1. ANALYTICAL CHEMISTRY ERT207By
DR. ZARINA ZAKARIA

2. What is Analytical Chemistry?
Task 1(Group 1): Prepare presentation on the topic based on page 2 to 9. Present on Monday 4/8/08

3. Classifying Separation Techniques
Separation based on….
size
mass or density
complexation reactions (masking)
Change of state
Partitioning between phases
Task 2 (Group 2): prepare presentation on separation classification. Present on Monday 4/8/08

5. terminology Sample : Analyte : Interferent : Chemical properties:
Physical properties:
(find out within 5 minutes)

6. Separation based on partitioning between phases
Terminology:
- partition coefficient:
- solute, S:
Selective partitioning of the analyte or interferent between two immiscible phases
Sphase 1 ↔ Sphase 2
The equilibrium constant would be:
KD = [Sphase 2]
[Sphase 1]

7. Separation based on partitioning between phases
2 techniques can be used to separate analyte and interferent:
Extraction
Chromatography
Will be discussed in chapter 6
Extraction is which a solute is transferred from one phase to a new phase

8. Type of extraction Liquid-liquid extraction Solid-phase extraction
Continuous extraction

9. Liquid-liquid extraction
Accomplished with a separatory funnel
Shaken to increase surface area between phases
When stop, denser phase settling to the bottom
Solid Phase Extraction
Solid-phase refer to solid adsorbent in the cartridge
many choice of adsorbent determined by the properties of the species being retain and
matrix in which it is found.

10. Replace liquid-liquid extraction due to ease of use, faster extraction time decreased volume of solvent and able to concentrate the analytes.
Solid-phase microextraction developed for less sample.
Gas-solid extraction.

11. A separatory funnel
Solid-phase extraction cartridge

12. Continuous Extraction
For component of interest that has unfavorable partition coofficient.
Extraction is accomplished by continuously passing the extracting phase through the sample until a quantitative extraction is achieved
Involving solid samples are carried out with a Soxhlet extractor

14. Modification for better extraction:
1. microwave-assisted extractions.
2. Purge and trap.
3. Supercritical fluids
Please read yourself, I may ask in test.
Group 1 and 2 presentation

15. Introduction:
Distribution of a solute between two immiscible, liquid phases.
Consider as rapid & clean
This technique includes:
Separations of both organic and inorganic substances.
Extraction of metal ions into organic solvents.
Multiple extractions for difficult separations and
Sequential multistep separations

16. The Partition Coefficient
If a solute is in an aqueous phase and is extracted into an organic phase.
A solute A will distribute itself between two phases.
The ratio of the cons. of the solute in the two phases will be a constant:
KD = [S]org
[S]aq
KD = distribution/partition coefficient
Subscript = solvent

17. To evaluate efficiency, consider solute’s total conc. In each phase
To evaluate efficiency, consider solute’s total conc. In each phase. Distribution ratio, D, is ratio of total solute’s in each phase.
D = [Sorg]tot
[Saq]tot
If solute exists in only one form in each phase;
KD = D

18. The Working Principle
The mixture is shaken for about a minute, and the phases are allowed to separate.
A solute is extracted from an aqueous solution into an immiscible organic solvent.
A solute is an organic compound, will distribute from water into organic solvents.
The principle is “like dissolves like”.
the bottom layer, the solvent is denser and will be drawn off after the separation is completed.

19. The Distribution Ratio
The ratio of the concentrations of all the species of the solute in each phase.
D = [HBz]e = in organic layer
[HBz]a + [Bz-]a in aqueous layer
Equation that relates D, KD and Ka is
D = KD
1 + Ka/[H+]a
the extraction efficiency will be independent of the original concentrations of the solute.

20. The Percent Extracted
Fraction of the solute extracted will depend on the volume ratio of the two solvents (what solvents?)
Fraction of solute extracted is equal to milimoles of solute in the organic layer divided by the total number of milimoles of solute.
The milimoles are the molarity times the mililiters.
Therefore, the percent extracted is given by,
%E = [S]oVo x %
[S]oVo + [S]aVa

21. Vo and Va are the volumes of the organic and aqueous phases, respectively.
Relation between %E and D is given by,
%E = D
D + (Va/Vo)
If Vo = Va, then
%E = 100D
D + 1
If D is less than 0.001, the solute can be considered quantitatively retained.
The percent extracted changes only from 99.5% to 99.9% when D is increased from 200 to 1000.

22. Example of calculation for extraction efficiency in one solute form.
In this case D and are KD equal.
D = [Sorg]tot = KD = [Sorg]
[Saq]tot [Saq]
(Moles aq)0 = (moles aq)1 + (moles org)1
- The derivation of equation to relate D and concentration of solute.

23. Tutorial 4 Explain example 7.14. page 217
2. Do question 24 in Chapter 7, page 229
(group 3 will explain the answer during tutorial class)
- Everybody has to submit answer for Q24 in a piece of paper. Marks will be given.