Solutions

Solutions

Solutions are homogeneous mixtures of two or more components, meaning their composition and properties are uniform throughout. The component present in the largest quantity is the solvent, which determines the physical state of the solution. Other components are called solutes. This unit primarily considers binary solutions (two components).

• Examples of Mixtures:
  • Brass (copper and zinc)
  • German silver (copper, zinc, and nickel)
  • Bronze (copper and tin)
  • Fluoride ions in water (1 ppm prevents tooth decay, 1.5 ppm causes mottling, high concentrations are poisonous)
  • Intravenous injections contain salts dissolved in water at specific ionic concentrations matching blood plasma.
• Biological Significance: Almost all processes in the body occur in liquid solutions.

Types of Solutions

Solutions can be formed by mixing components in various physical states, as summarised below:

Expressing Concentration of Solutions

Solution composition can be described qualitatively (dilute or concentrated) or quantitatively. Quantitative description is crucial for real-life applications.

Quantitative methods include:

• Mass percentage (w/w)

  • Defined as: (Mass of the component in the solution / Total mass of the solution) × 100.
  • Example: 10% glucose in water by mass means 10g glucose in 90g water, making 100g solution.
  • Commonly used in industrial chemical applications (e.g., commercial bleaching solution has 3.62 mass % sodium hypochlorite).

• Volume percentage (V/V)

  • Defined as: (Volume of the component / Total volume of solution) × 100.
  • Example: 10% ethanol solution in water means 10mL ethanol dissolved in water to make 100mL total solution.
  • Commonly used for liquid solutions (e.g., 35% (v/v) ethylene glycol antifreeze lowers water freezing point to 255.4K).

• Mass by volume percentage (w/V)

  • Mass of solute dissolved in 100 mL of the solution.
  • Commonly used in medicine and pharmacy.

• Parts per million (ppm)

  • Used for trace quantities of solute.
  • Defined as: (Number of parts of the component / Total number of parts of all components of the solution) × 10^6.
  • Can be expressed as mass to mass, volume to volume, or mass to volume.
  • Example: 6 × 10⁻³ g of dissolved oxygen in 1030 g of seawater is 5.8 ppm.
  • Pollutant concentrations in water or atmosphere are often expressed in mg mL⁻¹ or ppm.

• Mole fraction (x)

  • Defined as: (Number of moles of the component / Total number of moles of all the components).
  • Symbol x with a subscript for the component (e.g., x_A for component A).
  • For a binary mixture (A and B): x_A = n_A / (n_A + n_B).
  • For i components: x_i = n_i / Σ n_i.
  • The sum of all mole fractions in a given solution is unity (x_1 + x_2 + … + x_i = 1).
  • Useful for relating physical properties like vapour pressure to concentration and in gas mixture calculations.
  • Example 1.1: Mole fraction of ethylene glycol (C₂H₆O₂) in 20% by mass solution: 0.068 (ethylene glycol), 0.932 (water).

• Molarity (M)

  • Defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
  • Formula: Molarity = Moles of solute / Volume of solution in litre.
  • Units: mol L⁻¹ or M.
  • Example 1.2: Molarity of 5g NaOH in 450mL solution is 0.278 M.

• Molality (m)

  • Defined as the number of moles of the solute per kilogram (kg) of the solvent.
  • Formula: Molality (m) = Moles of solute / Mass of solvent in kg.
  • Units: mol kg⁻¹ or m.
  • Example 1.3: Molality of 2.5g ethanoic acid in 75g benzene is 0.556 mol kg⁻¹.

• Temperature Dependence

  • Mass %, ppm, mole fraction, and molality are independent of temperature.
  • Molarity is a function of temperature because volume changes with temperature, whereas mass does not.

Solubility

Solubility is the maximum amount of a substance that can be dissolved in a specified amount of solvent at a specified temperature. It depends on the nature of the solute and solvent, temperature, and pressure.

