CHEMISTRY 205 LECTURE EXAM IV Material - Section 1 Chapter 22 Chromatography refers to any separation method in which the components are distributed between a stationary phase and a moving (mobile) phase. Substances are separated by differences in their partition (distribution between a stationary phse and a mobile phase flowing through a column. I. Principle The Separation Process: A partitioning between two phases t0 t1 t2 t3 t4 Mobile Phase Mixture (Initial band with a mixture of three solutes) Stationary Phase (Column packing) To detector/ collector Page 1 2/12/2016 I. Classes A. Adsorption chromatography-The solute is adsorbed onto a solid stationary phase B. Partition chromatography-A stationary (liquid) phase is coated onto solid particles. The solid particles are called a solid support. The mobile phase is either a gas or liquid. C. Ion-exchange chromatography. Ions (either cations or anions) are bonded to a stationary solid phase (usually a resin). Ionic solute particles of the opposite charge are electrostaticaly attracted to the charged stationary phase. The mobile phase is a liquid. D. Molecular exclusion chromatography-This technique separates molecules by size. A mobile(dgas or liquid) passes through a very porous stationary phase (a gel). Small molecules pass through in the pores of the solid phase and while large molecules are excluded and quickly stream pass the porous stationary phase. E. Affinity chromatography-A solid support is coated with a solute molecule which is highly specific for only one type of molecule. Page 2 2/12/2016 III. Gas Chromatography Soap-bubble Flow Meter A. Experimental Set-up Flow Splitter Sample Injection Port Detector Pressure Gauge Bridge Carrier Gas Supply Pressure Regulator Flow Meter Recorder Column Thermostated Oven 1. Sample is injected at the injector A solute or solute mixture (a volatile liquid) is injected with a micro syringe through a rubber septum 2. A hot injector port rapidly volatiles the sample 3. The solute mixture passes through the column The sample is carried through the column by an inert carrier gas (He, N2, or H2). The column is a stainless steel or glass tubing packed with a stationary material-all thermostated at a temperature near or above the boiling point of the sample. 4. Separation Compounds are separated into bands in the column. 5. Elution Bands of the separated compounds moved along the column into a detector. 6. Detection Recording of peaks = chromatogram Page 3 2/12/2016 IV. Separation Process Page 4 2/12/2016 VI. Calculations Using the Ratio Method AREA XStandard AREA Ystandard AREA Xunknown AREA Yunknown Conc XStandard Conc Ystandard = Conc Xunknown Conc Yunknown Problem 1) Ethanol 2) n-Propanol 3) n-Butanol Peak Area Std. .373 .407 .425 Page 5 Peak Area Unknown .212 .428 .073 2/12/2016 EXAM IV Material - Section 2 Chapter 19 Spectrophotometry methods uses light to measure the concentrations of certain chemicals I. Background- Review!! A. Radiant Energy 1. 2. 3.IR, UV, X-ray, Microwave, and etc. 4. For all radiation: velocity = c = x where: c = 2.998 x 108 m sec-1 frequency (sec-1) wavelength (m) where: E = energy and E = h h = planck's constant a. b. c. Problem: If 500.0 nm, what is theand E? Page 6 2/12/2016 B. Electromagnetic Spectrum Radiant energy is characterized by its wavelength Wavelength (m) 10-11 Gamma Rays 1020 10-9 10-5 10-7 X-Rays Ultra Violet 1018 1016 Infra Red 10-3 10-1 Microwaves 1014 1012 Frequency (s-1) 101 103 Radio Frequency 1010 108 106 Visible Spectrum Violet Indigo 400 nm Blue Green