Notes-Part-1-Class-12-Chemistry-Chapter-2-Solutions-Maharashtra Board

Types of solutions :

Homogeneous solution : A solution in which solute and solvent form uniform homogeneous one phase due to attraction between their molecules/particles is called homogeneous solution.

• E.g. A solution of NaCl or sugar.

Heterogeneous solution : A solution consisting of two or more phases is called a heterogeneous solution.

• E.g. A colloidal solution of starch.

A solution consists of a solvent and a solute. Since the physical states of a solvent and a solute may be gaseous, liquid or a solid, there are nine types of solutions.

Factors on which the solubility of a substance (solute) depends :

The extent of dissolution of a substance (solute) depends upon the following factors

• Nature of a solute : A solute may be crystalline, amorphous, ionic or covalent. Hence accordingly its tendency to dissolve changes. The substances having similar intermolecular forces tend to dissolve in each other.
• Nature of solvents : Solvents are classified as polar and nonpolar. Polar solutes dissolve in polar solvents. For example, ionic compounds dissolve in polar solvent like water. A solvent other than water is called a non-aqueous solvent. For example, C₆H₆, CCl₄, etc. Solutions in these solvents are called nonaqueous solutions. The solvent may be a gas, a liquid or a solid.
• Amount of a solvent : More amount of a solvent will dissolve more quantity of the solute.
• Temperature : Depending on the nature of a solvent and a solute the solubility changes with temperature. The effect depends on the heat of solution, hydration energy, etc. Generally as the temperature increases, solubility of solid increases and that of gases decreases.
• Pressure : Pressure has no effect on the solubilities of solids and liquids since they are incompressible. The effect of pressure is important only for solutions which involve gases as solute. With the increase in pressure and decrease in temperature, the solubility of gases increases.

Effects of temperature on solubility of a gas in water :

• The gases are soluble in water and other liquids.
• According to Charles’ law, the volume of a given mass of a gas increases with the increase in temperature at constant pressure.
• Hence, the volume of the dissolved gas increases with the increase in temperature.

• This enormous increase in volume of the gases cannot be accommodated by the solvent molecules, hence excess of the gases escape out in the form of bubbles.
• Therefore, the solubility of gases in liquids decreases with temperature.

Illustration of Henry’s law : In case of aerated or carbonated drink beverage, the bottle is filled by dissolving CO₂ gas at high pressure and then sealed.

• Above the liquid surface there is air and undissolved CO₂. Due to high pressure, the amount of dissolved CO₂ is large.
• When the cap of the aerated bottle is removed, the pressure on the solution is lowered, hence excess of CO2 and air escape out in the form of effervescence. Thus by decreasing the pressure, solubility of CO₂ is decreased.

Exceptions to Henry's law :

The gases like NH₃ and CO₂ do not obey Henry’s law. This is because, these gases react with water,

• NH₃(g) + H₂O(l) --> NH₄⁺(aq) + OH^-(aq)
• CO₂(g) + H₂O(l)  —> H₂CO₃(aq)

Due to reactions of the gases like NH₃, CO₂(g), they have higher solubility than expected by Henry’s law.

Units of Henry’s law constant :

By Henry’s law, S = K_H x R where S is solubility of the gas in mol dm^-3, P is the pressure of the gas in atmosphere (or in bar) and K_H is Henry’s law constant.

S(mol dm-3)

K_H = \(\frac{S^{(mol/dm^3)}}{P_{atm}}\) mol dm^-3 atm^-1 or mol dm^-3 bar^-1

Hence the units of Henry’s law constant K_H are mol dm^-3 atm^-1

Variation of vapour pressure with mole fraction of a solute in a liquid mixture :

Because P⁰₁  and P⁰₂  are constants, a plot of P_T versus x₂ is a straight line as shown in the Fig The figure also shows the plots of P₁ versus x₁ and P₂ versus x₂ according to the equations 1. These are straight lines passing through origin.

When x₁= 1, x₂=0, P_T = P⁰₁ and when x₁=0, x₂ = 1, P_T = P⁰₂ as shown by lines I and II in Fig.

Composition of vapour phase above a liquid mixture.

Consider a liquid mixture of two liquids A and B having vapour pressures P⁰_A and P⁰_B and mole fractions x₁ and x₂ respectively in the liquid phase.

By Raoult’s law, the vapour pressures of two liquids will be,

P_A = x₁ P⁰_A and P_B = x₂ P⁰_B

The total vapour pressure of this liquid mixture is,

P_T = P_A + P_B

Pr = x₁P⁰_A + x₂P⁰_B

The vapour above liquid surface contains A and B.

If y₁ and y₂ are the mole fractions of A and B components respectively in the vapour phase, then by Dalton’s law of partial pressures,

P_A = y₁P_T and P_B = y₂P_T

and total vapour pressure is,

P_T =y₁P_T + y₂P_T

Characteristics of ideal solutions :

• The ideal solutions obey Raoult’s law over entire range of concentrations at constant temperature.
• In the formation of an ideal solution, heat is neither evolved nor absorbed and enthalpy change for mixing is zero, i.e. Δ_mixH = 0.
• In the formation of an ideal solution, there is no volume change on mixing two liquid components and the volume of solution is equal to the sum of volumes of two liquid components. Δ_mixV = 0
• In the ideal solution, solvent-solvent, solute-solute and solvent-solute interactions are comparable.

• The vapour pressure of an ideal solution lies between vapour pressures of two pure components.

