ANALYTICAL CHEMISTRY ERT207 By DR. ZARINA ZAKARIA
ANALYTICAL CHEMISTRYERT207
By
DR. ZARINA ZAKARIA
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
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
terminology
Sample :
Analyte :
Interferent :
Chemical properties:
Physical properties:
(find out within 5 minutes)
Separation based on partitioning between phases
Terminology: - partition coefficient:- solute, S:• Selective partitioning of the analyte or
interferent between two immiscible phasesSphase 1 ↔ Sphase 2
The equilibrium constant would be:KD = [Sphase 2]
[Sphase 1]
Separation based on partitioning between phases
2 techniques can be used to separate analyte and interferent:
1. Extraction
2. Chromatography
Will be discussed in chapter 6
Extraction is which a solute is transferred from one phase to a new phase
Type of extraction
1. Liquid-liquid extraction
2. Solid-phase extraction
3. Continuous extraction
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.
• 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.
A separatory funnelSolid-phase extraction cartridge
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
• 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
Introduction:
• Distribution of a solute between two immiscible, liquid phases.
• Consider as rapid & clean • This technique includes:
1. Separations of both organic and inorganic substances.
2. Extraction of metal ions into organic solvents.
3. Multiple extractions for difficult separations and
4. Sequential multistep separations
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 coefficientSubscript = solvent
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
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.
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.
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 100% [S]oVo + [S]aVa
• Vo and Va are the volumes of the organic and aqueous phases, respectively.
• Relation between %E and D is given by, %E = 100D
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.
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.
Tutorial 4
1. 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.