Solution and its types, Solubility

24/09/2020 Vinod 2 Comments

It is the homogeneous mixtures of two or more than two components .

When the composition and properties are uniform throughout the mixture then it is called homogeneous mixture .

It has following components –

  1. Solvent .
  2. Solute .

Solvent :-

The component that present in the largest quantity is known as Solvent .

Solvent can be solid , liquid and gas . It determines the physical state of the mixture .

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Solute :-

One or more components present in the Solvent is called Solute .

Saturated Solution :-

A mixture in which no more Solute can be dissolved under the given conditions I . e . temperature and pressure is called Saturated Solution .

Unsaturated Solution :-

A mixture in which more Solute can be dissolved under the given conditions I . e . temperature and pressure is called Unsaturated Solution .

Ideal Solution :-

The mixture which obey Raoult’s law over the entire range of concentration is known as Ideal Solution .

  • The enthalpy of mixing of the pure components to form the Ideal Solutions is Zero

I . e . ∆H = 0

  • Total volume of mixture is equal to sum if volumes of the components I . e . The volume of mixing is zero .

∆V = 0

  • If the intermolecular attractive forces between the molecules A – A and B – B are nearly equal to those between A – B . This leads to the formation of Ideal Solution .

A perfectly Ideal Solution is rare but some mixtures are nearly Ideal in behavior .

Examples – Solution of n – hexane and n – heptane , Bromoethane and Chloroethane , Benzene and Toluene etc .

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Fig 1 : – The plot of Vapour pressure and Mole fraction for an Ideal Solution at constant temperature .

Also read:- Colligative properties

Non – ideal Solution :-

When a mixture does not obey Raoult’s law over the entire range of concentration , then it is called Non – ideal Solution .

  • The enthalpy of mixing of both liquid components is changes .
  • Volume is increase after dissolution .
  • The intermolecular attractive forces between A – A and B – B are weaker than those between A – B and leads to decrease in Vapour pressure resulting in negative deviations .

The Vapour pressure of such Solution is either higher or lower than that predicted by Raoult’s law . If it is higher the Solution exhibits positive deviation and if it is lower , it exhibits negative deviation from Raoul’s law .

Examples – Solution of Acetone + Ethanol , Water + Ethanol , Water + Methanol , Acetone + Benzene etc .

Ideal and Non ideal Solutions
Fig 2 : – The plot if Vapour pressure and Mole fraction for a non – ideal Solution .

Vapour Pressure :-

It is defined as the pressure exerted by a Vapour in Thermodynamics equilibrium with solid or liquid at a given temperature in closed system .

  • The vapour pressure of any substance increase non-linearly with temperature .

Let us consider a binary solution of two volatile liquids and denote the two components as 1 and 2 . when taken in a closed vessel , both the components would evaporate and eventually an equilibrium . Let the total vapour pressure at this stage be P total , P₁ and P₂ be the partial pressure of the two components 1 and 2 respectively .

These partial pressure are related to the mole fraction X₁ and X₂ .

The French chemist Raoult ( 1886 ) gave the relationship between them . The relationship is known as the Raoult’s law .

Raoult’s law :-

The law states that for a solution of volatile liquids , the partial Vapour pressure of each component in the solution is directly proportional to its mole fraction present in the Solution .

If partial pressure for component 1 = P₁

Mole fraction for component 1 = X₁

then

P₁ proportional X₁

or P₁ = P₁⁰ X₁

Where P₁⁰ represent the Vapour pressure of component 1

If the partial Pressure for component 2 = P₂

Mole fraction for component 2 = X₂

then

P₂ proportional X₂

or P₂ = P₂⁰ X₂

Where P₂⁰ represent Vapour pressure of component 2

According to Dalton’s law of partial Pressure , the total pressure over the Solution phase in container will be the sum of the partial Pressures of the components .

Total Pressure

P = P₁ + P₂

= X₁ P₁⁰ + X₂ P₂⁰

since X₁ + X₂ = 1

hence

P = ( 1 – X₂ ) P₁⁰ + X₂ P₂⁰

or P = P₁⁰ + ( P₂⁰ – P₁⁰ ) X₂

Expressing Concentration of Solutions :-

Composition of solutions can be described by expressing its Concentration .

It can be expressed either qualitatively or quantitatively .

In qualitatively , Solutions are dilute ( relatively very small quantity of Solute ) while in quantitatively , Solutions are concentrated ( relatively very large quantity of Solute ) .

  • In real life we describe the Concentration of Solutions quantitatively in following ways –

1 . Mass percentage

2 . Volume percentage

3 . Mass by Volume percentage

4 . parts per million

5 . Mole fraction

6 . Molarity

7 . Morality

Mass percentage :-

The mass percentage of a component in a Solution is defined as

Mass % of component = mass of the component x 100 / Total mass of Solution

Volume percentage :-

The volume percentage is defined as –

Volume % of a component = Volume of the component X 100 / Total volume of Solution

Mass by Volume percentage ( W / V ) :-

It is defined as the mass of Solute dissolved in 100 ml of the Solution .

Parts of million ( ppm ) :-

When a Solute is present in trace amount , it is convenient to express concentration in parts per million ( ppm ) and is defined as –

Parts per million = Number of parts of the component X 10⁶ / Total no. of parts of all components of the Solution .

