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.