application of uv/vis spectroscopy common problems : a) mixtures :
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Application of UV/Vis Spectroscopy Common Problems : a) Mixtures : - blank samples often contain multiple absorbing species. - the absorbance is the sum of all the individual absorbencies A= A 1 + A 2 +A 3 + … = e 1 bc + e 2 bc + e 3 bc … - PowerPoint PPT PresentationTRANSCRIPT
Application of UV/Vis Spectroscopy
Common Problems:a) Mixtures:
- blank samples often contain multiple absorbing species.- the absorbance is the sum of all the individual absorbencies
A= A1 + A2 +A3 + … = 1bc + 2bc + 3bc …
- substances in both the blank and sample which absorb can be “blanked out” in both double and single beam spectrometers.
If blank absorbance too high:- dilute the sample- use a different where the analyte absorbs more relative to the interference.- use a different method of separation
But, If the blank absorbance is high, Po will decrease too much, the response will be slow and the results inaccurate
scan of substance
Large blank
absorbance
b) Instrumental Deviations from Beer’s Law:- stray light (already discussed).- polychromatic light (more then a single )
‚ since all instruments have a finite bandpass, a range of ’s are sent through the sample.
‚ may be different for each
Deviation’s from Beer’s law at high concentration
Illustration of Deviation from Beer’s law:
Let us say that exactly 2 wavelengths of light were entering the sample = 254 nm 254 =10,000
= 255 nm 255 = 5,000
let Po = 1 at both ’s
What happens to the Beer’s law plot as c increases?
A =A254 + A255
A = log Po/P (total) = log (Po254 + Po
255)/(P254 + P255)
At the individual ’s:
A254 = 254bc = log Po254/P254
10bc = Po254/P254
P254 = Po/10bc = Po10-bc
For both together:
A = log (Po254 + Po
255)/(Po25410-254bc + Po
25510-255bc)
Since Po254 = Po
255 =1:
A = 2.0/(Po25410-254bc + Po
25510-255bc)
A = 2.0/(10-10,000x1.0xc + 10-5000x1.0xc)
C A (actual) A(expected)
10-6M 0.0075 0.0075
10-5M 0.074 0.075
10-4M 0.068 0.75
10-3M 5.3 7.50
1
2
3
4
5
6
7
8
2 10-4 4 10-4 6 10-4 8 10-4 1 10-3
Concentration (M)
Negative Deviation
A = bc
The results are the same for more ’s of light. The situation is worse for greater differences in ’s (side of absorption peak, broad bandpass)
Always need to do calibration curve! Can not assume linearity outside the range of linearity curve!
HInKa
H+ + In-
Red, =600nm colorless
phenolphthalein:
If solution is buffered, then pH is constant and [HIn] is related to absorbance.
b) Chemical Deviations from Beer’s Law:- Molar absorptivity change in solutions more concentrated than 0.01M
‚ due to molecular interactions‚ Beer’s law assumes species are independent‚ electrolytes may also cause this problem‚ is also affected by the index of refraction
- association, dissociation, precipitation or reaction of analyte ‚ c in Beer’s law is the concentration of the absorbing species. ‚ commonly use the analytical concentration – concentration of all forms of the species.
CHIn [HIn] [In-] [HIn]/[In-]
10-5 8.5x10-7 9.2x10-6 0.0924
10-4 3.8x10-5 6.2x10-5 0.613
10-3 7.3x10-4 2.7x10-4 2.70
But, if unbuffered solution, equilibrium will shift depending on total analyte concentration
example: if Ka = 10-4 HInKa
H+ + In-
CHIn
Expected
Actual
“Apparent” deviation since can be accounted for by chemical equilibrium
Isosbestic point At the isosbestic point in spectra:
A = b([HIn] + [In-])
But, if unbuffered solution, equilibrium will shift depending on total analyte concentration
example: if Ka = 10-4 HInKa
H+ + In-
c) Non-constant b:- worse for round cuvettes- use parallel cuvettes to help
P0
A = log10 Po/P = bc
P1
B1
B2P2
d) Instrument noise:
noise - short term baseline fluctuations, which decrease the precision of the analysis
‚ can not measure A precisely ‚ the various sources of noise each cause some uncertainty in the absorbance measurement and can be treated as individual standard deviations.
1) 0%T noise:- noise when light beam is blocked- seldom important- typically 0.01%T
2) Readout Precision:- especially with a meter- typically 0.5%T 1-3% error in concentration
3) Shot Noise:- occurs when e- transfers across a junction (like the space between cathode & anode in PMT).- causes random fluctuations in current since individual e- arrive at random times- increases with increase current (%T). Especially bad above 95%T.
4) Flicker Noise:- noise from the lamp due to intensity changes- important at high transmittances.
