Download Analytical Chemistry CHEM 21122

2/9/2005
Course content & Exam
 Sampling and chemometrics (6 h)
 Analytical spectroscopy: (6 h)
 Analytical electrochemistry: (6 h)
 Analytical separation : (6 h)
 Analytical instrumentation : (6h)
Prof. K. A. S. Pathiratne(06 L)
Dr . Russel De Silva (12 L)
Dr. Sri Skandaraja (12 L)
Analytical Spectroscopy
End of semester exam : two hours, five Questions.
Expect in-class exams + take home assignments too.
Spectroscopy
 Latin: “spectrum” — ghost or spirit
 Greek: “σκοπειν” — to look carefully
 We do not look directly at the matter (atoms, molecules) but at its “ghost”
 We observe the interaction of electromagnetic radiation with different
properties of the matter.
 Each type of spectroscopy gives a different picture of the matter → the
spectrum.
Spectroscopy is the study of the interaction between
matter and radiated energy.
 The spectrum is the variation of the intensity of the radiation as a
function of the frequency or wavelength
Spectroscopy is any procedure that uses the interaction of
Electromagnetic Radiation (EMR) with matter to identify and/or to
estimate an analyte.
Quantitative
Analysis
Band spectra
molecules
ions
atoms
Mixtures
solid
liquid
gas
solutions
Qualitative
Analysis
line spectra
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Electromagnetic radiation (E-M radiation or EMR) is a form of energy that
exhibits wave-like behavior as it travels through space. (Dual nature of light)
EMR has both electric and magnetic field components, which oscillate in phase
perpendicular to each other and perpendicular to the direction of energy
propagation.
Properties of light waves
 Wavelength (): Distance from one wave peak to the
next.
Units: m, cm, m, nm
 Frequency (): Number of peaks that pass a given
point per second.
Units: Cycles/second or s-1 or Hertz (Hz)
 Wave number : Number of waves per cm.
Units: cm-1
 Electromagnetic radiation consists of discrete packets of
energy, which we call photons.
 Photons are the particles of light or the quanta of light and
each photon carries the energy, E (Joule).
E  hυ
where h is the Planck’s constant (=6.626x10-34 J s)
E = hυ = h
c
= hc υ
λ
Where c is the speed of light in vaccum ( = 2.998 x 108 m/s)
 The greater the energy, the higher the frequency and
wave number and the shorter the wavelength
Electromagnetic Spectrum
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Transmission and Absorption
Transmittance and Absorbance
Transmittance and Absorbance
 For a T = 0.7,
There are two quantities that relate the change in the intensity of EMR before
(I0) , and after, (I) interaction with matter.
1. Transmittance, T, is simply defined as “the fraction of light that reaches a detector after
passing through a sample”
T 
I
I
0≤T≤1
The percent transmittance, %T, is simply 100 T
I
%T 
x 100
0 < %T < 100
I
2. Absorbance, defined as:
I
0<A<2
A = log (  )
A=  log T
I
%T = 70%
then, A =-log T = 0.155
 When no light is absorbed ,
P=P0 ,
A=0, and %T = 100%
 If 90% of the light is absorbed, 10% is transmitted,
thus
T = I/I0 = 10/100 = 0.1 and
A = -log T = - log 0.1 = 1
Beer–Lambert law
the Beer–Lambert law, also known as Beer's law or the Lambert–
Beer law or the Beer–Lambert–Bouguer law (named after August
Beer, Johann Heinrich Lambert, and Pierre Bouguer) relates the
absorption of light to the properties of the material through which
the light is travelling.
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Beer-Lambert Law
Example

Absorbance of light of a particular wavelength is directly
proportional to:
Concentration, c, mol/L (M) of absorbing species in the sample
(A c)
Path length of light, (b or l), in cm through the sample
(A b )
Find the absorbance and transmittance of a 0.00240 M
solution of a substance with molar absorptivity of 313 M-1
cm-1 at  of 520 nm when measured in a cell of 2.00 cm
pathlength.
Thus,
A = bc
This relation is commonly known as Beer-Lambert law

 (M-1 cm-1 )is called the molar absorptivity or molar
absorption coefficient (or Extinction coefficient) “Absorbance of
1 M solution measured in a cell of 1 cm pathlength”

, is characteristic for each substance at a particular
wavelength, .

