This paper gives the copper solvent extraction simulation method based on the realistic theory. This method is simply to use for any complex configuration of copper solvent extraction (serie…Descripción completa
This paper gives the copper solvent extraction simulation method based on the realistic theory. This method is simply to use for any complex configuration of copper solvent extraction (serie…Full description
Solvent Solvent extraction extraction is defined defined as the extraction extraction or isolation isolation of a desired desired component or substance by extracting it with the help of a solvent having maximum affinity (solubility) (solubility) from the other immiscible liquid. For example, if we have a mixture of resorcinol and benzoic acid, both are soluble in ether one possibility is to add NaHCO3 in it, it will give sodium salts of both the compounds
These both give sodium salts and these salts are soluble in HCl, so it is a problem and hence need to separate them by solvent extraction with either and water. PARTITION COEFFICENT OR DISTRIBUTION LAW:
When a substance is dissolved in two immiscible solvents there is a defini definite te ratio ratio of the concentra concentratio tion n of that that substa substance nce in one solven solventt to the concentration of the substance in the second solvent, at a definite temperature or mathematically Kd i.e. Distribution coefficient or partition coefficient is equal to C1/C2. Kd or KD = C1/C2 Kd = Distribution constant D =
Distribution Ratio
This law is called distribution law or partition law and the constant is called distribution or partition partition constant or coefficient. coefficient. The major factors which which affects the distribution law 1
Temperature effect
2
Solvent effect
2 Temperature effect:
Temperature effect is the way that by the increase of temperature it increases the solubility of both the solvents and their ratio change. So the value of Kd change. Solvation effect:
Solvation effect change the value of Kd. Polymerization (Dimer):
When we add solute in solvent it changes by various reactions, for example by dimerization as benzoic acid in benzene, benzoic acid form dimer in benzene
Kd is generally taken for aqueous solution. One solvent is organic Kd =C1/C2 is the ratio for the very simple reaction if there is no side effect. But there are some side effects for example if the solvents are benzene and H2O and we add benzoic acid in this then the side effect are that in water benzoic acid ionizes and in benzene dimers are formed so we require some other way to avoid these side effects for this we have a distribution ratio D.
3
If there is no side reaction in the solvent then D
= Kd
=
C1/C2
=
Vo/Va
Vo =
Volume of organic phase
Va =
Volume of aqueous phase
One organic and one aqueous solvent is taken due to its immiscibility. Sometimes the volume of aqueous phase is above and organic phase is below and in some case the organic phase is above and aqueous phase is below. But in formula Vo is always above and Va is always below. The difference in Kd and D is that in case of D the concentration of all types of species in organic phase divided by concentration of all types of species
in
aqueous
phase.
Concentration
expressed
on
either
in
volume/volume or in milli moles/milli moles. If it is volume by volumeit is denoted by Dc and if it is in the form of millimoles it is denoted by Dm. Units of Extraction:
The extraction is always expressed in terms of %age % Extraction = 100[1 – 1/1(Dm+1)n] % Extraction =
For millimoles
100Dc/[Dc + (Va/Vo)]n
For Volume by Volume
n = is the number of extractions If Vo = Va then %E =
100 Dc/[Dc + (1)n
METHODS OF EXTRACTIONS:
Suppose there is 20 ml aqueous solution containing X millimoles of solute, extraction is with 20 ml of organic phase. There are two methods of extraction
4 1
One method is that these equal amounts are shaken together
and then separate. The other method is that in which solvent is divided into four parts then it is shaken in four installments with the other solvent having solute dissolved in it. In other words the “%age extraction depends upon the number of extractions, larger is the number of extraction more will be the %age of extraction provided the time and temperature of the extractions must be the same. NATURE OR TYPES OF EXTRACTION:
There are methods of extraction 1
Single short or Batch Method
2
Continuous Method
3
Multiple Extractions or Crag’s Method Batch Method:
It is an ordinary simple method usually used in labs. In this method the amount of the aqueous phase containing the specie is taken in a separating funnel to it added the whole amount or total volume of the extracting solvent then both are shaken profusely and after giving them rest the two immiscible layers are physically distinct from their organic layer or organic phase whether it form the upper or lower layer is separated from the aqueous layer is by draining from the separating funnel.
