1. Chem. 31 – 4/6 Lecture

2. Announcements I Exam 2 – Next Monday –Covering Ch. 6 (topics since exam 1), 7, 8-1, 17, and parts of 22 (parts in today’s lecture – even if we don’t get to them today) –Will review topics on Wednesday Lab Reports –Water Hardness lab report resubmissions due Wednesday –AA report due Wednesday 4/15 (delay possible)

3. Announcements II Homework Set 2 –Set 2.3 set shortened and posted Today’s Lecture –Chapter 17 Spectroscopy Beer’s Law/Basics on Instrumentation –Chapter 22: Chromatography Overview Partitioning Between Two Phases Partitioning and Retention in Chromatography

4. Spectroscopy Beer’s Law Light intensity in = P o Light intensity out = P Transmittance = T = P/P o Absorbance = A = -logT Light source Absorbance used because it is proportional to concentration A = εbC Where ε = molar absorptivity and b = path length (usually in cm) and C = concentration (M) b ε = constant for given compound at specific λ value sample in cuvette

5. Spectroscopy Beer’s Law Question Half of the 284 nm light is absorbed when benzoic acid at a concentration of 0.0080 M is in a cuvette with a path length of 0.5 cm. What is the molar absorptivity of benzoic acid at this wavelength?

6. Spectroscopy More on Beer’s Law Useful for determination of analyte concentrations Some limitations –Law not valid for high concentrations –Deviations to law appear to occur when multiple wavelengths of light used or when multiple species exist but absorb light differently –Uncertainties are lowest when 0.1 < A < 1 Example of deviations to Beer’s Law: Unbuffered Indicator with ε(In - ) = 300 M - 1 cm -1, ε(HIn) = 20 M -1 cm -1 ; pKa = 4.0 HIn ↔ H + + In -

7. Spectroscopy Spectrometers light source sample in cuvette light discriminator: monochromator (passes only a small range of wavelengths) light detector – measures light intensity by converting it to an electrical signal Data processor Components can look very different in different types of spectrometers, but spectrometers will have all of the major components (except other methods of wavelength discrimination may replace monochromators)

8. Spectroscopy Example Measurement: Ozone Ozone (O 3 ) is a pollutant (lower atmosphere) and in stratosphere provides UV protection Instrument is used for measurement at station or in airplane –compares absorbance through sample cell vs. –absorbance through reference cell Can also make measurements remotely (e.g. absorbance between two skyscrapers) light source ( = 254 nm) light detector air in O 3 scrubber sample cell reference cell chopper

9. Chromatography – Ch. 22 Introduction Purpose of Chromatography –To separate and detect components of a mixture –Analytical chemists are more interested in the detection part Advantages of Chromatography –Can handle more complex samples than typical spectroscopic methods –Also results in purification of mixtures (if desired) Disadvantage of Chromatography –Separation takes time (so generally not as fast as pure spectroscopic methods) Basis for Separation: –differential partitioning between a stationary and a mobile phase

10. Chromatography Partitioning – Ch. 22 Sect. 1 Covering to understand partitioning in chromatography Partitioning can occur between any two phases (as long as one phase is a fluid) Liquid-liquid is chosen as an example Partitioning governed by equilibrium equation X(org) X(aq) K = partition coefficient (a constant) note: technically, upper conc. is for “raffinate” phase while lower is for “extractant” phase

11. Chromatography Partitioning Partitioning coefficient depends on stability in solvents (related to solubility in solvents) Most common rule is likes dissolve likes Example of water – hexane partitioning Other Effects on Partitioning –K gives distribution if compound does not react further in either phase –However, compounds may react further (e.g. acid HA → H + + A - in aqueous phase) –Ions (e.g. A - ) will be found almost exclusively in aqueous phase –Distribution coefficient (D) gives ratio of total species concentration (only covering qualitatively) larger K Although both have K > 1, the OH group makes right molecule more polar (favors water more vs left)

