Solutions - Class 12th Chemistry

Introduction

A Solution is a homogeneous mixture of two or more components. It is defined by using the terms "Solute" and "Solvent".

Solvent

The component that is present in largest quantity is called Solvent. It determines the physical state of Solution.

Solute

One or more components present in Solution other than Solvent is called Solute.

Binary Solutions

Solution consisting of two components only.

Types of Solutions

According to the phase of Solvent, a binary Solution can be classified in following types:

Types of Solution	Solute	Solvent	Examples
Gaseous Solution	Gas	Gas	Mixture of O₂ and N₂
Gaseous Solution	Liquid	Gas	Chloroform mixed with N₂ Gas
Gaseous Solution	Solid	Gas	Camphor in nitrogen
Liquid Solutions	Gas	Liquid	Oxygen dissolved in water
Liquid Solutions	Liquid	Liquid	Ethanol dissolved in water
Liquid Solutions	Solid	Liquid	Glucose dissolved in water
Solid Solutions	Gas	Solid	Solution of H₂ and Pd
Solid Solutions	Liquid	Solid	Amalgam of Hg with Na
Solid Solutions	Solid	Solid	Alloys like brass

In this chapter we are mainly focusing on binary solutions (solution made up of two components) or liquid known as liquid solutions.

Gaseous Solutions

Solvent is a gas.
Small amounts of one gas are dissolved in a large amount of gas.
Examples: Air
- 78% nitrogen
- 21% oxygen
- 1% other gases

Different Methods of Expressing Concentration of Solutions

The Composition of Solution is defined via terms of concentration. There are several ways to define concentration of Solution as follows:

(A) Mass Percentage (ω/ω)

\[ \text{Mass %} \text{ of Component } = \frac{\text{Mass of the component in the Solution}}{\text{Total mass of Solution}} \times 100 \]

(B) Volume Percentage (v/v)

\[ \text{Volume %} \text{ of Component } = \frac{\text{Volume of the Component in the Solution}}{\text{Total volume of Solution}} \times 100 \]

(C) Mass by Volume percentage (ω/v)

\[ = \frac{\text{Mass of the component}}{100\text{mL Solution}} \]

(D) Part Per million (ppm)

\[ \text{ppm} = \frac{\text{Number of Parts of Component } \times 10^6}{\text{Total Number of Parts of all Components of Solution}} \]

Molarity and Molality

Molarity (M)

Molarity is the number of moles of solute present in 1 litre of solution.

\[ \text{Molarity} = \frac{\text{moles of Solute (n)}}{\text{Volume of Solution (V)}_L} = \frac{w_b \times 1000}{M_b \times V_{mL}} \] \[ \text{where } w_b = \text{mass of solute}, M_b = \text{molar mass of solute} \]

Question:

Calculate the molarity of a solution containing 5g of NaOH in 450 ml solution.

Solution:

\[ \text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \text{ g/mol} \] \[ \text{Molarity} = \frac{w_b \times 1000}{M_b \times V_{mL}} = \frac{5 \times 1000}{40 \times 450} = \frac{5000}{18000} = 0.278 \text{ M} \]

Molality (m)

Molality is the number of moles of solute present in 1 kg of solvent.

\[ \text{Molality} = \frac{\text{moles of Solute (n)}}{\text{mass of solvent (w}_a\text{) in kg}} = \frac{w_b \times 1000}{M_b \times w_a \text{ (in grams)}} \]

Question:

Calculate the molality of 2.5 g of ethanoic acid in 75 g of benzene.

Solution:

\[ \text{Molar mass of CH}_3\text{COOH} = 12 + 1 \times 3 + 12 + 16 + 16 + 1 = 60 \text{ g/mol} \] \[ \text{Molality} = \frac{2.5 \times 1000}{60 \times 75} = \frac{2500}{4500} = 0.556 \text{ m} \]

Normality

Normality is the number of gram equivalent of solute dissolved per litre of solution.

\[ \text{Normality} = \frac{\text{No. of gram equivalents of solute}}{\text{Volume of solution in litre}} \] \[ \text{No. of gram equivalents} = \frac{\text{mass of solute}}{\text{equivalent mass}} \] \[ \text{Equivalent mass} = \frac{\text{molar mass}}{\text{Basicity/Acidity}} \] \[ N = \frac{w_b \times 1000}{E_b \times V_{mL}} \]

Note: Normality = Molarity × basicity (for Acid) or Molarity × acidity (for Base)

Question:

250 ml solution is prepared by dissolving 1.325 g of Anhydrous Sodium Carbonate (Na₂CO₃) in water. Equivalent weight of Anhydrous Sodium Carbonate is 53. Calculate the Normality of Solution.