The amount of the specie is recovered by evaporating the
solvent, the sample or specie is weighed and calculated by usual measurement is done by using conventional / analytical method. Apparatus: Common glass separating funnel, it is Squibed pear shape
funnel or Squibed funnel. Continuous Method:
In continuous flow method the amount of the desired component is extracted with the help of extracting solvent in a way that it is added continuously as fresh installments in repeated or continuous manner.
Multiple Extractions, Counter current or Crag’s Method:
5 This method is used industrially or on big scale, it can’t be used in lab. By this method 99.9% extraction can be done. “In Crag’s method is consisted a unit contain desired number of Crag’s tube in which the fresh amounts of extracting solvent is added and the desired component is extracted in this process the material is transferred from tube to tube and in each tube the un-extracted material is shaken mechanically with extracting solvent and ultimately on reaching the 99th tube the total component is almost extracted by the extracting solvent. Applications:
1
It is mostly used for the solvent extraction of rare earths and the
metal ion present in minute amounts. 2
It is used as ion association complex technique. In ion association
complex method the desired metal ion is incorporated in some bulky molecules and it is then associated with some other ion to form in pair. The behavior of ion pair or that molecule is organic like e.g. tetra phenyl Arsonium molecule (C6H5)4As+. It make ion pair with Magnate ion MnO4 as (C6H5)4As+MnO4-. 3
The most important application of extraction is the extraction of
metals by chelation, e.g. formation of 8-hydroxy qunolinate complex of metals. It mostly reacts with divalent trivalent and tetra valent. They are pH dependent second example is of EDTA.
6
% Extrtaction:
Assume the extraction of a chemical specie from water to an organic solvent. Then the % extraction of specie ‘c’ extracted into the organic phase is
dividing the right hand side of the equation by Vo [c]w.
7
For the amount remaining in water after extractions in n number of extractions is
Vw & Vo W = Wa
=
Volume of water & volume of solvent
Amount of solute remaining =
Initial amount of solute in H2O
Example:
One gm of solute is contained in 100 ml of an aqueous solution. Calculate the amount remaining in the aqueous phase after (a)
A single extraction with 90 ml of an organic solvent.
(b)
A single extraction with 30 ml of an organic solvent.
(c)
Three extractions with 30 ml of an organic solvent. D = 10
(a)
Batch Extraction n = 1 Wa = 1 gm Vw = 100 ml Vo = 90 ml
8
(b)
Batch Method n = 1 Vw 100 ml Vo
30 ml
Wa 1 gm
9
(c)
Multiple Method: n = 3 Vw = 100 ml Vo = 30 ml Wa = 1 gm D = 10
As ‘C’ is a multiple method lets solve it stepwise. Ist Step: n = 1 Vw = 100 ml
10 Vo = 30 ml Wa = 1 gm D = 10
Now for sec step Wa2 will be Wa2 = W1 = 0.25 gm n = 1 Vw = 100 ml Vo = 30 ml D = 10
11
For third step Wa3 = W2 = 0.0625 gm n = 1 Vw = 100 ml Vo = 30 ml D = 10
12
Numerical: 20 ml of an aqueous solution of 0.1 M Buteric acid is shaken with 10 ml of ether. After the layers are separated. It is determined by titration that 0.5 millimoles of buteric acid remain in aqueous layer .
What is the distribution ratio and %
extraction. 20 ml contain buteric acid = 0.1 M For millimoles it is multiplied by volume Total concentration = 20 × 0.1 = 2.0 millimoles Millimoles of buteric acid in ether
=
[2.0 – 0.5]/10
=
1.5/10
=
0.15 millimoles
(Where 10 is the volume of ether in ml) Millimoles of buteric acid in aqueous layer = 0.5/20 (Where 20 is the volume of aqueous layer) D
