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EA-AC
Lecture W8
Today’s topics include:
February 26, 2003
Spectrophotometer Detector Noise and Slit widths
Analysis of Mixtures
Photometric titrations
Mole-ratio determinations of complexes
Last Week we talked a bit about errors in spectrophotometers.
Some sources of error are proportional to T. Cell positioning uncertainty falls in this class.
Another problem is fluctuations in source intensity. If the light source doesn’t have constant
output, then Io will vary with time and so will T and A measurements.
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Here are the results from the two different instruments we examined. The error of the Spec
20 is pretty much independent of transmittance. It is probable that the major source of
uncertainty lies in the limited resolution of the transmittance scale that is read by eye.
Errors that are independent of T include readout resolution, thermal detector noise – i.e. the
noise in the detector when no light falls on it, and amplifier and other electronic noise.
.
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The Cary 118 does has particularly low error at high A. The errors that exist have their
origin in the so-called shot noise that causes the outputs of photomultipliers and phototubes
to fluctuate randomly about a mean value.
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Shot noise is encountered whenever electrons cross a junction or a vacuum gap. In a
photomultiplier, light strikes a photoelectric surface in a vacuum and electrons are emitted.
These electrons are accelerated and strike a second plate that emits electrons – perhaps 6 of
them. Each of these electrons are attracted to a third electron emitting plate where each
produces 6 more electrons and so on. 69 = 10,077,696
The result is that a single photon produces a signal of about 107 electrons. The sum of all
the electrons the are collected by the last plate per second become the current out of the
detector and is proportional to the number of photons that strike the detector.
But there is a randomness about the number of electrons emitted at each step. Thus the
current is subject to statistical fluctuations , and these fluctuations vary with the square root
of the current.
All of this becomes unavoidable noise or error in the output of the spectrophotometer
detector.
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Now let’s revisit the idea of a slit width one more time. We said the wider the slit width the
greater the spectral power that passes through it, but the larger bandwidth results in reduced
spectral purity. So we have a design trade-off.
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Suppose that this is the true power response to molecules of a particular substance. This
theoretical curve is sometimes called the absorption envelope.
Now let’s see how the spectrum might change with different spectral slit widths. Note
especially the appearance of the narrow feature as the spectral slit width changes.
The spectral slit widths are indicated by the distance between pairs of parallel lines on the
wavelength axis as shown in this transparency.
These lines bracket the range of wavelengths that pass through the monochromator at any
nominal wavelength setting. The nominal wavelength is the center of the wavelength
range.
If the slit width is optimized, the spectrum will closely resemble the absorption envelope
of this transparency.
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If a narrower slit width is used, the decreased power arriving at the transducer means that
the contribution from noise begins to obscure the spectral features. Here we see such a
spectrum. All kinds of random noises now become visible.
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If you are stuck with this situation, one possible approach is to signal average. Signal
averaging simply involves obtaining the spectrum over and over again, adding the spectra
together and dividing A at each wavelength by the number of spectra.
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If the noise at each wavelength is truly random, then noise that is too high should balance
noise that is too low. In general the noise is reduced by a factor equal to the square root of
N the number of spectra taken.
The HP diode-array has this feature built in by letting you vary how long to continue signal
averaging spectra. The instrument takes a spectrum every 0.1 sec so collecting for 2 sec
signal averages 20 spectra.
Remember the transducer responds only to the total light power falling on it, with very
little discrimination of wavelength. As a result, the transducer averages the power arriving
at all the wavelengths passing through the monochromator.
Consequently, as the monochromator scans through wavelength, when the edge of the slit
reaches the spectral range of the sharp feature, the average power falling on the transducer
rises.
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This is seen as a rise in the power level before the monochromator's nominal wavelength
reaches the base of the sharp peak.
When the nominal wavelength of the sharp peak reaches the center of the slit, the highest
point (the highest power) of the peak appears to be reduced, since the transducer is
averaging the peak power with the points of lower power on either side.
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As the slit is made wider than the optimum, the spectral width becomes greater than that of
the sharp feature, and the sharp peak appears to be even broader and lower. Notice that
the broad peak appears relatively unchanged through all the slit-width changes.