A = bc
= (313 M-1 cm-1)(2.00 cm)(0.00240 M) = 1.50
- log T = A
log T = -A
T = 10logT = 10-A = 10-1.50 = 0.0316
Absorbance & Concentration
Molar absorptivity
(
Beer’s Law
A c)
Increasing Concentration
ε = 2212 (mol L-1)-1 cm-1
ε = 0.02 (mol L-1)-1 cm-1
Increasing absorbance
KMnO4
Lambert’s law
(A b )
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Transmittance and Concentration
Absorbance & concentration
T=10-A =10- bc
A = bc
= mx
Slope = b
1
Transmittance, T
Absorbance, A
y
Transmittance decreases
exponentially as concentration
increases
0.5
Solution with T = 50%
0
Concentration
12.5%
25%
P0 = 100%
32
50%
Concentration
not zero
Deviation from Beer’s law
 At high concentrations, the proximity of molecules to each others
alters their absorptivity. Scattering and changes in refractive index will
also contribute. the error by possible stray light is significant .
 At low concentrations, error due to sensitivity of the instrument is
significant .
 If the absorbing molecule participates in a concentration-
dependent chemical equilibrium:
HA
A- + H +
 The monochromator generally fails to choose only one λ, instead
it projects a very narrow band on the sample (recall that Beer’s law is
applied at one certain λ).
The range 0.1 - 1.0 absorbance(79.4 - 10%T) is considered ok for analytical purposes.
However the recommended best range to work is 0.2 -0.8. (63.1 – 15.8%T)
Absorption spectrum
 Shows the variation of
Absorbance with wave
length.
 We usually select the
wavelength of max
intensity for the analysis of
any compound.
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Instrumentation
Instrumentation (spectrophotometers)
1. Sources of light

Continuous source, that can emit radiation of all
wavelengths within the spectral region for which they are to
be used.
Tungsten Lamp (320-2500 nm) used for visible and near
IR spectrophotometers.
Deuterium Lamp (200-400 nm) used for UV
spectrophotometers.

Line sources, emit one or very narrow bands of radiation
whose wavelengths are known exactly.
Light emitting diodes (LED) used for fluorimeteric
measurements.
Laser and photodiode, that can be used in different
types of spectrophotometers according to selected
wavelength.
Hallow cathod lamps used in AAS.
A single beam spectrophotometer
• The spectrophotometer is composed of a polychromatic
light source separated into narrow band of wavelength by a
wavelength selector (Monochromator), passed through the
sample compartment and the transmitted intensity, P, after
the sample is measured by a detector .
• Single beam, means single light path through sample to
detector.
2. Wavelength selector – Filters
• Filters permit certain bands of wavelength (bandwidth of ~ 50
nm) to pass through.
2. Wavelength selector – Filters
• The simplest kind of filter is colored glass, in which the coloring
species absorbs a broad portion of the spectrum
(complementary color) and transmits other portions (its color).
• Disadvantage:
• They are not very good wavelength selectors and can’t be
used in instruments utilized in research.
• This is because they allow the passage of a broad
bandwidth which gives a chance for deviations from Beer’s
law and allows interference with impurities.39
Wavelength selectors (monochromator)
 A monochromator disperses light into its component
wavelengths and select a narrow band of wavelengths to
pass to the sample or detector.
 It can be in the form of prism or gratings.
Types of gratings
Echellette-type grating
Reflection followed by either constructive
or destructive interferences
Concave type grating
Collimating/focusing is not necessary so
its cost effective with more energy
throughput
Prisms are not preferable at long wave lengths as they don’t
offer enough resolution and the resolution power varies with
temperature (glass is affected by heat) thus they are now
replaced by different types of gratings.
Holographic grating
Two lasers with different angles – fringes
6000/mm for 50 cm possible
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3- Sample compartment (cells)
Importance of Gratings

