RI Planning Experiments 2 Tutorial - Ans to Self-check Questions

-15- Solutions to Self-Check Questions Question 1 (a) Cu(OH)2nCuCO3  (n+1) CuO + H2O + nCO2 [1] (b) 1. Using an analytical/weighing balance, we igh and record the mass of a clean, empty and dry crucible. 2. Weigh out accurately about 5.00 g of solid Cu(OH) 2nCuCO3 into the crucible. Record the total mass of the crucible and the solid Cu(OH)2nCuCO3. 3. Using a Bunsen burner, heat the crucible and its contents gently at first, and then heat strongly for 10 minutes. 4. Cool and weigh the crucible and its contents and record the total mass. 5. Repeat the heating-cooling-weighing pr ocess until constant mass is obtained. [1] weigh using an analytical balance and record mass of empty crucible, and total mass of crucible with solid [1] heat crucible and contents gently then strongly using a Bunsen burner [1] allow crucible and contents to cool down before weighing and recording the mass [1] repeat the heating-cooling-weighing process until constant mass is obtained. Tabulation of results Mass of empty crucible / g A Mass of crucible and Cu(OH)2nCuCO3 / g B Mass of crucible and its contents after first heating / g after second heating / g after third heating / g C D D Amount of CuO obtained = DA 63.5 + 16.0 = DA 79.5 mol [1] Amount of Cu in Cu(OH)2nCuCO3 = amount of CuO obtained % by mass of Cu = 63.5 × (DA) 79.5 × (BA) × 100% [1] Question 2 (a) ±0.05 x 2 23.50 x 100% = ±0.426% [1] (b) CO2(g) will dissolve / is soluble in water, resulting in a lower volume of CO2(g) measured. [1] (c) The percentage purity calculated will be lower than expected. [1] The smaller the volume of CO2(g), the smaller the amount of CO2 calculated, which in turn gives a smaller amount of Na2CO3 calculated in FA 1 and hence lower percentage purity. [1] [1] Table must be constructed with correct headers and units.

-16- (d)(i) [1] Collection of gas using a gas syringe [1] FA 1 in small test-tube in conical flask (d)(ii) With a gas syringe of 100 cm3 capacity, assume that 75.0 cm3 of CO2 gas is collected. Amount of CO2 gas = 75.0 / 24000 = 3.125 x 10–3 mol [1] Na2CO3(s) + 2HCl(aq)  CO2(g) + 2NaCl(aq) + H2O(l) Amount of Na2CO3 = 3.125 x 10–3 mol Molar mass of Na2CO3 = 2(23.0) + 12.0 + 3(16.0) = 106.0 g mol–1 Mass of Na2CO3 = (3.125 x 10–3)(106.0) = 0.3313 g Assuming that FA 1 contains 90% by mass of Na2CO3, mass of FA 1 to be used = 0.3313 / 90 x 100 = 0.368 g [1] Question 3 (a) Unlike HC l which is a strong acid, CH 3COOH is a weak acid and does not dissociate completely in aqueous solution. As energy is required to further ionise the unionised CH3COOH, this resulted in a less exothermic enthalpy change of neutralisation. (b) Pre-calculations Let volume of FA 1 or FA 2 used be 40 cm3. n(HCl) = 40/1000 x 0.50 = 0.0200 mol n(CH3COOH) = 40/1000 x 1.00 = 0.0400 mol n(Ba(OH)2) = ½ x n(HCl) = 0.0100 mol Volume of (Ba(OH)2) = 0.0100 / 0.5 x 1000 = 20 cm3 (Note: Volume of Ba(OH) 2 should not exceed 20 cm 3 so that Ba(OH) 2 will be the limiting reagent.) (Assuming capacity of Styrofoam cup is 200 cm3, volume of solution should be sufficient to submerge the bulb of the thermometer and should preferably not exceed more than half the capacity of the Styrofoam cup (i.e. up to 100 cm3)) Assume Vgas to be 50 – 90% of gas collection apparatus capacity.

