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Solutions — Notes

6 sections 10 min readBeta

1. Types of solutions and ways of expressing concentration

A solution is a homogeneous mixture of two or more components. The component in the larger amount is the solvent and decides the physical state; the others are solutes. A binary solution has two components. Nine types are possible (gas, liquid or solid solute in a gas, liquid or solid solvent), e.g. air, soda water, sugar syrup, brass, hydrogen in palladium.

Concentration can be stated in several ways:

  • Mass percentage =mass of componentmass of solution×100= \dfrac{\text{mass of component}}{\text{mass of solution}} \times 100; volume percentage similarly with volumes; mass by volume percentage = grams of solute per 100 mL of solution.
  • Parts per million (ppm) for traces: parts of component per 10610^{6} parts of solution (by mass, volume or mass/volume).
  • Mole fraction xB=nB∑nx_\text{B} = \dfrac{n_\text{B}}{\sum n}; all mole fractions add up to 1.
  • Molarity MM = moles of solute per litre of solution (changes with temperature because volume does).
  • Molality mm = moles of solute per kilogram of solvent (independent of temperature, like mass % and mole fraction).

When a percentage and a density are given, assume 100 g100\ \text{g} of solution; the volume is mass ÷ density.

Table of concentration units, what each is based on and whether it changes with temperature
UnitBased onTemperature dependent?
Mass %, ppm (by mass)MassesNo
Mole fractionMolesNo
MolalityMoles and mass of solventNo
MolarityMoles and volume of solutionYes
Volume %VolumesYes

Formulas

  • Mole fractions in a binary solution
    xA+xB=1x_\text{A} + x_\text{B} = 1
  • Molarity
    M=nsoluteVsolution (L)M = \dfrac{n_\text{solute}}{V_\text{solution}\,(\text{L})}
  • Molality
    m=nsolutewsolvent (kg)m = \dfrac{n_\text{solute}}{w_\text{solvent}\,(\text{kg})}

Key terms

Molarity

Moles of solute in one litre of solution; unit mol L−1\text{mol L}^{-1} (M).

Molality

Moles of solute per kilogram of solvent; unit mol kg−1\text{mol kg}^{-1} (m).

Mole fraction

Moles of one component divided by the total moles of all components.

ppm

Parts of a component per million parts of the solution.

2. Solubility of solids and gases; Henry’s law

Solubility is the maximum amount of a substance that dissolves in a specified amount of solvent at a specified temperature. A saturated solution is in dynamic equilibrium with undissolved solute.

Solids in liquids. Like dissolves like: ionic and polar solutes dissolve in polar solvents (water), non-polar solutes in non-polar solvents (benzene). If dissolving is endothermic, solubility increases with temperature; if exothermic, it decreases (Le Chatelier’s principle). Pressure has no significant effect.

Gases in liquids. Solubility increases with pressure and decreases with temperature. Henry’s law: at constant temperature, the partial pressure of a gas above a solution is proportional to its mole fraction in the solution, p=KH xp = K_\text{H}\,x. KHK_\text{H} depends on the gas, the solvent and the temperature; a higher KHK_\text{H} means lower solubility, and KHK_\text{H} rises with temperature.

Applications: carbonated drinks are sealed under high COX2\ce{CO2} pressure; divers breathing compressed air risk the bends (nitrogen bubbles in blood on rapid ascent), so tanks use air diluted with helium; at high altitude the low partial pressure of oxygen causes anoxia; aquatic life suffers in warm water, which holds less oxygen.

Schematic. Particles are drawn far larger than real molecules.

Formulas

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

Key terms

Saturated solution

A solution in which no more solute dissolves at that temperature; dissolving and crystallisation occur at equal rates.

Henry’s law

At constant temperature, p=KH xp = K_\text{H}\,x for a gas dissolved in a liquid.

Bends

A painful condition caused by nitrogen coming out of the blood as bubbles when a diver rises too quickly.

3. Vapour pressure, Raoult’s law, ideal and non-ideal solutions

Raoult’s law: for a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction in the liquid: p1=p1∘x1p_1 = p^\circ_1 x_1, p2=p2∘x2p_2 = p^\circ_2 x_2. By Dalton’s law ptotal=p1+p2=p1∘+(p2∘−p1∘)x2p_\text{total} = p_1 + p_2 = p^\circ_1 + (p^\circ_2 - p^\circ_1)x_2, a straight line against composition. The mole fraction in the vapour is y1=p1/ptotaly_1 = p_1 / p_\text{total}; the vapour is richer in the more volatile component.

For a non-volatile solute, only the solvent contributes: p=p1∘x1p = p^\circ_1 x_1. Raoult’s law is a special case of Henry’s law with KH=p∘K_\text{H} = p^\circ.

