CHEMISTY As LEVEL(FORM FIVE) NOTES – PHYSICAL CHEMISTRY – Two Components Liquid System(1)

PHYSICAL CHEMISTRY – Two Components Liquid System

IMMISCIBLE LIQUIDS

Immiscible liquids are liquids which do not mix to form a homogeneous mixture.

When there are two immiscible liquids, they form a so-called immiscible pair.

Immiscible liquids form a heterogeneous mixture.

For immiscible liquids, the intermolecular force of attraction is greater compared to intermolecular forces of attraction between the liquids; that’s why the liquids don’t mix.

Since the liquids do not mix, their total vapor pressure (P_T) is equal to the sum of the pure vapor pressures of the components.

In immiscible liquids, normally the denser component is found at the bottom (lower layer) while the less dense component floats on the denser component (upper layer).

Example

Consider the immiscible pair of components A and B in which A is denser than B.

The immiscible pair is kept in a separating funnel to observe them clearly.

Partition law states that

“When a solute that is soluble in both liquids is added, then it will dissolve.”

The ratio of concentration of solute added in a pair of immiscible liquids is constant, i.e.

The distribution of solute in a pair of immiscible liquids is governed by the Partition law or Distribution law.

PARTITION LAW

Partition law states that:

“In a pair of immiscible liquids, when a solute which is soluble in both is added, it will distribute itself in such a way that the ratio of its concentration between the two liquids is constant.”

The constant in distribution law is termed as Distribution coefficient, Distribution constant, or Partition constant. It is denoted by K_d or K_D.

Note

The solute added in a pair of immiscible mixtures can be in any of three states: liquid, gas, or solid.

APPLICATION OF PARTITION LAW

One application is the extraction of solute from one component by mixing the solution with a second liquid that has no solute at all.

The liquid component which removes the solute is called the Extracting component, and the one from which the solute is removed is called the Extracted component.

In terms of extractions:

Note

Concentration is the amount of substance per unit volume.

Also,

NOTE

During extraction of solute, the amount of solute in the extracted component will decrease while the amount in the extracting component will increase.

Example 1

a) State the partition law

Partition law states that:

“When a soluble solute is added in a pair of immiscible liquids, it will distribute itself in such a way that the ratio of its distribution between the two liquids is constant.”

b) What does the term “Partition coefficient” mean?

Partition coefficient is the concentration of solute in immiscible liquids.

c) An aqueous solution containing 10 g per litre of solute X was shaken with 100 cc of ether. On shaking, 6 g of X was extracted. Calculate the amount of X extracted from aqueous solution residue after shaking with 100 cc of ether.

Solution

Given

Extracted component = 10 g/l

Volume of extracting component = 100 cc

Mass of X in water (aqueous solution) = 10 g

Kd = 15

Let the amount extracted be ‘a’

60 – 15a = 10a

60 = 10a + 15a

60 = 25a

a = 60/25

a = 2.4 g/cc

∴ The amount of X extracted from the aqueous residue is 2.4 g/cc.

NOTE

If layers are not specified, then the word “between” shows the numerator and denominator of the formula.

Example 2

A solid X is added to a mixture of benzene and water. After shaking well and allowing to stand, 10 ml of benzene layer was found to contain 0.13 g of X and 100 ml of water layer contained 0.22 g of X.

Calculate the distribution coefficient of solute X between benzene and water layer.

Solution

Mass of solute X in benzene = 0.13 g

Volume of benzene = 10 ml

Example 3

In the distribution of succinic acid between ether and water at 15°C, 20 ml of the ethereal layer contains 0.092 g of the acid. Find out the weight of the acid present in 50 ml of the aqueous solution in equilibrium with it. If the coefficient Kb between water and ether is 1.196 g and Kd is 5.2.

Solution

Mass of succinic acid in ethereal = 0.092 g

Volume of ethereal = 20 ml

Conc. of succinic acid in ethereal = 0.092/20 = 4.6 × 10^-3 g/ml

Coefficient of succinic acid between water and ethereal = 5.2

Volume of water = 50 ml

Let X be the weight of succinic acid in water

Conc. of succinic acid in water = X/50

∴ The weight of the acid present in aqueous solution is 1.196 g.

