The method was discovered by Craig and Post. It is consists of 300-400 such chambers. The organic solvent and the aqueous solution is introduced to tube A and then it passes to B. It is shaken and is allowed to attain the equilibrium. Now, the apparatus is tilted so that the upper layer gets decanted through C and is collected in D. When the apparatus is again made vertical. The liquid passes through D into E in the next chamber of A and then to B. The process is repeated till the two liquids gets almost separated.

Application:

1. With the help of this method we can have accurate quantitative analysis of a single as well as the mixture of the components.

2. In this case apparatus required are very simple (separating funnel burette pipets conical flask etc.)

3. Time required for analysis is very small.

4. The method is very well used for detection of traces quantity of substance where precipitation method (Gravimetry) is not possible.

5. Fe⁺³ ferric ion can be easily extracted by ether from 6 molar HCl solution of the ferrous ally and iron ore.

6. The extraction can also be use in the extraction of metal as metal chelate where later has high solubility in an immiscible solvent such as chloroform and benzene.

7. In industrial and commercial field extraction by counter current extraction is frequently applied in the separation of components where the difference in the distribution coefficient are small.

8. The phenomenon is widely applied in drug analysis.

9. The solvent extraction is used in clinical laboratory.

10. Metal chelates are more soluble in non-polar solvents. Thus Ni(II) in its tetra co-ordinate complex with dimethyl glyoxime can be extracted into chloroform. In presence of citrate or tartrate the precipitation of Fe(III) and Cr(III) can be avoided.

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The method was discovered by Craig and Post. It is consists of 300-400 such chambers. The organic solvent and the aqueous solution is introduced to tube A and then it passes to B. It is shaken and is allowed to attain the equilibrium. Now, the apparatus is tilted so that the upper layer gets decanted through C and is collected in D. When the apparatus is again made vertical. The liquid passes through D into E in the next chamber of A and then to B. The process is repeated till the two liquids gets almost separated.

Application:

1. With the help of this method we can have accurate quantitative analysis of a single as well as the mixture of the components.

2. In this case apparatus required are very simple (separating funnel burette pipets conical flask etc.)

3. Time required for analysis is very small.

4. The method is very well used for detection of traces quantity of substance where precipitation method (Gravimetry) is not possible.

5. Fe⁺³ ferric ion can be easily extracted by ether from 6 molar HCl solution of the ferrous ally and iron ore.

6. The extraction can also be use in the extraction of metal as metal chelate where later has high solubility in an immiscible solvent such as chloroform and benzene.

7. In industrial and commercial field extraction by counter current extraction is frequently applied in the separation of components where the difference in the distribution coefficient are small.

8. The phenomenon is widely applied in drug analysis.

9. The solvent extraction is used in clinical laboratory.

10. Metal chelates are more soluble in non-polar solvents. Thus Ni(II) in its tetra co-ordinate complex with dimethyl glyoxime can be extracted into chloroform. In presence of citrate or tartrate the precipitation of Fe(III) and Cr(III) can be avoided.

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Though the multiple extraction is more beneficial than single step extraction. For analytical chemist it is easy to find out the amount extracted in single extraction, so he follow single extraction rather than multiple extraction because it is tedious. The relation between percentage extraction (E) for single extraction can be derived as follow: We know $$ { W }_{ 1 } = \left[ \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right] \times a \qquad ...(1) $$ where \({ W }_{ 1 }\) is the weight of solute remaining after 1st extraction.
\( { V }_{ W } \)= Volume of Aqueous Phase
\( { V }_{ o } \) = Volume of Organic Phase
\(D\) = Distribution Ratio
\(a\) = Total weight of Solute present initially.

The amount extracted will be a $$ { W }_{ 1 } = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \qquad ...(2) $$ $$ = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) $$ $$ \therefore Percentage \ extraction \ (E) = \left( \frac { a - { W }_{ 1 } }{ a } \right) \times 100 $$ Lets Substitute the values from equation (1) $$ E = \left( \frac { a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) }{ a } \right) \times 100 $$ $$ E = \left( 1 - \left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \right) \times 100 $$ $$ E = \left( \frac { { V }_{ o }D + { V }_{ W } - { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ $$E = \left( \frac { { V }_{ o }D }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ on dividing the numerator and denominator of RHS by \({ V }_{ o }\) we get $$ E = \left( \frac { D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } \right) \times 100 $$ $$ \therefore Percentage extraction (E) = \frac { 100D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } $$ Thus % extraction (E) depends on D as well as \(\frac { { V }_{ W } }{ { V }_{ o } } \).

Though the multiple extraction is more beneficial than single step extraction. For analytical chemist it is easy to find out the amount extracted in single extraction, so he follow single extraction rather than multiple extraction because it is tedious. The relation between percentage extraction (E) for single extraction can be derived as follow: We know $$ { W }_{ 1 } = \left[ \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right] \times a \qquad ...(1) $$ where \({ W }_{ 1 }\) is the weight of solute remaining after 1st extraction.
\( { V }_{ W } \)= Volume of Aqueous Phase
\( { V }_{ o } \) = Volume of Organic Phase
\(D\) = Distribution Ratio
\(a\) = Total weight of Solute present initially.

