Two hours of lab 8-10 am or 11 am - 1 pm or 2-4 pm One hour of discussion am or 1-2 pm or 4-5 pm. LAB GOALS E1 Session 2

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1 Experiment 1 Session 2 Electrons and Solution Color Two hours of lab 8-10 am or 11 am - 1 pm or 2-4 pm One hour of discussion am or 1-2 pm or 4-5 pm LAB GOALS E1 Session 2 Complete E 1 (Parts 1-4 and 5 B). Complete preparation for discussion. Complete team report and give to GSI or turn in within 48 hours -- see deadlines on page 221. Solution Color Solution color relates to metal ion electron configuration. The presence or absence of solution color is predictable based on the position of the metal ion s element in the periodic table. 1

2 Properties versus Periodic Table Position 1A 1 H 1s1 IIA IIIA IVA VA VIA VIIA 3 Li 2s 1 11 Na 3s 1 19 K 4s 1 37 Rb 5s 1 55 Cs 6s 1 87 Fr 7s 1 Pre-transition. Transition 4 Be 2s 2 12 Mg 3s2 IIIB IVB VB VIB VIIB VIIIB VIIIB IB IIB 20 Ca 4s 2 38 Sr 5s 2 56 Ba 6s 2 88 Ra 7s Sc Ti V Cr Mn Fe Co Ni 3d 1 4s 2 3d 2 4s 2 3d 3 4s 2 3d 5 4s 1 3d 5 4s 2 3d 6 4s 2 3d 7 4s 2 3d 8 4s Y Zr Nb Mo Tc Ru Rh Pd 4d 1 5s 2 4d 2 5s 2 4d 3 5s 2 4d 5 5s 1 4d 5 5s 2 4d 7 5s 1 4d 8 5s 1 4d La* 5d 1 6s 2 Post-transition VIIIA 2 He 1s B C N O F Ne 2s 2 2p 1 2s 2 2p 2 2s 2 2p 3 2s 2 2p 4 2s 2 2p 5 2s 2 2p Al Si P S Cl Ar 3s 2 3p 1 3s 2 3p 2 3s 2 3p 3 3s 2 3p 4 3s 2 3p 5 3s 2 3p 6 29 Cu 3d 10 4s 1 30 Zn 3d 10 4s 2 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 4s 4p 1 4s 4p 2 4s 4p 3 4s 4p 4 4s 4p 5 4s 4p 6 47 Ag 4d 10 5s 1 48 Cd 4d 10 5s 2 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 5s 5p 1 5s 5p 2 5s 5p 3 5s 5p 4 5s 2 5p 5 5s 5p 6 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 5d 2 6s 2 5d 3 6s 2 5d 4 6s 2 5d 5 6s 2 5d 6 6s 2 5d 7 6s 2 5d 9 6s 1 6s 2 6p 1 6s 2 6p 2 6s 2 6p 3 6s 2 6p 4 6s 2 6p 5 6s 2 6p 6 5d 10 6s 1 5d 10 6s Element synthesized, Ac # but no official name assigned 6d 1 7s 2 6d 2 7s 2 6d 3 7s 2 6d 4 7s 2 6d 5 7s 2 6d 6 7s 2 6d 7 7s 2 Energy and Electrons and Color Electrons can move to higher energy levels if available energy (heat or light) = exactly that needed for an electron energy level transition. If electrons move from a higher to a lower energy level, the difference in energy will be released. DEMO Wavelength Color Light is a form of energy and may cause electron energy level transitions or break bonds etc. λ 400 Violet - Blue - Green - Yellow - Orange - Red λ 800 Shorter wavelength Higher frequency Higher quantum energy 2

3 Wavelength and Energy Balloon containing H 2 and Cl 2 Light source DEMO 1. Expose the balloon to red light. 2. Expose the balloon to blue light. Color and Light Interaction The identity and color of a solution can be determined from its absorption (or transmission) spectrum. Color results from the selective absorption and transmission of visible wavelengths. DEMO Beer-Lambert Law A λ = ε c l at a given λ = (abs coefficient) (concentration) (path length) 3

