AJC N2014 H2 P2 solns

©2015AndersonJC/CHEM 1 H2 Chemistry 9647 2014 ‘A’ Level P2 Suggested Solutions 1 (a) H+(aq) + OH–(aq) H2O(l) [1] Comments: Students should not give the molecular equation. Many students forgot to include state symbols in their equations. (b) Justification for the volume of HNO3 required Assuming 20.00 cm3 of Ba(OH)2 was used for the reaction, No of moles of Ba(OH)2 used = 00.11000 00.20  = 0.0200 mol No of moles of HNO3 required = 0.0200 x 2 = 0.0400 (2 HNO3 ≡ 1 Ba(OH)2) Minimum volume of HNO3 required for complete neutralisation = 100050.1 0400.0  = 26.7 cm3 The maximum temperature change should occur when equ al number of moles of hydroxide and acid reacts. Hence, there should be at least 3 convenient volumes that are less than 26.7 cm3, and at least 3 convenient volumes that are more than 26.7 cm3, so as to obtain 3 points each to plot for the 2 straight line graphs. (accept other suitable volumes of Ba(OH)2 used, e.g. 10.00 cm3) [1] volume of Ba(OH)2 used = 20.00 cm3 initial burette reading / cm3 final burette reading / cm3 volume of HNO3 added / cm3 initial temperature / oC highest temperature of mixture / oC 0.00 8.00 a b 16.00 – 24.00 – 32.00 – 40.00 – 48.00 – [1]: appropriate volumes of HNO3 added. (Accept 3 convenient volumes before/after 26 cm3.) [1] Procedure 1. Using a 50.00 cm3 burette, transfer 20.00cm3 of Ba(OH)2 into a Styrofoam cup. 2. Place the Styrofoam cup in a 250 cm3 beaker to prevent it from toppling. 3. Fill up another 50.00 cm3 burette with HNO3. 4. Take the initial temperature of the solution in the Styrofoam cup. Record the value. 5. Take the initial burette reading and record the value into the table. 6. Add 8.00 cm3 of HNO3 from the burette into the Styrofoam cup. 7. Stir the mixture gently with the thermometer, measure and record the highest temperature reached. 8. Immediately, add another 8.00 cm3 of HNO3 from the burette into the Styrofoam cup. For every 8.00 cm3 of HNO3 added, stir the mixture gently, and measure and record its highest temperature reached. [3]

©2015AndersonJC/CHEM 2 [1]: appropriate sequence of procedure [S] [1]: appropriate apparatus and their respective capacities [T] [1]: ensuring reliability of results (i.e: stir gently, immediately) [R] How to recognise if the equivalence–point has passed The equivalence point has passed if the temperature of solution shows a decrease. [1] Graph expected to be obtained [1]: rough sketch of points, with Tmax at around 26.70 cm3. Explanation of the shape of the graph • before equivalence point, HNO3 is the limiting reagent. • with each addition of HNO3, n H2O increases and hence temperature of solution increases. • after equivalence point , Ba(OH) 2 has been completely neutrali sed and HNO 3 is in excess. • Temperature of solution decreases due to (i) cooling, (ii) increase in the total volume. [1] volume of HNO3 added / cm3 highest temperature reached / oC 10 20 30 40 50 32.5 35.0 37.5 40.0 30.0 0

