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Solutions and Their Concentration — Notes

5 sections 5 min readBeta

1. Types of solutions

A solution is a homogeneous mixture of two or more substances. The component in larger amount is the solvent; the others are solutes. By the states of solvent and solute there are nine types: gaseous (air; humid air; camphor in air), liquid (soda water; ethanol in water; salt solution) and solid solutions (hydrogen in palladium; sodium amalgam; alloys such as brass, bronze and carat gold). A binary solution has two components.

Table of gaseous, liquid and solid solutions with examples
SolventExamples
GasAir, humid air
LiquidSoda water, salt solution, ethanol in water
SolidBrass, sodium amalgam, H₂ in Pd

Key terms

Solution

A homogeneous mixture of two or more components.

Solid solution

A solution in which the solvent is a solid, e.g. an alloy such as brass.

2. Concentration terms

  • Mass % =mass of componentmass of solution×100= \dfrac{\text{mass of component}}{\text{mass of solution}} \times 100; volume %; w/v % (g per 100 mL).
  • ppm =mass of componentmass of solution×106= \dfrac{\text{mass of component}}{\text{mass of solution}} \times 10^{6} (1 ppm = 1 mg per kg).
  • Mole fraction xB=nB/(nA+nB)x_\text{B} = n_\text{B}/(n_\text{A} + n_\text{B}); ∑x=1\sum x = 1.
  • Molarity MM = mol of solute per L of solution; molality mm = mol of solute per kg of solvent.
  • Normality NN = gram equivalents of solute per L of solution; equivalent mass = molar mass ÷ n-factor (basicity, acidity, electrons transferred), so N=n×MN = n \times M. At an end point, N1V1=N2V2N_1V_1 = N_2V_2.
  • Mass %, ppm, mole fraction and molality are independent of temperature; molarity and normality fall slightly on heating because the volume of solution increases.
  • Conversions: assume 100 g of solution for a percentage; use density to get volume; molarity ↔ molality: m=1000M1000d−MMBm = \dfrac{1000M}{1000d - M M_\text{B}} (d in g mL⁻¹, MBM_\text{B} = molar mass of solute).
Table of concentration terms and whether they depend on temperature
TermTemperature dependent?
Mass %, ppmNo
Mole fractionNo
MolalityNo
Molarity, normalityYes

Formulas

  • Molarity
    M=nsoluteVsolution/LM = \frac{n_\text{solute}}{V_\text{solution}/\text{L}}
  • Molality
    m=nsolutemsolvent/kgm = \frac{n_\text{solute}}{m_\text{solvent}/\text{kg}}
  • Normality from molarity
    N=n×MN = n \times M
  • Normality equation
    N1V1=N2V2N_1V_1 = N_2V_2

Key terms

Molality

Moles of solute per kilogram of solvent; independent of temperature.

Normality

Number of gram equivalents of solute per litre of solution.

Equivalent mass

Molar mass divided by the n-factor of the substance in the reaction.

ppm

Parts of a component per million parts of the solution by mass.

3. Solubility of gases; Henry’s law

Solubility of a gas in a liquid increases with pressure and decreases with temperature (dissolving is usually exothermic).

Henry’s law: at constant temperature, the solubility of a gas is proportional to its partial pressure; p=KH xp = K_\text{H}\,x. Larger KHK_\text{H} → less soluble; KHK_\text{H} increases with temperature.

Applications: soft drinks bottled under high CO₂ pressure; the bends in divers (tanks use air diluted with helium); anoxia at high altitude; low oxygen in warm water.

Limitations: holds for low pressures, gases that do not react with or dissociate in the solvent (not NH₃ or HCl in water), and dilute solutions.

Schematic. Particles are drawn far larger than real molecules.

Formulas

  • Henry’s law
    p=KH xp = K_\text{H}\,x

Key terms

Henry’s law

At constant temperature the partial pressure of a gas over a solution is proportional to its mole fraction in the solution.

