Chapter 11 measuring solubility
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Week 1, Lesson 1. Chapter 11 – Measuring Solubility. Solubility. Solubility of a substance refers to the maximum amount of that substance that can be dissolved in a given quantity of solvent at a certain temperature.

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Chapter 11 – Measuring Solubility




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Chapter 11 measuring solubility

Week 1, Lesson 1

Chapter 11 – Measuring Solubility




  • A solution in which no more solute can be dissolved at that temperature is described as a saturated solution.

  • One way of measuring solubility is to determine the maximum mass of solute that can be dissolved in 100 grams of solvent at a particular temperature.

  • Solubility values allow us to compare the extent to which different solutes dissolve.

Solubility curves

Solubility Curves

  • The relationship between solubility and temperature can be represented by a solubility curve.

  • Each point on the solubility curve represents a saturated solution – no more solute can be dissolved at the particular temperature.

  • Any point below the curve represents an unsaturated solution for that solute.

  • Any unsaturated solution contains less solute than is needed to make a solution saturated at that temperature.

Solubility curve cont

Solubility Curve cont…



  • If a hot saturated solution of potassium nitrate is cooled, crystals of the solute will appear.

  • This happens because potassium nitrate also becomes less soluble as the temperature falls.

  • The potassium nitrate crystallises from solution.

  • This process is known as crystallisation.



  • With some substances, it is possible to produce an unstable solution that contains more dissolved solute than in a saturated solution.

  • Such a solution is said to be supersaturated.

  • In a solubility graph, any point above a solubility curve represents a supersaturated solution for that solute.

  • A supersaturated solution of potassium nitrate can be prepared by cooling a saturated solution very carefully so that crystallisation does not occur.

  • Adding a small seed crystal or knocking a supersaturated solution can cause the solute to crystallise or settle out until a saturated solution remains.

  • Crystallisation is often accompanied by the evolution of much heat energy.

Solubility of gases

Solubility of Gases

  • Gases such as oxygen and carbon dioxide are much less soluble in water than solids such as sodium chloride and sucrose.

  • The presence of oxygen and carbon dioxide is essential for the survival of aquatic life.

  • Aquatic plants and animals require a source of oxygen for respiration and although oxygen is a part of all water molecules, they cannot extract it from these molecules.

  • They must rely on the small amount of dissolved oxygen in the water.

  • Similarly, aquatic plants rely on dissolved carbon dioxide for photosynthesis.

  • The solubility of a particular gas in a liquid depends on the temperature of the liquid and the pressure of the gas.

Temperature and gas solubility

Temperature and Gas Solubility

  • Unlike most solids, gases become less soluble as the temperature increases.

  • Thermal pollution: a power station located next to a river or lake can release hot water as a waste material.

  • As the temperature of the water increases, less oxygen remains dissolved.

  • In addition, at increased temperatures, aquatic life becomes more active and so the demand for oxygen increases.

  • Competition for oxygen can become intense and can lead to large decreases in certain animal populations.

Pressure and gas solubility

Pressure and Gas Solubility

  • The solubility of gases increases with increasing pressure.

  • Carbon dioxide is forced into soft drink under high pressure to increase the amount that can be dissolved.

  • When the bottle top is removed, the carbon dioxide in the space at the top of the bottle escapes.

  • The gas pressure above the drink falls and so the solubility of carbon dioxide in the drink is lowered.

  • Dissolved carbon dioxide escapes from the liquid until only enough remains to saturate the solution at this pressure.

Solubility question examples

Solubility Question Examples…

Use the data in Table 11.2 to calculate the percentage of oxygen that would be lost from a pond, saturated with oxygen, if the temperature of water in the pond rose from 0°C to 20°C

Mass of oxygen, in 1kg saturated solution of oxygen;

= 0.069g at 0°C

= 0.043g at 20°C

So, mass of oxygen lost:

0.069 – 0.043 = 0.026g

% of oxygen lost =

= 38%

Concentration of solutions

Week 1, Lesson 2

Concentration of Solutions

  • The concentration of a solution describes the relative amounts of solute and solvent present.

  • A solution in which the ratio of the solute to solvent is high is said to be concentrated.

  • A solution in which the ratio of solute to solvent is low is said to be dilute.

Concentrations of solutions cont

Concentrations of Solutions cont…

  • Chemists use different measures of concentration depending on the particular situation.

  • Earlier, units of grams of solute per 100 grams of solvent were used to describe the concentration of saturated solutions.

  • Other ways of expressing concentration describe the amount of solute in a given solution.

  • They vary only in the units used to measure the amount of solute and the amount of solution.

Units of measurement

Units of Measurement

  • The most commonly used units of concentration are:

    • Mass of solute per litre of solution

    • Amount, in mol, of solute per litre of solution

Mass of solute per litre of solution

Mass of Solute per Litre of Solution

  • This unit expresses concentration in terms of the mass of solute present in 1 litre of solution.

  • Example: A 250mL glass of orange-flavoured mineral water contains 4.0mg of sulfate ions. What is the concentration (in mgL-1) of sulfate ions in the mineral water?

    Concentration = mass of sulfate ions(mg)

    volume of mineral water (L)

    = 4.0


    = 16mg/L or 16mg L-1



Amount in mol of solute per solution

Amount, in mol, of solute per solution

  • Expressing concentration in moles per litre of solution allows chemists to compare relative numbers of atoms, molecules or ions present in a given volume of solution.

  • This measure of concentration is known as molarity or molar concentration.

  • Molarity is defined as the number of moles of solute particles per litre of solution.

  • A one molar (1 M) solution contains one mole of solute dissolved in each litre of solution.

  • A concentration of such a solution is said to be one mole per litre, 1 mol L-1 or 1 M.

  • The term molarity can be used to mean concentration measured in moles per litre.

Amount of solute equation

Amount of Solute Equation

  • The amount of solute is linked to the concentration (molarity) and volume of the solution by the relationship:

    n = c x V

    Where n = amount, mol

    c = concentration, mol L-1

    v = volume, L



  • Calculate the molar concentration of a solution that contains 0.105 mol of potassium nitrate dissolved in 200mL of solution.

    STEP 1: Convert volume into L

    200mL = 200 x 10-3L

    STEP 2: Calculate the molar concentration

    c = n / v

    c = 0.105 / 200 x 10-3

    c = 0.525M

Unit conversion

Unit Conversion

grams M moles

litre x M litre

Molar Concentration and Solubility



  • The process adding more solvent to a solution is known as dilution.

  • When a solution is diluted, the solute particles are more widely spaced.

  • A dilution formula exists, which can be used to find the concentration or volume changes of a solution.

Dilution formula

Dilution Formula

C1V1 = C2V2

C1 = Initial concentration

V1 = Initial volume

C2 = Final concentration

V2 = Final volume

In this equation, the concentration unit is g L-1



  • The concentration of a seaweed extract in a bottle of seaweed fertiliser solution is 9.0gL-1. When used to fertilise plants, the seaweed fertiliser must be diluted. If 10mL of seaweed fertiliser is diluted with water to fill a 2.0L container, what is the new concentration?

    C1V1 = C2V2C1V1 = C2V2

    C1 = 9.09.0 x 0.010 = C2 x 2.0

    V1 = 0.010L0.09 = C2 x 2.0

    C2 = ?C2 = 0.09 / 2.0

    V2 = 2.0C2 = 0.045g L-1

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