Solutions and Colligative Properties

Fundamental Definition and Nature of Solutions

  • A solution is a homogenous mixture of two or more substances with variable compositions.
  • Mixtures exhibiting more than one phase are classified as heterogeneous (e.g., sand and water).
  • A solution consists of a solvent (component in large proportion) and a solute (component in small proportion).
  • Common examples include air (oxygen in nitrogen), soft drinks (CO2CO_2, sugar, and flavouring in water), and alloys.

Classification of Solutions

  • Solutions are classified by the number of components: binary (two), ternary (three), and quaternary (four).
  • Solutions exist in three physical states: Gaseous, Solid, and Liquid.
  • Gaseous solutions: All gas mixtures are homogeneous (e.g., Air). Components can be separated by phase changes such as low temperature distillation.
  • Solid solutions: Common in minerals and gemstones like emeralds, sapphires, rubies, and amethysts. Examples include alloys such as Brass (zinc-copper) and Bronze (tin and copper).
  • Liquid solutions: Formed by dissolving gases, liquids, or solids in a liquid solvent.

Factors Influencing Solubility of Gases in Liquids

  • Nature of substances: Highly soluble gases include Ammonia and Sulphur(iv)oxide in water.
  • Temperature: Solubility of gases decreases with an increase in temperature. This is relevant in degassing processes and aquatic ecology (e.g., eutrophication reducing dissolved oxygen).
  • Pressure: Henry's Law states that the amount of gas dissolved at a constant temperature is proportional to its pressure: Solubility∝Pressure\text{Solubility} \propto \text{Pressure}.

Methods of Expressing Concentration

  • Molarity (MM): Number of moles of solute per dm3dm^3 (litre) of solution.
  • Molality (mm): Number of moles of solute dissolved in 1000 g1000\,g (1 kg1\,kg) of solvent.
  • Normality (NN): Gramme-equivalent weight of solute per dm3dm^3 of solution. For H2SO4H_2SO_4, 1.0 g1.0\,g equivalent is calculated as 98.022=49.04 g\frac{98.02}{2} = 49.04\,g.
  • Mole Fraction (XX): Ratio of the moles of one component to the total moles in the solution. For component 1: X1=n1n1+n2X_1 = \frac{n_1}{n_1 + n_2}.
  • Weight Percent: Grams of solute per 100 g100\,g of solution.

Colligative Properties: Vapour Pressure and Raoult’s Law

  • Colligative properties depend solely on the number of solute particles, not their identity.
  • Vapour Pressure Lowering: Adding a non-volatile solute reduces the saturated vapour pressure because solute molecules exert an inward pull and block the escape route of solvent molecules.
  • Raoult's Law: The relative lowering of vapour pressure is proportional to the mole fraction of the solute: Po−PPo=nN+n=Xn\frac{P^o - P}{P^o} = \frac{n}{N + n} = X_n.
  • Vapour pressure of the solution (PP) is directly proportional to the mole fraction of the solvent (XNX_N): P=Po×XNP = P^o \times X_N.
  • Molecular weight (M2M_2) can be determined using Po−PPo=m2×M1m1×M2\frac{P^o - P}{P^o} = \frac{m_2 \times M_1}{m_1 \times M_2} for dilute solutions.

Boiling Point Elevation and Freezing Point Depression

  • Elevation of Boiling Point (ΔTb\Delta T_b): A solution boils at a higher temperature than the pure solvent (101325 Nm−2101325\,Nm^{-2}). The rise is proportional to molality: ΔTb=Ke×(1000×m2M2×m1)\Delta T_b = K_e \times (\frac{1000 \times m_2}{M_2 \times m_1}).
  • KeK_e is the molal elevation constant or ebullioscopic constant.
  • Depression of Freezing Point (ΔTf\Delta T_f): A solute lowers the freezing point of the solvent. The depression is proportional to the relative lowering of vapour pressure: ΔTf=Kf×(1000×m2M2×m1)\Delta T_f = K_f \times (\frac{1000 \times m_2}{M_2 \times m_1}).
  • KfK_f is the molal depression constant or cryoscopic constant.

Osmosis and Osmotic Pressure

  • Osmosis is the movement of solvent molecules through a semi-permeable membrane from a region of lower solute concentration (higher vapour pressure) to higher solute concentration (lower vapour pressure).
  • Osmotic Pressure (π\pi) is the hydrostatic pressure required to stop osmosis.
  • Van't Hoff Equation: π=nRTV=m2RTM2V\pi = \frac{nRT}{V} = \frac{m_2 RT}{M_2 V}, where R=8.314 JK−1mol−1R = 8.314\,JK^{-1}mol^{-1}.
  • Applications: Determination of molecular weights for polymers (polystyrene, Polypropylene, starch, proteins) and desalinization of sea water via reverse osmosis.
  • Relation to Vapour Pressure: For dilute solutions, P−P1P=Kπ\frac{P - P^1}{P} = K\pi.