Chemistry · Lesson

Solubility, temperature and gas dissolution

Describe solubility as an equilibrium property under named conditions. Separate how much can remain dissolved from how fast it dissolves, then use an explicitly defined Henry-law convention for a gas.

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By the end, you can

Distinguish unsaturated, saturated and supersaturated states.
Separate dissolution rate from equilibrium solubility.
Use gas partial pressure and the units of a defined Henry coefficient.

A saturation limit belongs to stated conditions

Imagine a hypothetical solute whose equilibrium solubility at a stated temperature is 10.0 g per 100 g of a specified solvent. If 12.0 g is added to 100 g of that solvent, a simple no-reaction equilibrium model can leave 10.0 g dissolved and 2.0 g as excess solid. The denominator is solvent mass, not final solution mass.

An unsaturated solution lies below the saturation composition. A saturated solution is at that equilibrium composition under the stated conditions. A supersaturated solution exceeds the equilibrium value relative to the relevant solid phase and can persist metastably before crystallization. Its persistence is not a new higher equilibrium solubility.

Stirring changes the approach, not the fixed-condition limit

Stirring or reducing solid particle size often speeds dissolution by improving contact and transport. These changes do not normally alter the equilibrium solubility of the same phases at unchanged temperature, pressure and composition. Faster dissolution and a larger equilibrium dissolved amount are separate claims.

Temperature can alter solubility, but “hotter always dissolves more” is too broad. Many solids become more soluble on heating, while some show other trends. Gas solubility often decreases with increasing temperature in familiar settings, yet the specific gas, solvent and reactions must be considered.

A gas equilibrium needs its own pressure

For a dilute physically dissolved gas at fixed temperature, one Henry-law convention is c = kp. Here c is dissolved-gas amount concentration and p is the gas’s partial pressure above the liquid. The coefficient k has units that make that equation consistent. It depends on the gas, solvent and temperature.

An original hypothetical case stipulates k = 1.00 × 10⁻⁵ mol L⁻¹ kPa⁻¹. Gas partial pressures of 20.0 and 40.0 kPa give 2.00 × 10⁻⁴ and 4.00 × 10⁻⁴ mol L⁻¹. The plotted concentrations are expressed as 200 and 400 µmol L⁻¹ for readable axis numbers. They are invented model values rather than tabulated gas data.

Defined partial pressure controls this dilute model

Dissolved concentration / µmol L⁻¹

Gas partial pressure / kPa

Calculated dilute equilibrium states
— Defined Henry relation: c = kp

Underlying data

Gas partial pressure / kPa	Dissolved concentration / µmol L⁻¹
10	100 · Calculated dilute equilibrium states
20	200 · Calculated dilute equilibrium states
40	400 · Calculated dilute equilibrium states

Original hypothetical gas/solvent case at fixed T: c = kp with k = 1.00 × 10⁻⁵ mol L⁻¹ kPa⁻¹. The same coefficient is 10.0 µmol L⁻¹ kPa⁻¹. Markers and dashed line are calculated values, not measurements or a claim about total pressure.

The equation defines the constant

Another convention writes p = K_H x, using dissolved-gas mole fraction x. That coefficient is not the same quantity or unit as k in c = kp. Under the latter convention larger k means greater concentration at fixed p; under p = K_H x, larger K_H means smaller x at fixed p. Read the equation before comparing a constant.

Henry behaviour describes a dilute equilibrium limit. Reactions, gas speciation, high concentrations or pressures outside the relevant range can require a more complete description. Likewise, “immiscible” liquids can have small mutual solubilities; the label need not mean exactly zero material crosses between the phases.

Worked examples

Separate dissolved and excess material

Use the hypothetical solubility 10.0 g per 100 g solvent at fixed T. After equilibration, how is 12.0 g added solute distributed in 100 g solvent under the no-reaction model?

The specified solvent amount matches the solubility basis.
At equilibrium with excess solid, at most 10.0 g remains dissolved under this model.
Subtract: 12.0 − 10.0 = 2.0 g excess solid.

10.0 g dissolved and 2.0 g undissolved. These are stipulated exercise values, not a named substance’s measured solubility.

Use the defined coefficient

For the hypothetical c = kp model, calculate c at gas partial pressure 40.0 kPa with k = 1.00 × 10⁻⁵ mol L⁻¹ kPa⁻¹.

Multiply k by the specified gas partial pressure.
The kPa units cancel.
c = 1.00 × 10⁻⁵ × 40.0 = 4.00 × 10⁻⁴ mol L⁻¹.

4.00 × 10⁻⁴ mol L⁻¹, equivalently 400 µmol L⁻¹ at the displayed precision.

Try it yourself

At unchanged conditions, does faster stirring normally raise equilibrium solubility?
Show the answer and reasoning
No.
It can change the approach rate without changing the equilibrium composition.
Must every solid become more soluble when temperature rises?
Show the answer and reasoning
No.
Temperature trends depend on the solute, solvent and equilibrium.
Which pressure belongs in the displayed c = kp gas relation?
Show the answer and reasoning
The gas’s partial pressure.
Total pressure can also include other gases that do not belong in this coefficient’s relation.
Are k in c = kp and K_H in p = K_H x interchangeable numerical constants?
Show the answer and reasoning
No.
They use different equations, concentration bases and units.

What to carry forward

Solubility is an equilibrium composition at stated conditions.
Rate and equilibrium extent differ.
Henry relations use a defined equation and gas partial pressure.
A metastable supersaturated state is not a new equilibrium limit.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.