Colligative Properties

Colligative properties:

• Depend on the number of solute particles present, not their identity
• Include properties like boiling point elevation, freezing point depression, vapor-pressure lowering, and osmotic pressure

Equation for Depression of Freezing Point

The equation for depression of freezing point is: ΔTf = i * Kf * m Where:

• ΔTf is the change in freezing point
• i is the van’t Hoff factor (number of particles formed when solute dissolves)
• Kf is the cryoscopic constant (specific for each solvent)
• m is the molality of the solution (moles of solute per kilogram of solvent)

Example: Calculating ΔTf

Let’s consider an example: A solution of 1.0 mol of glucose (C6H12O6) dissolved in 1.0 kg of water has a ΔTf of -1.86°C. Calculate the van’t Hoff factor (i). Solution: Given:

• ΔTf = -1.86°C
• Kf (for water) = 1.86 °C/m
• m (molality) = 1.0 mol/kg Substituting the values in the equation ΔTf = i * Kf * m, we can solve for i.

Solution to Example

ΔTf = i * Kf * m -1.86 = i * 1.86 * 1.0 i = -1.86 / (1.86 * 1.0) i = -1 Therefore, the van’t Hoff factor is -1.

Relationship Between Moles and Molality

The relationship between moles and molality is given by: m = (moles of solute) / (mass of solvent in kg) This equation allows us to calculate molality using the known values for moles and mass of solvent.

Example: Calculating Molality

Let’s consider an example: A solution contains 0.50 moles of sodium chloride (NaCl) dissolved in 1.5 kg of water. Calculate the molality (m) of the solution. Solution: Given:

• Moles of NaCl = 0.50 mol
• Mass of water = 1.5 kg Using the equation m = (moles of solute) / (mass of solvent in kg), we can calculate the molality.

Solution to Example

m = (moles of solute) / (mass of solvent in kg) m = 0.50 / 1.5 m = 0.33 mol/kg Therefore, the molality of the solution is 0.33 mol/kg.

Importance of Depression of Freezing Point

The depression of freezing point has several practical applications:

1. Antifreeze for car engines: Adding antifreeze to the radiator water lowers its freezing point, preventing it from solidifying in low temperatures.

2. Ice cream making: Salt is added to ice to lower its freezing point, creating a freezing environment for making ice cream.

3. Medical applications: Cryopreservation techniques use the depression of freezing point to preserve biological samples such as sperm, eggs, and tissues.

Conclusion

• Depression of freezing point is a colligative property that is based on the principle that the freezing point of a solvent is lowered when a non-volatile solute is added.
• The equation ΔTf = i * Kf * m is used to calculate the change in freezing point.
• The van’t Hoff factor (i) represents the number of particles formed when the solute dissolves.
• Molality (m) is the moles of solute per kilogram of solvent.
• Depression of freezing point has practical applications and is widely used in various fields.

Properties of Solutions

• Solutions are homogeneous mixtures of two or more substances.
• They have two components: the solvent and the solute.
• The solute is present in lesser amounts and is dissolved in the solvent.
• Solutions can be categorized into various types based on the state of the solvent and solute.

Types of Solutions

1. Solid Solution

• The solute is in the solid state, and the solvent is also usually a solid.
• Examples include alloys such as bronze (copper and tin) and brass (copper and zinc).

2. Liquid Solution

• The solute is a liquid, and the solvent is also a liquid.
• Example: Ethanol dissolved in water to make alcoholic beverages.

3. Gas Solution

• Both the solute and the solvent are in the gaseous state.
• Example: Air, which is a mixture of various gases.

Concentration of Solutions

• Concentration refers to the amount of solute present in a given amount of solvent or solution.
• It can be expressed in different ways, including:
  • Mass percent: Mass of solute / Mass of solution * 100%
  • Molarity (M): Moles of solute / Volume of solution (in liters)
  • Molality (m): Moles of solute / Mass of solvent (in kilograms)

Example: Calculating Concentration

Let’s consider an example: What is the molarity of a solution containing 0.5 moles of sodium hydroxide (NaOH) dissolved in 0.25 liters of water? Solution: Given:

• Moles of NaOH = 0.5 mol
• Volume of solution = 0.25 L Using the equation Molarity (M) = Moles of solute / Volume of solution, we can calculate the molarity.

Solution to Example

Molarity (M) = Moles of solute / Volume of solution M = 0.5 mol / 0.25 L M = 2 mol/L Therefore, the molarity of the solution is 2 mol/L.

Factors Affecting Solubility

Several factors influence the solubility of solutes in solvents:

1. Temperature:

• In general, the solubility of solid solutes increases with an increase in temperature.
• However, the solubility of gases typically decreases with an increase in temperature.

2. Pressure:

• For gases, an increase in pressure leads to an increase in solubility.
• This is described by Henry’s Law: Solubility (gas) ∝ Pressure (gas)

Solubility Curves

• Solubility curves represent the solubility of a particular solute at different temperatures.
• The curve shows how the solubility changes with temperature.
• Each solute has a unique solubility curve. Example: Solubility Curve of Potassium Nitrate (KNO3)
• The x-axis represents temperature (°C) and the y-axis represents grams of solute dissolved in 100 grams of water.
• Each point on the curve indicates the solubility of KNO3 at a specific temperature.

