Doc Brown's Advanced A level theoretical chemistry revision notes

Part  8.7 Explaining the solubility of covalent compounds in solvents, miscible and immiscible liquids - why mix or no mix?

[Author ©  Dr Phil Brown PhD: Doc Brown's Chemistry exam revision notes suitable for advanced A level chemistry students studying pre–university/college advanced level theoretical–physical chemistry courses:  solubility, miscible or immiscible?  [page updated May 2nd 2026 *]

8.7.1 Introduction to solubility

What factors control how soluble a covalent compound is in water or organic solvents i.e. will a solute dissolve in a particular solvent.

Polar molecules tend to be soluble and dissolve in polar solvents.

Non-polar molecules tend to be soluble and dissolve in non-polar solvents.

Highly polar liquids tend to be miscible with other highly polar liquids.

Why do some liquids completely mix (miscible) and others not (immiscible)?

In both situations the role of intermolecular attractive forces and intermolecular bonding is discussed.

8.7.2 Solubility of solid covalent compounds and miscible/immiscible liquids

An approach to the covalent case studies described below

(Solubility of inorganic salt like compounds is considered in Thermodynamics Part 2)

(solubility product is dealt with in Equilibria section 4.2 and partition calculations in section 4.1)

• The solubility of one material in another is a consequence of the physical and chemical interactions involved.

• A solvation process may involve intermolecular force interactions or a chemical change or both.

• When solids dissolve in liquids, a resulting saturated solution (maximum solubility) is another equilibrium situation.

• Similarly mixing two immiscible liquids is also an equilibrium situation even if only a trace of one dissolves in the other.

• It is important that you have read the section 8.2.1 on the nature of intermolecular forces because they all get mentioned again here, but not in as much detail, I will assume a working knowledge of them in this section.

• In considering whether A (liquid or solid solute) will dissolve in a liquid (e.g. the solvent) three possible interactions should be considered i.e. the intermolecular forces involved and solvation processes.

• The three 'interactions' will simply be A...A, A...B and B...B associations via intermolecular attractive forces.

• If either A...A or B...B associations are strong then one of them will dominate and A and B will be less likely to mix i.e. if two liquids, they are likely to be immiscible OR solid A won't dissolve in liquid B.

• * However if an A...B interaction is strong or comparable to A...A or B...B associations then two liquids are likely to be miscible and mix with each other OR solid A dissolves in liquid B.

• This conceptual way of thinking is a bit more sophisticated than 'like dissolves like', but the latter has much to commend it as a rough rule of thumb.

• If two liquids dissolve appreciably in each other but still two layers are formed above and below a meniscus, the two liquids may be referred to as partially miscible.

• If you can think like this, then most cases of what dissolves/mixes with what are usually quite understandable!

* For more details on miscible and immiscible liquids see Equilibria section 4.

Partition equilibrium between two phases, partition coefficient, immiscible liquids and solvent extraction

The case studies discussed below are:

Case study 8.7.3 Mixing hydrocarbons and water

Case study 8.7.4 The solubility of  alcohols in water and other polar homologous series

Case study 8.7.5 The solubility of carboxylic acids in water

Case Study 8.7.6 Why are pairs of some organic solvents miscible and others immiscible?

Case Study 8.7.7 Are dissolved molecules always in the same molecular state whatever the solvent?

Case study 8.7.8 The solubility of some gases in water

Case study 8.7.3 Mixing hydrocarbons and water

• The solubility of hydrocarbons in water is extremely low and for practical purposes it is virtually zero.

• It doesn't matter whether it is alkanes (lower–small or higher–large), alkenes, arenes (aromatic hydrocarbons like benzene, methyl benzene etc.) or even crude oil itself, non of them can associate with water.

• Hydrocarbon – hydrocarbon intermolecular forces are only the weak instantaneous dipole – induced dipole interactions.

• The polar water – oil interaction will be of a permanent dipole – induced dipole nature, but even if this is greater than the hydrocarbon – hydrocarbon attraction, what will dominate this situation is the strongest of the intermolecular forces, namely the hydrogen bonds between water molecules.

• ....^δ–:O–H^δ+ llll ^δ–:O–H^δ+....  rules ok!

• If a hydrocarbon molecule is 'shaken' into water, the hydrogen bonds will be disrupted and will tend to reform and quite literally squeeze out the hydrocarbon molecules.

• I would expect methane will be the most soluble hydrocarbon in water because it is the smallest possible hydrocarbon and its 'insertion' will cause the minimum disruption to water's hydrogen bonds.

• Selected solubility data of some hydrocarbons in water at ~room temperature.

• The values depend on ambient partial pressure and temperature, but all are very low solubilities.

• methane (22 mg/dm³), ethane (57 mg/dm³), propane gas (62 mg/dm³).

Case study 8.7.4 The solubility of alcohols in water and other homologous series

• The first three lower members of the alcohol homologous series are completely miscible with water.

• This is because all the A...A, B...B and B...B interactions are all similar, i.e. they can all hydrogen bond with each other.