Example 4

An aqueous solution of succinic acid at 15°C containing 0.07 g in 10 ml is in equilibrium with an ethereal solution which has 0.013 g in 10 ml. The acid has its normal molecular weight in both solvents. What is the concentration of the ethereal solution which is in equilibrium with aqueous solution containing 0.024 g in 10 ml? (Ans: 0.00044 g/ml)

Solution

Mass of succinic acid in aqueous solution = 0.07 g

Volume of aqueous solution = 10 ml

Concentration of succinic acid in ethereal = 0.013 g/ml

Concentration of succinic acid in solution = 0.07/10 = 7 × 10^-3 g/ml

Kd = 5.38

Now, for second extraction:

Mass of aqueous solution = 0.024 g

Volume of aqueous solution = 10 ml

Conc. of succinic acid in aqueous solution = 0.024/10 = 2.4 × 10^-3 g/ml

Volume of ethereal = 10 ml

Let X be the concentration of succinic acid in ethereal.

For more than one extraction, we normally use the following formula:

Where:

• Ar_e is the amount of solute remaining
• V_a is the volume of extracted solution
• V_b is the volume of extracting solution
• K is the constant of distribution
• W_o is the original weight of the solute

From the formula, the amount extracted can be calculated as (A_ex):

A_ex = W_o – Ar_e

If the amount extracted and amount remained are known, then their respective percentages can be calculated, i.e.

For extracted %:

Example 1

Solve the partition law:

Calculate the amount of solute extracted by shaking 1 litre of aqueous solution containing 11 g of Q with:

• i) 100 ml of ether (10 g)
• ii) Two successive volumes of 50 ml of ether (Kd = 100) (10.69 g)

Answer

The partition law states that:

“When a soluble solute is added in a pair of immiscible liquids, it will distribute itself in such a way that the ratio of its distribution in the liquids is constant.”

Data given

Mass of solute Q = 11 g

Volume of aqueous solution = 1 litre = 1000 cc

Conc. of Q in aqueous solution

Let X be the amount extracted

10x = 1100 – 100x

110x = 1100

x = 10 g.

∴ The solute extracted is 10 g.

b) (ii) Number of extractions = 2

Volume of ether (Vb) = 50 ml.

Volume of water (Va) = 1000 ml.

Kd = 100

Wo = 11 g

= 0.30556

Again

Aex = Wo – Are

= 11 – 0.3556

= 10.69 g

∴ The solute extracted is 10.69 g.

Example 2

a) Explain the principle of solvent extractions.

b) What is the condition necessary for solvent extraction?

Answer

b) The conditions of solvent extraction:

• (i) The liquids to be mixed must form an immiscible solution.
• (ii) The solute that is added to the extracted component must be soluble in the extracting component and vice versa.

a) The principles of solvent extraction:

• i) Division of the volume of extracting components: To be efficient, the extracting component can be divided into two or more portions.
• ii) When a liquid A (extracting component) is mixed with liquid B (extracted component), they must form immiscible liquids with a layer between them.
• iii) The solute should be soluble in both liquid components to allow its distribution in the pair of immiscible liquids.

Example 3

a) State the partition law

Two form five girls each were given a solution containing 10 g of solute A in 900 cc of solvent C. The first girl used 900 cc of solvent B to extract solute from C. The second girl decided to use 300 cc of B for three extractions.

The distribution coefficient of solute A between C and B is 8.

i) Calculate the percentage of A left in C by the first girl.

ii) Calculate the percentage of A left in C by the second girl.

iii) Comment on the result obtained by the two girls.

Solution

a) Partition law states that:

“When a soluble solute is added in a pair of immiscible liquids, it will distribute itself in such a way that the ratio of concentration in each component is constant.”

b) Solution

Mass of solute Wo = 10 g

Volume of extracted = 900 cc

Volume of extracting Vb = 900 cc

Distribution constant Kd = 8

= 1.11 g

% remained = 1.11/10 × 100 = 11%

ii) % remained by the second girl

= 2%

∴ The percentage of A left in C by the second girl is 2%.

iii) For a successful extraction, the extracting liquid component must be divided into as many small portions as possible.

Example 4

Find out the principles of solvent extraction. 100 cm³ of ether is available for extracting solute X from 100 cm³ of water. The partition coefficient of X between ether and water is 4.

i) Calculate the fraction of X extracted by using 100 cm³ of ether all at once.

ii) Calculate the fraction of X extracted by using 4 portions of 25 cm³ each. (If aqueous solution contains 8 g of X)

Answer

The principles of solvent extraction:

i) The solute added in the immiscible mixture must be soluble in both liquids.

To make extraction perfect, the extracting liquid must be divided into small portions.

ii) When the extracted liquid is mixed with the extracting liquid, they should form an immiscible mixture.