 Spectral purity
 Free of unwanted, scattered or stray radiation.
 Reflections of optical components.
 Mechanical imperfections
 Scattering by dust particles.
For Visible and UV spectroscopy, a liquid sample is usually
contained in a cell called a cuvet.
 Glass is suitable for visible but not for UV spectroscopy
because it absorbs UV radiation.
Quartz can be used in UV as well as in visible spectroscopy
Glass 400-3000 nm
(vis-near IR)
Opaque
Face
 Dispersion
 Ability to separate small wavelength differences.
1 cm
Cuvettes & other sample holders
Short pathlength
44
5. Detectors
 2. Photomultiplier tube
 It is a very sensitive device in which electrons emitted from the photosensitive cathode strike a second surface called dynode which is positive
with respect to the original cathode.
 Electrons are thus accelerated and can knock out more than one electrons
from the dynode.
h
 If the above process is repeated several times, so more than 106 electrons
are finally collected for each photon striking the first cathode.
 The electrons flow
through vacuum to an
anode to produce current
which is propor-tional to
radiation intensity.
Plastic containers
(VIS)
Examples of Detectors (cont.)
4-Examples of Detectors
1. Vacuum Phototube
 Phototube emits
electrons from a
photosensitive,-vely
charged cathode when
struck by visible or UV
radiation.
NaCl 200-15,000 nm
(UV-far IR)
Transparent
Face
 Light gathering
1 cm pathlength cuvet
Long pathlength
Silica/quartz 200-3000 nm
(UV-near IR)
eanode
-V
light
amplifier
dynodes
electrons
photochathode
anode
e-
Photosensitive cathode
47
voltage divider network
48
high voltage
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Double Beam spectrophotometers
Photodiode Array-multiplex
Good for quick (fraction of a second) scanning of a full spectrum
Uses semiconductor material:
Remember:
n-type silicon has a conduction electron – P or As doped
p-type silicon has a ‘hole’ or electron vacancy – Al or B doped
A diode is a pn junction:
under forward bias, current
flows from n-Si to p-Si
under reverse bias, no
current flows
 In double beam arrangement, the light alternately passes through
the sample and reference (blank), directed by rotating half-sector
mirror (chopper) into and out of the light path.
 When light passes through the sample, the detector measures the

Boundary is called a
depletion layer or region
radiant power P. When the chopper diverts the beam through the
blank solution, the detector measures P0.
 The beam is chopped several times per second and the electronic
circuit automatically compares P and P0 to calculate absorbance
and Transmittance.
Double Beam spectrophotometers (cont.)
Double Beam spectrophotometers (cont.)
Advantages of double beam over single beam
instruments:

The absorption in the sample is automatically corrected for
the absorption occurring in the blank, since the readout of the
instrument is the difference between the sample beam and the
blank beam.

Automatic correction for changes of the source intensity and
changes in the detector response with time or wavelength
because the two beams are compared and measured at the
same time.

Automatic scanning and continuous recording of spectrum
(absorbance versus wavelength).
Diode array Spectrophotometers
 It’s a double beam spectrophotometer having multiple
diodes (300-500) placed side by side and insulated from
one another in the form of a multi-layer sandwich.
 The common use of a diode array is to monitor light that
has passed through a liquid sensor cell as in a multiwavelength liquid chromatography detector.
 It needs, due to its high scanning rate, a data acquisition
unit and should be connected to computer and this allows
scanning of full spectrum at the same time with high
speed.
Single beam spectrophotometer is
inconvenient because:

The sample and blank must be placed alternately in the
light path.

For measurements at multiple wavelengths, the blank
must be run at each wavelength.