5) Cell positioning uncertainty:- not really noise, but affects precision- minor imperfections, scratches or dirt change %T- may be the major cause of imprecision
Category Characterized by Typical Sources Likely to be Important in
Case IST = k1
Limited Readout resolution Inexpensive photometers and spectrophotometers having small meter scales
Heat detector Johnson noise IR and near-IR spectrophotometers and photometers
Dark current and amplifier noise Regions where source intensity and detector are low
Case II ST = k2 T2 +T Photon detector shot noise High-quality UV-visible spectrophotometers
Case III ST = k3TCell positioning uncertainties High-quality UV-visible and IR
spectrophotometers
Source flicker Inexpensive photometers and spectrophotometers
Taken together, these noise sources indicate that the intermediate absorbance and transmittance ranges should be used.
- at low %T, 0%T, noise and readout precision are important - at high %T, shot and flicker noise are large.
Keep A in range of 0.1 – 1.5 absorbance units (80 -3%T)
Applications:A) Molar Absorptivities () in UV-Vis Range:
= 8.7 x 1019 PA
P – transition probability (ranges from 0.1 to 1, for likely transitions)A – cross-section area of target molecule (cm2)
- ~10-15 cm2 for typical organics- max = 104 to 105 L/mol-cm
- < 103 – low intensity (P 0.01)
Name Structure max
but-1-en-3-yne 219 7,600
cyclohex-2-enone 225 10,300
toluene 206 7,000
3,4-dimethylpent-3-en-2-one 246 5,300
O
CH
C CHH2C
CH3
OCH3
H3C
CH3
- For Compounds with Multiple Chromophores:‚ If greater then one single bond apart
- are additive- constant
CH3CH2CH2CH=CH2 max= 184 max = ~10,000
CH2=CHCH2CH2CH=CH2 max=185 max = ~20,000
‚ If conjugated- shifts to higher ’s (red shift)
H2C=CHCH=CH2 max=217 max = ~21,000
Example 6: The equilibrium constant for the conjugate acid-base pair
HIn + H2O H3O+ + In-
Calculate the absorbance at 430 nm for an indicator concentration of 3.00x10-4 M
K = 8.00x10-5
= 8.04x103 = 0.755x103
Applications:
B) Absorbing Species in UV/Vis:
1) Electronic transitions involving organic compounds, inorganic compounds, complexes, etc.
Basic process:
M + h M*
10-8 – 10-9s M* M + heat
(or fluorescence, light, or phosphorescence)
or
10-8 – 10-9s M* N
(new species, photochemical reaction)
Note: excited state (M*) is generally short and heat produced not generally measurable. Thus, get minimal disturbance of systems (assuming no photochemical reaction)
2) Absorption occurs with bonding electrons.
- E() required differs with type of bonding electron.- UV-Vis absorption gives some information on bonding electrons (functional groups in a compound.
- Most organic spectra are complex ‚ electronic and vibration transitions superimposed ‚ absorption bands usually broad ‚ detailed theoretical analysis not possible, but semi-quantitative or qualitative analysis of types of bonds is possible.
‚ effects of solvent & molecular details complicate comparison
- Single bonds usually too high excitation energy for most instruments (185 nm)
‚ vacuum UV‚ most compounds of atmosphere absorb in this range, so difficult to work with.‚ usually concerned with functional groups with relatively low excitation energies (190 850 nm).
- Types of electron transitions:
i) , , n electrons
Sigma () – single bond electron
Low energy bonding orbital High energy anti-bonding orbital
Pi () – double bond electron
Low energy bonding orbital High energy anti-bonding orbital
Non-bonding electrons (n): don’t take part in any bonds, neutral energy level.
Example: Formaldehyde
‚ * transition in vacuum UV‚ n * saturated compounds with non-bonding electrons
~ 150-250 nm ~ 100-3000 ( not strong)
‚ n *, * requires unsaturated functional groups (eq. double bonds)most commonly used, energy good range for UV/Vis ~ 200 - 700 nm n * : ~ 10-100 *: ~ 1000 – 10,000
Chromophore Example Solvent max (nm) max Type of transition
Alkene n-Heptane 177 13,000 *Alkyne n-Heptane 178
196225
10,0002,000
160
*__
Carbonyl n-Hexane
n-Hexane
186280
180293
1,00016
Large12
n*n*
n*n*
Carboxyl Ethanol 204 41 n*
Amido Water 214 60 n*
Azo Ethanol 339 5 n*
Nitro CH3NO2 Isooctane 280 22 n*
Nitroso C4H9NO Ethyl ether 300665
10020
_n*
Nitrate C2H5ONO2 Dioxane 270 12 n*
C6H13HC CH2
C5H11C C CH3
CH3CCH3
O
CH3CH
O
CH3COH
O
CH3CNH2
O
H3CN NCH3
Absorption Characteristics of Some Common Chromophores
Other Examples of Some Common Chromophores
ii) d/f electrons (transition metal ions)‚ Lanthanide and actinide series
- electronic transition of 4f & 5f electrons - generally sharp, well-defined bands not affected by associated ligands
‚ 1st and 2nd transition metal series- electronic transition of 3d & 4d electrons- broad peaks
Crystal-Field Theory
- In absence of external field d-orbitals are identical - Energies of d-orbitals in solution are not identical- Absorption involves e- transition between d- orbitals- In complex, all orbitals increase in energy where orbitals along bonding axis are destabilized
Magnitude of depends on:- charge on metal ion- position in periodic table- ligand field strength :
I- < Br- < Cl- < F- < OH- < C2O42- ~ H2O < SCN- < NH3
< ethylenediamine < o-phenanthroline < NO2- < CN-
increases with increasing field strength, so wavelength decreases
iii) Charge Transfer Complexes‚ Important analytically because of large (> 10,000)‚ Absorption of radiation involves transfer of e- from the donor to orbital
associated with acceptor- excited state is product of pseudo oxidation/reduction process
‚ Many inorganic complexes of electron donor (usually organic) & electron acceptor (usually metal)
- examples: Iron III thiocyanate Iron II phenanthroline
(deep red color)(colorless)
C) Qualitative Analysis:
1) Limited since few resolved peaks- unambiguous identification not usually possible.
