Transcript 7. Beer’s Law and It’s Implications for Instrument Construction 1. Derive Beer’s Law ASSUMPTIONS 1.

7. Beer’s Law and It’s Implications for
Instrument Construction
1. Derive Beer’s Law
ASSUMPTIONS
1. No light is emitted
M  h  M  kT
*
2. dx infinitesimal
3. Monochromatic light uniform on the surface, S
4. dn molecules in a section volume
V  dxS
5. Capture cross sectional area is
dS  dn
Set up Derivation
Photons impinging (capture area) P0 dS 
Photons captured 

total area
S
dS  dn
Consider a large # of boxes
Po
P2
P1
 dP 
P0 dn
S
This is an integration
dPx


P0 P
x
P
n
dn
0
S

dPx


P0 P
x
P
 ln Px
P
P0

n
dn
0
S

 n  n0 
S
 ln P  ln P0  
n
S
 P  n
 ln  
S
 P0 
 P
n
 log  
 P0  2.303S
 P
n
A   log    log T 
2.303S
 P0 
Substitutions
V
S
b
 1L 
n  MN a  3 3 V
 10 cm 
 1L 
  MN a  3 3 V
 10 cm 
 P
A   log  
 P0 
V 
2.303 
 b

 1L  
   N a  10 3 cm 3  
 Mb  bM
A
2.303




A   log T  bM
What is the absorbance when the
light transmitted is 50% of the initial
beam in a 2 cm path length cell for a
concentration of 10-3 M?
Deviations
1. Assumed each molecule was independent of the other
When will the assumptions fail?
Molecules not independent when:
Neighbors experience each other
1. High concentrations
2. High electrolyte
3. Large local fields due to large absorption probability (alpha)
Apparent Instrumental Deviations
**polychromatic radiation***
What is the source of polychromatic radiation?
1
2
Similarly
 P1 
    1bC
A1   log
 P01 
 P2 
A2   log
    2 bC
 P0 2 
Rearrange
P1  P01 10
Total absorbance
P2  P02 10
  1bC
Ameasured
Ameasured
   2bC
 P0 2  P0 2 
 P1  P 2 

  log
  log
 P0 2  P0 2 
 P1  P 2 


P0 2  P0 2

 log
   1bC
   2bC 
 P0 2 10
 P01 10

Consider several cases using this equation
Ameasured
1. Monochromatic light
 1   2  
Ameasured


P0 2  P0 2

 log
   1bC
   2bC 
 P0 2 10
 P01 10



P

P
0

2
0

2

 log


bC
 P  P 10

0 2
 01



 1 
Ameasured  log  bC 
 10



Ameasured  log 10bC  bC
Reality check ok
2. Case 2
Ameasured


P0 2  P0 2

 log
   1bC
   2bC 
 P0 2 10
 P01 10

Po, 1  P0, 2  P0
Ameasured
Ameasured
Ameasured


P0  P0
 log
   1bC
   2bC 
P
10

P
10
 0

0

2 P0
 log
 P 10   1bC  10    2bC
 0


2
 log
 10   1bC  10    2bC


Example Calculation
B=1
M=0.001
Molar absorptivity at 1=2000
at 2 = 200






Ameasured




2
 log

3
3

2000
1
10

200
1
10




   10
   
  10


Ameasured





2
 log
 10  2  10  0.2


.494
M


  0.494


When would this situation apply?
3. Stray Light
Ameasured


P0 2  P0 2

 log
   1bC
   2bC 
 P0 2 10
 P01 10

Po, 1  P0, 2 and   2  0
Ameasured


P0 1  P0 2
 log
   1bC
 0bC 
P
10

P
10
 0 1

0 2
Example Calculation
Stray light is 0.5% of total light
Ameasured


P0 1  P0 2
 log

   1bC
P
10

P
 0 1
0 2 
P0 2  0.005P0 1
What happens when light at 1 is strongly absorbed?
P0 110   1bC  P0 2
Ameasured
 P0 1  P0 2 
 log