Time consumption and possible errors due to stray light
or voltage changes during measurements.
Diode array Spectrophotometers (cont.)
 Advantages of Diode array over other double beam
spectrophotometers
 Fast spectral acquisition (in a fraction of a second).
 No wavelength selection needed, all wavelength are
recorded simultaneously while in others only one narrow
band reaches the detector at any time.
 No source intensity lost because we do not use slits as in
conventional spectrophotometers.
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Spectrophotometric Applications
Spectronic D20
Selection of wavelength
Selection of wavelength
Crystal Violet Absorption Spectrum
the wavelength should be
If the solution contains more than one absorbing species,
1.4
1.2
Absorbance
chosen to give the highest
1
the wavelength should be chosen, whenever possible, in a
0.8
possible sensitivity to be sure
0.6
0.2
concentration can be measured
region at which the other species does not absorb radiation
max
0.4
that the lowest sample
0
200
250
300
350
with fair accuracy.
400
450
500
wavelength, nm
550
600
650
700
750
measured with good accuracy
determination.
10-5 M
at max, while at other
wavelength (1), it may not be
detected
at all.
57
 max
1
sample
Absorbance
concentration (10-5 M) can be
10-3 M
10-4 M
5x10-5 M
Absorbance
For example, the lowest sample
or its absorbance is minimum. This will eliminate or at least
minimize the interference by the second species, in the
10-2 M
Reagent
(blank)
wavelength
wavelength
m
EXAMPLE
It is preferable to choose the wavelength at which the
absorbance will not significantly change if the wavelength is
slightly changed, i.e., A /  is minimum. At a wavelength
corresponding to broad horizontal band on the spectrum (band A), the
radiation is mainly absorbed to the same extent (A /  zero).
Absorbance
Broad horizontal
bands
Band B
A
shoulder
575 nm
Acidic
form
379 nm
Basic
form
Band A
A
445 nm
Absorbance
Selection of wavelength
isobestic point
503 nm
440 nm
Wavelength
Choose the proper  for detection of a weak acid (HA
following cases:
H+ + A-) in the
1- Determination of the acidic form in presence of the basic form: 445 nm
2- Determination of the basic form in absence of the acid form: 379 or 575nm
3- Determination of the basic form in presence of the acid form: 575 nm


m
wavelength
4- Determination of both species: 503 nm (isobestic point)
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Spectrophotometric Applications
1. Direct methods
Isosbestic point
 An isobestic point is a specific wavelength at which two
chemical species have the same molar absorptivity(ε) or -more
generally- are linearly related. The word derives from two Greek
words: isos: equal, and sbestos: extinguishable.
a. Single-point calibration method
 Standard with measured absorbance
Astd =  b Cstd
 Unknown with measured absorbance
Aunk =  b Cunk
 When an isosbestic plot is constructed by the superposition of
the absorption spectra of two species (whether by using molar
absorptivity for the representation, or by using absorbance and
keeping the same molar concentration for both species), the
isosbestic point corresponds to a wavelength at which these
spectra cross each other.
 Isosbestic point is very important in removing errors due to
compounds that can undergo chemical reactions as weak acids
or bases. ( This decreases deviation from beer’s law)
Ratio the two equations
Aunk/ Astd =  b Cunk /  b Cstd
 Solve for Cunk
Aunk/ Astd = Cunk /Cstd
 Note that, the standard solution of the species under test should
be prepared and measured under the same conditions of the
analyte (in the same cell and at the same ).
Calibration curve
b. By direct extrapolation
From a previously plotted calibration curve for the same
compound.
Series of standard solutions of the analyte is prepared
by successive dilutions.
Matrix effect, solvent effects and
interference
 In chemical analysis, matrix refers to the components
of a sample other than the analyte.
 Solvent, buffers, reagents, other ions & molecules
Blank solution.
• Solvent blank
• Reagent blank(a blank solution that contains the reagents
used in analysis)
• Method blank (a blank solution that has been handled
similar to a sample, and to which the same reagents have been added,
that had contact to the same type of vessels and that was treated by a
similar procedure.
Standard additions method
 Ideally the use of the Beer’s law plot of A vs conc. should be
sufficiently accurate for the evaluation of data.
 However, there are frequently what are known as matrix
effects where substances other than the analyte affect the
absorbance reading of the solution at the wavelength chosen
for the analysis.
 The method of standard additions is used to counteract the
matrix effect which we can not overcome or account for.
 Standard additions method usually involves adding one or
more increments of a standard solution to sample aliquots of
the same size (spiking).
interference
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Standard additions method (cont.)
c. Analysis of a mixture
Example
 A 2.00 mL urine specimen was treated with reagent to
generate a color with phosphate, following which the
sample was diluted to 100 mL. To a second 2.00 mL
sample was added exactly 5.00 mL of a phosphate solution
containing 0.0300 mg phosphate /mL, which was treated
in the same way as the original sample. The absorbance of
the first solution was 0.428 and that of the second 0.538.
Calculate the concentration in milligrams phosphate
per liter of the specimen.
Cx 
 Detector does not distinguish among absorbance readings
of individual analytes.
 Absorbance of analytes are additive
 For our example:
A λ mixture  ε λA b [A]  ε Bλ b [B]
 and more generically:
A  ε xb [X]  ε y b [Y]  ε z b [Z]  
0.428 x 0.0300 (mg PO 43- /ml) x 5.00ml
 0.292 mg PO 43- /ml sample
(0.538  0.428) x 2.00 ml sample
A1= c1b
A2= (c1-c2)b
Determination of  X , X ,  Y and Y
c. Analysis of a mixture (cont.)
 Prepare a standard solutions of pure X.
 Measure the absorbance of these solutions at ’ and ’’.
 Plot a calibration graph between absorbance readings and
A mixture contains two analytes, X and Y
 Measure absorbance of the mixture at max for X
 Measure absorbance of the mixture at max for Y
A