Solution

i) X extracted

Given:

Volume of extracting Vb = 100 cm³

Volume of extracted Va = 100 cm³

Distribution constant k = 4

Mass of solute X = 8 g

From:

∴ The fraction of X extraction is

ii) Fraction of X extracted by using two 50 cm³

= 8/9

Amount extracted = 8 – 8/9 = 64/9 g

∴ The amount of X extracted by using 2 portions of 50 cm³ is 64/9 g.

Example 5

The Mogul oil company is disturbed by the presence of impurity M in its four-star petrol. 1 litre of petrol contains 5 g of M. In an effort to decrease the concentration of M in petrol, Mogul has discovered the secret of solvent S and the partition coefficient of M between petrol and S is 0.01.

a) What is meant by the term partition?

b) Explain the principle of solvent extraction.

c) Calculate the total mass m of M removed by using:

• i) One portion
• ii) Two 50 cm³ portions of solvent

Solution

a) Partition is the distribution of solute in a pair of immiscible mixtures.

b) i. The principle of solvent extraction:

“When the two components are added (extracted and extracting), they should form an immiscible mixture.”

ii. The solute used must be soluble in both extracting and extracted components.

iii. The extracting component should be divided into small portions for the experiment to be effective.

c) Given:

Volume of petrol Va = 1 litre = 1000 cm³

Mass of M X = 5 g

Distribution constant kd = 0.05

Volume of solvent S Vb = 100 cc

i)

∴ The mass removed is 4.55 g

= 0.138 g

∴ The amount removed is 5 – 0.138 g = 4.86 g.

SEPARATION OF IMMISCIBLE LIQUIDS

Immiscible liquids are separated by the process of steam distillation.

Steam distillation

This is the process of separating immiscible liquids of different boiling points by passing superheated steam through them.

Conditions necessary for steam distillation

In order for steam distillation to be feasible, the following conditions must be met:

• i) The two liquids should have different boiling points.
• ii) The two liquids should be immiscible.
• iii) There should be no volume change.
• iv) The total vapor pressure of the liquids should be equal to the sum of the components’ vapor pressures.

APPLICATION OF STEAM DISTILLATION

Steam distillation can be used to determine the molar mass of an unknown liquid.

Let the two liquids A and B form an immiscible pair.

And

n_A = a

n_B = b

Since the two liquids are immiscible, their distillation process can be explained in terms of their proportions or compositions (mole fraction).

From the number of moles of components, the total number of moles can be obtained:

i.e. n_T = n_A + n_B = a + b

If n_T is known, then mole fraction or composition can be calculated, i.e.

For immiscible liquids, the ratio of their compositions is equal to the ratio of their vapor pressures.

Where:

• M_B is the molar mass of B
• M_A is the molar mass of A
• m_A is the mass of A
• m_B is the mass of B
• P_A is the vapor pressure of A
• P_B is the vapor pressure of B

Example 1

A solution of 6 g of substance X in 50 cm³ of aqueous solution is in equilibrium at room temperature with an ether solution of X containing 108 g of X in 100 cm³. Calculate the weight of X that could be extracted by shaking 100 cm³ of an aqueous liquid containing 10 g of X with:

• i) 100 cm³ ether
• ii) 50 cm³ of ether twice at room temperature

Solution

Concentration of X in H₂O = 6 g/50 cm³

Concentration of X in ethereal = 108/100

K = 9

Now

Example 2

What is steam distillation?

State the four conditions necessary for steam distillation.

An organic liquid distills in steam. The partial pressure of the two liquids at the boiling point are 5.3 kPa for organic liquid and 96 kPa for water. The distillate contains the liquids in the ratio of 0.48 g organic liquid to 1 g of water. Calculate the molar mass of organic liquid.

Answer

a) Steam distillation is the process of separating immiscible liquids of different boiling points by passing superheated steam through them.

b) The conditions necessary for steam distillation are:

• i) The liquids must form an immiscible solution.
• ii) The total vapor pressure is equal to the sum of the vapor pressures of the components.
• iii) There should be no change in volume.
• iv) The liquids should have different boiling points.

Solution

P_o = 5.3 kPa

m_o = 0.48 g

M_o = ?

P_w = 96 kPa

M_w = 18 g

m_w = 1 g

Example 4

An organic liquid Q which does not mix with water distills in steam at 96°C under the pressure of 1.01 × 10⁵ Pa. The pressure of water at 96°C is 8.77 × 10⁴ N/m². The distillate contains 51% by mass Q. Calculate the molar mass of Q.

