1. Instrumental
Analysis
PHC224
د. هالة زعــزع

2. Instrumental Analysis
3 hours lecture
spectrophotomtery
potentiometry
fluorimetry
conductometry
2 hour practical
Flam spectroscopy
polarography

3. Instrumental Analysis
PHC 224
Total Marks 100
Semester work 10
Fifth week 5
Mid term 15
Practical
25+5
Oral
Final Exam 30
Love Chemistry

4. Molecular Absorption Spectrophotometry
Contents
Light and radiation
Electromagnetic spectrum
Light as energy
Types of electromagnetic transitions
Interaction of Photons with matter
Absorption spectrum
Chromophores, and auxochromes
Factors affecting absorption spectrum (pH, solvent)
Beer’s law and quantitative analysis
Deviation from Beer’s law
Instrumentation (Basic component of spectrophotometer)
Analysis of some pharmaceutical drugs

5. What’s spectrophotometry
Spectroscopy is the study of interaction of spectrum of light with a substance to be analyzed, for its:
- Identification (qualitative analysis)
- Determination of its amount (quantitative analysis).
Nature of light
It is one of the different forms of radiant energy called electromagnetic radiations (EMR) due presence of both electric and magnetic components.

6. 1. Wave property Dual nature of light
Light exhibits wave property during its propagation and energy particle
during its interaction with matter.
The double nature of light (Waves and particles) is known as dualism.
1. Wave property
Light energy is transmitted in the form of waves
Wave length (λ)
It is the distance between two identical points in successive waves.
Its unit is nm = 10-9m=10 Ao 

7.  = C/ λ 2- Frequency () It is the number of waves per second
Its unit is sec-1 or (CPS) or
Hertz (Hz)
3-Velocity of light (C)
Light propagates at the highest known velocity= 300,000Km/sec.
 = C/ λ

8. Planck’s quantum theory
particles property
(light as energy)
Light consist of energy packets, known as photons.
Planck’s quantum theory
The atom or molecule can emit or absorb a discrete quantity of energy
i.e fixed amount of energy called “Quanta”.
The energy of a single quantum is proportional to the frequency i.e.
E α 
E = h 
= C/ λ
E = h C/ λ
h : is Plank’s constant = 6.63 x J.s
C: 3 x 108 m/s

9. Problem Solution - What is the energy emitted at 450 nm by CuCl2 ?
E = h  = h C/ 
=(6.63 x J.s) (3 x 108 m/s) / 450 x 10-9 m = 4.4 x J
It is so simple
If the energy difference between the ground state of an atom and its excited state is 4.4 x 10 -7J, what is the wavelength of the photon required to produce this transition? (h= 6.63 x J.s, C= 3 x 108 m/s)

10. Microwave
IR-radiation
Gamma-radiation
Radio wave
X-ray

11. E = h  = h C/ 
energy of a beam of EMR increase as wave length decreases.
The shorter the wave length, the greater the energy of the photons
According to  , EMR are classified into:

12. red, orange, yellow, green, blue, indigo and violet
For analytical purposes we use the region of I.R, visible and U.V radiations
x-rays
U.V
VIS
I.R
Microwaves
Radio waves
U.V range ( nm) which contain shorter  carrying more energy photon more than a beam of visible range( nm), and visible range more than I.R range (>800nm).
Visible radiation part (day light) consist of some colored radiations, which are,
red, orange, yellow, green, blue, indigo and violet UV
violet
indigo
blue
green
yellow
Orange
Red IR
If we observe the visible radiations of all wavelengths, we see white light
A beam containing several wavelengths is called polychromatic light.
A beam of light carrying radiation of only one discrete is said to be monochromatic

13. Interaction of a substance with EMR
Absorption of radiant energy by a substance molecule may occurs in three ways:
1- Raising electrons to a higher energy level (transitional energy), when molecule absorb visible and U.V region.
2- Raising the Vibration of the nuclei (Vibrational energy), when the molecule absorb I.R region.
3- Increasing rotation of the molecule around the axis (rotational energy), when the molecule absorb in F.I.R region.
The relative energies of transitional, vibrational and rotational are roughly in the order of 10000:100:1

14. Emission of radiant energy
The energy of transition is given by the following:
Excited state Es
 E = Es - Eg = h = h C/ 
Ground state Eg
Emission of radiant energy
Absorbed energy doesn’t accumulate in electronic system of a molecule.
If an excited electron returns to the ground state, it may lose absorbed energy in the form of heat, light or molecular collision.
When excited electron returns to the ground state via second excited state, light is emitted as fluorescence or phosphorescence.
Excited state Es
Ground state Eg