concentrations of the standards at both s. From the slopes of the
two graphs, the values of  X and  X can be determined. The values of
 Y and  can be determined in a similar way.
A  Xb[ X]  Yb[ Y]

Y
0.6
A  X b[ X]  Y b[ Y]
Y
By knowing the values of molar
absorptivities at the two wavelengths,
the above two equations can be solved
simultaneously for the two unknowns [X]
and [Y].

0.4
X
0.3
’’
0.2
0.1
 X
0
0
2
4
6
8
10
12
Concentration of standard soltion of X
69
Example
Example
 The concentration of Fe3+ and Cu2+ in a mixture can be determined
4-, which
following their reaction with hexacayano-ruthenate (II) Ru(CN)6
forma a purple blue complex with Fe3+ (max= 550 nm), and a pale green
complex with Cu2+ (max=396 nm). The molar absorbitivities (M-1cm-1) for
the metal complexes at the two wave lengths are summarized below.
 When a sample
’
0.5
Absorbance
X
()
()
Fe3+
550
9970
396
84
Cu2+
34
856
containing Fe3+ and
Cu2+ is
analyzed in a cell with pathlength of 1.00 cm, the absorbance at 550 nm is 0.183 and at 396 nm is 0.109.
 What are the molar concentrations of Fe3+ and Cu2+ in the