• However, although the hydroxyl O–H group is hydrophilic ('solvent liking') and readily hydrogen bonds with water, as well as itself, there is a counter trend of the hydrophobic ('solvent hating') alkyl group which disrupts the hydrogen bonds in water.

• Therefore, as the alkyl hydrocarbon chain increases in length, its greater hydrogen bond disruption results in a rapidly decreasing solubility so that by the time you reach octanol all successive alcohols are effectively insoluble.

• There are no compensating intermolecular forces to compensate this disruption of the strong hydrogen bonds in water.

• Note that if more than one hydroxyl group is present then the solubility will be increased e.g.

  • Liquid propane–1,2,3–triol (glycerol) has the structure HOCH₂CH(OH)CH₂OH (M_r = 92), so with its triple hydrogen bonding sites the glycerol – water interactions are strong and both liquids are completely soluble in each other i.e. fully miscible, despite being as bulky as pentan–1–ol on a molecular mass basis.

  • Sugars such as glucose (M_r = 180), despite the higher molecular mass, are highly soluble in water due to the multiple hydrogen bonding sites on the molecule (5 O–H groups).

  • Also note the increase in viscosity ('stickiness') of glucose (solid) > liquid propane–1,2,3–triol > liquid propane–1,2–diol > propan–1–ol,

  • which is accounted for in terms of increased hydrogen bonding as well as the increased instantaneous dipole – induced dipole forces from the increasing number of electrons.

• Although for linear ..1-ols, from nonan-1-ol onwards are ~insoluble, they will act as surfactants and form a monomolecular layer on the surface of water.

• The chemistry of ALCOHOLS

Case study 8.7.5 The solubility of carboxylic acids in water

• The solubility trend is similar to that of the linear alcohols in case study 8.7.4 as are the reasons for the trend.

• The first four lower members of this carboxylic acid homologous series are completely miscible with water and the solubilities, on a molar basis, tend to be a bit higher than for alcohols, presumably because they are more polar and have two sites for H bonding (C=O and O–H).

• As with alcohols, as the alkyl hydrocarbon chain increases in length, its greater hydrogen bond disruption results in a rapidly decreasing solubility so that by the time you reach decanoic acid all successive alcohols are effectively insoluble.

• There are no compensating intermolecular forces to compensate this disruption of the strong hydrogen bonds in water despite the multiple hydrogen bonding sites available via the carboxylic acid.

•  The chemistry of CARBOXYLIC ACIDS, ESTERS and other derivatives

Case 8.7.6 Why are some pairs of  organic solvents miscible and others immiscible?

(in these examples the cases are based on mixing equal volumes of each liquid)

• Propanone and benzene are miscible

  • (polar) propanone – propanone intermolecular forces = (instantaneous dipole – induced dipole) + (permanent dipole – permanent dipole)

  • (non–polar) benzene – benzene intermolecular forces =  (instantaneous dipole – induced dipole)

  • Although propanone is polar, most of its intermolecular force is still derived from instantaneous dipole – induced dipole attractive forces and the extra permanent dipole – permanent dipole force does not prevent miscibility.

• Propanone and butan–1–ol are miscible

  • (polar) propanone – propanone intermolecular forces = (instantaneous dipole – induced dipole) + (permanent dipole – permanent dipole)

  • (more polar) butan–1–ol – butan–1–ol intermolecular forces = (instantaneous dipole – induced dipole) + (hydrogen bonding)

  • Both molecules are polar and butan–1–ol can hydrogen bond with propanone as well as with itself, so miscibility is no problem.

• Glycerol (propan–1,2,3–triol) and propanone are immiscible

  • (polar) propanone – propanone intermolecular forces = (instantaneous dipole – induced dipole) + (permanent dipole – permanent dipole)

  • (much more polar) glycerol – glycerol intermolecular forces = (instantaneous dipole – induced dipole) + (multiple hydrogen bonding)

  • Glycerol has the structure HOCH₂CH(OH)CH₂OH, so with its triple hydrogen bonding sites the glycerol – glycerol attraction will outweigh any glycerol – propanone interactions.

• Butan–1–ol (1–butanol) dissolves a wide range of other organic liquids

  • Being highly polar with its hydroxyl group, butan–1–ol will readily dissolve many other polar organic compounds.

  • However, the alkyl hydrocarbon tail means the hydrogen bonding between the alcohol molecules doesn't totally dominate the situation and many relatively non–polar organic liquids will also dissolve in butan–1–ol.

Case 8.7.7 Are dissolved molecules always in the same molecular state whatever the solvent?

• For most organic compounds the molecular state is the same irrespective of the solvent e.g.

  • Sugars e.g. C₆H₁₂O₆ dissolving in water or alcohol.

  • Wax dissolving in other solvents such as hydrocarbon liquids or chloroalkane solvents.

  • Ethanol dissolving in water or other organic solvents such as propanone.

• Carboxylic acids

  • In water, typical dissolved carboxylic acids exist mainly as the solvated molecule, a few % ionised and virtually non of the dimer which can be formed by hydrogen bonding.