15. Types of electronic transition
The outer electrons in an organic molecule may occupy one of three different energy levels:
1- Sigma () electrons :They are bonding electrons posses the lowest energy level ( it is the most stable).
2- Pi () electrons: They are bonding electrons of higher energy than sigma electrons.
3- Non-bonding (n) electrons: They are of atomic orbital of hetero atoms (N,O, halogen or S) which don’t participate in bonding, they usually occupy the highest level of ground state.
In excited state  electrons occupy an antibonding energy level denoted as * and the transition is termed -* transition. -electrons occupy the antibonding * level, while n electrons occupy either * or *.

16. Electronic Transitions and UV-visible Spectra in Molecules
excited state
ground state
CH2 CH OH

17. Spectra-structure correlation
The absorbance of EMR in the UV-VIS regions depends on the structure of organic molecule
Absorbance of EMR by organic molecule is achieved by chromophoric groups(chromophores) assisted by auxochromes, where the electrons of absorbing molecule is excited, i.e it undergo transition from the ground state to the excited state.
Auxochrome
Chromophore
it is unsaturated group which is responsible for electronic absorption e.g
It is a saturated group which, when attached to a chromophore, alters both the  and intensity of absorption maximum e.g
-OH, - NH2, - Cl
All auxochromes have one or more non-bonding pair of electrons.

18. Absorption spectrum Absorption spectrum is obtained by plotting
Absorbance (A) as a function of wavelength ().
It is characteristic for each molecule, therefore it is used for identification of a chemical substance (qualitative analysis).
It has characteristic shape which show the  of maximum absorbance (max).
Absorption spectrum may show a shoulder or even no absorption characteristics.

19. different Organic compound
UV spectra of
different Organic compound

20. Why?  max is used for quantitative measurement in order to:
1-increase sensitivity .
2-minimize error of the analytical method

21. Saturated compounds containing heteroatoms (
Solvent Cut off
Compounds containing only -electrons are the saturated hydrocarbons which absorbs <170nm (i.e.) in the far UV. They are transparent in the near UV ( nm) making them ideal solvents for other compounds to be studied in this region.
However, their intense absorption usually extends to the edge of the near UV producing end absorption (Cut off wavelength) in the nm region.
What’s meant??
Cut off l l
220
200
lmax of solvent
Far UV
near UV UV
Visible
Absorbance
Wavelength, , generally in nanometers (nm) 0.
400
800
200

22. (showing high energy cutoffs)
Solvents for UV
(showing high energy cutoffs)
Solvent
Cut-off 
Water
205nm
CCl4
265 nm
Ethanol
210 nm
Chloroform
245 nm
Methanol
THF
220 nm
Ether
CH2Cl2
235 nm
Acetone
300 nm
Dioxane
Benzene
280 nm

23. Shifting of max 1- Bathochromic shift (or red shift)
Hyperchromic
Wavelength, nm
Absorbance
It is the shift of max to a longer wavelength due to substitution with certain functional groups
(e.g. –OH and –NH2) and conjugation
Hypsochromic
Bathochromic
Hyporchromic
2- Hypsochromic shift (or blue shift)
It is the shift of max to a shorter wavelength due to removal of conjugation
3- Hyperchromic effect (or shift)
4- Hypochromic effect (or shift)
This effect involves an increase in the intensity of absorption. It is usually brought about by introduction of an auxochrome.
It involves a decrease in the intensity of absorption

24. Factors affecting absorption spectrum
1- Effect of pH on absorption spectrum
The spectra of compounds containing acidic (phenolic-OH) or basic (-NH2) groups are dependent on the pH of the medium.
1- Phenol
The U.V spectrum of phenol in acid medium where benzenoid form is the predominant species, which is completely different from its spectrum in alkaline medium; where the predominant species is the corresponding phenate anion. The spectrum in alkaline medium exhibits bathochromic shift (red shift) due to delocalization of electrons that required lower energy for excitation so appear at longer wavelength
benzenoid
phenate anion

25. 2- Aniline
Aniline behaves like phenol, i.e. its spectrum exhibits bathochromic shift and hyperchromic effect in alkaline medium due to its conversion to the quinonoid species, while in acid medium its spectrum exhibit hypsochromic shift and hypochromic effect due to its conversion to the benzenoid species.
Give reason??

