NJC Data Processing Ans

1 Data Processing Exercises [Data Processing]: Volumetric Analysis Practical 1 1. 1. Determination of the number of molecules of H2O of crystallisation in borax In this experiment, you are to determine the number of moles of H2O in hydrated borax. Borax, also known as disodium tetraborate, reacts with hydrochloric acid according to the equation bxelow: Na2B4O7 (aq) + 2HCl (aq) + 5H2O (l) → 2NaCl (aq) + 4H3BO3 (aq) You are given the following: FA 1 is 109.30 g dm−3 borax, Na2B4O7.xH2O FA 2 is 0.100 mol dm−3 HCl Indicator Procedure 1. Using a burette, place between 17.80 cm3 and 18.20 cm3 of FA 1 into a 100 cm3 volumetric flask. Record your burette reading below. 2. Make up to the mark with deionised water and shake thoroughly. Label this solution FA 3. 3. Fill in a second burette with FA 2. 4. Pipette 25.0 cm3 of FA 3 into a conical flask and add a few drops of indicator provided. 5. Titrate the contents of the conical flask with FA 2 until the end point indicated by a colour change. 6. Repeat the titration as many times as necessary to obtain consistent results. 7. Tabulate your results in an appropriate manner. Results Table 1: Dilution of FA 1 Final burette reading / cm3 18.00 Initial burette reading / cm3 0.00 Volume of FA 1 used / cm3 18.00 Table 2: Titration of FA 3 against FA 2 Rough 1 2 Final burette reading / cm3 25.95 25.80 26.40 Initial burette reading / cm3 0.00 0.00 0.60 Volume of FA 2 used / cm3 25.95 25.80 25.80 Consistent results √ √

2 (a) (i) Using your titration results, calculate the number of moles of the Na2B4O7 in 25.0 cm3 of FA 3. no. of moles of HCl in 25.80 cm3 = 25.80 1000 x 0.100 = 2.58 x 10-3 mol no. of moles of Na2B4O7 in 25.0 cm3 = ½ (2.58 x 10-3) = 1.29 x 10-3 mol 1.29 x 10-3 mol number of moles of Na2B4O7 in 25.0 cm3 of FA 3 = ………………... [1] (ii) Calculate the number of moles of Na2B4O7 in 100 cm3 of FA 3. no. of moles of Na2B4O7 in 100 cm3 = 1.29 x 10-3 x 4 = 5.16 x 10-3 mol 5.16 x 10-3 mol number of moles of Na2B4O7 in 100 cm3 of FA 3 = ………………... [1] (iii) Hence calculate the concentration of Na2B4O7 in FA 1 in mol dm−3. No. of moles of Na2B4O7 in 100 cm3 of FA 3 = No. of moles of Na2B4O7 in 18.00 cm3 of FA 1 = 5.16 x 10-3 concentration = 5.16 x 10-3 / (18.00/1000) = 0.2866 = 0.287 mol dm−3 0.287 mol dm−3 concentration of Na2B4O7 in FA 1 = ………………... [1] (iv) Determine the relative molecular mass of Na2B4O7.xH2O. no. of moles of Na2B4O7.xH2O in 1 dm3 = no. of moles of Na2B4O7 in 1 dm3 Mr = 109.30 / 0.2866 = 381.4 381.4 Mr of Na2B4O7.xH2O = ……………………. Hence calculate the value of x in Na2B4O7.xH2O. Give your answer to the nearest whole number. [Ar: H, 1.0; B, 10.8; O, 16.0; Na, 23.0] x = (381.4 – 23.0 x 2 – 4 x 10.8 – 16.0 x 7) / 18.0 = 10 10 x is ………..……….[2]