Solution

Atmospheric pressure P_atm = 1.01 × 10⁵ N/m²

P_Q = P_atm – P_w = 1.01 × 10⁵ – 8.77 × 10⁴ = 1.033 × 10⁴ N/m²

M_Q = ?

M_w = 18 g/mol

Mass of Q m_B = 51 g

Mass of W m_w = 49 g

COLLIGATIVE PROPERTIES

What is colligative properties?

Definition

Colligative properties are properties of a liquid which change depending on the number of particles of solute added, but not on the nature of the solute.

Mainly there are 4 colligative properties:

• i) Lowering of vapor pressure.
• ii) Boiling point elevation.
• iii) Freezing point depression.
• iv) Osmotic pressure.

Assumptions of colligative properties:

• i) The solute should not react with solvent.
• ii) The solute should not be volatile compared to solvent.
• iii) The solute should not dissociate or associate in the solvent.

1. LOWERING OF VAPOR PRESSURE

Vapor pressure is the pressure exerted by vapor against the atmospheric pressure.

Lowering of vapor pressure: Is the difference between the original pressure of liquid solvent (P_o) and the pressure of the solution.

Effect of solute on the vapor pressure of the solvent

When solute particles are added to the solvent, the vapor pressure of the solution is lowered.

When solute particles dissolve in a given solvent, they collide with the solvent molecules and hence prevent or decrease the number of solvent molecules that escape from liquid phase to vapor phase. This causes a decrease in the amount of vapor above the solution and normally causes a decrease in vapor pressure.

Relative lowering of vapor pressure

Relative lowering of vapor pressure is the ratio of lowering vapor pressure to the vapor pressure of the solvent.

RAOULT’S LAW OF VAPOR PRESSURE

It states that “The relative lowering of vapor pressure is proportional to the mole fraction of the solute added.”

Mathematically

Let P_o be the vapor pressure of the solvent

P be the vapor pressure of the solution

X_s be the mole fraction of solute

From Raoult’s law of vapor pressure:

For mole fraction of solute:

Let n be number of moles of solute

N be number of moles of solvent

The molar mass of solute can be determined:

Example 1

What do you understand by the term “Colligative property”?

Colligative property is the property of a liquid which changes depending on the number of solute particles added but not on the nature of the solute.

A solution is prepared from 90 g of water and 10.6 g of non-volatile solute. If the vapor pressure of the solution at 60°C was found to be 0.1867 atm, calculate the relative molecular mass of solute. Given that vapor pressure of H₂O at 60°C was 0.1966 atm.

Solution

Mass of water (solvent) = 90 g.

Mass of non-volatile solute = 10.6 g.

Vapor pressure of solution (P) = 0.1867 atm.

Vapor pressure of solvent P_o = 0.1966 atm.

Recall:

Example 2

When 114 g of sucrose are dissolved in 1000 g of water, the vapor pressure was lowered by 0.11 mmHg. Calculate the relative molecular mass of sucrose if the vapor pressure of water at 20°C was 17.54 mmHg.

Solution

Mass of solute = 114 g

Mass of solvent = 1000 g

Lower vapor pressure = 0.11 mmHg (P_o – P)

Vapor pressure of solvent = 17.54 mmHg

Recall:

Example 3

Calculate the vapor pressure lowering caused by the addition of 100 g of sucrose of molar mass 342 g/mol to 1000 g of water if the vapor pressure of pure water at 25°C is 23.8 mmHg.

Solution

Mass of solute = 100 g

Mass of solvent = 1000 g

Molar mass of solute = 342 g/mol

Molar mass of solvent = 18 g/mol

Vapor pressure of solvent = 23.8 mmHg

Example 4

The vapor pressure of ether (molar mass 74 g/mol) is 442 mmHg at 293 K. If 3 g of compound A are dissolved in 50 g of ether at this temperature, the vapor pressure falls to 426 mmHg. Calculate the molar mass of A assuming that the solution of A in ether is very dilute.

Solution

Mass of solute = 3 g

Mass of solvent = 50 g

Vapor pressure of solution P = 426 mmHg

Vapor pressure of solvent P_o = 442 mmHg

Mr of solvent = 74 g/mol

Example 5

18.2 g of urea is dissolved in 100 g of water at 50°C. The lowering of vapor pressure produced is 5 mmHg. Calculate the molecular mass of urea if the vapor pressure of water at 50°C is 92 mmHg.

Solution

Mass of solute = 18.2 g

Mass of solvent = 100 g

Lowering vapor pressure (P_o – P) = 5 mmHg

Vapor pressure of solvent (P_o) = 92 mmHg