Ideal solutions obey Raoult’s law over the whole range; A–B interactions are about the same as A–A and B–B; ΔmixH=0\Delta_\text{mix}H = 0 and ΔmixV=0\Delta_\text{mix}V = 0 (n-hexane + n-heptane, bromoethane + chloroethane, benzene + toluene).

Non-ideal solutions:

  • Positive deviation: A–B attractions weaker; vapour pressure higher than predicted; ΔmixH>0\Delta_\text{mix}H > 0, ΔmixV>0\Delta_\text{mix}V > 0 (ethanol + acetone, carbon disulfide + acetone). Large positive deviation gives a minimum-boiling azeotrope (ethanol + water at about 95% ethanol by volume).
  • Negative deviation: A–B attractions stronger (often hydrogen bonding); vapour pressure lower; ΔmixH<0\Delta_\text{mix}H < 0, ΔmixV<0\Delta_\text{mix}V < 0 (chloroform + acetone, phenol + aniline). Large negative deviation gives a maximum-boiling azeotrope (nitric acid + water at about 68% nitric acid by mass).

An azeotrope boils at constant temperature with the same composition in liquid and vapour, so fractional distillation cannot separate it.

  • Partial pressure of A
  • Partial pressure of B
  • Total vapour pressure
Illustrative values: pure A 30 kPa, pure B 70 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 + Dalton
    ptotal=p1∘+(p2∘−p1∘) x2p_\text{total} = p^\circ_1 + (p^\circ_2 - p^\circ_1)\, x_2
  • Mole fraction in the vapour
    y1=p1ptotaly_1 = \dfrac{p_1}{p_\text{total}}

Key terms

Raoult’s law

The partial vapour pressure of a volatile component equals its pure vapour pressure times its mole fraction in the solution.

Ideal solution

A solution that obeys Raoult’s law at all compositions, with ΔmixH=0\Delta_\text{mix}H = 0 and ΔmixV=0\Delta_\text{mix}V = 0.

Azeotrope

A liquid mixture that boils at constant temperature with the same composition in the liquid and the vapour.

4. Colligative properties: vapour pressure, boiling point and freezing point

Colligative properties depend only on the number of solute particles in a given amount of solvent, not on their nature. They are measured for dilute solutions of non-volatile solutes.

Relative lowering of vapour pressure equals the mole fraction of the solute: p1∘−p1p1∘=x2≈w2M1M2w1\dfrac{p^\circ_1 - p_1}{p^\circ_1} = x_2 \approx \dfrac{w_2 M_1}{M_2 w_1} for dilute solutions.

Elevation of boiling point. A solution’s vapour pressure reaches atmospheric pressure at a higher temperature: ΔTb=Kb m\Delta T_\text{b} = K_\text{b}\, m. KbK_\text{b} (ebullioscopic constant) for water is 0.52 K kg mol−10.52\ \text{K kg mol}^{-1}.

Depression of freezing point. The solution’s vapour pressure equals that of the solid solvent at a lower temperature: ΔTf=Kf m\Delta T_\text{f} = K_\text{f}\, m. KfK_\text{f} (cryoscopic constant) for water is 1.86 K kg mol−11.86\ \text{K kg mol}^{-1}; for benzene 5.12 K kg mol−15.12\ \text{K kg mol}^{-1}.

Molar mass from either: M2=K×1000×w2ΔT×w1M_2 = \dfrac{K \times 1000 \times w_2}{\Delta T \times w_1} with masses in grams. A temperature difference has the same value in K and °C.

Schematic: fewer solvent particles at the surface means fewer escape per second.
  • Pure water
  • Solution
Illustrative curves; the gap is exaggerated so that it can be seen.

Formulas

  • Relative lowering of vapour pressure
    p1∘−p1p1∘=x2\dfrac{p^\circ_1 - p_1}{p^\circ_1} = x_2
  • Elevation of boiling point
    ΔTb=Kb m\Delta T_\text{b} = K_\text{b}\, m
  • Depression of freezing point
    ΔTf=Kf m\Delta T_\text{f} = K_\text{f}\, m

Key terms

Colligative property

A property of a solution that depends only on the number of solute particles, not on their identity.

Ebullioscopic constant

KbK_\text{b}: elevation of boiling point for a solution of molality 1 mol kg−11\ \text{mol kg}^{-1}.

Cryoscopic constant

KfK_\text{f}: depression of freezing point for a solution of molality 1 mol kg−11\ \text{mol kg}^{-1}.

5. Osmosis and osmotic pressure

Osmosis is the net flow of solvent through a semipermeable membrane (one that passes solvent but not solute) from pure solvent or a dilute solution into a more concentrated solution. The osmotic pressure π\pi is the excess pressure that must be applied to the solution to just stop osmosis.

For dilute solutions π=CRT=n2RTV\pi = CRT = \dfrac{n_2 RT}{V}, so M2=w2RTπVM_2 = \dfrac{w_2 RT}{\pi V}. Osmotic pressure is preferred for the molar masses of proteins and polymers: it is measured at room temperature and uses molarity, and its values are large enough to measure even for very dilute solutions.

