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3. In Part C, you prepare four solutions (starting concentrations are given blow) in test tubes and you will measure the Absorbance after the solutions have reached equilibrium. Shown below is an example of just ONE solution, with absorbance data shown as well. Using this data and the standard curve from question 2 on the previous page, calculate Kf. SHOW ALL WORK BELOW! Starting Concentrations Test Tube 0.00200 M Fe(NO3)3 [mL] 0.0020 M KSCN [mL] H20 [mL] 1 3.0 2.0 5.0 Absorbance Data [FESCN2"] at equilibrium Test Tube Absorbance at Amax Kf 1 0.118 2.88 x 10-4 M

3. In Part C, you prepare four solutions (starting concentrations are given blow) in test tubes and you will measure the Absorbance after the solutions have reached equilibrium. Shown below is an example of just ONE solution, with absorbance data shown as well. Using this data and the standard curve from question 2 on the previous page, calculate Kf. SHOW ALL WORK BELOW! Starting Concentrations Test Tube 0.00200 M Fe(NO3)3 [mL] 0.0020 M KSCN [mL] H20 [mL] 1 3.0 2.0 5.0 Absorbance Data [FESCN2"] at equilibrium Test Tube Absorbance at Amax Kf 1 0.118 2.88 x 10-4 M

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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3. In Part C, you prepare four solutions (starting concentrations are given blow) in test tubes and you will measure the Absorbance after the solutions have reached equilibrium. Shown below is an example of just ONE solution, with absorbance data shown as well. Using this data and the standard curve from question 2 on the previous page, calculate Kf. SHOW ALL WORK BELOW! Starting Concentrations Test Tube 0.00200 M Fe(NO3)3 [mL] 0.0020 M KSCN [mL] H20 [mL] 1 3.0 2.0 5.0 Absorbance Data (FESCN2*] at equilibrium Test Tube Absorbance at Amax Kf 1 0.118 2.88 x 10-4 M Given absorbance: y = 0.118 y - 0.0483 X =: 2417.5 0.118 – 0.0483 2417.5 x = 2.883 x 10$ M (FESCN2*] = 2.88 x 10$ M MFex VFe 0.002 x 3 (Fe**) = 0.006 M Total Volume MSCN X V SCN 10 0.002 x 2 (SCN'] = = 0.0004 M %3D Total Volume 10 ICE Table: Fe3+ SCN FESCN2+ 0.006 0.0004 -X -X +x 0.006 -x 0.004 - x [FESCN2+] Kf = [Fe3+][SCN-] (0.006 – x)(0.004 - x) 0.118 = 2417.5 + 0.0483 2417.5x = 0.334 - 0.0483 = 0.0697 0.0697 X = = 2.88 x 10-5 M 2417.5 At equilibrium, (FESCN2*] = 2.88 x 10S M (Fe3*] = 0.0060 - x = 0.0060 –- 2.88 x 105 = 0.0059712 [SCN'] = 0.0040 – x = 0.0040 - 2.88 x 10S = 0.0039712 %3D not sure what wernt wrong -. 2.88 x 10-5 Kf = (0.0059712 x 0.0039712) Kf = 121
Transcribed Image Text:3. In Part C, you prepare four solutions (starting concentrations are given blow) in test tubes and you will measure the Absorbance after the solutions have reached equilibrium. Shown below is an example of just ONE solution, with absorbance data shown as well. Using this data and the standard curve from question 2 on the previous page, calculate Kf. SHOW ALL WORK BELOW! Starting Concentrations Test Tube 0.00200 M Fe(NO3)3 [mL] 0.0020 M KSCN [mL] H20 [mL] 1 3.0 2.0 5.0 Absorbance Data (FESCN2*] at equilibrium Test Tube Absorbance at Amax Kf 1 0.118 2.88 x 10-4 M Given absorbance: y = 0.118 y - 0.0483 X =: 2417.5 0.118 – 0.0483 2417.5 x = 2.883 x 10$ M (FESCN2*] = 2.88 x 10$ M MFex VFe 0.002 x 3 (Fe**) = 0.006 M Total Volume MSCN X V SCN 10 0.002 x 2 (SCN'] = = 0.0004 M %3D Total Volume 10 ICE Table: Fe3+ SCN FESCN2+ 0.006 0.0004 -X -X +x 0.006 -x 0.004 - x [FESCN2+] Kf = [Fe3+][SCN-] (0.006 – x)(0.004 - x) 0.118 = 2417.5 + 0.0483 2417.5x = 0.334 - 0.0483 = 0.0697 0.0697 X = = 2.88 x 10-5 M 2417.5 At equilibrium, (FESCN2*] = 2.88 x 10S M (Fe3*] = 0.0060 - x = 0.0060 –- 2.88 x 105 = 0.0059712 [SCN'] = 0.0040 – x = 0.0040 - 2.88 x 10S = 0.0039712 %3D not sure what wernt wrong -. 2.88 x 10-5 Kf = (0.0059712 x 0.0039712) Kf = 121
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