Experiment 5 Don't Get Stressed

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Feb 20, 2024

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Experiment 5 Don’t Get Stressed in the Lab Total Points: 43 Student’s Name: Adam Miller Lab Section: 804 NOTE: You must show your work for all calculations; no work, no credit. Equilibrium Reactions of Iron with Thiocyanate Table Test Tube Initial [Fe 3+ ] Initial [SCN - ] Absorbance Equilibrium [Fe 3+ ] Equilibrium [SCN - ] Equilibrium [FeSCN 2+ ] 1 .001 M .0004 M .353 9.496E-4 M 3.496E-4 M 5.039E-5 M 2 .001 M .0006 M .415 9.408E-4 M 5.408E-4 M 5.924E-5 M 3 .001 M .0008 M .463 9.339E-4 M 7.339E-4 M 6.610E-5 M 4 .001 M .001 M .544 9.223E-4 M 9.223E-4 M 7.766E-5 M 5 .18 M .0002 M 1.401 .1798E-4 M 0 M .0002 M Initial Temperature = 22 ° C (5 pts) 1. (2 pts) Express the equilibrium constant (K c ) for the iron complex formed in this investigation. (For instance: K c = x/y) K c = [FeSCN 2+ ] eq / ([Fe 3+ ] eq [SCN - ] eq ) 2. (2 pts) Calculate the initial concentration of Fe 3+ ([Fe 3+ ] i ) for all the test tubes . This is based on the dilution that results from adding the KSCN and H 2 O to the original 0.0020 M Fe(NO 3 ) 3 solution. Test Tube 1: .002(5) = 10x x = .001M Test Tube 2: .002(5) = 10x x = .001M Test Tube 3: .002(5) = 10x x = .001M Test Tube 4: .002(5) = 10x x = .001M Test Tube 5: .2(18) = 20x x = .18M 3. (4 pts) Calculate the initial concentration of SCN - ([SCN - ] i ) for all the test tubes . It was diluted by Fe(NO 3 ) 3 and H 2 O. Test Tube 1: .002(2) = 10x x = .0004M Test Tube 2: .002(3) = 10x x = .0006M Test Tube 3: .002(4) = 10x x = .0008M Test Tube 4: .002(5) = 10x x = .001M Test Tube 5: .002(2) = 20x x = .0002M

Experiment 5 Don’t Get Stressed in the Lab Total Points: 43 4. (2 pts) Calculate the concentration of FeSCN 2+ in the standard solution , test tube 5. (The conversion of SCN - to FeSCN 2+ is essentially 100% because of the large excess of Fe 3+ ; thiocyanate is the limiting reagent.) SCN - = Limiting Reagent [SCN - ] in test tube = [FeSCN 2+ ] Therefore [FeSCN 2+ ] = .0002M 5. (4 pts) Using the following formula, calculate the [FeSCN 2+ ] eq for test tubes 1-5 . [FeSCN 2+ ] eq = (A eq /A std ) x [FeSCN 2+ ] std A eq = absorbance values for the equilibrium solutions A std = absorbance values for the standard test tube Test Tube 1: .0002 x (.353/1.401) = 5.039E-5 M Test Tube 2: .0002 x (.415/1.401) = 5.924E-5 M Test Tube 3: .0002 x (.463/1.401) = 6.610E-5 M Test Tube 4: .0002 x (.544/1.401) = 7.760E-5 M Test Tube 5: .0002 x (1.401/1.401) = .0002 M 6. (4 pts) Calculate the concentration of Fe 3+ at equilibrium ([Fe 3+ ] eq ) for test tubes 1-5 . (Hint: [Fe 3+ ] eq = [Fe 3+ ] I – [FeSCN 2+ ] eq ) Test Tube 1: .001 - 5.039E-5 = 9.496E-4 M Test Tube 2: .001 - 5.929E-5 = 9.408E-4 M Test Tube 3: .001 – 6.610E-5 = 9.339E-4 M Test Tube 4: .001 – 7.769E-5 = 9.223E-4 M Test Tube 5: .18 - .0002 = .1798 M

Experiment 5 Don’t Get Stressed in the Lab Total Points: 43 7. (4 pts) Calculate the concentrations of SCN - at equilibrium ([SCN - ] eq ) for test tubes 1-5 , including the standard. (Hint: [SCN - ] eq = [SCN - ] I – [FeSCN 2+ ] eq ) Test Tube 1: .0004 - 5.039E-5 = 3.496E-4 M Test Tube 2: .0006 - 5.929E-5 = 5.408E-4 M Test Tube 3: .0008 – 6.610E-5 = 7.339E-4 M Test Tube 4: .001 – 7.769E-5 = 9.223E-4 M Test Tube 5: .0002 - .0002 = .0 M 8. (4 pts) Calculate the K c values for test tubes 1-4 ; show their expression before you calculate. Test Tube 1: [5.039E-5] / ([9.496E-4][3.496E-4]) = 151.79 M -1 Test Tube 2: [5.924E-5] / ([9.408E-4][5.408E-4]) = 116.43 M -1 Test Tube 3: [6.610E-5] / ([9.339E-4][7.339E -4]) = 96.442 M -1 Test Tube 4: [7.766E-5] / ([9.223E-4][9.223E-4]) = 91.296 M -1 9. (2 pts) Using the K c values you calculated in the above calculations, determine the average value for K c and the standard deviation. (NOTE: Standard deviation can be calculated using Excel or a calculator.) (151.79 + 96.442 + 116.43 + 91.296) / 4 = 113.9895 (average) Used STDEV.P in Excel Standard Deviation = 23.8 Post-lab Questions NOTE: You must show your work for all calculations; no work, no credit. 1. (4 pts) Balance the following reaction of iron(II) with 1,10-phenanthroline and provide the K eq expression. 1Fe 2+ (aq) + 3C 12 H 8 N 2 (aq)  1Fe(C 12 H 8 N 2 ) 3 2+ (aq) K c = [Fe(C 12 H 8 N 2 ) 3 2+ ] / ([Fe 2+ ][C 12 H 8 N 2 ] 3 )

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Experiment 5 Don’t Get Stressed in the Lab Total Points: 43 2. (2 pts) Why in the crystal violet experiment did you set the colorimeter to 565 nm and in the iron thiocyanate experiment use 470 nm? Make sure there is an explanation as to the different wavelengths with regard to color. (NOTE: The color wheel on page 62 can be useful for this question.) The wavelength was changed to absorb a different color for this week. This week the color needing to be absorbed was blue. Blue light has a different wavelength than any other color, and therefore needs to be accounted for. 3. (4 pts) Find the absorbance and transmittance of a 0.300 M solution of a substance with a molar absorptivity of 0.333 M -1 cm -1 in a cell with a 1.50 cm path length. A = Ebc A = .333 * 1.5 * .3 A = .14985 A = -logT .14985 = -logT T = .70819