Substituting known values into equation given in slide 22 gives:
A550 = 0.183 = 9970CFe + 34CCu
A396= 0.109 = 84CFe + 856CCu
To determine the CFe and CCu, we solve the 1st equation for CCu
CCu = 0.183-9970CFe/34
And substitute the result into the 2nd equation,
0.109 = 84CFe + 856 (0.183-9970CFe/34)
=4.607-(2.51x105)CFe
Thus Concentration of Fe3+ = 1.80x10-5 M and by substituting
back in equation of mixture’s absorbance at 396 , concentration
of Cu2+ = 1.26x10-4M.
sample?
71
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2- Spectrophotometric titrations
2- Spectrophotometric titrations
radiation in the UV-Vis. regions, absorbance measurements at
fixed wavelength can be used to locate the endpoint of the
titration:
At + T
P
 The absorbance of the analyte solution is measured after each
addition of the titrant and the end point is located from the plot
of the absorbance readings and the volume of the titrant added.
 Let’s consider the analysis of hydrogen peroxide with potassium
permanganate in an acidic solution.
 The potassium permanganate or MnO4- is the only colored
substance in the reaction.
 How would the absorbance change as titrant was added?
5H2O2 + 2MnO4- + 6H+  5O2 (g) +
(Analyte)
Absorbance
 If the analyte (At), the titrant, or the reaction product absorbs
(Titrant)
purple
2Mn2+ + 8H2O
(colorless products)
Equivalence point
MnO4- reacting,
color disappears
MnO4accumulates
Volume of titrant (mL KMnO4)
To determine the concentation of the analyte
2 (MV)H2O2 = 5 (MV)MnO4-
Stoichiometry of complex ions
Examples
Determination of metal
ligand ratio
 Solutions of cation and
ligand with identical formal
concentrations are mixed in
varying volume ratios; but
V T = const.
 A plot of A vs volume ratio
(volume ratio = mole
fraction) gives maximum
absorbance when there is a
stoichiometric amount of
the two.
76
Example (cont.)
Determination metal ligand ratio (cont.)
Example
 To determine the formula
of a the complex between
Fe2+ and 0-phenathroline, a
series of solutions was
prepared in which the total
concentration of metal and
ligand was held constant at
3.15x10-4M. The absorbance
of each solution was
measured at a wavelength of
510nm. Using the data
given in table determine
the formula of the
complex
XL
Absorbance
0.0
0.000
0.1
0.116
0.2
0.231
0.3
0.347
0.4
0.462
0.5
0.578
0.6
0.693
0.7
0.809
0.8
0.693
0.9
0.347
1.0
0.000
Determination metal ligand ratio (cont.)
Thus, formula of the metal-ligand complex is
Fe(o-phenthroline)32+
77
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Absorption spectra of organic molecules
Determination of acid-base dissociation constant
 Let’s consider, as a simple example, an acid–base reaction of
* or *
molecular orbital
(anti-bonding)
the general form
HIn + H2O
H3O+ + In–
where HIn and In– are the conjugate weak acid and weak base
forms of a visual acid–base indicator. The equilibrium constant
for this reaction is
s or p
atomic
orbital,
Atom X
 The following formulae can be used for calculating Ka and pKa
of weak acids and bases:
s or p
atomic
orbital,
Atom Y
non-bonding
orbital
 or 
molecular orbital
(bonding)
79
Vacuum UV or Far UV
(λ<190 nm )
Absorption spectra of organic molecules (cont.)

(1)
(2)
(3)
(4)
Electrons in molecule can be classified into four different
types:
Closed shell electrons that are not involved in bonding.
These have very high excitation energies and do not contribute
to absorption in UV-visible region.
Covalent single bond electrons (, or sigma electrons).
These are firmly held and their excitation required energies
corresponding to vacuum UV region, thus they are not widely
used for analytical purpose because of experimental
difficulties to work in this region.
Paired non bonding outer-shell electrons( n electrons)
such as those on N, S, O and halogens. These are less tightly
held than  electrons and can be excited in UV-visible
region.
Electrons in (pi) orbital, as in double and triple bonds. These
are the most readily excited and are responsible for the
majority of electronic species in UV-Visible region.
UV/VIS
82
Electronic Spectra and Molecular Structure
The electronic transition that takes place in the UV-Vis regions of spectrum
is due to the absorption of radiation by:
Specific types of functional groups or bonds (called CHROMOPHORS)
that contains valence electrons with relatively low excitation energy.
Absorption characteristics of some
common chromophors
Chromophore
Example
C=C
Alkenes
Excitation
λmax
(nm)
ε
π
π*
190
15,000
C≡C
Alkynes
π
π*
190
10,000
C=O
Ketones
n
π
π*
π*
290
195
15
10,000
N=O
Nitro-compounds
n
π
π*
π*
275
200
17
5,000
C-X X=Br
X=I
Alkyl halides
n
n
σ*
σ*
205
255
200
360
The high value
of ε for 
*
indicates that
these transition
are the most
easily occurring
and thus the
most probable
transition
occurs between
these levels.
Note that all the above transitions are observed in
the ultraviolet region. That is the reason why most
simple organic compounds are colorless
83
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Effect of Conjugation (cont.)
Effect of Conjugation
-electrons are considered to be
further delocalized by conjugation as
this delocalization leads to decrease in
the energy gap between  and *.
*
*
*