Orange Yellow 600 nm 500 nm Red 700 nm 750 nm 1. Absorption by species (molecules, hydrated ions,complexes etc.) in solution. a. Energy changes (1) Electronic (2) Vibrational (3) Rotational b. Absorption of light by a Na atom Page 7 2/12/2016 104 Absorption cont'd a) Tuv b) Tvis c) TIR Page 8 2/12/2016 2. Absorption and color 3. Absorption spectra C. Spectrophotometers red 1. Single beam 420 P Meter nm Radiant Source Prism or Grating Phototube Detector Amplifier blue Cell Page 9 2/12/2016 2. Double beam Light Source Sample Cuvet Scanning Monochromator P Detector Rotating Mirror (Beam chopper) Mirror Amplifier Semitransparent Mirror Reference Cuvet Page 10 P0 Mirror 2/12/2016 Recorder D. Measurements 1. Po, initial radiant power of the light beam 2. P, final radiant power of the light beam P 3. % T = P o x 100 4. Absorbance, A or abs Po Psolvent A = log P P solution E. Beer - Lambert Law 1. %T and Length of Cell Log %T Length 1_ %T Length 1 a. %T length b. A length c. Causes: 1. Reflection losses 2. Scattering losses in solution Page 11 2/12/2016 F. % T and sample concentration Log %T 1_ %T [X] [X] 1. A [X] Log 1_ %T %T %T 1 2. % T [X] G. Absorbance is dependant on [X] and length of the cell H. Beer's law - equation Po log P = bc Where Po = %T, blank P = T, sample = molar absorptivity in also: Po log P = abc L cm-1 mol-1 a = absorpitivity used when c is not in moles b = sample cell length c = sample concentration Page 12 2/12/2016 I. A 100 = log %T = abc Aunk Cx Problem: 1. A known conc. of Cr(III) has an abs = 0.315 at 570 nm. The [Cr3+] = 4.00 x 10-3 M 2. An unknown concentration of Cr(III) transmits 35.5 % T in the same cuvette at 570nm. What is the [Cr+3] ? Page 13 2/12/2016 J. Limitations to Beer's Law 1. 2. K. Analysis 1. 2. 3. 4. L. Wavelength selection for analysis Key: Use the maximum absorbance for analysis of compounds 1. Spectrum-%T vs absorbance 2. Scan the compound Page 14 2/12/2016 M. Analysis of multicomponent systems {Mixtures} Key: Absorbances at one are additive: ATotal = A1 + A2 + A3 + A4 + A5 +....+....An = 1bc1 + 2bc2 + 3bc4 +....+nbcn EXPT: A mixture containing "X" and "Y" is analyzed. Problem: Both absorb at 500 and 700 nm Part. 1 Determine the absorptivity constants, ax and ay Assume b = 1 cm Absorptivity constants for "X" Absorptivity constants for "Y" Page 15 2/12/2016 Part. B Analysis for concentrations for "X" and "Y" Page 16 2/12/2016 N. Problems 1. Textbook problem 2. Textbook problem 3. A 1.200 g sample containing 1.00% Mn is oxidized and diluted to 250 mls: (1) 5 mls is diluted to 100 mls A = 0.100 (2) 10 mls is diluted to 100 mls A = 0.180 (3) 20 mls is diluted to 100 mls A = 0.340 (4) 30 mls is diluted to 100 mls A = 0.400 An 1.500g unknown sample is oxidized and diluted to 500 mls and a ten ml aliquot is diluted to 100 mls. This solution gives an A of 0.260. Calculate the % Mn in the sample. Page 17 2/12/2016 4. The equilibrium constant for the reaction: 2 CrO42- + 2 H+ Cr2O72- + H2O is 4.2 x 1014. The molar absorptivities at 345 nm for the two principal species in a solution of K2CrO4 are: = 1.84 x 103 for chromate and = 10.7 x 102 for dichromate. What is the theoretical absorbance (in a 1.00 cm cell)for this solution? Page 18 2/12/2016 EXAM IV Material - Section 3 Chapter 14,15, & 16 I. BACKGROUND/REVIEW Electrochemistry - The interchange of chemical and electrical energy A. Redox reaction B. Balancing redox equations Page 19 2/12/2016 D. Electrical measurements 1. Charge 2. Current 3. Cell potential a. Definitions Page 20 2/12/2016 b. Table of standard