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How do substances interact with water?
the polar nature of the water molecule, the intermolecular forces (hydrogen bonding, dipole-dipole and dispersion) that operate between water molecules and between water and solute particles, and the use of these forces to predict relative solubility of substances in water
A focused VCE Chemistry Unit 2 answer on water polarity and intermolecular forces in aqueous solutions. Covers the bent shape and dipole of water, hydrogen bonding versus dipole-dipole versus dispersion forces, the like-dissolves-like rule, and how to rank the relative solubility of polar, ionic and non-polar substances in water.
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What this dot point is asking
VCAA wants you to use the polar bent shape of the water molecule and the three intermolecular forces (hydrogen bonding, dipole-dipole, dispersion) to predict which substances dissolve in water and how strongly. The framing is the rule "like dissolves like" applied to specific named substances.
The answer
Why water is polar
Water has the molecular formula . Oxygen is far more electronegative than hydrogen (3.44 vs 2.20 on the Pauling scale), so each bond is polar covalent: the oxygen carries a partial negative charge (), each hydrogen a partial positive charge ().
The shape matters as much as the bond polarity. Oxygen has two bonding pairs and two lone pairs, giving a bent () geometry. The two bond dipoles do not cancel; they add to a net dipole pointing from the two H atoms toward the oxygen. Water is therefore a polar molecule.
The three intermolecular forces in water and aqueous solutions
| Force | Where it acts | Relative strength |
|---|---|---|
| Hydrogen bonding | When H is bonded to N, O or F and another N, O or F lone pair is nearby | Strongest IMF (about 10 to 40 kJ per mol) |
| Dipole-dipole | Between any two permanent dipoles | Moderate (about 5 to 25 kJ per mol) |
| Dispersion (London) | Between any two species; the only force in non-polar substances | Weakest per pair; grows with electron count and surface area |
In pure water, hydrogen bonding dominates because each water molecule can donate two H and accept two via its lone pairs.
In aqueous solutions the solute brings its own forces. The relevant question is whether the solute-water interactions can replace the water-water hydrogen bonds the solute disrupts.
Predicting solubility in water
Apply this checklist to any solute:
- Is it ionic? Then ion-dipole attractions form a hydration shell. Most ionic compounds dissolve well unless their lattice energy is unusually high (see the solubility-rules dot point for the exceptions).
- Is it a polar molecule with an O-H, N-H or F-H bond? Then it can hydrogen bond directly with water (alcohols, ammonia, carboxylic acids, sugars). Small ones (methanol, ethanol) are miscible; longer chains (butanol onwards) are less soluble because the non-polar tail does not fit the water network.
- Is it polar but cannot hydrogen bond? Then it relies on dipole-dipole with water. Soluble but less so than H-bonders (acetone, propanone are good examples).
- Is it non-polar? Then only dispersion forces with water. Dissolving it would force water to give up hydrogen bonds with nothing to replace them. Essentially insoluble (oils, hexane, dinitrogen).
The shorthand: like dissolves like. Polar/ionic in polar solvents (water), non-polar in non-polar solvents (hexane).
Comparing solubilities of similar molecules
A common VCAA pattern asks you to rank a family. Use the balance of polar head vs non-polar tail.
Alcohols: methanol and ethanol are miscible with water; propan-1-ol is miscible; butan-1-ol is partially soluble; pentan-1-ol and longer are essentially insoluble. The single group cannot keep a long alkyl chain in solution.
Carboxylic acids: ethanoic acid is miscible; longer-chain fatty acids are insoluble.
Sugars: glucose is highly soluble because it has five hydroxyl groups per molecule, all able to hydrogen bond with water.
Worked example
Predict whether iodine (), ammonia () and potassium chloride () dissolve in water. Justify using IMFs.
: non-polar diatomic. Only dispersion with water. Insoluble in water (very low solubility; iodine is soluble in non-polar solvents like hexane or cyclohexane instead).
: polar bent molecule with bonds. Hydrogen bonds with water (donor through , acceptor through the N lone pair). Highly soluble; aqueous ammonia is a standard reagent.
: ionic. Hydration shell forms around and ions. Soluble at about at room temperature.
Common traps
Calling water non-polar because the bond dipoles cancel. They do not cancel in a bent molecule. They cancel in (linear) but not in .
Putting hydrogen bonding in molecules that lack N-H, O-H or F-H. , and do not hydrogen bond, even though each has H. The H must be on N, O or F.
Confusing the intramolecular covalent bond with the intermolecular hydrogen bond. The first is inside the water molecule and is broken only by chemical reaction. The second is between molecules and breaks during boiling and dissolving.
Saying a long alcohol is insoluble because it is non-polar. It is amphiphilic. The end is polar; the alkyl chain is not. Long alcohols are insoluble because the non-polar end wins.
In one sentence
Water is a polar bent molecule that hydrogen-bonds with itself, so a solute dissolves only when it offers comparable interactions (ion-dipole for ionic solutes, hydrogen bonding for or molecules, dipole-dipole for other polar molecules), while non-polar species cannot replace the water-water hydrogen bonds they would have to break and so stay out of solution.
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE4 marksMethanol (CH3OH), ethane (C2H6) and sodium chloride (NaCl) are added separately to water. (a) Rank the three substances from most to least soluble in water. (b) Identify the dominant intermolecular force operating between each solute and water. (c) Use bonding to justify your ranking.Show worked answer →
A 4-mark answer needs the ranking, the forces and the justification.
(a) Ranking, most to least soluble in water:
NaCl > CH3OH > C2H6.
(b) Forces between each solute and water:
NaCl with water: ion-dipole attraction (the Na+ and Cl- ions are surrounded by water dipoles).
CH3OH with water: hydrogen bonding (the O-H of methanol both donates and accepts H-bonds with water).
C2H6 with water: only weak dispersion forces (ethane is non-polar; only induced-dipole interactions are possible).
(c) Justification:
NaCl dissolves because the ion-dipole attractions, summed over the whole hydration shell, release energy comparable to or greater than the NaCl lattice energy.
CH3OH dissolves freely (in fact it is miscible with water) because each methanol can hydrogen bond with several water molecules, releasing energy that compensates for breaking water-water H-bonds.
C2H6 has no dipole and cannot hydrogen bond. Dissolving it would require breaking water-water hydrogen bonds with nothing of comparable strength to replace them. The process is unfavourable, so ethane is essentially insoluble in water.
Related dot points
- the explanation of the properties of water (including high boiling point, high specific heat capacity, surface tension and the density of ice relative to liquid water) and the role of water as a solvent for polar and ionic substances, including the use of solubility rules to predict precipitation reactions and write ionic equations
A focused VCE Chemistry Unit 2 answer on the chemistry of water. Covers hydrogen bonding and how it explains water's anomalous physical properties, how water dissolves ionic and polar molecular substances, the use of solubility rules to predict precipitation, and writing balanced ionic and net ionic equations.
- expressing the concentration of solutions (mol L^-1, g L^-1, %m/v, %m/m, %v/v and ppm) including dilution calculations, and the Brønsted-Lowry model of acids and bases including conjugate acid-base pairs, the distinction between strong and weak (and concentrated and dilute) acids and bases, and the calculation of pH from [H+]
A focused VCE Chemistry Unit 2 answer on concentration and acid-base chemistry. Covers concentration units (mol L^-1, g L^-1, %m/v, %m/m, %v/v, ppm) and dilution calculations, the Brønsted-Lowry model with conjugate acid-base pairs, strong vs weak and concentrated vs dilute, and the calculation of pH from [H+].