26. On running U.V spectrum of known concentration of phenol as a function of pH (at different pH). The spectrum will be shifted to different max by changing the pH, but all spectra intersect at certain  which is known as isosbestic point.
Isosbestic point
At isosbestic point, the same absorbance is given for the same concentration at different pH (i.e absorbance is not pH dependent but concentration dependent).
Analytical measurements of pH- sensitive compounds are done at isosbestic point

27. Isosbestic points
Single clear point, can exclude intermediate state, exclude light scattering and Beer’s law applies

28. 2- Effect of conjugation
Generally, extending conjugation leads to red shift

29. N.B. 3- Effect of redox reaction on absorption spectrum.
Oxidation of diphenylamine will convert the benzenoid spectrum of diphenylamine to the quinonoid spectrum of the oxidised form, as represented by the following equation.
N.B.
Increase in conjugation, increases absorbance of light by a compound which appears colored and exhibits hyperchromic effect.

30. Laws of light absorbance
Scattering losses
in solution
Reflection losses
at interfaces
When a monochromatic light having intensity (Io) is allowed to pass through absorbing medium, some is absorbed (Ia), reflected (Ir), transmitted (It), refracted (If) and scattered (Is). I0
(radiant intensity) It
(transmitted intensity)
Refraction losses
Reflection, Refraction
and scattering losses
Io = Ia + It + Ir + If + Is

31. Therefore To minimize refraction, reflection and scattering loss
Is = zero for clear solution, while If and Ir may be canceled by means of control cuvette (Blank) containing the solvent in which the substance to be analyzed is dissolved.
Therefore
Io = Ia It or
Ia (A) = Io - It

32. where K is proportionality constant.
Log
1- lambert’s law
2- Beer’s Law:
When a monochromatic light enters an absorbing medium, its intensity is decreased exponentially with the increase of the concentration of the absorbing medium, when (b) is constant.
When a monochromatic light enter absorbing medium, its intensity is decreased exponentially with the increase of thickness of the absorbing medium (i.e solution) (b) at constant concentration (C)
log Io / It  b or
log Io / It = K b
where K is proportionality constant.
log Io / It  C or
log Io / It = K C b C C b
Log Io/It

33. Beer-lambert’s law A = abc
log Io/It  b (thickness or path length) log Io/It  c (Concentration)
log Io/It  bc or log Io/It = abc or
A = abc
where: A = log Io/It = absorbance
a: is a constant, known as absorptivity which is:
The absorbance, when thickness of solution is unity ( 1cm) and concentration is unity.
- Molar absorptivity or epsilon ()
A (1% - 1cm):
-If the unit of concentration is 1M, (a) is known as molar absortivity or epsilon () or molar extinction coefficient. (Unit of  is L mol-1 cm-1)
If unity of concentration is 1%, (a) is known as A (1%, 1cm).

34. A/C =a= slope of the curve
Both  and A (1%, 1cm) are characteristic for each substance and are used for qualitative purpose.
Absorptivity (a), can be calculated from the slope of the curve produced on plotting (A) as a function of (C) at fixed (b). When (b) is 1 cm b C A x
A/C =a=
slope of the curve
A = aC or a = A/C Or
From the curve
A/C =a= slope of the curve

35. R –C– OEt + H2N – OH R –C– NH – OH + EtOH
Colorimetry
In colorimetry we are measuring A of visible radiation by colored sample.
Requirements for substances to be measured colorimetrically:
Substance must be colored e.g CuSO4, organic dyes,…etc.
2- If the substance to be analysed is colorless, it must react with certain reagent (known as chromogen) to produce equivalent colored product. e.g orthophenanthrolene which reacts with ferrous (Fe2+) in buffered medium(acidic pH) to produce intense red color.
3- If the substance to be analyzed is colorless and there is no suitable chromogen, it must be converted to a certain derivative which has a suitable chromogen. e.g esters, which is first converted to hydroxamic acid derivative through the reaction with hydroxylamine. Hydroxamic acid derivative gives purple color on addition of ferric (Fe3+) due to the formation of iron chelate.
R –C– OEt + H2N – OH R –C– NH – OH + EtOH
Hydroxamic acid derivative

36. *Requirements for ideal chromogen
1- Should be colorless or easily separated.
2- Should be selective.
3- The full development of color must be rapid.
4- Produce only one color of specified max.
5- Its reaction to produce colored product, should be of known mechanism and proceed stoichiometrically.
*Requirements for colored product
1- Should be of intense color, to increase the sensitivity.
2- Should be unaffected by pH or the pH must be specified and maintained by suitable buffer or the measurement is carried out at  of isosbestic.
3- Should be stable with time.
4- The reaction of its formation, must be rapid and quantitative.
5- The colored product, should obey Beer-lambert’s law, i.e on plotting A versus C at fixed b, we obtain straight line passing through the origin.