The difference in the slit's effect on the broad and narrow features illustrates a general
rule. The effect depends on the ratio of the linewidth of each spectral feature compared
with the spectral slit width.
As a general rule, in order to measure a spectral feature precisely, spectral bandwidth need
to be < 1/10th the linewidth of the feature.
As an example of the effect of slit width on the absorption spectrum of a real substance,
here is the spectrum of reduce cytochrome c as the bandpass is varied from 20 nm to 1 nm.
Here we see both peaks improve as we go to narrower slits.
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Here is an example of the use of a spectrophotometer in analysis of a nickel
complex.
A 5.00 x 10-4 M solution of a nickel complex is put into a sample cuvette with a
pathlength of 1.000 cm. The absorbance at 592 nm is found to be 0.446. What is
the molar absorptivity of the nickel compound at that wavelength?
To find ε592:
A = ε592 b c
0.446 = ε592 (1.000 cm) (5.00 x 10-4 M)
ε592 = 0.446/(5.00 x 10-4) = 8920
If a solution of unknown concentration of the nickel complex has an absorbance of
0.125 at the same wavelength, what is its concentration?
A = ε592 b c
0.125 = 8920 x 1.000 c
c = 0.125/8920 = 1.40 x 10-5 M
So Beer’s law is based on a linear relationship between A and c. We can take
advantage of this fact to analysis multiple chromophores in the same solution.
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The total absorbance of a solution at any given wavelength is equal to the sum of the
absorbances of the individual components in the solution.
Atotal = A1 + A2 + …… + An = 1bc1 + 2bc2 +…… + nbcn.
This relationship makes it possible in principle to determine the concentrations of the
individual components of a mixture even if total overlap in their spectra exists.
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For example, this transparency shows the spectrum of a solution containing a mixture of
species M and species N as well as absorption spectra for the individual components.
Clearly, no wavelength exists at which the absorbance is due to just one of these
components.
At each point in the mixture the absorbance is
Atotal = 1bc1 + 2bc2
To analyze the mixture, molar absorptivities for M and N are first determined at
wavelengths 1 and 2 with enough standard to be sure that Beer's law is obeyed over an
absorbance range that encompasses the absorbance of the sample.
Note that the wavelengths selected are ones at which the two spectra differ significantly.
Thus, at 1 the molar absorptivity of component M is much larger than that for component
N. The reverse is true for 2.
Once we know  values for each substance at each of the two wavelengths then the
absorbance of the mixture is determined at the same two wavelengths.
A1 = 1bc1 + 2bc2
A2 = 3bc1 + 4bc2
We know everything except the two concentration terms, so we have two equations and
two unkowns to solve for. You worked one of these problems in the homework and they
are pretty straight forward.
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Here is a real world example of a two component solution. The top spectrum is of a CoEDTA solution while the middle spectrum is that of a Ni-EDTA solution, The bottom
spectrum is from a mixture of the two solutions but not a 1:1 mixture.
Using this method, you do not have to do an EDTA titration in order to do 3 sig fig analysis
of a mixture of Co and Ni. Once you have determined the two ’s, you can do this analysis
rapidly.
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Our book presents in Chapter 19 an Excel spreadsheet method for solving these multicompoent mixture probems. Solving sets of linear equations is the basis for the math field
called linear algebra. The book mentions both matrix and determinant methods for these
analyses as well.
Here is a graphical approach that is based on using more wavelengths to improve the
statistics of the measurement. At any wavelength
Am =
A1s
c2 A2s
c2s A1s
+ c1
c1s
Where the s’s represent values of the standard solutions and the rest are for the mixture.
Notice that the equation has the form y = mx + b. This example is based on measurements at
five wavelengths.
Finding the slope and intercept then allow you to solve for c1and c2 because
c1 = intercept * c1s and c2 = slope * c2s
The advantage of this method is that you can use least squares analysis to get the best straight
line from the data and use least squares statistics to determine the errors in the values of
c1and c2.
We have been talking about how instruments work and how we use spectroscopy to
determine the concentration of analytes in samples. Now let’s look at how
spectrophotometers are used in titrations.