• Solubility of a Solid in a Liquid

  • “Like dissolves like”: Polar solutes dissolve in polar solvents, and non-polar solutes dissolve in non-polar solvents, because intermolecular interactions are similar.
  • Dissolution: Solute dissolves, increasing its concentration.
  • Crystallisation: Solute particles in solution collide with solid solute and separate out.
  • Dynamic Equilibrium: When dissolution and crystallisation occur at the same rate, solute concentration becomes constant. (Solute + Solvent ⇌ Solution).
  • Saturated Solution: A solution in dynamic equilibrium with undissolved solute, containing the maximum amount of solute at given temperature and pressure.
  • Unsaturated Solution: More solute can be dissolved at the same temperature.
  • Effect of Temperature:
    • If dissolution is endothermic (Δsol H > 0), solubility increases with temperature.
    • If dissolution is exothermic (Δsol H < 0), solubility decreases with temperature.
    • This follows Le Chatelier’s Principle.
  • Effect of Pressure: Pressure has no significant effect on the solubility of solids in liquids because solids and liquids are highly incompressible.

• Solubility of a Gas in a Liquid

  • Effect of Pressure: Solubility of gases increases with an increase in pressure. Increased pressure over the solution leads to more gaseous particles striking and entering the solution phase until a new equilibrium is reached.

  • Henry’s Law: Quantitative relation between pressure and solubility of a gas.

    • States that at a constant temperature, the 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.
    • Most commonly used form: p = K_H x.
      • p = partial pressure of the gas in vapour phase.
      • x = mole fraction of the gas in the solution (measure of solubility).
      • K_H = Henry’s law constant.
    • Higher K_H value at a given pressure means lower solubility of the gas in the liquid.
    • K_H is a function of the nature of the gas.
    • Effect of Temperature (from K_H values): K_H values for N₂ and O₂ increase with temperature, indicating that solubility of gases decreases with increasing temperature. This is why aquatic species prefer colder waters. Dissolution of gas is considered an exothermic process.

  • Applications of Henry’s Law:

    • Soft drinks/soda water: Sealed under high pressure to increase CO₂ solubility.
    • Scuba diving: Divers breathe air at high pressure underwater, increasing dissolved gases in blood. As they ascend, pressure decreases, releasing N₂ bubbles (bends – painful and dangerous). To avoid this, scuba tanks are filled with air diluted with helium (11.7% He, 56.2% N₂, 32.1% O₂).
    • High altitudes (Anoxia): Lower partial pressure of oxygen at high altitudes leads to low O₂ concentration in blood and tissues, causing weakness and impaired thinking (anoxia).

  • Example 1.4: Calculating millimoles of N₂ dissolved in water using Henry’s Law.

    • Given: p(N₂) = 0.987 bar, K_H (N₂) = 76.48 kbar (76480 bar).
    • x(Nitrogen) = p(N₂) / K_H = 0.987 / 76480 = 1.29 × 10⁻⁵.
    • Moles of N₂ in 1 litre (55.5 mol) of water ≈ 7.16 × 10⁻⁴ mol or 0.716 mmol.

Vapour Pressure of Liquid Solutions

This section discusses solutions where the solvent is a liquid, and the solute can be a gas, liquid, or solid.

• Vapour Pressure of Liquid-Liquid Solutions (Raoult’s Law)

  • For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
  • For component 1: p_1 = p_1⁰ x_1.
    • p_1 = partial vapour pressure of component 1.
    • p_1⁰ = vapour pressure of pure component 1 at the same temperature.
    • x_1 = mole fraction of component 1 in solution.
  • Similarly for component 2: p_2 = p_2⁰ x_2.
  • Dalton’s Law of Partial Pressures: The total vapour pressure (p_total) over the solution is the sum of the partial pressures: p_total = p_1 + p_2.
  • Combining: p_total = x_1 p_1⁰ + x_2 p_2⁰.
  • The total vapour pressure over the solution varies linearly with the mole fraction of any one component.
  • Composition of Vapour Phase: If y_1 and y_2 are mole fractions in the vapour phase:
    • p_i = y_i p_total.
    • The vapour phase will be richer in the more volatile component.
  • Example 1.5: Calculating vapour pressure and vapour phase composition for chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) mixture.
    • Molar masses: CH₂Cl₂ = 85 g mol⁻¹, CHCl₃ = 119.5 g mol⁻¹.
    • Moles: CH₂Cl₂ = 0.47 mol, CHCl₃ = 0.213 mol.
    • Mole fractions in liquid: x_CH₂Cl₂ = 0.688, x_CHCl₃ = 0.312.
    • Given pure vapour pressures: p⁰_CHCl₃ = 200 mm Hg, p⁰_CH₂Cl₂ = 415 mm Hg.
    • Total Vapour Pressure: p_total = 200 + (415 - 200) × 0.688 = 347.9 mm Hg.
    • Partial pressures: p_CH₂Cl₂ = 285.5 mm Hg, p_CHCl₃ = 62.4 mm Hg.
    • Mole fractions in vapour phase: y_CH₂Cl₂ = 0.82, y_CHCl₃ = 0.18.