2) Solvent can affect position and shape of curve.- polar solvents broaden out peaks, eliminates fine structure.
Loss of fine structure for acetaldehyde when transfer to solvent from gas phase
Also need to consider absorbance of solvent.
Loss of fine structure for 1,2,4,5-tetrazine as solvent polarity increases
(a) Vapor
(b) Hexane solution
(c) Aqueous
2) Solvent can affect position and shape of curve.- polar solvents broaden out peaks, eliminates fine structure.
3) Solvent can also absorb in UV-vis spectrum.
3) Can obtain some functional group information for certain types of compounds..- weak band at 280-290 nm that is shifted to shorter ’s with an increase in polarity (solvent) implies a carbonyl group.
‚acetone:in hexane, max = 279 nm ( = 15)in water, max = 264.5 nm
- solvent effects due to stabilization or destabilization of ground or excited states, changing the energy gap.
‚ since most transitions result in an excited state that is more polar than the ground state- 260 nm with some fine structure implies an aromatic ring.
Benzene in heptane
More complex ring systems shift to higher ’s (red shift) similar to conjugated alkenes
UV Spectral Nomenclature
C) Quantitative Analysis (Beer’s Law):1) Widely used for Quantitative Analysis Characterization
- wide range of applications (organic & inorganic)- limit of detection 10-4 to 10-5 M (10-6 to 10-7M; current)- moderate to high selectivity- typical accuracy of 1-3% ( can be ~0.1%)- easy to perform, cheap
2) Strategiesa) absorbing species- detect both organic and inorganic compounds containing any of these species (all the previous examples)
b) non- absorbing species- react with reagent that forms colored product- can also use for absorbing species to lower limit of detection- items to consider:
, pH, temperature, ionic strength- prepare standard curve (match standards and samples as much as possible)
Complex(red)
As Fe3+ continues to bind proteinred color and absorbance increases.
When all the protein is bound to Fe3+,no further increase in absorbance.
Standard Addition Method (spiking the sample)
- used for analytes in a complex matrix where interferences in the UV/Vis for the analyte will occur: i.e. blood, sediment, human serum, etc..
- Method:(1) Prepare several identical aliquots, Vx, of the unknown sample.
(2) Add a variable volume, Vs, of a standard solution of known concentration, cs, to each unknown aliquot.
Note: This method assumes a linear relationship between instrument response and sample concentration.
(3) Dilute each solution to an equal volume, Vt.
(4) Make instrumental measurements of each sample to get an instrument response, IR.
Where:S = signal or instrument responsek = proportionality constantVs = volume of standard addedcs = concentration of the standardVx = volume of the sample aliquotcx = concentration of the sampleVt = total volume of diluted solutions
V s
Inst
rum
ent R
espo
nse
(S
) m = y/ x
b = y-intercept
(V s ) 0
t
xx
t
ss
VckV
VckVS
bmVS s
x
sx mV
bcc
VsSScVSC ss
x)( 12
1
(5) Calculate unknown concentration, cx, from the following equation.
Note: assumes a linear relationship between instrument response and sample concentration.
c) Analysis of Mixtures- use two different ’s with different ’s
A1 = 1MbcM + 1
NbcN (1)
A2 = 2MbcM + 2
NbcN (2)
Note: need to solve simultaneous equations
d) Photometric titration- can measure titration with UV-vis spectroscopy.- requires the analyte (A), titrant (T) or titration product (P) absorbs radiation
Example 7: Given:
Calculate the concentrations of A and B in solutions that yielded an absorbance of 0.439 at 475 nm and 1.025 at 700 nm in a 2.50-cm cell.
Absorbance (1.00 cm cell)Species 475 nm 700 nm
A (7.50x10-5 M) 0.155 0.755
B (4.25x10-5 M) 0.702 0.091