Solutions · Lesson 3 of 11
Solubility and Henry’s Law
“Explain dissolution through competing interactions and dynamic equilibrium, then quantify gas solubility using partial pressure.”
• Explain saturation as a dynamic equilibrium. • Relate solubility to solute–solvent interactions and temperature. • Use Henry’s law with a clearly specified constant convention. • Calculate dissolved gas amounts and explain pressure-dependent applications.
Why some substances dissolve and others do not
Water dissolves sugar and many salts, but does not dissolve substantial amounts of naphthalene. Benzene dissolves naphthalene much more readily. Dissolution requires separating solute particles, separating some solvent particles, and forming solute–solvent contacts. The balance of these changes, together with the tendency of particles to spread out, determines whether mixing is favourable. “Like dissolves like” is a useful guide to compatible interactions, not a law that guarantees every polar solid dissolves in water.
| Pair | Main attractive interaction | Meaning for dissolution |
|---|---|---|
| n-Hexane and n-octane | London dispersion | Both are non-polar; their interactions are similar. |
| I₂ and CCl₄ | London dispersion | Both are polarizable, non-polar molecular substances. |
| NaClO₄ and water | Ion–dipole attraction | Water hydrates separated ions. |
| Methanol and acetone | Hydrogen bonding, with dipolar contributions | Methanol can donate an O–H hydrogen bond to acetone oxygen. |
| Acetonitrile and acetone | Dipole–dipole, with dispersion also present | Both have permanent dipoles; neither supplies an O–H donor. |
Water’s polarity stabilises ions through hydration. Sugar dissolves because its many O–H groups can hydrogen-bond with water even though it does not split into ions. A non-polar hydrocarbon lacks those strong contacts with water. In an alcohol–water mixture, both components can form hydrogen bonds, but mixing reorganises existing networks; the presence of hydrogen bonding alone does not prove an ideal solution.
Problem
Compare formic acid, ethylene glycol, phenol, pentanol, toluene and chloroform in water.
- 1.Formic acid and ethylene glycol interact strongly with water and are highly soluble; their small structures and hydrogen-bonding groups support mixing.
- 2.Phenol and pentanol have polar O–H groups but also substantial hydrocarbon regions, giving limited or partial solubility rather than the extensive mixing of a short-chain alcohol.
- 3.Toluene is very poorly soluble and chloroform is only sparingly soluble. “Insoluble” in a qualitative classification means very low solubility, not rigorously zero molecules dissolved.
- 4.Temperature and the exact isomer matter; a functional group is one factor, not the entire explanation.
Saturation is an equilibrium, not the end of molecular motion
The maximum equilibrium amount of a substance that dissolves in a specified amount of solvent at a stated temperature and pressure. Its units and reference amount must be given.
Initially, particles leave an added solid and disperse through the liquid. As dissolved concentration rises, particles also return to the solid. When dissolution and crystallisation rates become equal, the dissolved amount stops changing although particles continue to move between phases. With excess solid present, that dynamic equilibrium is a saturated solution. An unsaturated solution can dissolve more under the same conditions. Stirring usually changes how quickly equilibrium is reached, rather than the final equilibrium solubility.
For a solid whose dissolution is endothermic, increasing temperature generally favours dissolution and increases solubility. For exothermic dissolution, heating can decrease solubility. This follows the equilibrium response to added heat. It is therefore incorrect to say every solid becomes more soluble when heated. Pressure has little effect on most solid–liquid solubilities under ordinary conditions because their volumes change only slightly with pressure.
Problem
In a simplified model, CuS(s) ⇌ Cu²⁺(aq) + S²⁻(aq) has Ksp = 6 × 10⁻¹⁶. Find its molar solubility in pure water, neglecting other equilibria.
- 1.Let s mol L⁻¹ dissolve. One formula unit supplies one ion of each kind, so [Cu²⁺]=s and [S²⁻]=s.
- 2.The solid is omitted from the concentration product. Ksp = [Cu²⁺][S²⁻] = s².
- 3.s = √(6 × 10⁻¹⁶) = 2.44949 × 10⁻⁸ mol L⁻¹ ≈ 2.45 × 10⁻⁸ M.
- 4.This is the stipulated simple model. Sulfide protonation, solution pH, complex formation and activities can affect a real CuS solubility. Do not present the square-root result as an unrestricted material constant.
Gas pressure changes the dissolution equilibrium
A gas above a liquid exchanges molecules with dissolved gas. At equilibrium, entry and escape rates are equal. Raising the gas partial pressure at fixed temperature increases its tendency to enter the liquid. More dissolves until a new equilibrium is reached. Partial pressure means the pressure contributed by that gas; in an ideal gas mixture it equals gas-phase mole fraction times total pressure.
At a fixed pressure, larger K_H means smaller dissolved mole fraction, since x_g = p_g/K_H. Different references use reciprocal constants or concentrations in place of mole fractions. Read the defining equation and units before using a constant. A pressure-only constant is consistent with the dimensionless mole-fraction form; a constant multiplying molality must include reciprocal-molality units. Henry’s simple linear law works best at low dissolved concentrations when the gas does not undergo substantial reaction with the solvent.
Problem
At 293 K, p(N₂)=0.987 bar and K_H=76.48 kbar. Estimate dissolved N₂ in 1.00 L water, taking water mass as 1000 g.
- 1.Convert 76.48 kbar to 76480 bar. x(N₂)=0.987/76480=1.290533 × 10⁻⁵.
- 2.Water amount =1000/18=55.55556 mol. Use n = xn₁/(1−x) to obtain 7.16972 × 10⁻⁴ mol.