P


0 2
Ameasured
 P0 1  0.005P0 1 
 log

0
.
005
P


0 1
.
 1005

Ameasured  log
  2.303
 0.005
The maximum absorbance the
Instrument is capable of measuring is
2.303
Comparison of Instruments
Instrument
%stray light
maxA
Spect 20
0.5
2.3
McPherson
0.1
3
McPherson +filter 0.01
4
Double monochromator 0.001
5
Physical Dimensions:
89.1 mm x 63.3 mm x 34.4 mm
Weight:
190 grams
Detector:
Sony ILX511 linear silicon CCD array
Detector range:
200-1100 nm
Pixels:
2048 pixels
Pixel size:
14 μm x 200 μm
Pixel well depth: ~62,500 electrons
Sensitivity:
75 photons/count at 400 nm; 41 photons/count at 600 nm
Design:
f/4, Symmetrical crossed Czerny-Turner
Focal length:
42 mm input; 68 mm output
Entrance aperture:
Czerny-Turner
construction
5, 10, 25, 50, 100 or 200 µm wide slits or fiber (no slit)
Grating options: 14 different gratings, UV through Shortwave NIR
Detector collection lens option:
Yes, L2
OFLV filter options:
OFLV-200-850; OFLV-350-1000
Other bench filter options:
Longpass OF-1 filters
Collimating and focusing mirrors:
Standard or SAG+
UV enhanced window:
Yes, UV2
Fiber optic connector:
SMA 905 to 0.22 numerical aperture single-strand optical fiber
Spectroscopic Wavelength range:
Grating dependent
Optical resolution:
~0.3-10.0 nm FWHM
Signal-to-noise ratio:
250:1 (at full signal)
A/D resolution:
12 bit
Dark noise:
3.2 RMS counts
Dynamic range:
2 x 10^8 (system); 1300:1 for a single acquisition
Integration time: 3 ms to 65 seconds
Stray light:
<0.05% at 600 nm; <0.10% at 435 nm
Corrected linearity:
What would be
Electronics Power consumption:
The maximum
Data transfer speed:
A this could measure?
>99.8%
90 mA @ 5 VDC
Full scans to memory every 13 ms with USB 2.0 or 1.1 port, 300 ms with serial port
What is the maximum amount of absorbance
you can measure if the stray light in an
instrument is 8%?
If it is 0.05% at 600 nm as for the Ocean
Optics?
1. Where does stray light come from?
2. Is stray light likely to be more important for 200 or
for 900 nm light?
3. Is stray light likely to be more or less important
near a region where solvent interferes?
Double Dispersion Reduces the Stray Light
Comparison of Instruments
Name
$
∆
range
Ps/Po%
Spect 20
Double Beam
PE-57
Double dispersive
Multichannel Array
2-4k
4-15k
>5k
2-8
190-1000
195-850
190-750
185-3125
200-920
0.5
0.1
<0.1
0.0008
7-9k
0.2
0.07
Beer’s Law and Standard Additions
QUANTITATION
1. Wide chromophore range (universality)
-extended by color forming reactions
for example complexation
2. Good sensitivity
3. Selectivity
4. Accuracy
5. Ease
1. Standard Curves
Choose a wavelength where the molar absorptivity does not change
where would this be?
why choose this wavelength region?
Need clear cells and no greasy fingers. Why?
Need to control: temperature; pH; electrolyte/solvent. Why?
2. Standard addition method
is useful when matrix (the solution containing the
sample analyte) effects complicate matters
Overcoming Matrix Effects in Calibration Curves
Solvent
Signal
Matrix
Sample
Signal
Matrix
Example: Flame
Atomic Absorption for
Pb in SeaWater, PbCl2
Is lost lowering the signal
Matrix Effect
If we don’t have a
Clear idea what
The matrix effect is
Then we drastically
Misjudge the conc
Of the sample from
The measured signal
Ppm Metal
Our standards
Suggest this
Sample conc.
Standards made
Up in the matrix of
The sample would
Suggest this sample
Conc.
Overcoming Matrix Effects in Calibration Curves
0
 total moles 
Ameasured  b

total
volume


10
20
Anew,measured
 nunknown  nadded 
 b 

V

V
 sample of unknown
added 

30
40
Ameasured
 Vunkown M unknown  Vstamdard M s tan dard 
 b

Vunkown  Vs tan dard


V
M
 Vstamdard M s tan dard 
Ameasured  b unkown unknown

V
total


Ameasured
 Vunkown M unknown 
 M s tan dard 
 b
  b
Vstamdard
V
V
total
total




50
intercept
slope
y
Ameasured
x
 Vunkown M unknown 
 M s tan dard 
 b
  b
Vstamdard
V
V
total
total




V

M
int ercept  b unkown unknown 
Vtotal


 M s tan dard 
slope  b

V
total


int ercept

slope
 Vunkown M unknown 

V
total


b 
 M s tan dard 

V
total


b 
 int ercept   M s tan dard 
  M unknown


 slope   Vunkown 

Vunkown M unknown
M s tan dard
M=slope=0.03912
B=intercept=0.2422
Vunknown= 10 ml
Mstandard=11.1ppm
 total moles 
Ameasured  b

 total volume 
 int ercept   M s tan dard 
  M unknown


 slope   Vunkown 
. ppm 
 0.2422   111


  7.01 ppm
 0.03812   10 
You did standard addition for the flame lead
analysis. You found:
A  00.0521  0.433( ppb)
Your unknown volume is 10 mL and the standard
you added is 20 ppb.
What is the unknown concentration?
Two Component Spectra
A1   M 1bC M   N 1bC N
A2   M 2 bC M   N 2 bC N
measure
Must be known
Result is two equations in two unknowns – can be solved