37. Instrumentation

38. Instrumentation A- Visual methods:
which are applied only in case of colored samples.
B- Photo-electric methods:
which are applied in both colored and Colourless samples.
A. Visual methods
using polychromatic light
1-Standard series method. (using polychromatic light)
This method, based on the comparison of the colored sample with standard series of colors.
Standard colors may be freshly prepared or a permanent colored system.
For freshly prepared standards, matched Nessler tubes are used.
For permanent colored system e.g colored disc, lovibond comparator or sealed ampoules are used.

39. b) variable depth or balancing method
According to Beer-lambert’s Law:
Log I0/It = A = a b1 C1 = a b2C2 or
b1 C1 = b2 C2 or C1/C2 = b2/b1
Comparing a sample with a solution of known concentration of the same sample (standard solution) till the intensity of It emerging from sample and standard solution are the same, we can calculate the concentration of the unknown sample.

40. B- Photo-electric methods
Using monochromatic light
These methods depend on photoelectric phenomenon, where, the intensity of EMR is measured through the intensity of electric current produced by electrons liberated from a photosensitive metal under the influence of incident EMR.
The instrument used, usually consists of 5 basic components
1- Radiant energy source.
2- Dispersing system (or monochromator).
3- Sample compartment (cuvette).
4- Detector.
5- Recorder (meter).
Light source
Monochromator
Sample
Compartment
Detector
Recorder

41. 1. Radiant energy source (source of light)
It must be of sufficient intensity and must cover the desired spectral range. In visible range, we must use tungesten lamp, in U.V range we must use deutrium lamp (D2) (or hydrogen lamp.)
2. Dispersing system (monochromator)
It convert polychromatic light to monochromatic light, i.e of definite range or .
Monochromatic light may be obtained by the use of:
act by selective absorption of unwanted  and transmit the complementary color
a) Filters
b) Prisms
Act by refraction of light
C) Grating
Act by diffraction and interference

42. a) Filters
Filters act by selective absorption of unwanted  and transmit the complementary color, which is needed to be absorbed by the sample to be analyzed.
N.B
If a substance absorbs all visible light it will appear black.
If a substance doesn’t absorb visible light, it will appear colorless.
* Filters may be
1-gelatin and tinted glass:
These types of filter transmit a wide band of nm. which is not exactly monochromatic.
2-interfering filters
It consist of multiple layers. Interference result in a narrow band of nm.

43. Act by refraction of light. In visible range we use glass prism.
b) Prisms
Act by refraction of light.
In visible range we use glass prism.
In U.V range we use prism made of quartz or fused silica.
ROYGBIV
C) Grating
Act by diffraction and interference
Grating consists of a large number of parallel lines ruled very close to each other on a highly polished surface e.g aluminum or aluminized glass (600 line/mm).
Each ruled groove functions as a scattering center for light rays falling on its edge. Through diffraction and interference, the grating make the light beam into almost single .

45. 3.Sample compartment (Cuvette)
Transparent surface
Obaque surface
It is made of glass for visible range and quartz or fused silica for U.V range.
Its standard path length is 1cm (10mm) some times it is ½ cm.
4. Detector
:There are two types of detectors
a) photocell (Photovoltaic cell) e.g Barrier layer cell
Light falling on cell
Transparent metal layer of Ago
(Collecting electrode)
Photosenitive semiconductor of selenium
Metal base Plate of iron - +
EMR falling upon a semiconductor surface, where electrons are excited and produce EMF (current) proportional to the intensity of incident light.
This type requires no external source of power.

46. :
b) Phototube (photomultiplier or photoemissive tube)
In order to obtain greater sensitivity (through magnification of EMF produced) multiplication of the initial photoelectrons by secondary emission, using several anodes arranged in gradually increasing potential (i.e this type requires external source of power).
5. Recorder (meter)
Electric signal produced in detector is fed to a sensitive galvanometer, its scale is graduated in absorbance or/and transmittance units.