4 Beer-Lambert Law Concentration (mm) is proportional to concentration at a given wavelength (λ ). Concentration and Light Calibration Curve The Beer-Lambert law becomes less and less accurate as solution concentration increase Never extrapolate a linear line of a calibration curve beyond tested absorbance- concentration values. Concentration and Light A change in sample concentration will alter the absorbance readings proportionately across all wavelengths of an absorption spectrum; the pattern will not alter. 4

5 Beer-Lambert Law Path length and light absorbance are directly proportional at a fixed wavelength and concentration. DEMO Error caution: Spectrophotometers/sample holders have different path lengths! Don t change spectrophotometers in the middle of an analysis! and Path Length The path length (and λ) must be fixed when plotting a calibration curve or absorbance readings will be in error DEMO = 1/2path length Beer-Lambert Law A λ = ε c l at a given λ = (abs coefficient) (concentration) (path length) 5

6 Beer-Lambert Law Spectrum of 0.10 M differences across wavelengths are due to? 1. Differences in the absorptivity coefficient ( ε ) 2. Differences in the concentration of the sample. 3. Differences in the path length of the sample holder. 4. All the above. Violet Blue Green Yellow Orange Red Wavelength λ (nm) Which statement below is correct? 1. Color of Abs λ max = blue-purple. 2. The sample is green. 3. E is greater at λ 500 than at λ 400. Part 4. Concentration and Light Prepare a calibration curve graph (Abs vs. concentration plot) at a team chosen and fixed λ. Use known concentrations of the assigned sample of unknown concentration Concentration (mm) 6

7 Preparation of Calibration Curve 1. Prepare a set of solutions of known concentration (e.g., 0.08 M, 0.06 M, 0.04 M, 0.02M) from your assigned 0.10 M solution. 2. Choose an appropriate wavelength for your calibration curve graph. 3. Plot a calibration curve graph. 4. Determine the slope (εl) of the calibration line. Preparation of Solutions #M = Molarity of Solution # = mmoles per ml of solution or moles per 1000 ml of solution 1. Preparation of calibration curve solutions. Reminder: M 1 V 1 = M 2 V 2 If V= milliliters: M x V = mmol x ml = mmol ml Example: 20.0 ml of a 0.07 M solution contains 1.4 mmol. 7

8 Q. What volume of 0.10 M Ni(NO 3 ) 2 and water do you use to prepare 20.0 ml of 0.07 M Ni(NO 3 ) 2? M 1 V 1 = M 2 V 2 ml 0.10 M Ni(NO 3 ) 2 + ml H 2 O? 2. Wavelength Choice for Calibration Curve? The wavelength of maximum absorbance is typically chosen so that changes in absorbance with changes in concentration are maximum and the calibration graph line has a maximum slope. Wavelength Choice for Calibration Curve? Wavelength (nm) Absorption Spectrum: 0.16 mm Plastocyanin Calibration curve at: 600nm 550nm? 8

9 Wavelength of Calibration Graph?(F 05 exam) 1.2 Purple Blue Green Yellow Orange Red 0.8 Absorption Absorption λ (nm) [M + ] (Molar) Q. A 0.4M solution of M + has the absorption spectrum on the left. Circle the wavelength of its calibration graph: Determine the slope of the calibration curve Slope in Abs/ M? Part 5B. What is the sample concentration? Calibration Curve λ Q. A sample of plastocyanin has Abs = 0.65 at λ 600. What is its concentration (mm)? 9

10 Part 5B. What is the sample concentration? Calibration Curve λ Eyeball the graph only to determine an approximate concentration Use the Beer-Lambert law to determine an exact concentration The slope of the calibration graph line is = εl in the Beer- Lambert equation A λ = εlc Substitute the slope value in the Beer-Lambert equation to solve for the unknown concentration. Calibration Curve λ Q. A sample of plastocyanin has Abs = 0.65 at λ 600. What is its concentration (mm)? 10

11 Questions? Contact 11

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