• Raoult’s Law as a Special Case of Henry’s Law

  • Comparing Raoult’s law (p_i = x_i p_i⁰) and Henry’s law (p = K_H x), it is seen that the partial pressure is proportional to mole fraction.
  • Raoult’s law is a special case of Henry’s law where the Henry’s law constant (K_H) becomes equal to the vapour pressure of the pure component (p_i⁰).

• Vapour Pressure of Solutions of Solids in Liquids

  • When a non-volatile solute is added to a solvent, the vapour pressure of the solution is lower than that of the pure solvent at the same temperature.
  • This is because the surface of the solution has both solute and solvent molecules, reducing the fraction of the surface covered by volatile solvent molecules, thus reducing the number of solvent molecules escaping.
  • The decrease in vapour pressure depends only on the quantity of non-volatile solute, irrespective of its nature (e.g., 1.0 mol sucrose or 1.0 mol urea lowers vapour pressure of 1kg water similarly).
  • General form of Raoult’s Law: For any solution, the partial vapour pressure of each volatile component is directly proportional to its mole fraction.
  • For a non-volatile solute (component 2) in a liquid solvent (component 1): p_1 = x_1 p_1⁰.

Ideal and Non-ideal Solutions

Solutions are classified based on their adherence to Raoult’s law.

• Ideal Solutions

  • Obey Raoult’s law over the entire range of concentration.
  • Enthalpy of mixing is zero (Δ_mix H = 0): No heat is absorbed or evolved during mixing.
  • Volume of mixing is zero (Δ_mix V = 0): The total volume of the solution is the sum of the volumes of components.
  • Molecular Interactions: Intermolecular attractive forces between A-A, B-B, and A-B types are nearly equal.
  • Examples: Solutions of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene. Perfectly ideal solutions are rare.

• Non-ideal Solutions

  • Do not obey Raoult’s law over the entire range of concentration.

  • Vapour pressure is either higher or lower than predicted by Raoult’s law.

  • Positive Deviation from Raoult’s Law

    • Vapour pressure is higher than predicted.
    • A-B interactions are weaker than A-A or B-B interactions. Molecules find it easier to escape into the vapour phase.
    • Examples:
      • Ethanol and acetone: Acetone breaks hydrogen bonds in pure ethanol, weakening interactions.
      • Carbon disulphide and acetone: Dipolar interactions between solute-solvent molecules are weaker.
    • Δ_mix H > 0 (Endothermic mixing): Heat is absorbed. (Implied by the source’s discussion on molecular interactions and deviation, common knowledge often links positive deviation to endothermic mixing, though not explicitly stated for Δ_mix H in this section for positive/negative deviation)
    • Δ_mix V > 0 (Volume expands): Components occupy more volume than expected. (Implied by weaker interactions, common knowledge)

  • Negative Deviation from Raoult’s Law

    • Vapour pressure is lower than predicted.
    • A-B interactions are stronger than A-A or B-B interactions. Decreased escaping tendency of molecules.
    • Examples:
      • Phenol and aniline: Stronger intermolecular hydrogen bonding between phenol and aniline molecules.
      • Chloroform and acetone: Chloroform forms hydrogen bonds with acetone.
    • Δ_mix H < 0 (Exothermic mixing): Heat is evolved. (Implied by the source’s discussion on molecular interactions and deviation, common knowledge)
    • Δ_mix V < 0 (Volume contracts): Components occupy less volume than expected. (Implied by stronger interactions, common knowledge)

• Azeotropes

  • Binary mixtures having the same composition in liquid and vapour phase and boiling at a constant temperature.
  • Cannot be separated by fractional distillation.
  • Minimum boiling azeotrope: Formed by solutions showing large positive deviation from Raoult’s law at a specific composition.
    • Example: Ethanol-water mixture (approx. 95% ethanol by volume).
  • Maximum boiling azeotrope: Formed by solutions showing large negative deviation from Raoult’s law at a specific composition.
    • Example: Nitric acid and water (approx. 68% nitric acid, 32% water by mass, boiling point 393.5 K).

