 2(aq) will also take place to produce the product shown OH OH Br Br Br OR lone pair of electrons on O in phenol group delocalises into benzene ring, hence making benzene ring more electron rich towards attack by electrophile. Step II: Nucleophilic substitution requires nucleophile (CN –) which is in very low concentration as HCN is a weak acid. Alternative method: Heat with ethanolic KCN Step V: Esterification between –CO2H and phenol group cannot take place under such conditions. The lone pair of electrons on O in phenol group is not nucleophilic enough to attack the –CO2H as it is delocalized into the benzene ring. Alternative method: Add PCl5 / SOCl2 [6] (c) No. of moles of compound A = 10/146 = 0.06849 mol No. of moles of 2-vi