If two charges \({ q }_{ 1 }\) and \({ q }_{ 2 }\) are separated by distance \(r\) in vacuum then electrostatic force between them will be given by the Coulomb law as below:

$${ F }_{ v } = \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(1)$$ Now if these charges are placed in a medium of absolute permittivity \(\epsilon\) , then electrostatic force between these charges will be $${ F }_{ m } = \frac { 1 }{ 4\pi \epsilon } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(2)$$ We know that $$\epsilon = { \epsilon }_{ 0 } { \epsilon }_{ r } =K{ \epsilon }_{ 0 }$$ where \(K\) is dielectric constant or Relative permittivity or specific inductive capacity. or $$K = \frac { \epsilon }{ { \epsilon }_{ 0 } }$$ Now, eqn. (2) becomes, $${ F }_{ m } = \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(3)$$ Dividing eqn. (1) by eqn. (3), $$\frac { { F }_{ v } }{ { F }_{ m } } = \frac { \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) }{ \left( \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) } = K$$ Hence, dielectric constant (or relative permittivity) of a medium is defined as the ratio of the electrostatic force between two point charges when placed a certain distance apart in vacuum to the electrostatic force between the same charges when placed same distance apart in a medium.

This means a dielectric reduces the force between two charges.

If two charges \({ q }_{ 1 }\) and \({ q }_{ 2 }\) are separated by distance \(r\) in vacuum then electrostatic force between them will be given by the Coulomb law as below:

$${ F }_{ v } = \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(1)$$ Now if these charges are placed in a medium of absolute permittivity \(\epsilon\) , then electrostatic force between these charges will be $${ F }_{ m } = \frac { 1 }{ 4\pi \epsilon } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(2)$$ We know that $$\epsilon = { \epsilon }_{ 0 } { \epsilon }_{ r } =K{ \epsilon }_{ 0 }$$ where \(K\) is dielectric constant or Relative permittivity or specific inductive capacity. or $$K = \frac { \epsilon }{ { \epsilon }_{ 0 } }$$ Now, eqn. (2) becomes, $${ F }_{ m } = \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \qquad ...(3)$$ Dividing eqn. (1) by eqn. (3), $$\frac { { F }_{ v } }{ { F }_{ m } } = \frac { \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) }{ \left( \frac { 1 }{ 4\pi K{ \epsilon }_{ 0 } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \right) } = K$$ Hence, dielectric constant (or relative permittivity) of a medium is defined as the ratio of the electrostatic force between two point charges when placed a certain distance apart in vacuum to the electrostatic force between the same charges when placed same distance apart in a medium.

This means a dielectric reduces the force between two charges.

Permittivity is the property of a particular medium which affects the magnitude of the force existing between two point charges.

We know that the greater the value of the permittivity of the medium placed between the two charged bodies the lesser the value of force existing between them.

Absolute permittivity \({ \epsilon }_{ 0 }\)

Relative permittivity \({ \epsilon }_{ r }\)

The absolute permittivity of air or vacuum is minimum and its value is \(8.854 \times { 10 }^{ -12 }\) F/m (farad/metre) whereas the value of absolute or (actual) permittivity \( \epsilon \) of all other insulating medium is more than \({ \epsilon }_{ 0 }\).

The ratio of these two permittivities i.e. absolute permittivity (\( \epsilon \)) of the insulating medium to the absolute permittivity (\({ \epsilon }_{ 0 }\)) of the air or vacuum is known as relative permittivity of that medium and is denoted by \({ \epsilon }_{ r }\). i.e. $$\quad { \epsilon }_{ r } = \frac { \epsilon }{ { \epsilon }_{ 0 } } $$

Permittivity is the property of a particular medium which affects the magnitude of the force existing between two point charges.

We know that the greater the value of the permittivity of the medium placed between the two charged bodies the lesser the value of force existing between them.

Absolute permittivity \({ \epsilon }_{ 0 }\)

Relative permittivity \({ \epsilon }_{ r }\)

The absolute permittivity of air or vacuum is minimum and its value is \(8.854 \times { 10 }^{ -12 }\) F/m (farad/metre) whereas the value of absolute or (actual) permittivity \( \epsilon \) of all other insulating medium is more than \({ \epsilon }_{ 0 }\).

The ratio of these two permittivities i.e. absolute permittivity (\( \epsilon \)) of the insulating medium to the absolute permittivity (\({ \epsilon }_{ 0 }\)) of the air or vacuum is known as relative permittivity of that medium and is denoted by \({ \epsilon }_{ r }\). i.e. $$\quad { \epsilon }_{ r } = \frac { \epsilon }{ { \epsilon }_{ 0 } } $$

Coulomb performed a number of experiments to see the effect of placing two small charges near each other. On his experimental researches, he established a law which is known as Coulombs Law of Electrostatics.