Isotonic solutions have the same osmotic pressure (0.9% w/v sodium chloride is isotonic with blood cells). In a hypertonic solution cells lose water and shrink; in a hypotonic solution they gain water and swell. Applying a pressure greater than π\pi reverses the flow: reverse osmosis, used to desalinate sea water.

At equilibrium the pressure of the extra column of height h equals the osmotic pressure of the solution.

Formulas

  • Osmotic pressure
    π=CRT\pi = CRT
  • Molar mass from osmotic pressure
    M2=w2RTπVM_2 = \dfrac{w_2 R T}{\pi V}

Key terms

Semipermeable membrane

A membrane that allows solvent molecules but not solute particles to pass.

Osmotic pressure

The excess pressure that must be applied to a solution to prevent the entry of solvent through a semipermeable membrane.

Isotonic solutions

Solutions with the same osmotic pressure at a given temperature.

Reverse osmosis

Flow of solvent out of a solution through a semipermeable membrane when a pressure greater than the osmotic pressure is applied.

6. Abnormal molar masses and the van’t Hoff factor

If a solute dissociates in solution (electrolytes in water), the number of particles increases and the molar mass calculated from a colligative property is lower than the true value. If it associates (ethanoic acid dimerising in benzene through hydrogen bonds), the number of particles decreases and the calculated molar mass is higher.

The van’t Hoff factor i=normal molar massabnormal molar mass=observed colligative propertycalculated colligative propertyi = \dfrac{\text{normal molar mass}}{\text{abnormal molar mass}} = \dfrac{\text{observed colligative property}}{\text{calculated colligative property}}. Corrected equations: ΔTb=iKbm\Delta T_\text{b} = iK_\text{b}m, ΔTf=iKfm\Delta T_\text{f} = iK_\text{f}m, π=iCRT\pi = iCRT, Δpp1∘=i n2n1\dfrac{\Delta p}{p^\circ_1} = i\,\dfrac{n_2}{n_1}.

  • Dissociation into nn ions with degree α\alpha: i=1+(n−1)αi = 1 + (n - 1)\alpha, so α=i−1n−1\alpha = \dfrac{i - 1}{n - 1}.
  • Association of nn molecules into one with degree α\alpha: i=1−α+αni = 1 - \alpha + \dfrac{\alpha}{n}; for dimers i=1−α2i = 1 - \dfrac{\alpha}{2}.
  • Non-electrolytes (glucose, urea, sucrose): i=1i = 1. Complete dissociation: NaCl\ce{NaCl} i=2i = 2, CaClX2\ce{CaCl2} and KX2SOX4\ce{K2SO4} i=3i = 3.
Table of urea, potassium chloride, potassium sulfate and ethanoic acid with their behaviour in solution and approximate van’t Hoff factors
SoluteSolventBehaviourii
UreaWaterNo change11
Potassium chlorideWaterDissociates into 2 ionsclose to 22
Potassium sulfateWaterDissociates into 3 ionsclose to 33
Ethanoic acidBenzeneForms dimersclose to 0.50.5

Formulas

  • Dissociation
    i=1+(n−1)αi = 1 + (n - 1)\alpha
  • Association
    i=1−α+αni = 1 - \alpha + \dfrac{\alpha}{n}

Key terms

Abnormal molar mass

A molar mass from a colligative property that differs from the formula value because of dissociation or association.

van’t Hoff factor

The ratio of the observed colligative property to the value calculated assuming no dissociation or association.

Degree of dissociation

The fraction α\alpha of the solute formula units that split into ions.

Exam tips

Common mistakes

  • Dividing by the mass of the solution instead of the solvent when finding molality.
  • Using millilitres instead of litres in molarity or in π=CRT\pi = CRT, or a gas constant whose unit does not match the pressure unit.
  • Adding 273 to a temperature difference: ΔT\Delta T is the same in K and °C.
  • Saying a gas with a larger KHK_\text{H} is more soluble (it is less soluble).
  • Mixing up the deviations: positive deviation = weaker A–B attraction, higher vapour pressure, endothermic mixing, minimum-boiling azeotrope.
  • Forgetting the van’t Hoff factor for electrolytes, or using the number of atoms instead of the number of ions (for KX2SOX4\ce{K2SO4}, n=3n = 3).

How marks are usually awarded

  • 1-mark items: correct option, term or value with unit.
  • Numericals (2–3 marks): formula (1/2–1) + substitution with units (1) + answer with correct unit and sensible significant figures (1/2–1).
  • Explanations (2–3 marks): the law or principle named (1) + the reason in terms of particles or attractions (1) + an example or consequence (1).
  • 5-mark answers: usually a definition, a derivation or graph, and a numerical; label graphs with axes and the pure-component points.