In order to do a titration of course we need to be able to locate the equivalence point. The
application of absorption measurements obviously requires that one or more of the reactants
or products absorb radiation or that an absorbing indicator be present.
A photometric titration curve is a plot of absorbance as a function of titrant volume. The
absorbance value must be corrected for solution volume. That is, the solution becomes more
and more dilute with added titrant. So the Absorbance has to be corrected back to simulate
the initial volume.
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If conditions are chosen properly, the curve consists of two straight-line regions with
different slopes, one occurring at the outset of the titration and the other located well beyond
the equivalence-point region.
The end point is taken as the intersection of extrapolated linear portions of the two lines.
This transparency shows typical photometric titration curves.
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The graph in the upper left is the curve for the titration of a nonabsorbing species with an
absorbing titrant that is decolorized by the reaction. An example is the titration of thiosulfate
ion with triiodide ion. As the letters indicate, s = 0 p = 0 and the titrant is colored t >0.
2 S2O32+ == > S4O62+ + 2 eI3- + 2 e- == > 3 IThe titration curve for the formation of an absorbing product from colorless reactants is
shown in upper-middle graph (b). The product is colored but the analyte and titrant are not.
An example is the titration of iodide ion with a standard solution of iodate ion to form
colored triiodide.
The remaining figures illustrate the curves obtained with various combinations of absorbing
analytes, titrants, and products.
Notice that the equivalence points are estimated by extrapolation of each of the volume
corrected lines. Volume correction is achieved by multiplying each observed absorbance by
(V + v)/V, where V is the original volume of the solution and v is the volume of added
titrant.
In order to obtain titration curves with linear portions that can be extrapolated, the absorbing
system(s) must obey Beer's law.
Photometric titrations are ordinarily performed with a spectrophotometer or a photometer that
has been modified so that the titration vessel is held in the light path.
The instrument is set to a suitable wavelength and with radiation passing through the analyte
solution to the detector. The instrument is adjusted to a convenient absorbance reading by
varying the source intensity or the detector sensitivity.
Ordinarily, no attempt is made to measure the true absorbance since relative values are
perfectly adequate for end-point detection. Titration data are then collected without
alteration of the instrument settings.
The power of the radiation source and the response of the detector must remain constant
during a photometric titration.
Cylindrical cells are ordinarily used and stirring provided to get complete mixing. Care must
be taken to avoid any movement of the vessel that might alter the length of the radiation path.
Photometric titrations often provide more accurate results than a direct photometric
determination because the data from several measurements are pooled in determining the end
point.
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When there are other absorbing species in the matrix during direct abosrbance measurements
we are many time forced to use the method of standard addition.
However other absorbing species do not usually interfere with a photometric titration since
only a change in absorbance is being measured.
One advantage of a photometric end point is that the experimental data are taken well away
from the equivalence-point region.
Consequently, the equilibrium constant for the reaction need not be as favorable as that
required for a titration that depends upon observations near the equivalence point such as is
required for indicator end points.
For the same reason, more dilute solutions can be titrated.
Photometric titrations has been applied to all types of reactions. For example, most standard
oxidizing agents have characteristic absorption spectra and thus produce photometrically
detectable end points.
Although standard acids or bases do not absorb, the introduction of colored acid/base
indicators permits photometric neutralization titrations.
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The photometric end point has also been used to great advantage in titrations with EDTA and
other complexing agents. This transparency illustrates the application of this technique to the
successive titration of bismuth(III) and copper(II).
Bismuth binds EDTA tighter (large formation constant) and so binds first to EDTA, then
copper binds.
At 745 mn, the cations, the reagent, and the bismuth complex formed in the first part of the
titration do not absorb, but the copper complex does. Thus, the solution exhibits no
absorbance until essentially all the bismuth has been titrated.
With the first formation of the copper complex, an increase in absorbance occurs. The
increase continues until the copper equivalence point is reached. Further reagent additions
cause no further absorbance change. Clearly, two well-defined end points result.
Now let’s turn to the use of spectrophotometry for determining the composition of complex
ions in solution and for determining their formation constants. A major advantage of the
spectrophotometric method is that you do not have to isolate the complex as a pure
compound.
