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 1