12. Chromatography Partitioning Example of Effect of Aqueous Reactions on Compound Distribution Compound A is nearly as polar as B However, acidity affects distribution between water and organic layer Compound B will undergo dissociation in water: HA ↔ H + + A - Distribution of B given by: D = [HA] org /{[HA] + [A - ]} aq Compound A Compound B K = 7.59K = 6.17 pK a = 4.62 Not very acidic If aqueous phase is buffered at pH > pK a (e.g. pH = 6), most of B will be in anion form and very little of B will be in organic phase With a low pH buffer, D ~ K

13. Chromatography Partitioning - Questions 1.A compound with an octanol water partition coefficient of 52 is placed in a separatory funnel with water and octanol and shaken. The concentration of it in octanol is found to be 0.150 M. What is its concentration in water? 2.It is desired to separate the following two compounds: CH 3 (CH 2 ) 3 OH and CH 3 (CH 2 ) 3 NH 2. The two compounds have similar K OW values (around 11) but the second compound is basic. What can be done to separate the two? 3.It is desired to transfer butanol (left compound in #2) from water to an organic phase. Would it be transferred most efficiently using 1-octanol, a less polar solvent (e.g. octane), or a more polar solvent (e.g. 1-hexanol) as the organic solvent?

14. Chromatography Partitioning in Chromatography Separation Occurs in Column Partitioning Requires Two Phases: –Mobile phase fluid flowing through the column type of fluid determines type of chromatography fluid = gas means gas chromatography (GC) liquid chromatography (high performance liquid chromatography or HPLC) supercritical fluid (SFC) [supercritical fluid = fluid at high temperature and pressure with properties intermediate between liquid and gas] –Stationary phase (solid or liquid within column) most commonly liquid-like substance on solid support

15. Chromatography More on Stationary Phases Open Tubular – in GC (end on, cross section view) Column Wall Mobile phase Stationary phase (wall coating) Packed column (side view) (e.g. Silica in normal phase HPLC) Packing Material (solid) Stationary phase is surface (larger area than shown because its porous) Bonded phase (liquid-like)Expanded View Stationary Phase Chemically bonded to packing material Packing Material

16. Chromatography Flow – Volume Relation Relationship between volume (used with gravity columns) and time (most common with more advanced instruments): V = t·u V V = volume passing through column part in time t at flow rate u V Also, V R = t R ·u V where R refers to retention time/volume (time it takes component to go through column or volume of solvent needed to elute compound)

17. Chromatography More on Volume Hold-up volume = V M = volume occupied by mobile phase in column Stationary phase volume = V S Calculation of V M : V M = t M ·u V, where t M = time needed for unretained compounds to elute from column

18. Chromatography Partition and Retention Partition Coefficient = K = [X] S /[X] M K is constant for X if T and/or solvent remain constant K is not used that frequently in chromatography Retention Factor = k = main measure of partioning/retention in column k = (moles X) S /(moles X) M = K(V S /V M ) Retention Factor is more commonly used because of ease in measuring, and since V M /V S = constant, k = constant · K (for a given column) Note: k Column1 ≠ k Column2 (if V M /V S changes)

19. Chromatography Definition Section – Partition and Retention Since the fraction of time a solute molecule spends in a given phase is proportional to the fraction of moles in that phase, k = (time in stationary phase)/(time in mobile phase) Experimentally, k = (t R – t M )/t M Note: t ’ R = t R – t M = adjusted retention time

20. Chromatography Definition Section – Relative Retention NOT ON EXAM 2 For a separation to occur, two compounds, A and B must have different k values The greater the difference in k values, the easier the separation Relative Retention =  = k B /k A (where B elutes after A) = measure of separation ease = “ selectivity coefficient ”  value close to 1 means difficult separation

21. Chromatography Reading Chromatograms Determination of parameters from reading chromatogram (HPLC example) t M = 2.374 min. (normally determined by finding 1 st peak for unretained compounds – contaminant below) V M = u V · t M = (1.0 mL/min)(2.37 min) = 2.37 mL 1 st peak, t R = 4.958 min.; k = (4.958 - 2.374)/2.374 = 1.088  (for 1 st 2 peaks) = k B / k A = t RB ’ / t RA ’ = (5.757 – 2.374)/(4.958 – 2.374) = 1.31 [NOT ON EXAM 2]