Colligative Properties and Determination of Molar Mass

Colligative properties depend on the number of solute particles irrespective of their nature relative to the total number of particles in solution. These properties are connected to the decrease of vapour pressure when a non-volatile solute is added to a volatile solvent.

The four colligative properties are:

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

• Relative Lowering of Vapour Pressure

  • When a non-volatile solute is added, the vapour pressure of the solvent (p₁) is less than that of the pure solvent (p₁⁰).
  • Raoult’s law: p₁ = x₁ p₁⁰.
  • Lowering of vapour pressure (Δp₁): Δp₁ = p₁⁰ - p₁ = p₁⁰ (1 - x₁) = x₂ p₁⁰.
  • Relative lowering of vapour pressure: (p₁⁰ - p₁) / p₁⁰ = x₂.
    • This expression equals the mole fraction of the solute (x₂).
    • For dilute solutions (n₂ << n₁), this simplifies to: (p₁⁰ - p₁) / p₁⁰ ≈ n₂ / n₁.
    • In terms of masses and molar masses: (p₁⁰ - p₁) / p₁⁰ = (w₂/M₂) × (M₁/w₁).
  • Used to calculate the molar mass of the solute (M₂).
  • Example 1.6: Molar mass of a solid calculated from lowering of benzene vapour pressure.
    • M₂ = (K*b × 1000 × w₂) / (ΔT_b × w₁) (Typo in source, this is formula for elevation of boiling point. For relative lowering, use the one derived above: M₂ = (w₂ * M₁) / (w₁ _ ((p₁⁰ - p₁) / p₁⁰))).
    • Using the correct formula from source 1.28: M₂ = (w₂ _ M₁ _ p₁⁰) / (w₁ * (p₁⁰ - p)).
    • M₂ = (0.5 g × 78 g mol⁻¹ × 0.850 bar) / (39 g × (0.850 bar - 0.845 bar)) = 170 g mol⁻¹.

• Elevation of Boiling Point

  • The boiling point of a solution containing a non-volatile solute is always higher than that of the pure solvent. This is because the vapour pressure of the solution is lower, so a higher temperature is needed for it to equal atmospheric pressure.
  • Elevation of boiling point (ΔT_b): ΔT_b = T_b - T_b⁰.
    • T_b⁰ = boiling point of pure solvent.
    • T_b = boiling point of solution.
  • For dilute solutions, ΔT_b is directly proportional to the molal concentration (m) of the solute.
    • ΔT_b = K_b m.
    • K_b = Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).
    • Unit of K_b is K kg mol⁻¹.
  • Molar mass calculation: M₂ = (K_b × 1000 × w₂) / (ΔT_b × w₁).
  • Example 1.7: Boiling point of water with glucose dissolved.
    • Molality of glucose = 0.1 mol kg⁻¹.
    • ΔT_b = K_b × m = 0.52 K kg mol⁻¹ × 0.1 mol kg⁻¹ = 0.052 K.
    • Boiling point of solution = 373.15 K (water) + 0.052 K = 373.202 K.
  • Example 1.8: Molar mass from boiling point elevation.
    • ΔT_b = 354.11 K - 353.23 K = 0.88 K.
    • M₂ = (2.53 K kg mol⁻¹ × 1.80 g × 1000 g kg⁻¹) / (0.88 K × 90 g) = 58 g mol⁻¹.