3 (b) (i) Calculate the oxidation state of boron in Na2B4O7 and in H3BO3. +3 oxidation state of B in Na2B4O7 is ……………………………….. [1] +3 oxidation state of B in H3BO3 is ……………………………….. [1] (ii) Hence suggest the type of reaction between Na2B4O7 and HCl. acid base reaction / neutralisation ………………………………………………………………………………….. [1] (iii) Hence or otherwise, suggest a identity for the indicator used in this titration. methyl orange / phenolphthalein / thymolphthalein ………………………………………………………………………………….. [1] (c) Student A repeated the experiment, however in step 1, he transferred FA 1 to a 250 cm3 graduated flask by mistake and top up with deionised water to the mark accordingly. He then obtained the titration results as shown below. 1 2 3 Final burette reading / cm3 10.00 20.80 30.80 Initial burette reading / cm3 0.10 10.50 20.40 volume of FA 2 / cm3 9.90 10.30 10.40 (i) With reference to his titration results, explain how the mistake affected the reliability of his results. The FA 3 solution prepared become 2.5 times more dilute, hence the titre values become 2.5 times smaller. This decreases the reliability of his results as percentage error due to measurement is higher with a smaller titre volume. (ii) Identify any anomalous titre value in the student’s result. Suggest an error that could have lead to the anomalous result. Experiment 1 as it is not within ±0.10 cm3 of experiment 2 and 3. The student did not rinse the pipette with FA 3 after rinsing with deionised water, the residual water in the pipette diluted the aliquot transferred to the conical flask. Hence lesser amount of Na2B4O7 is used, lowering the amount of HCl needed to neutralise it, resulting in a smaller titre volume.

4 [Date Processing] : Volumetric Analysis Practical 2 2. To calculate the values of a and b in the formula of the ammonium salt (NH4)aFe(SO4)b.6H2O Iron(II) ammonium sulfate, (NH4)aFe(SO4)b.6H2O dissociates into iron(II) sulfate and ammonium ions when in solution. In this experiment, you will perform a redox titration of (NH4)aFe(SO4)b.6H2O with potassium manganate(VII). You are given the following: FA 1 is iron(II) ammonium sulfate crystals FA 3 is 0.0100 mol dm-3 potassium manganate (VII) sulfuric acid solution You are to record all the mass measurements and titration results in the space provided. Preparation of FA 2 by dissolving FA 1 in sulfuric acid 1. Weigh accurately about 5.00 g of FA 1. Record your readings in an appropriate manner. 2. Using a 50 cm3 measuring cylinder, transfer 30.0 cm3 of sulfuric acid to dissolve the crystals in a small beaker. 3. Transfer the solution and washings into a 250 cm3 graduated flask and make up to the mark with deionised water. Shake the solution thoroughly and label this FA 2. Titration with potassium manganate(VII) solution 4. Pipette 25.0 cm3 of FA 2 into a conical flask. 5. Titrate the mixture against FA 3 until the solution changes from yellow to permanent orange. 6. Repeat the titration as many times as necessary to obtain consistent results. (a) Results Table 1: Mass readings Mass of weighing bottle and iron(II) ammonium salt /g 9.256 Mass of weighing bottle and residual iron(II) ammonium salt /g 4.253 Mass of iron(II) ammonium salt /g 5.003 Table 2: Titration with FA 3 Rough 1 2 Final burette reading / cm3 26.50 27.50 25.50 Initial burette reading / cm3 0.00 2.00 0.00 Volume of KMnO4 used /cm3 26.50 25.50 25.50 Consistent results √ √

5 (b) (i) Give the balanced chemical equation for the reaction taking place during titration. MnO4− + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O ………………………………………………………………………….. [1] (ii) Using your titration results, calculate the amount of KMnO4 used in the reaction. average titre volume = 25.50+25.50 2 = 25.50 cm3 no. of moles of KMnO4 used = 25.50 1000 x 0.01 = 2.55 x 10-4 mol 2.55 x 10-4 mol amount of KMnO4 = ……………………..………… [1] (iii) Calculate the amount of Fe2+ in 25.0 cm3 of FA 2. Hence, calculate the amount of Fe2+ in the mass of FA 1 used. No. of moles of Fe2+ in 25.0 cm3 = 5 x 2.55 x 10-4 = 1.275 x 10-3 mol No. of moles of Fe2+ in 250 cm3 = 250 25 x 1.275 x 10-3 = 1.28 x 10-2 mol 1.28 x 10-2 mol amount of Fe2+ = ……………………..………… [2] (c) Upon futher analysis, the amount of NH4+ ion in FA 2 is found to be 2.55 x 10-2 mol. Using your answer to (b)(iii), calculate the values of a and b in the formula (NH4)aFe(SO4)b.6H2O. nNH4+ / nFe2+ = a 2.55 x 10−2 / 1.275 x 10−2 = a = 2 Looking at charges: (NH4)2Fe(SO4)b 2(+1) + (+2) + b(−2) = 0 b = 2 2 2 a is ………………… and b is ……………….... [2]

6 (d) (i) Explain the purpose of using sulfuric acid, instead of deionised water, to dissolve FA 1 in step 2. This is to provide an acidic medium fo