Lec_08 - Today 1. Report for Experiment 6 is due (40 pts)...

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Unformatted text preview: Today 1. Report for Experiment 6 is due (40 pts) 2. Post-lab for Experiment 6 is due (10 pts) 3. Pre-lab for Experiment 7 is due (5 pts) 4. Experiment 7. Synthesis and Analysis of a Complex Iron Compound. Part 3 have a write-up in your notebook Next week 1. Post-lab for Experiment 7 is due (10 pts) Spring 2007 Experiment 7. Synthesis and Analysis of a Complex Iron Compound. Part 3 PURPOSE AND LEARNING OBJECTIVES To synthesize and isolate potassium oxalatoferrate coordination compound To use redox titration to determine the oxalate content of a prepared sample To use visible spectrophotometry to determine the iron content of a prepared sample TODAY: To determine molar absorptivity of Fe(phen)32+ To determine molecular stoichiometry of a synthesized sample using redox titration data in combination with the spectrophotometric data Kx[Fey(C2O4)x] zH2O Potassium content h Iron content Oxalate content h Water content Spring 2007 Absorption of light Visible Spectrum 400 500 Wavelength (nm) Energy 600 700 E = h = hc Any substance that absorbs visible light appears colored to human eye when white light is transmitted through it or reflected from it. White light contains all the colors in the visible spectrum. When substance absorbs certain wavelengths (colors) of white light, human eye detects the wavelengths (colors) that are not absorbed (are transmitted). Example: Chlorophyll absorbs blue and red appears green Spring 2007 Fundamentals Transmittance fraction of the original light that passes through the sample I0 c I Transmittance T = I I0 I 100% I0 Percent Transmittance %T = l I0 the intensity of the incident light I the intensity of the light transmitted through the sample Spring 2007 Fundamentals Absorbance A = - log T Transmittance T = e -cl A = lc Absorbance Beer's Law (Also known as Beer-Lambert law) Concentration A absorbance (no units) - molar absorptivity (M1cm 1) is a constant for a particular substance, characteristic of how much light is absorbed at a particular wavelength. l pathlength (cm) is a width of the sample (or the cuvette) c concentration (M) Spring 2007 Genesys 20 Light source selector Sample Blank Light detector Amplifier Readout Wavelength range 325 to 1100 nm Wavelength accuracy 2.0 nm Wavelength precision 0.5 nm Use the SAME spectrophotometer for the entire experiment! Spring 2007 Experimental Procedure Part 1. Next week Part 2. Standard solution preparation Part 3. Calibration curve using standard solution Part 4. Next week Spring 2007 Standard Solution preparation Objective: Fe2+ Fe(phen)32+ 1,10-phenanthroline Start with Fe2+ (aq) solution of known concentration Add excess of 0.3% solution of 1,10-phenanthroline Fe2+ + 3 phen Fe(phen)32+ Add 10% hydroxylamine, NH2OH, to preserve your Fe2+ standard: 4Fe3+ + 2 NH2OH 4Fe2+ + N2O (g) + 4 H+ + H2O Add 1 M sodium acetate, NaCH3COO, to buffer the solution Let the solution sit for at least 20 minutes to insure the reaction goes to completion Spring 2007 Calibration curve Objective: Determine molar absorptivity,, for Fe(phen)32+ Prepare 5 standard solutions of varying concentration 0.9 0.8 0.7 Least squares fit y=ax Absorbance 0.6 0.5 0.4 0.3 0.2 0.1 0 0.0E+00 2.0E-05 4.0E-05 6.0E-05 8.0E-05 A = const c const = l (slope) Note: l = 1 cm Concentration, M Spring 2007 Dilutions 1. M1 V1 = M 2 V2 # of moles before = # of moles after Adding up to a volume (volumetric flasks are usually used) Example: 20.0 ml of 0.50 M solution was transferred into a 100.0 ml volumetric flask and diluted to the mark. Determine the molarity of the resulting solution. M1 = 0.50 M V1 = 20.0 ml M2 = ? V2 = 100.0 ml Answer: 0.10 M f5 factor o n Dilutio 2. Adding volumes together (mixing solutions) Example: 20.0 ml of 0.50 M solution were mixed with 100.0 ml of deionized water. Determine the molarity of the resulting solution. M1 = 0.50 M V1 = 20.0 ml M2 = ? V2 = 120 ml Answer: 0.083 M ctor of 6 Dilution fa Spring 2007 Two Step Dilutions M1 V1 = M 2 V2 # of moles before = # of moles after Example: 20.0 ml of 0.20 M solution was transferred into a 100.0 ml volumetric flask and diluted to the mark. Then 15.00 ml of the prepared solution was mixed with 65.0 ml of water. Determine the molarity of the final solution. Dilution 1. M1 = 0.20 M V1 = 20.0 ml M2 = ? V2 = 100.0 ml M2 = 0.040 M Dilution 2. M1 = 0.040 M V1 = 15.0 ml M2 = ? V2 = (15.0 ml + 65.0 ml) = 80.0 ml M2 = 0.0075 M NOTE: M2 from Dilution 1 becomes M1 in Dilution 2. f5 factor o n Dilutio of 5.3 ution factor Dil Spring 2007 Miscellaneous Tips Most common source of error DILUTION ERRORS Make sure you know how to use a volumetric flask Make sure you know how to handle spectrophotometer cuvettes: Don't forget to prime Handle from the top Fill the cuvette at least 3/4 of the way up DO NOT discard any of your solutions until AFTER you plot your data and OK it with your TA Spring 2007 ...
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This note was uploaded on 11/10/2008 for the course CH 204 taught by Professor Leytner during the Spring '08 term at University of Texas at Austin.

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