According to Coulombs Law :The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges \({ q }_{ 1 } \) and \({ q }_{ 2 }\) have distance d between them, then force F between the charges can be mathematically expressed as: $$ F \propto { q }_{ 1 }{ q }_{ 2 }\qquad ...(1)$$ $$ F \propto \frac { 1 }{ { d }^{ 2 } } \qquad ...(2)$$ Combining (1) and (2), $$ F \propto \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } or F = K\frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \qquad ...(3) $$ Where, \(K\) is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used.

In SI units force is measured in newton, charge in coulomb, distance in metre and the value of \(K\) is given as, $$ K = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \qquad ...(4) (in SI system)$$ where, \({ \epsilon }_{ 0 }\) = Absolute permittivity of vacuum or permittivity of free space = \(8.854 \times { 10 }^{ -12 }\) farad/metre
\( { \epsilon }_{ r }\) = Relative permittivity of the medium w.r.t. vacuum in which the charges are placed (e.g. for air \({ \epsilon }_{ r }\) = 1)

Putting the value of \(K\) in equation (3), we get, $$ \boxed { F = \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } \qquad ... (5)$$ Unit Charge or one coulomb charge in SI system can be defined as the amount of charge which when placed at a distance of one metre from an equal and similar charge in air, is repelled with a force of \(9 \times { 10 }^{ 9 } \)newton from it.

As per this defination, $$ { q }_{ 1 } = { q }_{ 2 } = q, $$ $$F = 9 \times { 10 }^{ 9 } newton,$$ $$ { \epsilon }_{ 0 } = 8.854 \times { 10 }^{ -12 } farad/metre,$$ \({ \epsilon }_{ r } = 1\) (Assuming two charges to be placed in air)
d = 1m

Putting these values in the above relation, we get $$ 9 \times { 10 }^{ 9 } = \frac { q \times q }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 \times { 1 }^{ 2 } } $$ $$ 9 \times { 10 }^{ 9 } = \frac { { q }^{ 2 } }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) } $$ $$ { q }^{ 2 } = \left( 9 \times { 10 }^{ 9 } \right) \times 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) = 1.00086 $$ $$ \therefore q = \pm 1.00043 \ coulomb $$ Hence, unit charge.

Determine the force between two charges, each of one coulomb when they are separated at one metre distance in air.

Solution:We have, $${ q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = 1m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \frac { 1 \times 1 }{ { \left( 1 \right) }^{ 2 } } $$ $$ F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } $$ $$ F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } $$ $$ F = 8.992 \times { 10 }^{ 9 } N $$

An electron and a proton are at a distance of 10^-9 m from each other in a free space. Compute the force between them.

Solution:We have, The charge on an electron, \({ q }_{ 1 } = -1.6 \times { 10 }^{ -19 } \) C
and the charge on a proton, \({ q }_{ 2 } = +1.6 \times { 10 }^{ -19 }\) C. $$ { q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = \({ 10 }^{ -9 }\) m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \left( \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } $$ $$ \therefore F = \left( \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \right) \left( \frac { \left( -1.6 \times { 10 }^{ -19 } \right) \times \left( +1.6 \times { 10 }^{ -19 } \right) }{ { \left( { 10 }^{ -9 } \right) }^{ 2 } } \right) $$ $$ \therefore F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \left( -2.56 \times { 10 }^{ -20 } \right) $$ $$ \therefore F = -23.063 \times { 10 }^{ -11 } N $$ The negative sign indicates that the force is attractive in nature.

Coulomb performed a number of experiments to see the effect of placing two small charges near each other. On his experimental researches, he established a law which is known as Coulombs Law of Electrostatics.

According to Coulombs Law :The force of attraction or repulsion between two charges is directly proportional to the product of magnitude of the two charges and inversely proportional to the square of distance between them.

If two point charges \({ q }_{ 1 } \) and \({ q }_{ 2 }\) have distance d between them, then force F between the charges can be mathematically expressed as: $$ F \propto { q }_{ 1 }{ q }_{ 2 }\qquad ...(1)$$ $$ F \propto \frac { 1 }{ { d }^{ 2 } } \qquad ...(2)$$ Combining (1) and (2), $$ F \propto \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } or F = K\frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } \qquad ...(3) $$ Where, \(K\) is a constant of proportionality and its value depends upon the medium in which the charges are placed and the system of units used.

In SI units force is measured in newton, charge in coulomb, distance in metre and the value of \(K\) is given as, $$ K = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \qquad ...(4) (in SI system)$$ where, \({ \epsilon }_{ 0 }\) = Absolute permittivity of vacuum or permittivity of free space = \(8.854 \times { 10 }^{ -12 }\) farad/metre
\( { \epsilon }_{ r }\) = Relative permittivity of the medium w.r.t. vacuum in which the charges are placed (e.g. for air \({ \epsilon }_{ r }\) = 1)

Putting the value of \(K\) in equation (3), we get, $$ \boxed { F = \left( \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } \qquad ... (5)$$ Unit Charge or one coulomb charge in SI system can be defined as the amount of charge which when placed at a distance of one metre from an equal and similar charge in air, is repelled with a force of \(9 \times { 10 }^{ 9 } \)newton from it.

