Experiment 5 Don't Get Stressed

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University of Nebraska, Lincoln *

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Chemistry

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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
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