Mole fraction :-

It is defined as mole fraction if a component = Number of moles of the component / Total no. of moles of all the components .

Example – If the number of moles if A and B are nA and nB respectively .

The mole fraction of A will be

XA = nA / nA + nB

The mole fraction of B will be

XB = nB / nA + nB

XA + XB = ( nA / nA + nB ) + ( nB / nA + nB ) = 1

Q 1. Calculate the mole fraction benzene containing 30 % by mass in Carbon tetrachloride .

Solve –

moles if Carbon tetrachloride ( CCl₄ ) = 70 gm / 154 gm mol⁻¹ = 0 ·45 mol .

Moles of benzene = 30 gm / 78 gm mol⁻¹ = 0 · 38 mol .

or XB = nB / nA + nB

= 0 · 38 / 0 · 45 + 0 · 38

= 0 · 38 / 0 · 83 = 0 · 457

XA = 1 – 0 · 457 = 0 · 543

Molarity ( M ) :-

It is defined as number of moles of Solute dissolved in one litre of liquid .

Molarity = Moles of Solute / Volume of Solutions in litre

I . e . M = n / V

Molarity is a function of temperature . This is because Volume depends on temperature and the mass does not .

Molality ( m ) :-

It is defined as the number of moles of the Solute per kilogram ( Kg ) of the Solvent and us expressed as

Molality = Moles of Solute / mass of Solvent in Kg

I . e . m = n / W

Note :- Mass % ,ppm , mole fraction and Molality are independent of temperature .

Q 2 . Calculate molality of 2 · 5 gm of ethanoic acid ( CH₃COOH ) in 75 gm of benzene .

solve –

moles of ethanoic acid ( n ) = 2 · 5 gm / 60 gm mol⁻¹ = 0 · 0417 mol⁻¹

mass of benzen in kg ( W ) = 75 x 10⁻³ kg

Molality of ethanoic acid ( m ) =?

m = n / W

m = 0 · 0417 x 1000 / 75 = 0 · 556 mol kg⁻¹

Solubility :-

It is defined as the amount of substance ( solute ) that dissolves in a unit volume of a liquid substance ( solvent ) to form a saturated solution under specified conditions of temperature and pressure .

Solubility = mass of Solute in gm X 100 / mass of Solvent .

  • Solubility of one substance into another substance depends on the following factors –

1 . Nature of the substances .

2 . Temperature .

3 . Pressure .

Nature of the Substances :-

Strong solute – solvent attractions equate to greater solubility while weak solute – solvent attraction equate to lesser solubility .

Polar solute tend to dissolve best in polar solvent . non – polar solute tend to dissolve best in non – polar solvent .

Example – NaCl and sugar dissolve in water .

Effect of temperature :-

The solubility of a solid in a liquid is affected by temperature changes .

In general , if in a nearly saturated solution ,the dissolution process is endothermic , the solubility should increase with rise in temperature and if it is exothermic the solubility should decrease .

  • Solubility of gases in liquids decreases with rise in temperature . As dissolution is an exothermic process , the solubility should decrease with increase of temperature .

Effect of Pressure :-

Pressure does not have any significant effect on solubility of solids in liquids .

it is so because solids and liquids are highly incompressible .

Henry’s law :-

At constant temperature , Solubility of gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid .

which is known as Henry’s law .

If we use the mole fraction of a gas in the liquid as a measure of its solubility , then it can be said that the mole fraction of a gas in the liquid is proportional to the partial pressure of the gas over the liquid .

I . e . P proportional X

Or

P =K X

Where P – partial pressure of the gas , X – mole fraction of the gas and K is constant called Henry’s law constant .

The formula is called Henry’s law constant .

Henrys law 1
Fig . 3 . Graph between Solubility and Partial Pressure .

Different gases have different K values at the same temperature . So K is a function of nature of the gas .

Q 3 . If N₂ gas is bubbled through water at 273 K . How many milli moles if N₂ gas would dissolve in 1 litre of water . Assume that N₂ exerts a partial pressure of 0 · 987 bar . Give that Henry’s law constant for N₂ at K is 76 · 48 k bar .

solve –

X = P / K = 0 · 987 bar / 76480 bar = 1 · 29 x 10⁻⁵

As 1 litre of water contains 55 · 5 mol of it therefore if n represent number of moles of N₂

X = n / n + 55 ·5 = n / 55 · 5 = 1· 29 x 10⁻⁵

( n in denominator is neglected as it is << 55 · 5 )

Thus n = 1 · 29 x 10 ⁻⁵ x 55 · 5 mol = 7 · 16 x 10⁻⁴ mol = 0 · 716 m mol.

Applications of Henry’s law :-

1 . Increase the solubility of CO₂ in cold drinks and soda water , the bottle is sealed under high pressure .

2 . At high altitudes the partial pressure of oxygen at high altitudes less than the ground level . This results in low concentration of oxygen in the blood and tissues of the Peoples .

3 . In scuba diving – during scuba diving when the diver breaths in compressed air from the supply tank , more Nitrogen dissolves in the blood because the pressure at that Depth is far greater the surface atmospheric pressure .

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