©2015AndersonJC/CHEM 3 Determination of the concentration of nitric acid The equivalence point can be obtained by finding the volume of HNO 3 added at the intersection of the 2 lines. Let the equivalence volume be x cm3 [HNO3] = 1000 x 21000 00.2000.1  (or x 200.2000.1  ) = x 40 mol dm–3 [1] Determination of the ∆Hn for the reaction The maximum temperature, Tmax, corresponds to the temperature at the intersection of the 2 lines in the graph. maximum temperature rise, ∆T = Tmax – initial temperature = y oC number of moles of H2O formed = 2 x number of moles of Ba(OH)2 = 21000 00.2000.1  = 0.0400 mol Taking density of reaction mixture to be that of water (1.0 g cm-3) m = (volume of Ba(OH)2 + volume of HNO3 used at equivalence pt) = (20.00 + x) Hc = O2Hn Tmc− = )1000(0400.0 )y)(18.4)(x00.20( +− kJ mol–1 [1] Comments: Students were expected to describe how to use the appropriate apparatus of the correct precision for a given step, rather than just outlining generally what had to be done. For thermometric experiments, the use of some type of insulated container is usually necessary. In this experiment, the subsequent addition of acid had to be done in small enough portions to give sufficient points on the graph to draw two straight lines. Ideally, the titrant ( HNO3) should be continuously added and the temperatures are taken at the specified volumes to prevent heat loss (NOT add-stir-read cycle). Students are required to show clearly on the graph that the maximum temperature reached corresponded with the correct volume of HNO 3 added (i.e. around 26.70 cm 3, depending on the initial volume of Ba(OH) 2 used). Subsequently, they need to indicate how T was obtained from their graph.

©2015AndersonJC/CHEM 4 (c) Hc = O2Hn Tmc− If the volume of acid and hydroxides are used was doubled, volume of reaction mixture will double while the amount of heat evolved from the reaction will also double [since now twice the n(H2O) will be formed]. Hence, there will be no change in maximum temperature rise. [1] FYI Experimental set-up 20.00cm3 HNO3 Ba(OH)2 styrofoam cup supported in a beaker retord stand burette containing HNO3 is added continuously while temperatures are taken at specified volumes.

©2015AndersonJC/CHEM 5 2 (a) (i) Mg2+ and SO42− [1] (ii) Fe2+, Mn2+, Cu2+ [1] Comments: Hint for (i): concentration of ions should decrease to 0.0 in the hydrothermal vent water. Hint for (ii): there should be an increase in concentration of ions from 0.0 in normal sea water. As zinc is not a transition metal, Zn2+ should not be included. (b) In normal seawater, [H+] = 10−7.8 = 1.58 x 10−8 mol dm−3 In hydrothermal vent water, [H+] = 10−4.3 = 5.01 x 10−5 mol dm−3 [1] [1] (c) SO42− + 10H+ + 8e− H2S + 4H2O [1] (d) (i) oxidised = sulfur (from –2 in H2S to –1 in FeS2) reduced = hydrogen (from +1 in H2S to 0 in H2) [1] (ii) [1]: correct dot-and-cross [1]: clear distinction of electrons from the two sulfur atoms [2] (iii) Electronic configuration of Fe2+ is [Ar]3d6 and it has partially filled 3d–orbitals. In the presence of ligands, the partially filled 3d–orbitals of Fe2+ are split into two (non–degenerate) levels with a small energy gap (E) (d orbitals splitting) between them. When energy is absorbed from the visible light region, an electron is promoted from the d orbital of a lower energy level to a d orbital of higher energy level (d -d transition) The colour of Fe 2+ is the complement of the colour absorbed. This energy corresponds to violet light from the visible light region. The yellow colour observed is the complementary colour of the violet light absorbed. [1]: partially filled 3d–orbital; small energy gap [1]: d–d transition [1]: absorption of violet (complementary to yellow) [3] Comments: For (iii), clarity must be shown in the description of d –d transition. Electrons are being promoted to a higher energy level d orbital with the absorption of the photon, and not the other way around. OR

©2015AndersonJC/CHEM 6 (e) (i) Relative atomic mass of an element is the weighted average of the mass of its isotopes, relative to 12 1 the mass of one atom of 12C. [1] (ii) Let the percentage of 3He be x%. So the percentage of 4He is (100 – x)%. Ar of He = 100 )x100(0026033.4x0160293.3 −+ 400.25959 = 3.0160293x + 400.26033 – 4.0026033x x = 7.50 x 10−4 Percentage of 3He is 7.50 x 10−4 % [2] Comments: For (i), students are to state that it is an average mass of the isotopes, and that it is relative to 1/12 t