Solution:

\[ \text{Normality} = \frac{1.325 \times 1000}{53 \times 250} = \frac{1325}{13250} = 0.1 \text{ N} \]

Mole Fraction

Mole fraction of a component = No. of moles of the component / Total No. of moles of all components

Denoted by a Greek letter 'X' (chi)

\[ X_B = \frac{n_B}{n_A + n_B} \] \[ X_A = \frac{n_A}{n_A + n_B} \] \[ X_A + X_B = 1 \]

Henry's Law

Solubility of a gas in a liquid is directly proportional to the Partial Pressure of the gas present above the Surface of Liquid or Solution.

\[ P \propto X \] \[ P = k_H \cdot X \]

where:

\( P \) = Partial Pressure
\( X \) = Mole Fraction
\( k_H \) = Henry's Law Constant

k_H Value for O₂ increases with Increase in Temperature, So Solubility decreases with Increase in Temperature.

Henry's Law Applications:

Soda bottles
Scuba divers

Question:

Henry's law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 ml of Soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Solution:

\[ P_{CO_2} = 2.5 \text{ atm} \times 1.01325 \times 10^5 \text{ Pa/atm} = 2.533 \times 10^5 \text{ Pa} \] \[ P_{CO_2} = k_H \times X_{CO_2} \] \[ 2.533 \times 10^5 = 1.67 \times 10^8 \times X_{CO_2} \] \[ X_{CO_2} = 0.001517 \]

Mass of H₂O = 500 g (assuming density of water = 1 g/mL)

Molar mass of water = 18 g/mol

\[ n_{H_2O} = \frac{500}{18} = 27.78 \text{ mol} \] \[ X_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{H_2O}} \approx \frac{n_{CO_2}}{n_{H_2O}} \text{ (since } n_{CO_2} \ll n_{H_2O}\text{)} \] \[ n_{CO_2} = X_{CO_2} \times n_{H_2O} = 0.001517 \times 27.78 = 0.042 \text{ mol} \]

Molar Mass of CO₂ = 44 g/mol

\[ \text{Mass of CO}_2 = n_{CO_2} \times \text{Molar Mass of CO}_2 = 0.042 \times 44 = 1.848 \text{ g} \]

Vapour Pressure of Liquid Solution

Pressure exhibited by Vapour present above a liquid surface is known as Vapour Pressure.
Only in a closed container.

Raoult's Law

For a solution of a Volatile Liquid, the Partial Vapour Pressure of each component of the solution is directly proportional to the mole fraction present in the solution.

\[ P_1 \propto x_1 \Rightarrow P_1 = P_1^0 x_1 \] \[ P_2 \propto x_2 \Rightarrow P_2 = P_2^0 x_2 \]

where:

\( P_1 \) = Partial Pressure of 1st liquid substance
\( P_2 \) = Partial Pressure of 2nd liquid substance
\( x_1 \) = mole fraction of 1st liquid substance
\( x_2 \) = mole fraction of 2nd liquid substance
\( P_1^0, P_2^0 \) = Vapour Pressure of Pure Components at the same temperature

Dalton's Law of Partial Pressures

Total Pressure is equal to the sum of the partial pressures of the components.

\[ P_{\text{total}} = P_1 + P_2 \text{ [when two components]} \]

Total Vapour Pressure: Calculation

Raoult's Law: \( P_1 = P_1^0 x_1 \), \( P_2 = P_2^0 x_2 \)

For Binary Solution: \( x_1 + x_2 = 1 \) ⇒ \( x_1 = 1 - x_2 \)

Dalton's Law: \( P_{\text{total}} = P_1 + P_2 \)

\[ = P_1^0 x_1 + P_2^0 x_2 \] \[ = P_1^0 (1 - x_2) + P_2^0 x_2 \] \[ = P_1^0 - P_1^0 x_2 + P_2^0 x_2 \] \[ P_{\text{total}} = P_1^0 + (P_2^0 - P_1^0) x_2 \]

Total Vapour Pressure depends upon only one component mole fraction.

Ideal and Non-ideal Solutions

(A) Ideal Solutions

An Ideal Solution is one that obeys Raoult's Law over the entire range of concentration and temperature.

Characteristics of Ideal Solution:

Obeys Raoult's Law \( P = P_1 + P_2 \) ⇒ \( P = P_1^0 x_1 + P_2^0 x_2 \)
Volume change on mixing is zero ⇒ \( \Delta V = 0 \)
The Enthalpy remains unchanged when the components are mixed ⇒ \( \Delta H = 0 \)
No change in intermolecular forces. The intermolecular forces A-A, B-B & A-B are nearly equal.

Examples of Ideal Solution:

Benzene & Toluene
Mixture of Methanol & Ethanol

(B) Non-Ideal Solution

A non-ideal solution is one that does not obey Raoult's law over the entire range of concentration and temperature.

Characteristics of Non-Ideal Solution:

Does not obey Raoult's Law (\( P_1 \neq P_1^0 x_1 \) & \( P_2 \neq P_2^0 x_2 \))
These solutions show change in Volume when components are mixed ⇒ \( \Delta V \neq 0 \)
These solutions show change in Enthalpy when mixed ⇒ \( \Delta H \neq 0 \)
The intermolecular forces A-A & B-B are different from A-B resulting in deviation.

Types of Non-Ideal Solutions:

Positive Deviation: Partial Vapour Pressure is more than predicted by Raoult's law. It occurs when intermolecular forces (A-B force) are weaker than solute-solute (A-A) and solvent-solvent (B-B) forces.

Colligative Properties

The Properties which depend upon the number of Solute Particles present in a definite amount of solvent, but not on the chemical nature of Solute.

The Important Colligative Properties are:

Relative Lowering of Vapour Pressure (RLVP)
Elevation of boiling Point
Depression of freezing Point
Osmotic Pressure

1. Relative Lowering of Vapour Pressure

When a Non-Volatile Solute is added to the solvent, the V.P. of the resulting Solution is lower than that of pure Solvent. This phenomenon is called lowering of V.P.

Why it happens: The Solute particles occupy space at the surface of the liquid. Thus fewer solvent molecules can escape into vapour Phase. Hence, the V.P. decreases.

\[ \text{Lowering of vapour pressure} = P^0 - P \] \[ \text{Relative lowering of vapour pressure} = \frac{P^0 - P}{P^0} \text{ (RLVP)} \]

The Relative lowering of vapour pressure is equal to the Ratio of moles of Solute present in the Solution to the total No. of moles present in the Solution (mole fraction of the Solute).

According to the Raoult's Law:

\[ \frac{P^0 - P}{P^0} = X_b = \frac{n_B}{n_A + n_B} \] \[ \frac{n_B}{n_A + n_B} \approx \frac{n}{N} \text{ (for dilute solutions)} \]

Calculation of Molecular Mass of Non Volatile Compound from Relative Lowering of Vapour Pressure:

\[ \frac{P^0 - P}{P^0} = \frac{w_b \times M_A}{M_b \times w_A} \]

Question:

The V.P. of pure Benzene at a certain temperature is 640 mm/Hg. A Non-Volatile Solid weighing 2.75 g is added to 39 g of benzene. Vapour Pressure of Solution obtained is 600 mm/Hg. Calculate the molecular weight of solid substance.

Solution:

\[ P^0 = 640 \text{ mm/Hg}, P = 600 \text{ mm/Hg} \] \[ w_b = 2.75 \text{ g}, w_A = 39 \text{ g} \] \[ M_A \text{ (benzene)} = 78 \text{ g/mol} \] \[ \frac{640 - 600}{640} = \frac{2.75 \times 78}{M_b \times 39} \] \[ \frac{40}{640} = \frac{2.75 \times 2}{M_b} \] \[ \frac{1}{16} = \frac{5.5}{M_b} \] \[ M_b = 5.5 \times 16 = 88 \text{ g/mol} \]

Elevation of Boiling Point

The increase in the boiling point of the solvent on addition of solute to it is called the Elevation of boiling point of solvent. It is represented by \( \Delta T_b \).

\[ \Delta T_b = T_b - T_b^0 \]

where:

\( T_b \) = boiling point of solution
\( T_b^0 \) = boiling point of pure solvent

From experiment it is observed that the elevation of boiling point is proportional to the molality of solute in the dilute solution.

\[ \Delta T_b \propto m \] \[ \Delta T_b = k_b \cdot m \]

where \( k_b \) is molal Elevation constant or ebullioscopic constant.

\[ \Delta T_b = k_b \cdot \frac{w_b \times 1000}{M_b \times w_A \text{ (in grams)}} \]

Question:

The boiling point of benzene is 353.23 K. When 1.80 g of a Non-volatile Solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of Solute (K_b for benzene: 2.53 K kg/mol).

Solution:

\[ \Delta T_b = T_b - T_b^0 = 354.11 - 353.23 = 0.88 \text{ K} \] \[ \Delta T_b = k_b \cdot m = k_b \cdot \frac{w_b \times 1000}{M_b \times w_A} \] \[ 0.88 = 2.53 \times \frac{1.80 \times 1000}{M_b \times 90} \] \[ M_b = \frac{2.53 \times 1.80 \times 1000}{0.88 \times 90} \] \[ M_b = \frac{4554}{79.2} = 57.5 \text{ g/mol} \]

Depression of Freezing Point

The decrease in the freezing point of the solvent on adding of non-volatile solute is known as depression of freezing point.

\[ \Delta T_f = T_f^0 - T_f \]

where:

\( T_f^0 \) = freezing point of pure solvent
\( T_f \) = freezing point of solution

From experiment it is observed that depression of freezing point \( \Delta T_f \) is directly proportional to the molality of the solute in the dilute solution.

\[ \Delta T_f \propto m \] \[ \Delta T_f = k_f \times m \] \[ \Delta T_f = k_f \times \frac{w_b \times 1000}{M_b \times w_A \text{ (in grams)}} \]

where \( k_f \) is molal depression constant or cryoscopic constant.

Osmosis and Osmotic Pressure

(A) Osmosis

The spontaneous flow of the solvent through a semi-permeable membrane from a pure solvent to a solution or from a dilute solution to a concentrated solution is called osmosis.

Difference between Osmosis and Diffusion:

Osmosis	Diffusion
In osmosis, the solvent molecules move from the lower concentrated solution to higher concentrated solution.	In diffusion, the molecules move from the higher concentrated region to the region of lower concentration.
In osmosis, semi-permeable membrane is necessary.	No membrane is required in diffusion.
Osmosis takes place only in solutions.	Diffusion takes place in gases and liquids.
In osmosis, the flow of solvent particles is in one direction only.	In diffusion both the solute as well as solvent particles flow in opposite directions.

Osmotic Solutions Concepts

IDENTIFICATION

ISOTONIC SOLUTION

Two solutions having same osmotic pressure
Example: 0.9% solution of NaCl is isotonic with blood plasma

HYPERTONIC SOLUTION

Having higher osmotic pressure than reference solution
Causes water to move out of cells (plasmolysis)

HYPOTONIC SOLUTION

Having lower osmotic pressure than reference solution
Causes water to move into cells (cytolysis)

OSMOSIS PROCESSES

EXOSMOSIS

Outward flow of water or solvent from a cell through semi-permeable membrane (occurs in hypertonic solutions)

ENDOSMOSIS

Inward flow of water or solvent into a cell through semi-permeable membrane (occurs in hypotonic solutions)

Law of Osmotic Pressure (van't Hoff's Law)

Definition: Osmotic pressure (π) is the minimum pressure required to prevent osmosis when a solution is separated from pure solvent by a semipermeable membrane.

1. van't Hoff Equation for Dilute Solutions

\[ \pi = C R T \]

Where:

\( \pi \) = Osmotic pressure (atm)
\( C \) = Molar concentration (mol/L)
\( R \) = Universal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹)
\( T \) = Absolute temperature (K)

ABNORMAL MOLECULAR MASS

Observed colligative properties deviate from normal calculated values due to:

Association of molecules (e.g., dimerization of acetic acid in benzene)
Dissociation of molecules (e.g., ionization of NaCl in water)

Higher values observed in case of dissociation

Lower values observed in case of association

4. Van't Hoff Factor (i)

Definition: Accounts for dissociation/association of solutes.

\[ i = \frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}} \] \[ i = \frac{\text{Normal Molecular Mass}}{\text{Observed Molecular Mass}} \] \[ i = \frac{\text{Number of particles after association/dissociation}}{\text{Number of particles initially}} \]

Examples:

NaCl → \( i \approx 2 \) (dissociates into Na⁺ + Cl⁻)
Glucose → \( i = 1 \) (no dissociation)

Example Calculation:

Calculate osmotic pressure of 0.1M glucose solution at 300K.

\[ \pi = (0.1 \text{ mol/L})(0.0821 \text{ L·atm·K⁻¹·mol⁻¹})(300 \text{ K}) \] \[ \pi = 2.463 \text{ atm} \]