- 3.This is 0.716972 mmol, about 0.717 mmol. Neglecting x in the denominator has a very small effect here.
- 4.The tiny mole fraction justifies the dilute approximation; it does not justify using total air pressure in place of nitrogen partial pressure.
| Gas in water | Temperature / K | K_H / kbar |
|---|---|---|
| He | 293 | 144.97 |
| H₂ | 293 | 69.16 |
| N₂ | 293 | 76.48 |
| N₂ | 303 | 88.84 |
| O₂ | 293 | 34.86 |
| O₂ | 303 | 46.82 |
| Ar | 298 | 40.3 |
| CO₂ | 298 | 1.67 |
| Formaldehyde, apparent equilibrium value | 298 | 1.83 × 10⁻⁵ |
| Methane, representative dilute value | 298 | Approximately 40 |
| Vinyl chloride | 298 | 0.611 |
Compare constants at the same temperature and under the same convention. The increase in K_H for oxygen and nitrogen on warming indicates lower solubility at the same partial pressure. Many common gas dissolutions are exothermic, so heating shifts equilibrium toward escape. This is a common trend, not a universal theorem for every reactive gas–solvent system. Highly reactive gases, including formaldehyde in water, can have apparent constants that include dissolved chemical reactions. A representative methane constant is approximately 40 kbar at 298 K in the pressure–mole-fraction convention; constants expressed using mass concentration must be converted before comparison.
Interpreting real situations
A sealed carbonated drink holds more CO₂ under high CO₂ pressure. Opening lowers that partial pressure and dissolved gas escapes as the liquid approaches a new equilibrium. At high altitude, the oxygen fraction of dry air is broadly similar but total pressure is lower; oxygen partial pressure falls. Henry’s law explains part of the physical dissolution effect, while oxygen binding to haemoglobin makes blood more complex than an inert solvent.
During a dive, breathing gas is supplied at greater pressure, increasing the amount of inert gas that can dissolve in tissues. A rapid pressure decrease can create bubbles. Helium-containing mixtures are used in some forms of diving, but no single composition is suitable for every depth. The chemistry illustrates pressure-dependent dissolution; it does not specify a diving procedure.
Problem
Water equilibrates with air at 10 atm and 298 K. Take oxygen and nitrogen fractions as 0.20 and 0.79, with K_H values 3.30 × 10⁷ and 6.51 × 10⁷ mm Hg. Find their aqueous mole fractions.
- 1.Partial pressures are 2.0 atm O₂ and 7.9 atm N₂, or 1520 and 6004 mm Hg. The small remaining fraction belongs to other gases.
- 2.x(O₂)=1520/(3.30×10⁷)=4.60606×10⁻⁵. x(N₂)=6004/(6.51×10⁷)=9.22273×10⁻⁵.
- 3.These fractions include water in their denominator. Among these two dissolved gases alone, oxygen fraction ≈0.3331 and nitrogen ≈0.6669.
- 4.The dissolved gas ratio differs from the gas-phase ratio because their Henry constants differ.
Quick Quiz
Quiz
A saturated solution in contact with excess solid has constant dissolved concentration. Why?
Explanation: Dynamic equilibrium involves continuing forward and reverse processes at equal rates.
At fixed pressure, what does a larger K_H imply in p=K_Hx?
Explanation: Rearranging gives x=p/K_H. The direction depends on the chosen constant convention.
For endothermic dissolution of a solid, what is the usual equilibrium effect of heating?
Explanation: Adding heat favours the endothermic dissolution direction, increasing equilibrium dissolved concentration under the model.
Which quantity belongs in Henry’s law for oxygen in air?
Explanation: It is the pressure contribution of the gas whose dissolved concentration is being calculated.
Which interaction most directly stabilises Na⁺ in water?
Explanation: Polar water molecules orient around the ion, producing ion–dipole hydration interactions.
Practice Problems
- At the same temperature and solvent amount, ethane pressure is 1 bar when 6.56 × 10⁻³ g is dissolved. Estimate the pressure for 5.00 × 10⁻² g in the dilute regime.
- CO₂ has K_H=1.67 × 10⁸ Pa at 298 K. Estimate its amount in 500 mL water at CO₂ pressure 2.5 atm. Take density of water as 1 g mL⁻¹.
- The molality of H₂S in water is 0.195 mol kg⁻¹ at gas pressure 1 atm. Estimate K_H for p=K_Hx, using water 18 g mol⁻¹.
- Rank KCl, methanol, acetonitrile and cyclohexane qualitatively by increasing solubility in n-octane. Explain the limitation of the middle ordering.
- A methane-in-benzene constant is stated as 4.27 × 10⁵ mm Hg. At 760 mm Hg gas pressure, calculate the mole fraction under p=K_Hx. Why does this not uniquely specify molality?
- Explain why stirring does not generally increase equilibrium solubility and why “gases always become less soluble on heating” is too broad.
For dilute solutions at unchanged solvent amount, dissolved amount and mole fraction are approximately proportional. Thus p₂/p₁≈w₂/w₁. p₂=1×0.0500/0.00656=7.62195 bar≈7.62 bar. The constant temperature and solvent amount are essential assumptions.
Key Takeaways
• Solubility depends on the balance of interactions and the specified equilibrium conditions. • Saturation is a dynamic equality of dissolution and crystallisation rates. • Heating often increases endothermic solid dissolution and decreases common exothermic gas dissolution. • Henry’s law uses gas partial pressure, rather than automatically using total pressure. • For p=K_Hx, K_H has pressure units and larger values mean lower solubility. • A gas mole fraction must be converted to moles using the solvent amount. • Constant conventions, reactive gases and ambiguous units require explicit interpretation.