-17- Procedure 1. Using a 50 cm 3 measuring cylinder, add 20 cm 3 of Ba(OH) 2 into a clean and dry Styrofoam cup supported in a 250 cm3 beaker. 2. Using a thermometer, measure and record the steady initial temperature of Ba(OH) 2. 3. Using another 50 cm3 measuring cylinder, measure 40 cm3 of FA 1. Ensure that both FA 1 and Ba(OH)2 are at the same initial temperature. (Note: A burette/pipette cannot be used for measurement of FA1 as FA1 must be added to Ba(OH)2 ‘at one go’. Time is required for solution to run down the burette/pipette and is hence not suitable for use here) 4. Add the FA 1 into the cup containing Ba(OH) 2. Stir gently with the thermometer and record the highest temperature reached. 5. Repeat steps 1 to 4 by replacing FA 1 with FA 2. 6. The acid which gave a lower temperature rise will be CH 3COOH while the acid which gave a higher temperature rise will be HCl. Key points:  Use of appropriate apparatus & capacity (e.g. 50 cm 3 measuring cylinders) to measure the volumes of solutions, thermometer to measure temperatures, Styrofoam cup supported in 250 cm 3 beaker.  Record steady initial temperature of Ba(OH) 2 & ensuring both solutions are at the same initial temperature.  Record highest temperature reached.  Determine identity of solutions based on temperature measurements. (c) H2SO4 is a dibasic acid and HCl is a monobasic acid. Compared to HCl, per mole of H2SO4 reacts with NaOH to form 1 more mole of H 2O, releasing greater heat of neutralisation. Hence the enthalpy change of reaction per mole of H2SO4 is more exothermic than that of HCl. Question 4 (a) H+(aq) + OH–(aq)  H2O(l) [1] (b) Pre-calculations [1] Assuming that 20.00 cm3 of Ba(OH)2 was used, Amount of OH– = 20 1.00 21000 = 0.04 mol Amount of HNO 3 used = 0.04 mol Assuming [HNO3] = 1.5 mol dm–3, volume of HNO3 required = 30.04 x 1000 26.67 cm1.5  Thus, HNO3 will be added in 5.00 cm3 portions until 50.00 cm3 has been added. Procedure [3] 1. Using a 50.00 cm 3 burette, transfer 20.00 cm 3 of Ba(OH) 2 into a polystyrene cup supported in a 250 cm3 beaker. 2. Place a thermometer into the polystyrene cup and measure the initial steady temperature of the Ba(OH)2 solution in the cup. 3. Fill a separate 50.00 cm 3 burette with HNO 3 and place the polystyrene cup with the beaker under the burette. 4. Add 5.00 cm3 of HNO3 from the burette into the polystyrene cup. Stir the mixture with the thermometer. Measure and record the maximum temperature reached, Tmax, attained and record it in the following table. Note: Since extrapolation of a straight line is required, aim for  5 points before and after the estimated end-point to allow for good extrapolation. The use of 5 cm3 portions will allow 5 points before the end-point and ending at 50.00 cm3 will allow 5 points after the end-point.

-18- volume of HNO3 added / cm3 Tmax / °C 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 5. Repeat step 4 until a total of 50.00 cm3 of HNO3 has been added. Key points:  Burette to add HNO 3 + burette for measurement of Ba(OH)2  Styrofoam/polystyrene cup (for insulation)  Thermometer + measurement of initial tempetaure & max temperature upon each addition of HNO3  Tabulation of data How you would recognise that the equivalence-point has been passed [1] Neutralisation is an exothermic reaction and the maximum temperature is reached at the end-point. The temperature of the mixture incr eases until the end-point is reached. After the end-point, the temperature will decrease as excess acid is added, because no further reaction is taking place and the excess HNO 3 added is at a lower temperature than the reaction mixture. Sketch of the expected graph [1] (Plot a graph of Tmax against volume of HNO3. Draw best-fit lines for the points before and after the maximum Tmax has been reached and extend them until they intersect. The point at which the two best-fit lines meet corresponds to the volume of HNO3 required for neutralisation and the maximum Tmax.) (Alternatively, graph of T against volume of HNO3 can be drawn  Refer to Y5 Expt 7)

-19- An explanation of the shape of your graph [2] Before equivalence point has been passed: Ba(OH)2 is in excess while HNO3 is limiting. As more HNO3 is added, more Ba(OH