The bends

Pain caused by nitrogen bubbles forming in the blood when a diver surfaces too quickly.

4. Raoult’s law; ideal and non-ideal solutions

Raoult’s law: for a solution of volatile liquids, the partial vapour pressure of each component equals its pure vapour pressure times its mole fraction: pA=pA∘xAp_\text{A} = p^\circ_\text{A}x_\text{A}; ptotal=pA+pBp_\text{total} = p_\text{A} + p_\text{B} (Dalton). Vapour composition: yA=pA/ptotaly_\text{A} = p_\text{A}/p_\text{total}; the vapour is richer in the more volatile component. For a non-volatile solute, p=p1∘x1p = p^\circ_1 x_1 (vapour pressure lowered). Raoult’s law is a special case of Henry’s law with KH=p∘K_\text{H} = p^\circ.

IdealPositive deviationNegative deviation
A–B forces≈ A–A, B–Bweakerstronger
Vapour pressureas Raoulthigherlower
ΔmixH\Delta_\text{mix}H0> 0< 0
ΔmixV\Delta_\text{mix}V0> 0< 0
Exampleshexane + heptane; benzene + tolueneethanol + water; acetone + CS₂chloroform + acetone; HNO₃ + water
  • Partial pressure of A
  • Partial pressure of B
  • Total vapour pressure
Illustrative values: pure A 24 kPa, pure B 64 kPa at the same temperature.
  • Ideal (Raoult)
  • Positive deviation
  • Negative deviation
Illustrative curves. A large enough deviation gives a maximum or a minimum in total pressure: an azeotrope.

Formulas

  • Raoult’s law
    pA=pA∘ xAp_\text{A} = p^\circ_\text{A}\,x_\text{A}
  • Total vapour pressure
    ptotal=pA∘+(pB∘−pA∘) xBp_\text{total} = p^\circ_\text{A} + (p^\circ_\text{B} - p^\circ_\text{A})\,x_\text{B}

Key terms

Ideal solution

A solution that obeys Raoult’s law at all compositions, with zero enthalpy and volume of mixing.

Positive deviation

Vapour pressure higher than Raoult’s law predicts because A–B attractions are weaker.

5. Azeotropes and fractional distillation

An azeotrope is a liquid mixture that boils at constant temperature and distils with unchanged composition (liquid and vapour have the same composition).

  • Minimum-boiling azeotrope: from large positive deviation; e.g. ethanol + water (about 95% ethanol by volume, about 351 K).
  • Maximum-boiling azeotrope: from large negative deviation; e.g. nitric acid + water (about 68% HNO₃, about 393.5 K), HCl + water (about 20% HCl, about 383 K).

Boiling point–composition diagrams: the liquid and vapour curves meet at the azeotrope (minimum or maximum). Fractional distillation of a mixture gives one pure component and the azeotrope — never both pure components — because at the azeotrope the vapour has the same composition as the liquid.

  • Liquid (boiling) curve
  • Vapour (condensing) curve
Illustrative curves. At the minimum the liquid and vapour have the same composition.

Key terms

Azeotrope

A constant-boiling mixture whose vapour has the same composition as the liquid.

Fractional distillation

Separation of volatile liquids by repeated evaporation and condensation in a fractionating column.

Exam tips

Common mistakes

  • Dividing by the mass of solution instead of the mass of solvent for molality.
  • Treating 1 M H₂SO₄ as 1 N (it is 2 N).
  • Saying a larger Henry’s constant means a more soluble gas.
  • Using liquid mole fractions for the vapour instead of y = pA/ptotal.
  • Linking positive deviation with a maximum-boiling azeotrope.

How marks are usually awarded

  • Numericals: formula, substitution with units, and a final answer with correct units (full working expected).
  • Deviation questions: the change in intermolecular forces plus the signs of ΔmixH and ΔmixV, with an example.
  • Azeotrope questions: type, cause, an example, and a labelled boiling point–composition sketch.