• Depression of Freezing Point

  • The freezing point of a solution containing a non-volatile solute is lower than that of the pure solvent. This is because the vapour pressure of the solution becomes equal to that of the pure solid solvent at a lower temperature.
  • Freezing point (T_f): Temperature at which vapour pressure of liquid phase equals vapour pressure of solid phase.
  • Depression in freezing point (ΔT_f): ΔT_f = T_f⁰ - T_f.
    • T_f⁰ = freezing point of pure solvent.
    • T_f = freezing point of solution.
  • For dilute solutions, ΔT_f is directly proportional to the molality (m) of the solution.
    • ΔT_f = K_f m.
    • K_f = Freezing Point Depression Constant or Molal Depression Constant (Cryoscopic Constant).
    • Unit of K_f is K kg mol⁻¹.
  • Molar mass calculation: M₂ = (K_f × 1000 × w₂) / (ΔT_f × w₁).
  • Constants K_f and K_b can be calculated from properties of the solvent:
    • K_f = (R M₁ T_f² ) / (1000 Δ_fus H).
    • K_b = (R M₁ T_b² ) / (1000 Δ_vap H).
      • R = gas constant.
      • M₁ = molar mass of solvent.
      • T_f, T_b = freezing/boiling point of pure solvent in Kelvin.
      • Δ_fus H, Δ_vap H = enthalpies of fusion/vaporisation of solvent.
  • Example 1.9: Freezing point of ethylene glycol solution.
    • Molality of ethylene glycol = 1.2 mol kg⁻¹.
    • ΔT_f = 1.86 K kg mol⁻¹ × 1.2 mol kg⁻¹ = 2.2 K.
    • Freezing point of solution = 273.15 K (water) - 2.2 K = 270.95 K.
  • Example 1.10: Molar mass from freezing point depression.
    • M₂ = (5.12 K kg mol⁻¹ × 1.00 g × 1000 g kg⁻¹) / (0.40 K × 50 g) = 256 g mol⁻¹.

• Osmosis and Osmotic Pressure

  • Semipermeable Membranes (SPM): Membranes with submicroscopic holes that allow small solvent molecules to pass through but hinder larger solute molecules.
  • Osmosis: The process of flow of solvent molecules through a semipermeable membrane from the pure solvent side to the solution side.
    • Flow continues until equilibrium.
    • Solvent molecules always flow from lower concentration to higher concentration of solution.
  • Osmotic Pressure (P or Π): The excess pressure that must be applied to a solution to prevent osmosis (to stop solvent flow into the solution through an SPM).
    • It is a colligative property as it depends on the number of solute molecules, not their identity.
    • For dilute solutions, osmotic pressure is proportional to the molarity (C) of the solution at a given temperature (T).
    • P = C R T (or Π = C R T).
    • Can also be written as: P = (n₂/V) R T.
      • n₂ = moles of solute.
      • V = volume of solution in litres.
      • R = gas constant.
    • Molar mass calculation: M₂ = (w₂ R T) / (P V).
      • w₂ = mass of solute.
      • M₂ = molar mass of solute.
    • Advantages for molar mass determination:
      • Pressure measurement is typically done around room temperature, beneficial for biomolecules unstable at higher temperatures.
      • Uses molarity, not molality, simplifying measurements.
      • Its magnitude is large even for very dilute solutions.
      • Useful for polymers due to their poor solubility.
  • Isotonic Solutions: Two solutions with the same osmotic pressure at a given temperature; no osmosis occurs between them when separated by an SPM.
    • Example: 0.9% (mass/volume) sodium chloride solution (normal saline) is isotonic with blood plasma.
  • Hypertonic Solution: Solution with higher salt concentration (e.g., >0.9% NaCl). Water flows out of cells placed in it, causing them to shrink. (Raw mangoes shrivel in brine).
  • Hypotonic Solution: Solution with lower salt concentration (e.g., <0.9% NaCl). Water flows into cells placed in it, causing them to swell. (Wilted flowers revive in fresh water, blood cells swell).
  • Edema: Swelling caused by water retention in tissue cells due to osmosis, often from high salt intake.
  • Biological applications: Water movement in plants, food preservation (salting meat, sugaring fruits).
  • Example 1.11: Molar mass of a protein from osmotic pressure.
    • P = 2.57 × 10⁻³ bar, V = 0.200 L, T = 300 K, R = 0.083 L bar mol⁻¹ K⁻¹.
    • M₂ = (1.26 g × 0.083 L bar mol⁻¹ K⁻¹ × 300 K) / (2.57 × 10⁻³ bar × 0.200 L) = 61,022 g mol⁻¹.