Using Solubility Curves

• Solubility curves can be used to determine whether a given solution is saturated, unsaturated, or supersaturated.
• If the concentration of a solute lies on the curve, the solution is saturated.
• If the concentration is below the curve, the solution is unsaturated.
• If the concentration is above the curve, the solution is supersaturated. Example:
• If the solubility curve of KNO3 indicates that 50 grams are soluble in 100 grams of water at a specific temperature, and the solution contains 60 grams of KNO3, it is supersaturated.

Colligative Properties

Colligative properties are properties that depend on the number of solute particles present, not their identity. They include:

1. Boiling Point Elevation

• The boiling point of a solvent is increased when a non-volatile solute is added.
• This is used in the antifreeze properties of solutions.

2. Freezing Point Depression (discussed earlier)

• The freezing point of a solvent is lowered when a non-volatile solute is added.

3. Vapor-Pressure Lowering

• The vapor pressure of a solvent is decreased when a solute is added.
• This affects the rate of evaporation and boiling point.

Osmosis and Osmotic Pressure

• Osmosis is the movement of solvent molecules across a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration.
• Osmotic pressure is the pressure required to prevent osmosis and is dependent on the concentration of solute particles.
• Osmosis plays a vital role in various biological systems and has applications in medicine and industry.

Applications of Depression of Freezing Point

• Depression of freezing point is utilized in the production of artificial snow for winter sports.
• It is also used in the preservation of food by freezing, which helps to extend its shelf life.
• Depression of freezing point is employed in the production of ice creams and frozen desserts.

Example: Calculating Molality

Let’s consider an example: A solution contains 0.20 moles of urea (CH4N2O) dissolved in 250 grams of water. Calculate the molality (m) of the solution. Solution: Given:

• Moles of urea (CH4N2O) = 0.20 mol
• Mass of water = 250 g Using the equation m = (moles of solute) / (mass of solvent in kg), we can calculate the molality.

Solution to Example

m = (moles of solute) / (mass of solvent in kg) m = 0.20 mol / 0.250 kg m = 0.80 mol/kg Therefore, the molality of the solution is 0.80 mol/kg.

Example: Calculating ΔTf

Let’s consider an example: A solution of 2.5 moles of potassium chloride (KCl) dissolved in 500 grams of water has a ΔTf of -3.2°C. Calculate the van’t Hoff factor (i). Solution: Given:

• ΔTf = -3.2°C
• Kf (for water) = 1.86 °C/m
• m (molality) = 2.5 mol / 0.500 kg Substituting the values in the equation ΔTf = i * Kf * m, we can solve for i.

Solution to Example

ΔTf = i * Kf * m -3.2 = i * 1.86 * (2.5 / 0.500) -3.2 = i * 1.86 * 5 i = -3.2 / (1.86 * 5) i = -0.345 Therefore, the van’t Hoff factor is approximately -0.345.

Properties of Solutions

• Solutions have physical properties that differ from those of the pure solvent or solute.
• The properties of solutions include:
  • Density
  • Viscosity
  • Conductivity
  • pH (in the case of acidic or basic solutions)
  • Color (in the presence of colored solutes)

Factors Affecting Solubility

Various factors influence the solubility of solutes in solvents:

1. Nature of the solute and solvent:

• Polar solvents dissolve polar solutes, while nonpolar solvents dissolve nonpolar solutes.
• Like dissolves like.

2. Temperature:

• Generally, an increase in temperature increases the solubility of solid solutes.
• However, for some solutes, solubility may decrease with temperature.

3. Pressure:

• For solids and liquids, pressure has a negligible effect on solubility.
• But for gases, increased pressure leads to increased solubility.

Solubility Product (Ksp)

• The solubility product (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt.
• It is the product of the concentrations of the ions raised to their stoichiometric coefficients in the balanced equation for the dissolution.
• The Ksp expression does not include the solid salt. Example: The solubility product expression for the dissolution of AgCl is given by Ksp = [Ag⁺][Cl⁻].

Example: Calculating Solubility Product (Ksp)

Let’s consider an example: Given the solubility product (Ksp) of calcium fluoride (CaF2) is 3.4 × 10⁻¹¹. If the concentration of calcium ions is 1.2 × 10⁻⁴ M in a saturated solution of CaF2, calculate the concentration of fluoride ions. Solution: Given:

• Ksp = 3.4 × 10⁻¹¹
• [Ca²⁺] = 1.2 × 10⁻⁴ M Using the Ksp expression, we can solve for [F⁻].

Solution to Example

Ksp = [Ca²⁺][F⁻]

3.4 × 10⁻¹¹ = (1.2 × 10⁻⁴)[F⁻] [F⁻] = (3.4 × 10⁻¹¹) / (1.2 × 10⁻⁴) [F⁻] = 2.83 × 10⁻⁷ M Therefore, the concentration of fluoride ions in the saturated solution is 2.83 × 10⁻⁷ M.