Characteristics of nonideal solutions :

• Nonideal solutions do not obey Raoult’s law over the entire range of concentrations.
• The vapour pressures of these solutions may be higher or lower than ideal solutions.
• These solutions exhibit two types of deviations from Raoult’s law namely (a) positive deviation and (b) negative deviation.
• These solutions give azeotropic mixtures.

Solutions with positive deviations from Raoult’s law :

• A solution or a liquid mixture which has higher vapour pressure than theoretically calculated by Raoult’s law or higher than those of pure components is called a nonideal solution with positive deviation.
• In these solutions, solute-solvent intermolecular attractions are weaker than those between solvent-solvent and solute-solute interactions.
• For example, solutions of acetone and ethanol, carbon disulphide and acetone, etc.

Solutions with negative deviations from Raoult’s law :

• A solution or a liquid mixture which has lower vapour pressure than theoretically calculated by Raoult’s law or lower than those of pure components is called a nonideal solution with negative deviation.
• In these solutions, the intermolecular interactions between solvent and solute molecules are stronger than solvent-solvent or solute-solute interactions.
• For example, solutions of phenol and aniline, chloroform and acetone, etc.

Vapour pressure of a liquid :

• If a volatile liquid is placed in an open vessel, the liquid molecules have a tendency to escape in a gaseous state forming vapour and diffuse into surroundings. Hence evaporation takes place continuously and no equilibrium is attained.
• If a liquid is placed in a closed vessel, the vapour molecules get accumulated on its surface. These vapour molecules are in continuous random motion. They collide with each other, with walls of a container and surface of the liquid, and return to the liquid state. This reverse phenomenon is called condensation.
• After some time, rates of evaporation and condensation become equal and an equilibrium is established between liquid and vapour phases. At this stage the vapour exerts a constant pressure called vapour pressure on liquid surface at constant temperature.
• Vapour pressure: The pressure exerted by the vapour of a liquid (or solid) when it is in equilibrium with the liquid (or solid) phase at a constant temperature is called the vapour pressure of the liquid (or solid).
• The vapour pressure of a liquid increases with the increase in temperature.

Explanation : Let P₀ and P be the vapour pressures of a pure solvent and a solution respectively. If x₁ is the mole fraction of the solvent then Raoult’s law can be represented as, P = x₁P₀

For a binary solution containing one solute, if x₁ and x₂ are mole fractions of a solvent and a solute respectively then,

x₁ + x₂ = 1

 x₁ = 1 - x₂

 P = x₁P₀

= (1 - x₂)P₀

= P₀ - x₂P₀

 P₀ - P = x₂P₀

∴ x₂ = \(\frac{P_0-P}{P_0}\)

P₀ - P = ΔP is the lowering of vapour pressure

∴ x₂  = \(\frac{ΔP}{P_0}\)

In the equation,  is called relative lowering of vapour pressure.

Hence Raoult’s law can also be stated as the relative lowering of vapour pressure is equal to mole fraction of the solute.

Variation of vapour pressure with mole fraction of a solvent in solution. OR

Variation of vapour pressure with the concentration of a solution :

The vapour pressure of a pure solvent decreases when a nonvolatile solute is dissolved in it.

Consider a pure solvent with vapour pressure P₀ and mole fraction x₁.

For a solution containing a nonvolatile solute, if x₁ and x₂ are the mole fractions of a solvent and a solute respectively, then x₁ + x₂ = 1 and x₁ < 1.

By Raoult’s law, P = x₁P_0.  As the mole fraction of a solvent in the solution increases the vapour pressure increases as shown in the above Fig. When x₁ becomes equal to 1, the vapour pressure becomes P₀, ie the vapour pressure of a pure solvent.

If at any mole fraction of a solvent, the vapour of the solution is P, then the lowering of vapour pressure will be, ΔP = P₀ - P.

Relation between relative lowering of vapour pressure and molar mass of nonvolatile solute :

Consider a solution in which W₁ gram of a solvent of molar mass (or molecular weight) M₁ contains W₂ gram of a solute of molar mass M₂.

Then number of moles of a solvent = n₁ = W₁/ M₁

Number of moles of a solute = n₂ = W₂/ M₂

Total number of moles = n = n₁ + n₂

Mole fraction of the solvent = x₁ = n₁/n

Mole fraction of the solute = x₂ = n₂/n

∴ x₁ + x₂ = \(\frac{n_1}{n}+\frac{n_2}{n}\)=\(\frac{n_1+n_2}{n}=\frac{n}{n} =1\)

∴ x₁ + x₂ =1

In the case of an ideal solution which is a dilute solution, the concentration and the number of moles of the solute are very low, i.e. n₂ << n₁

∴ n₁ + n₂ ≅ n₁

Now x₂ = n₂/n = \(\frac{n_2}{n_1+n_2}≅ \frac{n_2}{n_1}\)

∴ x₂ = \(\frac{W_2/M_2}{W_1/M_1}\)= \(\frac{W_2M_1}{W_1M_2}\)

If P₀ and P are the vapour pressures of a pure solvent and a solution respectively, then

relative lowering of vapour pressure = \(\frac{P_0-P}{P_0}\)

By Raoult’s law,

\(\frac{ΔP}{P_0}\)=\(\frac{P_0-P}{P_0}\)  = x₂

∴\(\frac{P_0-P}{P_0}\)=\(\frac{W_2M_1}{W_1M_2}\)

or

\(\frac{ΔP}{P_0}\)=\(\frac{W_2M_1}{W_1M_2}\)

Hence by measuring the vapour pressure of a pure solvent and a solution, the molar mass of the dissolved nonvolatile substances can be determined. .