• Reverse Osmosis (RO) and Water Purification

  • Reverse Osmosis: The phenomenon where pure solvent flows out of the solution through a semipermeable membrane when a pressure larger than the osmotic pressure is applied to the solution side.
  • Practical Utility: Used in desalination of seawater.
  • Membrane: A porous membrane like cellulose acetate is used, which is permeable to water but impermeable to impurities and ions.

Abnormal Molar Masses

The molar mass of a solute determined experimentally using colligative properties can be different from its normal (theoretical) molar mass due to dissociation or association in solution.

• Dissociation: When solute particles break into more particles (ions) in solution.

  • Example: KCl dissociates into K⁺ and Cl⁻ ions.
  • Results in a lower experimentally determined molar mass than the true value because the number of particles in solution increases.

• Association: When solute particles combine to form larger particles (e.g., dimers).

  • Example: Ethanoic acid (acetic acid) dimerises in benzene due to hydrogen bonding, especially in low dielectric constant solvents.
  • Results in a higher experimentally determined molar mass than the true value because the number of particles in solution decreases.

• van’t Hoff Factor (i)

  • Introduced by van’t Hoff (1880) to account for the extent of dissociation or association.

  • Defined in three ways:

    1. i = Normal molar mass / Abnormal molar mass.
    2. i = Observed colligative property / Calculated colligative property.
    3. i = Total number of moles of particles after association/dissociation / Number of moles of particles before association/dissociation.

  • Interpretation of i:

    • i < 1 for association (e.g., ethanoic acid in benzene, i ≈ 0.5).
    • i > 1 for dissociation (e.g., aqueous KCl solution, i ≈ 2).
    • i = 1 for non-dissociating/non-associating solutes (ideal behaviour).

  • Modified Colligative Property Equations (with i):

    • Relative lowering of vapour pressure: (p₁⁰ - p₁) / p₁⁰ = i n₂ / (n₁ + i n₂) or simplified to i n₂ / n₁ for dilute solutions.
    • Elevation of Boiling point: ΔT_b = i K_b m.
    • Depression of Freezing point: ΔT_f = i K_f m.
    • Osmotic pressure of solution: P = i n₂ R T / V.

  • Example 1.12: Percentage association of benzoic acid in benzene.

    • Experimental molar mass (M₂) from ΔT_f = 241.98 g mol⁻¹.
    • Normal molar mass of benzoic acid (C₆H₅COOH) = 122 g mol⁻¹.
    • Calculated i = Normal Molar Mass / Experimental Molar Mass = 122 / 241.98 = 0.504.
    • For dimerization (2A ⇌ A₂), if x is degree of association, total moles = 1 - x/2.
    • i = 1 - x/2 (from total moles).
    • Solving for x: 0.504 = 1 - x/2 => x/2 = 0.496 => x = 0.992.
    • Percentage association = 99.2%.

  • Example 1.13: van’t Hoff factor and dissociation constant of acetic acid.

    • Calculated ΔT_f (assuming no dissociation) = K_f × molality = 1.86 K kg mol⁻¹ × 0.0106 mol kg⁻¹ = 0.0197 K.
    • Observed ΔT_f = 0.0205 °C (or K).
    • van’t Hoff factor (i) = Observed ΔT_f / Calculated ΔT_f = 0.0205 K / 0.0197 K = 1.041.
    • For dissociation (CH₃COOH ⇌ H⁺ + CH₃COO⁻), if x is degree of dissociation, total moles = n(1+x).
    • i = n(1+x) / n = 1+x.
    • Solving for x: 1.041 = 1 + x => x = 0.041.
    • Dissociation constant (K_a) = ([H⁺][CH₃COO⁻]) / [CH₃COOH] = (nx * nx) / (n(1-x)).
    • K_a = (0.0106 × 0.041)² / (0.0106 × (1 - 0.041)) = 1.86 × 10⁻⁵.