Experiment 1 Session 2 Electrons and Solution Color Three hours of lab Complete E1 (Parts 1, 2, 3, 4, and 5B) Prepare discussion presentation Prepare team report. Course theme There are structure and property and periodic table relationships 1A 1 H 1s1 IIA IIIA IVA VA VIA VIIA 3 Li 2s 1 1 1 Na 3s 1 1 9 K 4s 1 3 7 Rb 5s 1 5 5 Cs 6s 1 8 7 Fr 7s 1 4 Be 2s 2 1 2 Mg 3s2 IIIB IVB VB VIB VIIB VIIIB! VIIIB IB IIB 2 0 Ca 4s 2 3 8 Sr 5s 2 Properties versus Periodic Table Position Pre-transition. Transition 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 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 2 3 9 4 0 4 1 4 2 4 3 4 4 4 5 4 6 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 10 Post-transition VIIIA 2 He 1s 2 5 6 7 8 9 1 0 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 6 1 3 1 4 1 5 1 6 1 7 1 8 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 2 9 Cu 3d 1 0 4s 1 3 0 Zn 3d 1 0 4s 2 3 1 Ga 3 2 Ge 3 3 As 3 4 Se 3 5 Br 3 6 Kr 4s 4p 1 4s 4p 2 4s 4p 3 4s 4p 4 4s 4p 5 4s 4p 6 3 4 7 Ag 4d 1 0 5s 1 4 8 Cd 4d 1 0 5s 2 4 9 In 5 0 Sn 5 1 Sb 5 2 Te 5 I 5 4 Xe 5s 5p 1 5s 5p 2 5s 5p 3 5s 5p 4 5s 2 5p 5 5s 5p 6 5 6 Ba 5 7 7 2 La* Hf 7 3 Ta 7 4 W 7 5 Re 7 6 Os 7 7 Ir 7 8 Pt 7 9 Au 8 0 Hg 8 1 Tl 8 2 Pb 8 3 Bi 8 4 Po 8 5 At 8 6 Rn 6s 2 5d 1 6s 2 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 1 0 6s 1 5d 1 0 6s 2 8 8 8 9 1 0 4 1 0 5 1 0 6 1 0 7 1 0 8 1 0 9 + Element synthesized, Ra Ac # + + + + + + but no official name assigned 7s 2 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 1
Part I Preparation and Color of Solutions Can you answer the following questions? 1) Is solution color predictable from the position of the metal ion s element in the periodic table? If so, how? 2) Is solution color predictable based on the metal ion s electron configuration? If so, how? 3) Is solution color predictable based on the metal ion s ionic radius? If so, how? Light Energy and Electrons Light is a form of energy and may cause electron energy level transitions λ 400 Violet - Blue - Green - Yellow - Orange - Red λ 800 Shorter wavelength Higher frequency Higher quantum energy Energy and Electrons and Color Electrons can move to higher energy levels if available energy (e.g., 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 2
Wavelength and Energy Light is a form of energy and may break bonds λ 400 Violet - Blue - Green - Yellow - Orange - Red λ 800 Shorter wavelength Higher frequency Higher quantum energy 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. Part 3. Solution Color and Light Interaction The identity of a solution can be determined from its absorption (or transmission) spectrum. 3
Part 3. Solution Color and Light Interaction Solution color results from the selective absorption and transmission of visible wavelengths. DEMO Part 5B. What is the sample concentration? 1. Prepare a calibration curve graph (Abs vs. concentration plot) using known sample concentrations at a chosen and fixed λ (Part 4) 2. Determine the Abs of the sample at the calibration curve graph λ (Part 5 B) 3. Use the Beer-Lambert Law to calculate the unknown sample concentration (Part 5 B) Beer-Lambert Law A λ = ε c l at a given λ = (abs coefficient) (concentration) (path length) 4
Beer-Lambert Law and Concentration 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 Concentration (mm) is proportional to concentration at a given wavelength (λ ). DEMO Concentration and Calibration Curve Calibration Curve The Beer-Lambert law becomes less and less accurate as solution concentration increases Never extrapolate a linear line of a calibration curve beyond tested absorbance- concentration values. Concentration and Absorption Spectra A change in sample concentration will alter the absorbance readings proportionately across all wavelengths of an absorption spectrum; the pattern will not alter. 5
Beer-Lambert Law and Path Length 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. 2.5 2.0 1.5 1.0 0.0 0.0 0.1 0.2 0.3 0.4 [Plastocyanin], mm = 1/2path length Beer-Lambert Law A λ = ε c l at a given λ = (abs coefficient) (concentration) (path length) 6
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 0.7 0.6 0.4 0.3 0.2 0.1 0 400 450 500 550 600 650 700 Wavelength! (nm) Which statement below is correct? 1. Color of Abs λ max = blue-purple. 2. The sample is green. 3. ε 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 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 7 8 Concentration (mm) 7
Preparation of Calibration Curve Graph 1. Prepare a set of diluted solutions of known concentration (e.g., 0.08 M, 0.06 M, 0.04 M, 0.02M) of the sample of unknown concentration. 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. 1. Prepare solutions for Calibration Curve Graph Prepare a solution of known molarity of the same identity as the solution of unknown concentration. #M = Molarity of Solution # = mmoles per ml of solution or moles per 1000 ml of solution Prepare diluted solutions for Calibration Curve Graph 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. 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 Wavelength Choice for Calibration Curve? 1.20 2.5 0.80 0.40 0.00 250 350 450 550 650 750 0.0 0.0 0.1 0.2 0.3 0.4 Wavelength (nm) [Plastocyanin], mm Absorption Spectrum: 0.16 mm Plastocyanin 2.0 1.5 1.0 Calibration curve at: 600nm 550nm 2. Choose the Calibration Curve Wavelength The wavelength of maximum absorbance is typically chosen so that changes in absorbance with changes in concentration are maximum and the calibration curve line has a maximum slope. 9
Wavelength of Calibration Graph?(F 05 exam) 1.2 Purple Blue Green Yellow Orange Red 0.8 Absorption 1 0.8 0.6 0.4 0.2 Absorption 0.7 0.6 0.4 0.3 0.2 0.1 0 400 450 500 550 600 650 700! (nm) 0 0 0.1 0.2 0.3 0.4 [M + ] (Molar) A 0.4M solution of M + has the absorption spectrum on the left. Circle the wavelength of its calibration graph: 425 500 550 600 650 3. Determine the slope of your calibration curve Slope in Abs/ M? y = x Part 5B. What is the unknown concentration? 2.5 2.0 1.5 1.0 Calibration Curve λ 600 0.0 0.0 0.1 0.2 0.3 0.4 [Plastocyanin], mm Q. A sample of plastocyanin has Abs = 0.65 at λ 600. What is its concentration (mm)? 10
Part 5B. What is the unknown concentration? 2.5 2.0 1.5 1.0 Calibration Curve λ 600 0.0 0.0 0.1 0.2 0.3 0.4 [Plastocyanin], mm Eyeball the graph ONLY to determine the approximate concentration of the unknown. Use the Beer-Lambert law to determine an exact concentration. 2.5 2.0 1.5 1.0 0.0 0.0 0.1 0.2 0.3 0.4 [Plastocyanin], mm 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. 2.5 2.0 1.5 1.0 Calibration Curve λ 600 0.0 0.0 0.1 0.2 0.3 0.4 [Plastocyanin], mm Q. A sample of plastocyanin has Abs = 0.65 at λ 600. What is its concentration (mm)? Abs= εlc 11
Questions? Contact nkerner@umich.edu 12