The power of the technique lies in the fact that quantitative absorption measurements can be
performed without disturbing the equilibria under consideration. Although most
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spectrophotometric studies of complexes involve systems in which a reactant or a product
absorbs, nonabsorbing systems can also be investigated successfully.
The most common techniques employed for complex-ion studies are the method of
continuous variations given in the book and the mole-ratio method.
In the method of continuous variations, cation and ligand solutions with identical analytical
concentrations are mixed in such a way that the total volume (and hence the total moles) of
reactants in each mixture is constant but the mole ratio of reactants ( XL) varies
systematically (for example, 1:9, 8:2, 7:3, and so forth).
The absorbance of each solution is then measured at a suitable wavelength and corrected for
any absorbance the mixture might exhibit if no reaction had occurred.
The corrected absorbance is plotted against the volume fraction of one reactant, that is,
Vm/(Vm + VL), where Vm is the volume of the cation solution and VL that of the ligand.
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A typical plot is shown in this transparency. A maximum (or minimum if the complex
absorbs less than the reactants) occurs at a volume ratio Vm/VL corresponding to the
combining ratio of cation and ligand in the complex.
In the Figure, Vm/(Vm + VL) is 0.33 and VL/(Vm + VL) is 0.66; thus, Vm/VL is 0.33/0.66,
which suggests that the complex has the formula ML2.
The curvature of the experimental lines in the Figure is the result of incompleteness of the
complex-formation reaction. A formation constant for the complex can be evaluated from
measurements of the deviations from the theoretical straight lines.
The Mole-Ratio Method is a third method for determine combining ratios of ligands and
metals.
In the mole-ratio method, a series of solutions is prepared in which the analytical
concentration of one reactant (usually the cation) is held constant while that of the other is
varied.
A plot of absorbance versus mole ratio of the reactants is then prepared. If the formation
constant is reasonably favorable, two straight lines with different slopes are obtained. The
two intersect at a mole ratio that corresponds to the combining ratio in the complex.
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Typical mole-ratio plots are shown in this transparency. Note that the ligand of the 1:2
complex absorbs at the wavelength selected so that the slope beyond the equivalence point is
greater than zero.
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We deduce that the uncomplexed cation involved in the 1:1 complex absorbs because the
initial point has an absorbance greater than zero. Formation constants can be evaluated from
the data in the curved portion of mole-ratio plots.
A mole-ratio plot may reveal the stepwise formation of two or more complexes as successive
slope changes, provided the complexes have different molar absorbtivities and provided the
formation constants are sufficiently different from each other.
The Slope-Ratio Method
This approach is particularly useful for weak complexes but is applicable only to systems in
which a single complex is formed.
The method assumes (1) that the complex-formation reaction can be forced to completion by
a large excess of either reactant and (2) that Beer's law is followed under these
circumstances.
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Let us consider the reaction in which the complex MxLy is formed by the reaction of x moles
of the cation M with y moles of a ligand L:
x M + y L < == > MxLy
Mass-balance expressions for this system are
CM = [M] + x [MxLy]
CL = [L] + y [MxLy]
where cM and cL are the molar analytical concentrations of the two reactants.
We now assume that at very high analytical concentrations of L, the equilibrium is shifted
far to the right and [M] << x [MxLy]. Under this circumstance, the first mass-balance
expression simplifies to
cM = x [MxLy]  [MxLy] = cM/x
So Beer's law becomes,
A1 = b[MxLy] = bcM/x
A plot of absorbance as a function of cM becomes linear whenever sufficient L is present to
satisfy the assumption that [M] << x[MxLy]. The slope of this plot is b/x.
When cM is made very large, we assume that [L] << y[MxLy], whereupon the second
mass-balance equation reduces to
CL = y[MxLy] and
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A2 = b[MxLy] = bcL/y
Again, if our assumptions are valid, a linear plot of A versus cL is observed at high
concentrations of M. The slope of this line is b/y.
The ratio of the slopes of the two straight lines gives the combining ratio between M and L:
b/x = y/x
b/y
These are some of the other uses of photometric methods in analytical chemistry besides
straight analyte concentration determination.
Homework
Graphing – zero not a point
Exam
Classes on Wed Mar 5 will be in 1039. Read Chapter 21 on AA.