The amount extracted will be a $$ { W }_{ 1 } = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \qquad ...(2) $$ $$ = a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) $$ $$ \therefore Percentage \ extraction \ (E) = \left( \frac { a - { W }_{ 1 } }{ a } \right) \times 100 $$ Lets Substitute the values from equation (1) $$ E = \left( \frac { a - a\left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) }{ a } \right) \times 100 $$ $$ E = \left( 1 - \left( \frac { { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \right) \times 100 $$ $$ E = \left( \frac { { V }_{ o }D + { V }_{ W } - { V }_{ W } }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ $$E = \left( \frac { { V }_{ o }D }{ { V }_{ o }D + { V }_{ W } } \right) \times 100 $$ on dividing the numerator and denominator of RHS by \({ V }_{ o }\) we get $$ E = \left( \frac { D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } \right) \times 100 $$ $$ \therefore Percentage extraction (E) = \frac { 100D }{ D + \frac { { V }_{ W } }{ { V }_{ o } } } $$ Thus % extraction (E) depends on D as well as \(\frac { { V }_{ W } }{ { V }_{ o } } \).

If a solution contains two or more solutes say A and B, it is observed that when A is extracted, some amount of B is also extracted. The extent of the seperation can be expressed in terms of one factor called Seperation Factor \(\beta\) . This is related to the distribution ratio of A and B. $$ \boxed { \beta = \frac { D_{ A } }{ { D }_{ B } } = \frac { { { C }_{ o(A) } }/{ { C }_{ a(A) } } }{ { { C }_{ o(B) } }/{ { C }_{ a(B) } } } } $$ It is ratio therefore no unit and no dimension.

The larger value is always placed in the numerator. \beta must be made as large as possible by choice of extractant and by adjusting the volume ratio.

When \(D_{ A }\) = 10 and \(D_{ B } \)= 0.1. The \(\beta = \frac { 10 }{ 0.1 } = 100%\)

single extraction in case will remove 91% of A and 9% of B. It can be obtained for A, $$ E = \left[ \frac { 100{ D }_{ A } }{ { D }_{ A } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ for $${ V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 10 }{ 10 + 1 } \right] = \left[ \frac { 1000 }{ 11 } \right] = 90.9% $$ $$ Similarly \ for \ B, \ E = \left[ \frac { 100{ D }_{ B } }{ { D }_{ B } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ $$ for \ { V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 0.1 }{ 0.1 + 1 } \right] = \left[ \frac { 10 }{ 1.1 } \right] = 9.1% $$ The seperation of A is almost complete from B if the seperation factor B is high. It can be only in the case when \({ D }_{ A }\) is large and \({ D }_{ B }\) is small. For a given value of \({ D }_{ A }\) and \({ D }_{ B }\), the seperation effect can be increased by adjusting the volume ratio given by Nush Densen Equation which says. $$ \boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } } $$

If a solution contains two or more solutes say A and B, it is observed that when A is extracted, some amount of B is also extracted. The extent of the seperation can be expressed in terms of one factor called Seperation Factor \(\beta\) . This is related to the distribution ratio of A and B. $$ \boxed { \beta = \frac { D_{ A } }{ { D }_{ B } } = \frac { { { C }_{ o(A) } }/{ { C }_{ a(A) } } }{ { { C }_{ o(B) } }/{ { C }_{ a(B) } } } } $$ It is ratio therefore no unit and no dimension.

The larger value is always placed in the numerator. \beta must be made as large as possible by choice of extractant and by adjusting the volume ratio.

When \(D_{ A }\) = 10 and \(D_{ B } \)= 0.1. The \(\beta = \frac { 10 }{ 0.1 } = 100%\)

single extraction in case will remove 91% of A and 9% of B. It can be obtained for A, $$ E = \left[ \frac { 100{ D }_{ A } }{ { D }_{ A } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ for $${ V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 10 }{ 10 + 1 } \right] = \left[ \frac { 1000 }{ 11 } \right] = 90.9% $$ $$ Similarly \ for \ B, \ E = \left[ \frac { 100{ D }_{ B } }{ { D }_{ B } + { { V }_{ W } }/{ { V }_{ o } } } \right] $$ $$ for \ { V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1 $$ $$ \therefore E = \left[ \frac { 100 \times 0.1 }{ 0.1 + 1 } \right] = \left[ \frac { 10 }{ 1.1 } \right] = 9.1% $$ The seperation of A is almost complete from B if the seperation factor B is high. It can be only in the case when \({ D }_{ A }\) is large and \({ D }_{ B }\) is small. For a given value of \({ D }_{ A }\) and \({ D }_{ B }\), the seperation effect can be increased by adjusting the volume ratio given by Nush Densen Equation which says. $$ \boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } } $$

In choosing a proper solvent for solvent extractions, the following factors must be taken into consideration.

1. The solvent selected should be such that, the solubility of the solute to be extracted is more. This means seperation factor (B) must be large.

2. The two phases used must be totally immiscible.

3. The solvent must not react chemically with the solute present in aqueous solution.

4. The solvent must be recoverable without much expenditure.

5. The density of the solvent must be different than aqueous phase it will help in quick settling of the two phases.

6. The solvent should not be toxic.

7. The solvent should be cheap.

8. The solvent should possess low inflammability.

9. The solvent should have low viscosity, low vapour pressure and low freezing point.

10. The solvent should be readily available.

In choosing a proper solvent for solvent extractions, the following factors must be taken into consideration.

1. The solvent selected should be such that, the solubility of the solute to be extracted is more. This means seperation factor (B) must be large.

2. The two phases used must be totally immiscible.

3. The solvent must not react chemically with the solute present in aqueous solution.

4. The solvent must be recoverable without much expenditure.

5. The density of the solvent must be different than aqueous phase it will help in quick settling of the two phases.

6. The solvent should not be toxic.

7. The solvent should be cheap.

8. The solvent should possess low inflammability.

9. The solvent should have low viscosity, low vapour pressure and low freezing point.

10. The solvent should be readily available.