47. Commercial instruments
1. Filter photo-electric colorimeter

48. 2. Compensating two-photocell colorimeter

49. Deviation from Beer-Lambert's law
1) Real deviations A C
Solute molecule in concentrated solution doesn’t absorb radiant energy in the same manner as does the same molecule in dilute solution, due to:
molecular interaction, leading to non-linear response when A is plotted against C.
This deviation decreases or disappear in very dilute solution.
2) Instrumental deviations
It may be irregular or regular :
a. Irregular deviations
may result from:
1- The use of unmatched cuvette (due to industrial defect)
2- Unclean handling (e.g finger print on the cuvette)
3- Unclean optics (lenses, mirrors or lamp)

50. b. Regular deviations
Regular deviations may result from:
1- Error in  scale.
2- Error in potentiometric reading of absorbance
3- Stray light, which is any radiations or  other than that absorbed by the sample or any light reaches the detector without passing through the sample. It may result from incorrect choice of filters, the presence of fluorescent impurities and light source.
3) Chemical deviations
These include; effect of pH (which leads to shifting of max.), temperature effect and time factor of colored solution (color may fade by time due to deterioration of organic dye by oxidation, reduction, hydrolysis, etc….).

51. Application

52. 1- Qualitative analysis 2- Quantitative analysis
3- Determination of some physical constants.
Applications
1- Qualitative analysis
Absorptivity ( or A (1% - 1cm), max, U.V and visible absorption spectrum usually give finger print of the sample to be analyzed.
2- Quantitative analysis
a. Quantitative analysis of a single component
The substance to be analyzed is dissolved in a suitable solvent, e.g methanol, ether, water, etc….
The solvent is used as blank.
Absorbance readings are taken in the expected range ( nm for colorless sample or nm for colored sample).
To get absorption spectrum of the substance to be analyzed, absorbance readings are taken at suitable intervals, usually 5 nm or less.

53. For quantitative determination of a single component we have to:
1- Detect max of the substance to be analyzed after dissolving in a suitable solvent.
2- Construct a calibration curve by plotting A against C at fixed b using standard series of the same chemical present in the sample, at the characteristic max..
3- The Absorbance of the sample to be analyzed is determined under the conditions adopted during construction of the calibration curve.
4- The concentration of the sample can be determined from the calibration curve. A C or
A = abc
i.e c = A / (a b)

54. b. Quantitative analysis of multicomponent mixture
Consider a binary mixture of X and Y. For quantitative analysis of this mixture, the following requirements must be fulfilled:
1-The absorption spectrum of X and Y should not show sever overlap.
2-Beer-Lambert’s law must be obeyed for X and Y at their characteristic max.
3-X and Y must be chemically inert to each other.

55. 3- Determination of some
physical constants
(figure A)
Determination of pKa of weak acid or pkind
Consider the weak acid HB during its dissociation
HB H B-
OH- H+
Ka = [H+][B-] / [HB]
- log Ka = - log [H+] – log [B-] / [HB]
pKa = pH – log [B-] / [HB]
(figure B)
To determine pKa, the absorption spectra of a fixed concentration of HB as function of pH is constructed (figure A), then absorbance is measured as a function of pH at 1,(max of HB) and 2 (max of B-) (figure B). The pH at which the two curves are intersected represents pKa of weak acid.

56. 4.Determination of complexation ratio (Job’s method).
During study of complexation reaction and its stoichiometry, we have to prepare aserial dilutions of Metal ion (Mn+) ranging from 0 to 10 (i.e. 0, 1, 2, ….., 10) and ranging from 10 to 0 (i.e. 10, 9, 8,…….0) for complexing agent (L).
Depending on additive property of absorbance, we prepare complementary mixtures of Mn+ and by measuring A of these mixtures.
if there is no change in A, this indicates that, there is no complexation reaction.
If A of mixtures are varied, this indicates that complexation reaction takes place.
The ratio of Mn+: L can be detected by their ratio at the point of maximum or minimum absorbance.

57. 5-Detection of impurities
Absorbing impurities, cause distortion of the absorption curve of the mother substance,leading to over estimation (i.e higher absorbance), also it may lead to shifting of max.
a) Impurity index (I.I.)
The absorption spectra of tested sample and reference standard of the same substance contained in the tested sample are constructed.
I.I. = A`1 /A`2 – A1/A2
it must be equal to zero if the sample is 100% pure as the reference.
b) Spectrophotometric purity index (S.P.I)
Ā and A refer to the absorbance of the sample and reference standard respectively.
S.P.I. = (A`1 / A`2) / (A1 / A2)
this value equals 1 in case of pure sample.