  • However they dimerise in most organic solvents and are quite soluble in many of them.

  • 

  • This because most organic solvents such as hydrocarbons, chloroalkanes are not as polar as the acid itself (or water!).

  • Therefore the strongest intermolecular force that can operate is the hydrogen bonding between two carboxylic acid molecules.

  • In other words A...A solute interactions are stronger than A...R solute – solvent OR B...B solvent – solvent intermolecular forces.

• Iodine

  • Non–polar iodine has a very low solubility in water because the strongest interactions possible in the mixture are H₂O...H₂O via hydrogen bonding.

  • Explaining the partition of the solute iodine dissolving in a mixture of the immiscible liquids water and hexane by considering the C₆H₁₄ - I₂ and H₂O - I₂ interactions in terms of intermolecular bonding and solvation.

  • So when the iodine molecules try to dissolve, its interaction with water is so weak that it fails to significantly disrupt the hydrogen bonds between water molecules, hence the solubility is low.

  • Iodine is much more soluble in less polar solvents like hydrocarbons and chloroalkanes where the A...A, A...B and B...B intermolecular attractive forces are more comparable.

  • In both pure water and organic solvents iodine exists as the iodine molecule I_2(aq) or I_{2(organic solvent)}.

  • Although iodine has a low solubility in pure water, it is much more soluble in potassium iodide solution which is a much more convenient for delivering a useful solution of iodine.

  • What actually happens when you add solid iodine to potassium iodide solution is that a chemical reaction takes place that more than compensates for the disruption of hydrogen bonding in water. The iodine forms the 'tri–iodide' ion.

    • (i)  I_2(s) I_2(aq)  and then  (ii) I_2(aq) + I–_(aq) [I₃]^–_(aq) ....

    • .... and the 2nd equilibrium shifts the first equilibrium to the right, i.e. an increase in solubility,

    • but remember, this is due to a chemical reaction, NOT a change in the intermolecular forces.

• Hydrogen chloride gas

  • Hydrogen chloride gas is a small polar covalent molecule readily soluble in organic hydrocarbon solvents and will retain its molecular character.

  • It is also very soluble in water, with which it reacts to form hydrochloric acid, a solution of hydrated ions, a very non-molecular state!

  • HCl(g)  +  H₂O(l)  ===>  H₃O⁺(aq)  +  Cl_-(aq)

8.7.8 The solubility of some gases in water

• The maximum solubility is quoted as the volume of gas dissolved in 1 cm³ of water at 0^oC at 1 atm pressure of the gas (101325 Pa).

  • This corresponds to the equilibrium X_(g) X_(aq) where X is the gas dissolving in water.

  • This is only quoted specifically where the gas reacts with water moving this equilibrium further to the right i.e. increasing solubility.

• From Avogadro's Law, equal volumes of gases contain the same number, the data represents solubility on a molar basis.

• Solubility of gases decreases on increase in temperature. and increases with increase in pressure of the gas.

• In most cases the solubility is low because the dissolving molecule is disrupting the hydrogen bonds between water molecules.

• * However, in a few cases, there is a chemical reaction with water OR strong hydrogen bonding occurs between solute and the solvent water, and in such cases the solubility is considerably increased.

• When comparing any two of the molecules in the list, you can produce reasonable arguments to explain either the difference or similarity in solubility and I've quoted a few such arguments.

Henry's Law and Gas Solubility

• Henry's Law states that the mass of slightly soluble gas that dissolves in a definite mass of liquid at a given temperature is very nearly proportional to the partial pressure of that gas (in contact with the liquid).

• This law holds for gases that do NOT react with the liquid solvent (e.g. hydrogen chloride in water) OR have a very strong intermolecular attraction with the liquid (eg. ammonia in water).

• One mathematical expression of Henry's Law is: p_x = K_H c

• where p_x = the partial pressure of the dissolved gas X (i.e. p_x above the liquid), K_H is Henry's Constant (which depends on the gas solute, the liquid solvent and the ambient temperature, and c is the concentration of the dissolved gas solute.

• For relatively non–polar molecules of low solubility you do get a 'reasonably' linear graph of concentration (x) versus partial pressure (y)

• For a graph with the axis origin 0,0 and this corresponds to the equilibrium

  • X_(g) X_(aq) where X is the gas dissolving in water.

WHAT NEXT?

Part 8 sub–index: 8.1 Vapour pressure, nature, origin and examples explained * 8.2.1 Introduction to the types of intermolecular forces and examples explained (index) * 8.2.2 Detailed comparative discussion of boiling points of 8 organic molecule of similar molecular mass * 8.3 Boiling point plots of six organic homologous series - graphs and explanation * 8.4 Other case studies of boiling points related to intermolecular forces * 8.5 Steam distillation – theory and practice * Evidence and theory for hydrogen bonding in simple covalent hydride * 8.7 Solubility of covalent compounds, miscible and immiscible liquids

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. K_c and K_p equilibrium expressions and calculations * Part 3. Equilibrium and industrial processes * Part 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series