CH2=CH2
Absorption maxima are thus shifted to
CH2=CH-CH=CH2
conjugated system
longer wavelengths (Bathochromic
shift Or red shift). The higher the
extent of conjugation, the longer the
wavelength of transition.
BLUE shift
RED shift
Also, the molar absorptivity (ε)
roughly doubles with each new
conjugated double bond
(Hyperchromic effect).
85
86
Effect of Conjugation (cont.)
Some highly conjugated organic compounds such
as carotene and lycopene absorb radiations at longer
wavelength enough to be observed in the visible
region i.e., they are colored.
Effect of solvents
 The solvents used to obtain UV-VIS spectra can also influence
the band shape and wavelength of the measured absorption
bands.
 Non-polar, non-hydrogen bonding solvents approximate the
-CAROTENE
conditions in the vapor state and often give spectra with more fine
structure.
 Polar and/or hydrogen bonding solvents tend to give more
band broadening.
 Because the ground and excited states often have different charge
LYCOPENE
87
distributions the interactions with polar solvents can shift the
transitions of n→π* to shorter wavelengths (hypsochromic shift,
blue shift); and the π →π* transitions to longer wavelengths
(bathochromic shift, red shift)
88
The effects of substitution
Effect of solvents (cont.)
89
Auxochrome
function group
Auxochrome is a functional group that does not absorb in UV
region but has the effect of shifting chromophore peaks to longer
wavelength (bathochromic effect) as well as increasing their
intensity (hyperchromic effect).
90
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Analysis of non absorbing species
•
Many non-absorbing analytes can be determined
spectrophotometrically by causing them to react with
chromophoric reagents to give products that absorb strongly in
the UV-Vis regions (products of high ).
•
Most of the spectrophotometric clinical tests are based on
such kind of interactions .
to the colorless solution containing Fe2+ add buffer and
excess ferrozine (chromophoric reagent) where a purple
complex is formed. Measure the absorbance of at 562 nm
(max for the complex)
Fe2+ + 3ferrozine
colorless
[(ferrozine)3Fe]4purple complex
emission can be induced or spontaneous
Emission intensity
I = kP0C
Where p0 is the incident radiant power
C is concentration of emitting species
k is a constant
An isolated atom scatters light because the electric field of the incident light wave forces
the electrons in the atom to oscillate back and forth about their equilibrium position.
By the laws of electromagnetism, when a charge changes its velocity, it emits radiation.
Inelastic scattering is important for gases. In liquids, random
molecular motions broaden Rayleigh scattering.
Light is emitted uniformly in all directions in the plane ⊥ to oscillation, but decreases in
amplitude as the viewing angle shifts away from that plane.
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Determination of acid-base dissociation constant (cont.)
Determination of acid-base dissociation constant (cont.)
Example 1
 Example 2
The acidity constant for an acid–base indicator was determined by preparing
three solutions, each of which has a total indicator concentration of 5.00x10–5
M. The first solution was made strongly acidic with HCl and has an
absorbance of 0.250. The second solution was made strongly basic and has
an absorbance of1.40. The pH of the third solution was measured at 2.91,
with an absorbance of0.662. What is the value of Ka for the indicator?
 The Absorbance of a fixed concentration of a pharmaceu-tical
compound (phenylephrine) at 292 nm is found to be 1.224 in
0.1M NaOH and 0.02 in 0.1M HCl. Its absorbance in a buffer
of pH 8.5 is found to be 0.349. Calculate the pKa value of its
acidic phenolic hydroxyl group.
 The value of Ka is determined as
 [H3O]+ = -antilog 2.91 = 1.23x10-3M
pKa  8.5  log
1.224  0.349
 8.5  0.402  8.902
0.349  0.02
99
100
Derivative spectrophotometry
Derivative spectrophotometry (cont.)
 Two closely spaced wavelengths are alternately passed through the sample. To