As per this defination, $$ { q }_{ 1 } = { q }_{ 2 } = q, $$ $$F = 9 \times { 10 }^{ 9 } newton,$$ $$ { \epsilon }_{ 0 } = 8.854 \times { 10 }^{ -12 } farad/metre,$$ \({ \epsilon }_{ r } = 1\) (Assuming two charges to be placed in air)
d = 1m

Putting these values in the above relation, we get $$ 9 \times { 10 }^{ 9 } = \frac { q \times q }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 \times { 1 }^{ 2 } } $$ $$ 9 \times { 10 }^{ 9 } = \frac { { q }^{ 2 } }{ 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) } $$ $$ { q }^{ 2 } = \left( 9 \times { 10 }^{ 9 } \right) \times 4\pi \times \left( 8.854 \times { 10 }^{ -12 } \right) = 1.00086 $$ $$ \therefore q = \pm 1.00043 \ coulomb $$ Hence, unit charge.

Determine the force between two charges, each of one coulomb when they are separated at one metre distance in air.

Solution:We have, $${ q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = 1m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \frac { 1 \times 1 }{ { \left( 1 \right) }^{ 2 } } $$ $$ F = \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } $$ $$ F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } $$ $$ F = 8.992 \times { 10 }^{ 9 } N $$

An electron and a proton are at a distance of 10^-9 m from each other in a free space. Compute the force between them.

Solution:We have, The charge on an electron, \({ q }_{ 1 } = -1.6 \times { 10 }^{ -19 } \) C
and the charge on a proton, \({ q }_{ 2 } = +1.6 \times { 10 }^{ -19 }\) C. $$ { q }_{ 1 } = { q }_{ 2 } = 1 \ coulomb $$ $$ { \epsilon }_{ 0 } = absolute \ permittivity = 8.854 \times { 10 }^{ -12 }F/m $$ $$ { \epsilon }_{ r } = 1 \qquad \qquad \left[ \because medium \ is \ air \right] $$ d = distance between the charges = \({ 10 }^{ -9 }\) m

The magnitude of force between two charges is given by $$ \boxed { F = \frac { 1 }{ 4\pi { \epsilon }_{ 0 }{ \epsilon }_{ r } } \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } } $$ Lets Substitute the values, $$ \therefore F = \left( \frac { 1 }{ 4 \times 3.14 \times \left( 8.854 \times { 10 }^{ -12 } \right) \times 1 } \right) \frac { { q }_{ 1 }{ q }_{ 2 } }{ { d }^{ 2 } } $$ $$ \therefore F = \left( \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \right) \left( \frac { \left( -1.6 \times { 10 }^{ -19 } \right) \times \left( +1.6 \times { 10 }^{ -19 } \right) }{ { \left( { 10 }^{ -9 } \right) }^{ 2 } } \right) $$ $$ \therefore F = \frac { 1 }{ 0.111 \times { 10 }^{ -9 } } \left( -2.56 \times { 10 }^{ -20 } \right) $$ $$ \therefore F = -23.063 \times { 10 }^{ -11 } N $$ The negative sign indicates that the force is attractive in nature.

The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as a whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. However, if from such a neutral body, electrons are removed, there occurs a shortage of electrons in the body.

Consequently the body no longer remains neutral. The result is that the body attains positive charge.

Thus, when a body is having shortage of electrons, it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge on the body.

To give a negative charge to any body, extra electrons must be supplied to it.

To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy.

This makes the charged body capable of doing work.

The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C).

1 coulomb of charge = The charge on \(625 \times {10}^{16}\) electrons.

Therefore, practical unit of charge is coulomb.

Like charges repel each other while unlike charges attract each other, means when two like charges are brought near to each other, they will repel each other and if two unlike charges [a +ve and a -ve charge] are brought near to each other, they will start attracting each other.

The number of electrons in an atom is equal to the number of protons, therefore, atom is neutral as a whole. A body consists of atoms, therefore, the body is neutral under ordinary conditions. However, if from such a neutral body, electrons are removed, there occurs a shortage of electrons in the body.

Consequently the body no longer remains neutral. The result is that the body attains positive charge.

Thus, when a body is having shortage of electrons, it is said to be positively charged.

On the other hand, a negatively charged body has excess of electrons from its normal due share.

Total deficiency or excess of electrons in a body is known as charge on the body.

To give a negative charge to any body, extra electrons must be supplied to it.

To supply these extra electrons, work will have to be done, which is stored in the body in the form of energy.

This makes the charged body capable of doing work.

The charge on an electron is so small that it is not possible and convenient to take it as the unit. In practice, the charge is measured in coulomb (C).

1 coulomb of charge = The charge on \(625 \times {10}^{16}\) electrons.

Therefore, practical unit of charge is coulomb.

Like charges repel each other while unlike charges attract each other, means when two like charges are brought near to each other, they will repel each other and if two unlike charges [a +ve and a -ve charge] are brought near to each other, they will start attracting each other.

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. $$ \boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } } $$