Vapour Pressure of Solutions

When a solvent is pure, its surface is entirely occupied by solvent molecules. These molecules possess kinetic energy and can escape from the liquid surface into the gaseous phase, a process known as vaporisation. This leads to the establishment of a dynamic equilibrium between the liquid and vapour phases, where the rate of vaporisation equals the rate of condensation. The pressure exerted by the vapour in this equilibrium state at a given temperature is called the vapour pressure of the pure solvent.

Now, consider a solution formed by dissolving a non-volatile solute in a volatile solvent. The surface of this solution will have both solvent molecules and solute molecules. Since the solute is non-volatile, its molecules do not enter the vapour phase. Consequently, the number of solvent molecules at the surface is reduced compared to the pure solvent. This reduction in the surface area available for solvent molecules to escape leads to a lower rate of vaporisation.

At equilibrium, the rate of vaporisation of the solvent will equal the rate of condensation. The pressure exerted by the solvent vapour in this equilibrium will be the vapour pressure of the solution. Crucially, the presence of a non-volatile solute lowers the vapour pressure of the solvent. This phenomenon is a direct consequence of the solute molecules occupying some of the surface area, thereby hindering the escape of solvent molecules into the vapour phase. The extent of this lowering depends on the concentration of the solute.

Factors Affecting Vapour Pressure

The vapour pressure of a solution is primarily influenced by two factors:

• Temperature: Like pure liquids, the vapour pressure of a solution increases with an increase in temperature. This is because higher temperatures provide more kinetic energy to the solvent molecules, increasing their tendency to escape into the vapour phase.
• Concentration of Solute: For solutions containing a non-volatile solute, the vapour pressure decreases as the concentration of the solute increases. This is due to the reduced surface area available for solvent molecules to vaporise, as explained earlier.

Raoult's Law

Raoult's law is a fundamental principle that quantitatively describes the relationship between the vapour pressure of a solution and the composition of the solution, particularly for ideal solutions. It was formulated by the French chemist François-Marie Raoult in 1886.

Statement of Raoult's Law

Raoult's law states that for a solution of volatile components, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction in the solution. The proportionality constant is the vapour pressure of the pure component.

Mathematically, if we consider a binary solution composed of component 1 and component 2, with mole fractions $x_1$ and $x_2$ respectively, and their respective pure vapour pressures as $P_1^\circ$ and $P_2^\circ$, then the partial vapour pressure of component 1 ($P_1$) and component 2 ($P_2$) in the solution are given by:

$P_1 = x_1 P_1^\circ$

$P_2 = x_2 P_2^\circ$

The total vapour pressure ($P_{total}$) of the solution is the sum of the partial vapour pressures of all components, according to Dalton's law of partial pressures:

$P_{total} = P_1 + P_2 = x_1 P_1^\circ + x_2 P_2^\circ$

Since $x_1 + x_2 = 1$, we can also write:

$P_{total} = x_1 P_1^\circ + (1 - x_1) P_2^\circ$

Or,

$P_{total} = (1 - x_2) P_1^\circ + x_2 P_2^\circ$

Raoult's Law for Non-Volatile Solute

A special case of Raoult's law applies when a non-volatile solute is dissolved in a volatile solvent. In this scenario, the solute does not contribute to the vapour pressure. Let the solvent be component 1 and the solute be component 2. The vapour pressure of the solution ($P_{solution}$) is then solely due to the solvent.

According to Raoult's law:

$P_{solution} = P_1 = x_1 P_1^\circ$

Where $x_1$ is the mole fraction of the solvent and $P_1^\circ$ is the vapour pressure of the pure solvent.

Since $x_1 + x_2 = 1$, where $x_2$ is the mole fraction of the solute, we have $x_1 = 1 - x_2$. Substituting this into the equation:

$P_{solution} = (1 - x_2) P_1^\circ = P_1^\circ - x_2 P_1^\circ$

Rearranging this equation, we get:

$P_1^\circ - P_{solution} = x_2 P_1^\circ$

The term $(P_1^\circ - P_{solution})$ represents the lowering of the vapour pressure. The term $\frac{P_1^\circ - P_{solution}}{P_1^\circ}$ represents the relative lowering of vapour pressure.

Therefore, Raoult's law can also be stated as: The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.

$\frac{P_1^\circ - P_{solution}}{P_1^\circ} = x_2$

Ideal and Non-Ideal Solutions

Solutions can be broadly classified into ideal and non-ideal solutions based on their behaviour with respect to Raoult's law.

Ideal Solutions

An ideal solution is defined as a solution that obeys Raoult's law over the entire range of concentration at all temperatures. In an ideal solution, the intermolecular forces between the solute-solute molecules, solvent-solvent molecules, and solute-solvent molecules are almost identical.

Key characteristics of ideal solutions:

• Obeys Raoult's Law: $P_A = x_A P_A^\circ$ and $P_B = x_B P_B^\circ$.
• No Heat Change on Mixing: The enthalpy of mixing ($\Delta H_{mix}$) is zero. This means that the process of mixing the solute and solvent is neither exothermic nor endothermic. Energy is neither released nor absorbed because the bond breaking and bond forming energies are equal.
• No Volume Change on Mixing: The volume of mixing ($\Delta V_{mix}$) is zero. This means that the final volume of the solution is exactly equal to the sum of the volumes of the individual components before mixing.
• Intermolecular Forces: The forces of attraction between A-A, B-B, and A-B molecules are nearly equal.

Examples of Ideal Solutions: While perfectly ideal solutions are rare, some mixtures approximate ideal behaviour. Common examples include:

• Mixtures of benzene and toluene.
• Mixtures of n-hexane and n-heptane.
• Mixtures of chlorobenzene and bromobenzene.

These examples consist of components that are chemically similar and have similar molecular sizes and structures.

Non-Ideal Solutions

A non-ideal solution is a solution that does not obey Raoult's law over the entire range of concentration. The deviation from Raoult's law arises because the intermolecular forces between the components are not equal.

Non-ideal solutions can exhibit two types of deviations:

1. Positive Deviation from Raoult's Law

In solutions showing positive deviation, the partial vapour pressures of the components are greater than what is predicted by Raoult's law.

• Partial Vapour Pressures: $P_A > x_A P_A^\circ$ and $P_B > x_B P_B^\circ$.
• Total Vapour Pressure: $P_{total} > x_A P_A^\circ + x_B P_B^\circ$.
• Enthalpy of Mixing: $\Delta H_{mix} > 0$ (Endothermic process). Energy is absorbed during mixing because the solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions. More energy is required to break the existing bonds than is released when new bonds are formed.
• Volume of Mixing: $\Delta V_{mix} > 0$. The final volume of the solution is greater than the sum of the volumes of the individual components. This is because the weaker interactions between unlike molecules lead to a slight expansion.
• Intermolecular Forces: Solute-solvent interactions are weaker than solute-solute and solvent-solvent interactions.

Examples of Non-Ideal Solutions with Positive Deviation:

• Mixture of ethanol and water: Ethanol molecules form hydrogen bonds with each other, and water molecules form hydrogen bonds with each other. When mixed, the hydrogen bonds between ethanol and water molecules are weaker than the original hydrogen bonds within pure ethanol and pure water. This leads to a higher tendency for molecules to escape into the vapour phase.
• Mixture of acetone and carbon disulphide ($CS_2$): The $CS_2$ molecules are non-polar, while acetone has a polar carbonyl group. The interaction between acetone and $CS_2$ is weaker than the interactions within pure acetone or pure $CS_2$.

2. Negative Deviation from Raoult's Law

In solutions showing negative deviation, the partial vapour pressures of the components are less than what is predicted by Raoult's law.

• Partial Vapour Pressures: $P_A < x_A P_A^\circ$ and $P_B < x_B P_B^\circ$.
• Total Vapour Pressure: $P_{total} < x_A P_A^\circ + x_B P_B^\circ$.
• Enthalpy of Mixing: $\Delta H_{mix} < 0$ (Exothermic process). Energy is released during mixing because the solute-solvent interactions are stronger than the solute-solute and solvent-solvent interactions. More energy is released when new bonds are formed than was required to break the existing bonds.
• Volume of Mixing: $\Delta V_{mix} < 0$. The final volume of the solution is less than the sum of the volumes of the individual components. This is due to stronger interactions causing the molecules to pack more closely.
• Intermolecular Forces: Solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions.

Examples of Non-Ideal Solutions with Negative Deviation:

• Mixture of phenol and aniline: Both phenol and aniline can form hydrogen bonds. When mixed, they form stronger hydrogen bonds with each other than they do within their pure states.
• Mixture of hydrochloric acid (HCl) and water: The strong attraction between HCl molecules and water molecules (due to ionisation and hydration) leads to a decrease in vapour pressure.
• Mixture of acetic acid and pyridine: Pyridine is a base, and acetic acid is an acid. They react to form a salt, leading to strong intermolecular forces.

Colligative Properties

Colligative properties are properties of solutions that depend solely on the number of solute particles present in a given amount of solvent, and not on the nature or identity of the solute. These properties arise due to the presence of solute particles disrupting the solvent's behaviour.

There are four main colligative properties:

1. Relative lowering of vapour pressure.
2. Elevation of boiling point.
3. Depression of freezing point.
4. Osmotic pressure.

The magnitude of these properties is directly proportional to the number of solute particles. For solutions of electrolytes, which dissociate into ions, the number of particles is greater than the number of formula units of the solute, and this must be accounted for using the van't Hoff factor.

1. Relative Lowering of Vapour Pressure

As discussed earlier under Raoult's law for non-volatile solutes, the relative lowering of vapour pressure is given by:

$\frac{P_1^\circ - P_{solution}}{P_1^\circ} = x_2$

If the solute is non-electrolyte (like glucose or urea), $x_2$ is the mole fraction of the solute.

$x_2 = \frac{n_2}{n_1 + n_2}$

Where $n_1$ is the moles of solvent and $n_2$ is the moles of solute.

For dilute solutions, $n_1 \gg n_2$, so $n_1 + n_2 \approx n_1$.

Therefore, $x_2 \approx \frac{n_2}{n_1}$.

$\frac{P_1^\circ - P_{solution}}{P_1^\circ} = \frac{n_2}{n_1}$

This equation shows that the relative lowering of vapour pressure is directly proportional to the moles of solute and inversely proportional to the moles of solvent, hence dependent on the number of solute particles.

2. Elevation of Boiling Point ($\Delta T_b$)

The boiling point of a liquid is the temperature at which its vapour pressure equals the external atmospheric pressure. When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases. To reach the atmospheric pressure, the solution needs to be heated to a higher temperature than the pure solvent. This increase in boiling point is called the elevation of boiling point.

The elevation of boiling point ($\Delta T_b$) is directly proportional to the molality ($m$) of the solution.

$\Delta T_b \propto m$

$\Delta T_b = K_b \cdot m$

Where:

• $\Delta T_b = T_{b, solution} - T_{b, solvent}^\circ$ is the elevation in boiling point.
• $K_b$ is the ebullioscopic constant (or molal boiling point elevation constant) of the solvent. It is a characteristic property of the solvent. Its units are K kg/mol.
• $m$ is the molality of the solution, defined as moles of solute per kilogram of solvent ($m = \frac{n_2}{w_1(kg)}$).

For a non-electrolyte solute, $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.

If the molar mass of the solute is $M_2$ and the mass of the solute is $w_2$, then moles of solute $n_2 = \frac{w_2}{M_2}$.

So, $m = \frac{w_2}{M_2 \cdot w_1(kg)}$.

Substituting this into the $\Delta T_b$ equation:

$\Delta T_b = K_b \cdot \frac{w_2}{M_2 \cdot w_1(kg)}$

This equation can be used to determine the molar mass ($M_2$) of an unknown non-volatile, non-electrolyte solute.

Example: Adding 1 mole of glucose (a non-electrolyte) to 1 kg of water will raise the boiling point of water by $K_b$ (for water, $K_b = 0.52$ K kg/mol). So, $\Delta T_b = 0.52$ K. The boiling point of pure water is 373.15 K (100°C), so the boiling point of the solution will be $373.15 + 0.52 = 373.67$ K.

3. Depression of Freezing Point ($\Delta T_f$)

The freezing point of a liquid is the temperature at which its solid and liquid phases are in equilibrium. When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lowered. This also lowers the freezing point of the solvent. The decrease in freezing point is called the depression of freezing point.

The depression of freezing point ($\Delta T_f$) is directly proportional to the molality ($m$) of the solution.

$\Delta T_f \propto m$

$\Delta T_f = K_f \cdot m$

Where:

• $\Delta T_f = T_{f, solvent}^\circ - T_{f, solution}$ is the depression in freezing point. (Note the order: pure solvent freezing point minus solution freezing point).
• $K_f$ is the cryoscopic constant (or molal freezing point depression constant) of the solvent. It is a characteristic property of the solvent. Its units are K kg/mol.
• $m$ is the molality of the solution.

Similar to boiling point elevation, if the molar mass of the solute is $M_2$ and its mass is $w_2$, and the mass of the solvent is $w_1(kg)$:

$\Delta T_f = K_f \cdot \frac{w_2}{M_2 \cdot w_1(kg)}$

This equation is also used to determine the molar mass of non-volatile, non-electrolyte solutes.

Example: Adding 1 mole of sugar (a non-electrolyte) to 1 kg of water will lower the freezing point of water by $K_f$ (for water, $K_f = 1.86$ K kg/mol). So, $\Delta T_f = 1.86$ K. The freezing point of pure water is 273.15 K (0°C), so the freezing point of the solution will be $273.15 - 1.86 = 271.29$ K. This is why salt is spread on icy roads; it lowers the freezing point of water, preventing ice formation or melting existing ice.

4. Osmotic Pressure ($\Pi$)

Osmosis is the spontaneous movement of solvent molecules from a region of higher solvent concentration (lower solute concentration) to a region of lower solvent concentration (higher solute concentration) through a semipermeable membrane. A semipermeable membrane allows solvent molecules to pass through but not solute particles.

Osmotic pressure ($\Pi$) is the minimum pressure that needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane. It can be thought of as the "pressure" that drives osmosis.

The osmotic pressure of a solution is directly proportional to the molarity ($C$ or $M$) of the solution.

$\Pi \propto C$

$\Pi = C \cdot R \cdot T$

Where:

• $\Pi$ is the osmotic pressure (usually in atm or Pa).
• $C$ is the molarity of the solution (moles of solute per litre of solution, $C = \frac{n_2}{V(L)}$).
• $R$ is the universal gas constant (e.g., 0.0821 L atm/mol K or 8.314 J/mol K).
• $T$ is the absolute temperature in Kelvin.

This equation is known as the van't Hoff equation for osmotic pressure. It is analogous to the ideal gas law, $PV = nRT$, where $\Pi V = nRT$, and $\frac{n}{V} = C$.

If the molar mass of the solute is $M_2$ and its mass is $w_2$, then $C = \frac{w_2}{M_2 \cdot V(L)}$.

So, $\Pi = \frac{w_2 R T}{M_2 V(L)}$.

This form of the equation is particularly useful for determining the molar mass of unknown solutes.

Isotonic, Hypotonic, and Hypertonic Solutions:

• Isotonic solutions: Solutions having the same osmotic pressure. If two solutions are isotonic, there is no net movement of solvent across a semipermeable membrane separating them.
• Hypotonic solution: A solution with a lower osmotic pressure compared to another solution.
• Hypertonic solution: A solution with a higher osmotic pressure compared to another solution.

Osmotic pressure is the most widely used colligative property for determining molar masses of biomolecules like proteins and polymers because these substances are often soluble only in small amounts, and using molality or molarity can be difficult.

van't Hoff Factor (i)

The colligative properties discussed above are based on the assumption that the solute does not dissociate or associate in the solution. However, many solutes, particularly electrolytes, dissociate into ions when dissolved in a solvent, while others may associate to form larger molecules. This changes the effective number of solute particles in the solution, thereby altering the colligative property.

The van't Hoff factor ($i$) is introduced to account for the extent of dissociation or association of a solute in a solution. It is defined as the ratio of the observed colligative property to the calculated colligative property assuming no dissociation or association. Alternatively, it is the ratio of the number of particles after dissociation/association to the number of formula units initially dissolved.

Definition

$i = \frac{\text{Observed colligative property}}{\text{Calculated colligative property (assuming no dissociation/association)}}$

Or,

$i = \frac{\text{Number of moles of particles in solution after dissociation/association}}{\text{Number of moles of solute dissolved}}$

Effect on Colligative Properties

The general equations for colligative properties need to be modified by multiplying by the van't Hoff factor ($i$) when dealing with electrolytes or associated solutes:

• Relative lowering of vapour pressure: $\frac{P_1^\circ - P_{solution}}{P_1^\circ} = i \cdot x_2$
• Elevation of boiling point: $\Delta T_b = i \cdot K_b \cdot m$
• Depression of freezing point: $\Delta T_f = i \cdot K_f \cdot m$
• Osmotic pressure: $\Pi = i \cdot C \cdot R \cdot T$

Dissociation of Electrolytes

Consider an electrolyte $AB$ that dissociates into $n$ ions in solution. For example, $NaCl$ dissociates into $Na^+$ and $Cl^-$ ions ($n=2$). $CaCl_2$ dissociates into $Ca^{2+}$ and $2Cl^-$ ions ($n=2$). $Al_2(SO_4)_3$ dissociates into $2Al^{3+}$ and $3SO_4^{2-}$ ions ($n=5$).

Let $\alpha$ be the degree of dissociation. Initially, we have 1 mole of the electrolyte. After dissociation, we have $(1-\alpha)$ moles of undissociated electrolyte and $n\alpha$ moles of ions.

Total moles of particles in solution = $(1-\alpha) + n\alpha = 1 + \alpha(n-1)$

So, the van't Hoff factor for dissociation is:

$i = \frac{\text{Total moles of particles}}{\text{Initial moles}} = \frac{1 + \alpha(n-1)}{1} = 1 + \alpha(n-1)$

Where $n$ is the number of ions produced per formula unit of the electrolyte.

Example: For $NaCl$, $n=2$. If $\alpha=1$ (complete dissociation), $i = 1 + 1(2-1) = 1 + 1 = 2$. This means 1 mole of $NaCl$ effectively gives 2 moles of particles ($Na^+$ and $Cl^-$). If $\alpha=0.9$, $i = 1 + 0.9(2-1) = 1.9$.

Association of Solutes

Some solutes associate in solution to form larger molecules. For example, carboxylic acids like acetic acid can dimerise in non-polar solvents due to hydrogen bonding.

Consider a solute $A$ that associates to form $n$ molecules of $(A)_n$. Let $\alpha$ be the degree of association. Initially, we have 1 mole of $A$. After association, we have $(1-\alpha)$ moles of undissociated $A$ and $\frac{\alpha}{n}$ moles of associated particles.

Total moles of particles in solution = $(1-\alpha) + \frac{\alpha}{n}$

So, the van't Hoff factor for association is:

$i = \frac{\text{Total moles of particles}}{\text{Initial moles}} = 1 - \alpha + \frac{\alpha}{n}$

Note that for association, $i$ is always less than 1.

Example: Acetic acid in benzene dimerises. $2CH_3COOH \rightleftharpoons (CH_3COOH)_2$. Here $n=2$. If $\alpha=0.8$, $i = 1 - 0.8 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6$.

Determination of Molecular Mass using van't Hoff Factor

When calculating the molar mass of an electrolyte or an associating solute using colligative properties, the van't Hoff factor must be incorporated into the formulas.

• From $\Delta T_b = i \cdot K_b \cdot \frac{w_2}{M_2 \cdot w_1(kg)}$, the molar mass $M_2 = \frac{i \cdot K_b \cdot w_2}{w_1(kg) \cdot \Delta T_b}$.
• From $\Delta T_f = i \cdot K_f \cdot \frac{w_2}{M_2 \cdot w_1(kg)}$, the molar mass $M_2 = \frac{i \cdot K_f \cdot w_2}{w_1(kg) \cdot \Delta T_f}$.
• From $\Pi = i \cdot C \cdot R \cdot T = i \cdot \frac{w_2 R T}{M_2 V(L)}$, the molar mass $M_2 = \frac{i \cdot w_2 R T}{V(L) \cdot \Pi}$.

These modified formulas allow for accurate determination of molar masses even when the solute undergoes dissociation or association.