 Derivative spectrphotometry can be
used to clarify bands in more complex
UV spectra.
 The main effect of derivatization is
background correction and better
selectivity than normal spectrophotometry for resolving binary
mixtures and some ternary
mixtures.
 Derivative spectrophotometry have
vast applications in analysis of
pharmaceutical compounds that
contain more than one active
ingredient at the same time.
101
record spectra, one wavelength is set so as to lag behind the other by a fixed
amount during the wavelength scanning process.
 Three points on the derivative curve are of special significance.
 The point where the curve intercepts the zero line corresponds to the
wavelength of the conventional peak's maximum . The negative portion of
the derivative curve has a minimum value, and the positive has a maximum
value.
 The horizontal distance between the two is a measure of peak bandwidth
This derivatization is done
automatically through
spectrophotometer.
and the vertical distance from the maximum to the minimum relates to the
amplitude of the conventional spectral peak and is dependent on
chromophore concentration.
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Derivative spectrophotometry (cont.)
 Example, UV spectrum of
pseudoephedrine,
dextromethorphan and
triprodline, pesudoephedrine lies on
top of a large background because of
the other two compounds, which have
much stronger chromophoric groups
than pesudoephedrine.
 By special calculations, done
through softwares, the amount of
pseudoephedrine can be tested by
plotting a calibration graph using
information obtained from 1st and 2nd
derivative absorption spectra.
Application of UV spectrophotometry in dissolution testing
 In recent years, dissolution testing has become increasingly important in
the pharmaceutical industry. Dissolution testing provides information
on batch homogeneity and conformity, and is routinely used for quality
control purposes. In addition, conclusions about the in vivo behavior of
the product can be drawn from its in-vitro dissolution behavior.
Therefore, the dissolution test has become an indispensable tool in the
development of new solid oral dosage forms and the assessment of the
physical stability of the formulation.
 An appropriate in-vitro-dissolution test method should meet the
following criteria:
- Sufficient discriminatory power Identification of critical manufacturing
variables
- Low variability of the method
- Proof of batch homogeneity
103
Application of UV spectrophotometry in dissolution testing (cont.)
Application of UV spectrophotometry in dissolution testing (cont.)
 Traditionally, dissolution testing has been carried out on solid dosage forms
using a standard dissolution test apparatus.
 Samples of the dissolving active within the dosage form are taken for estimation
from the dissolution tester either manually, or using more automated
means such as a pump which has been connected to a suitable measurement
device, usually a UV-VIS Spectrophotometer.
 The amount dissolved is usually calculated by comparison to a standard
calibration curve.
105
106
In-class exam 1
Take home exam 1
 Calculate the absorbance of an organic dye (C= 7 x 10 -4 mol L-1),
knowing that the molar absorbtivity is 650 mol-1 L cm-1 and that
the length of the optical path of the cell used is 2 x 10 -2 m.
 Calculate the absorbance of an organic dye (C= 7 x 10 -4 mol L-1),
knowing that the molar absorbtivity is 650 mol-1 L cm-1 and that
the length of the optical path of the cell used is 2 x 10 -2 m.
 What would happen to the (i) absorbance & (ii) transmittance, if
the cell used was double its present thickness?
 What would happen to the (i) absorbance & (ii) transmittance, if
the cell used was double its present thickness?
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In-class exam 2
 Calculate the energy of a mole of protons corresponding
to a wavelength of 300 nm.
the speed of light in vaccum ( = 2.998 x 108 m/s)
the Planck’s constant (=6.626x10-34 J s)
19