contents

Chemistry

8.1 - Solubility & Solubility Curves

5 min

note 1 of 6

Solute, solvent, solution and saturated solution

When a solid dissolves in a liquid, the liquid is the solvent and the dissolved solid is the solute.
Together they form a solution - a mixture of solute and solvent that doesn't separate back out on its own.
Soluble describes a substance that dissolves in a given solvent; insoluble describes one that won't. Common salt is soluble in water, but sand is not.

Saturated solutions

A saturated solution contains the maximum amount of dissolved solute possible at that temperature; no more solute will dissolve.

Three beakers in a row: blue particles labelled solvent with orange solute dropped in; the two mixed, labelled solution; then saturated solution, with extra orange solute piled on the bottom labelled undissolved solute.

note 2 of 6

Solubility and solubility curves

The solubility of a solid is the mass of it that dissolves in 100 g of water at a stated temperature to form a saturated solution.
A solubility curve is a graph plotting solubility against temperature.

How temperature affects solubility

For most solids, solubility increases as temperature increases.
For gases, solubility decreases as temperature increases.

Two graphs of solubility against temperature: solids solubility curve, where solid 1 rises steeply and solid 2 gently, and gases solubility curve, where gas 1 and gas 2 both fall. — Solubility changes much more with temperature for some substances than for others.

note 3 of 6

Calculating solubility from experimental data

Add excess solid to a known mass of water at a set temperature and stir until no more dissolves.
Filter the mixture, then dry and weigh the undissolved solid left on the filter paper.
Mass of solute dissolved = mass added − mass of dried excess.

Formula for solubility

$Solubility = \frac{mass of solute dissolved}{mass of water} \times 100$

Where:

Solubility is in grams of solute per 100 g of water
Mass of solute dissolved is in grams (g)
Mass of water is in grams (g)

note 4 of 6

Worked example - Solubility from experimental data

30 g of sodium nitrate is added to 25 g of water at 20 °C and stirred well. The mixture is filtered, and the solid left on the filter paper is dried and weighed at 8 g. Calculate the solubility of sodium nitrate at 20 °C.

Step 1 - Find the mass of solute that dissolved

Mass dissolved = 30 - 8 = 22 g

Step 2 - Apply the solubility formula

$\begin{matrix} Solubility & = \frac{22}{25} \times 100 \\ = 88 g per 100 g of water \end{matrix}$

note 5 of 6

Mass of solute deposited on cooling

When a saturated solution is cooled, solubility falls and the solute it can no longer hold crystallises out.
The mass deposited is the difference between the solubility at the starting temperature and the cooled temperature.

Formula for mass deposited

$Mass deposited = \frac{solubility change \times mass of water}{100}$

Where:

Mass deposited is the mass of solute that crystallises out on cooling, in grams (g)
Solubility change is the solubility at the starting temperature minus the solubility at the cooled temperature, in g per 100 g of water
Mass of water is the mass of water in the solution, in grams (g)

A rising solubility curve with dashed lines from 70 °C and 30 °C; the gap between their solubilities is labelled mass of solute deposited, and a leftward arrow is labelled the solution is cooled.

note 6 of 6

Worked example - Solute deposited on cooling

A saturated solution of ammonium sulfate containing 100 g of water is cooled from 80 °C to 20 °C. From the solubility curve, the solubility is 95 g per 100 g of water at 80 °C and 75 g per 100 g of water at 20 °C. Calculate the mass of solid that crystallises out.

Step 1 - Find the difference in solubility between the two temperatures

95 - 75 = 20 g per 100 g of water

Step 2 - State the mass deposited

The solution contains exactly 100 g of water, so 20 g of ammonium sulfate crystallises out.

Step 3 - Interpretation

If the solution had contained 40 g of water instead, less solid comes out in the same proportion:

$20 \times \frac{40}{100} = 8 g$

key terms

TermDefinition

SolventThe liquid in which a solute dissolves to form a solution.

SoluteThe substance that dissolves in a liquid to form a solution.

SolutionThe mixture formed when a solute dissolves in a solvent.

Saturated solutionA solution that contains the maximum amount of dissolved solute possible at a given temperature.

SolubilityHow much of a substance dissolves in a solvent. For a solid, it is usually given as the mass that dissolves in 100 g of water at a stated temperature to form a saturated solution.

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contents

Chemistry

8.1 - Solubility & Solubility Curves

5 min

note 1 of 6

Solute, solvent, solution and saturated solution

When a solid dissolves in a liquid, the liquid is the solvent and the dissolved solid is the solute.
Together they form a solution - a mixture of solute and solvent that doesn't separate back out on its own.
Soluble describes a substance that dissolves in a given solvent; insoluble describes one that won't. Common salt is soluble in water, but sand is not.

Saturated solutions

A saturated solution contains the maximum amount of dissolved solute possible at that temperature; no more solute will dissolve.

note 2 of 6

Solubility and solubility curves

The solubility of a solid is the mass of it that dissolves in 100 g of water at a stated temperature to form a saturated solution.
A solubility curve is a graph plotting solubility against temperature.

How temperature affects solubility

For most solids, solubility increases as temperature increases.
For gases, solubility decreases as temperature increases.

note 3 of 6

Calculating solubility from experimental data

Add excess solid to a known mass of water at a set temperature and stir until no more dissolves.
Filter the mixture, then dry and weigh the undissolved solid left on the filter paper.
Mass of solute dissolved = mass added − mass of dried excess.

Formula for solubility

$Solubility = \frac{mass of solute dissolved}{mass of water} \times 100$

Where:

Solubility is in grams of solute per 100 g of water
Mass of solute dissolved is in grams (g)
Mass of water is in grams (g)

note 4 of 6

Worked example - Solubility from experimental data

Step 1 - Find the mass of solute that dissolved

Mass dissolved = 30 - 8 = 22 g

Step 2 - Apply the solubility formula

$\begin{matrix} Solubility & = \frac{22}{25} \times 100 \\ = 88 g per 100 g of water \end{matrix}$

note 5 of 6

Mass of solute deposited on cooling

When a saturated solution is cooled, solubility falls and the solute it can no longer hold crystallises out.
The mass deposited is the difference between the solubility at the starting temperature and the cooled temperature.

Formula for mass deposited

$Mass deposited = \frac{solubility change \times mass of water}{100}$

Where:

Mass deposited is the mass of solute that crystallises out on cooling, in grams (g)
Solubility change is the solubility at the starting temperature minus the solubility at the cooled temperature, in g per 100 g of water
Mass of water is the mass of water in the solution, in grams (g)

note 6 of 6

Worked example - Solute deposited on cooling

Step 1 - Find the difference in solubility between the two temperatures

95 - 75 = 20 g per 100 g of water

Step 2 - State the mass deposited

The solution contains exactly 100 g of water, so 20 g of ammonium sulfate crystallises out.

Step 3 - Interpretation

If the solution had contained 40 g of water instead, less solid comes out in the same proportion:

$20 \times \frac{40}{100} = 8 g$

key terms

TermDefinition

SolventThe liquid in which a solute dissolves to form a solution.

SoluteThe substance that dissolves in a liquid to form a solution.

SolutionThe mixture formed when a solute dissolves in a solvent.

Saturated solutionA solution that contains the maximum amount of dissolved solute possible at a given temperature.

SolubilityHow much of a substance dissolves in a solvent. For a solid, it is usually given as the mass that dissolves in 100 g of water at a stated temperature to form a saturated solution.

end of notes

Key points to remember

Take a quiz View flashcards

9.1 - The Reactivity SeriesNext up