potentials (at standard conditions) Standard. conditions: Solutes at1 M (solns) { Actually, activity = 1} Gasses at 1 atm, 760 torr,760 mm Metals are pure T° = 25 °C Table of E° Values E° (V) Na+ + e- Na° - 2.71 Zn° - .76 Cu2+ + 2e- Cu° +.34 Fe° +.77 Zn2+ + 2eFe2+ + 2e- °E species is measured against: 2H+ + 2e- H2 °E = 0 ie. .....all other potentials are measured relative to the H reduction reaction c. Nerst equations - E for nonstandard conditions Use the nerst equation if nonstandard conditions exist....ie [species] 1M E = E° - A Cc A Dd 0.059 log A aA b n A B E = E° - [C]c[D]d 0.059 log [A]a[B]b n (1) Using the nerst equation \ Page 21 2/12/2016 II. Cells A. Diagram B. Cell equations: Page 22 2/12/2016 C. Danielle cell John Danielle invented the first battery to generate electricity Zn(s) + Cu2+ Zn2+ + Cu(s) D. Galvanic cells Spontaneous reactions/ Reactions that go as written: Oxidation (spontaneous) || Reduction (spontaneous) D. Electrolytic cells Electricity is required for the reaction to proceed as written: Oxidation || Reduction Page 23 2/12/2016 E. Galvanic and Electrolytic examples Page 24 2/12/2016 F. E calculations 1. Calculate E when the following reaction is half-over chemically: Cu2+ + Zn° Zn2+ + Cu° 2. Calculate the electrode potential for platinum electrodes immersed in a solutions that are 0.075 M in Iron (III) sulfate and 0.060 M in Iron (II) sulfate 3. Calculate the electrode potential for the following cell: Ag° | AgBr(sat'd) , Br-(0.04M) || H+(1.00 x 10-4) | H2 (0.90 atm), Pt Page 25 2/12/2016 G. Equilibrium G. Equilibrium calculations 1. Calculate the Keq for the following reaction: Sn2+ + Cr2+ Sn° + Cr3+ 2. Calculate the Ksp of Ag2C2O4 if the cell has a potential of + 0.277 3. The Ksp for ZnCO3 is 1.26 x 10-10 Calculate the E° for the following process: 2e - + ZnCO3 Zn° + CO32- 4. Textbook problems: Page 26 2/12/2016 Chapter 16 I. PROBLEMS: Key: 1. Balance the redox equation first! - Use Ap34 The Sb(III) in a 1.080 g sample of stibnite ore required a 41.67 ml titration with 0.03134 M I2 2. . Calculate the percentage stibnite as Sb2O3 in the sample A 4.971 g sample containing the mineral tellurite was dissolved and then treated with 50.00 ml of 0.03114 M K2Cr2O7. 3 TeO2 + Cr2O72- + 8 H+ 3 H2TeO4 + 2 Cr3+ + H 2O Upon completion of the reaction, the excess dichromate required a 10.05 ml back-titration with 0.1135 M Fe2+. Calculate the percentage of TeO2 in the sample. Page 27 2/12/2016 II. A REDOX TITRATION CURVE Problem 1: Derive a titration curve for the titration of 30.00 ml of 0.100 M Fe2+ titrated with 0.1000M Ce4+ Construction of Titration Curve Calculation of the volume of base needed to reach the equivalence point Region 1 = BEFORE THE EQUIVALENCE POINT Page 28 2/12/2016 Region 2 = THE EQUIVALENCE POINT Page 29 2/12/2016 Region 3 = AFTER THE EQUIVALENCE POINT Graph Page 30 2/12/2016 III. REDOX INDICATORS Redox indicator changes color from one oxidation state to another Ex. Ferroin (Phen)3Fe3+ + e- (Phen)3Fe2+ A. Theory E = E° - [Inred] 0.05916 log n [Inox] Typically, a color change from the oxidized form of the indicator to the reduced form requires a in the concentration ratio from: [Inred] 1 [Inox] < 10 to [Inred] [Inox] 10 < 1 B. Types of indicators 1. Auto indicators: a. b. c. 2. Specific Indicators: a. b. C. Titrants 1. Cerium 2. Potassium permanganate Page 31 2/12/2016 change 3. Sodium thiosulfate 4. Iron 5. Potassium dichromate D. IODINE TITRATIONS 1. Solubility of I2 2. Reduction reaction 3. Direct titrations 4. Starch indicator Page 32 2/12/2016