Year 11 Chapter 11:~ Measuring Solubility
11.1 Measuring Solubility
The solubility of a substance refers to the maximum amount of that substance that can
be ____________ in a given quantity of a solvent at a certain temperature.
A solution that no more solute can be dissolved at that temperature is described as a
____________ solution. To measure solubility you need to determine the maximum
amount of solute that can be dissolved in ______ grams of solvent.
Solubility values allow you to compare the extent to which different solutes dissolve.
Example:
A maximum of 6g of solute can be dissolved in 20g of water at 20°C. What is the
solubility?
Solubility Curves – The relationship between solubility and temperature can be
represented by these curves (see Figure 11.1).
Solubility (g/100g)
Supersaturated
Saturated
Unsaturated
Temperature (°C)
Example:
An 80g sample of NaNO3 is added to 200g of H2O at 20°C. Use the solubility curve in
Fig 11.1 to calculate how much more NaNO3 needs to be added to make the solution
saturated with NaNO3 at 20°C.
Crystallisation
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.
Example:
What happens if a solution containing 50g of potassium nitrate in 100g of water is
allowed to cool from 40°C to 20°C?
Supersaturation
With some substances, it is possible to produce an unstable solution that contains more
dissolved solute that in a saturated solution, this is a supersaturated.
A supersaturated solution of potassium nitrate can be prepared by cooling a saturated
solution very carefully so that crystallisation does not occur.
Questions: 1, 2, 3, 4 & 5.
Solubility of Gases
Gases such as oxygen and carbon dioxide are much less soluble in water then solids such
as sodium chloride and sucrose. Aquatic life relies on dissolved oxygen in the water and
aquatic plants rely on dissolved _______ ________ .
The solubility of a particular gas in a liquid depends on the _____________ of the
liquid and the pressure of the gas.
Temperature and gas solubility
Unlike most solids, gases become _______ soluble as the temperature increase. You may
have notice when you heat water small bubbles of air form and escape from the water.
Pressure and gas solubility
The solubility of gases increases with ____________ pressure. When you open a can or
bottle of soft drink you hear the sound of the gas escaping.
Example: Using table 11.2
Calculate the percentage of oxygen that would be lost from a pond, saturated with
oxygen, if the temperature of the water in the pond rose from 0°C to 20°C.
Mass of Oxygen in 1kg:
at 0°C =
_____ g
at 20°C =
_____ g
So, mass of oxygen lost:
Therefore % of oxygen lost
= ______ X 100
0.069
= ______ %
QUESTIONS: 6, 7, & 15.
11.2 Concentration of Solutions
The concentration of a solution describes the relative amounts of solute and solvent
present.
 A solution in which a ratio of solute to solvent is high is said to be
________________ .
 A solution in which a ratio of solute to solvent is low is said to be
______________ .
You have measured concentration in g/100g so far, there are other ways of expressing
concentration where they measure the amount of solute in a given amount of solution.
 Mass of solute per litre of solution: g/L (g L-1) or mg/L
 Amount, in mol, of solute per litre of solution: mol/L (mol L-1) or M
*Example of cordial
Symbol or Unit
Mass of solute per litre of Solution
This unit expresses concentration in terms of the mass solute present in 1 litre of
solution (g/L)
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.
1mL of water = 1g of water
m(SO42-) = 4.0mg
m(solvent) = 250mL = 0.250L
mass of solute (mg)
Concentration =
volume of solution (L)
Microl
Millil
Litre
L
itre
itre
Amount, in mol, of Solute per
litre
of
Solution
mL
L
Kiloli
tre
kL
This allows chemists to compare relative numbers of atoms, molecules or ions present in
a given volume of solution. The measure of concentration is known as molarity (M) or
molar concentration.
Molarity if defined as the number of moles of solute particles per _________ of
solution. A one molar (1M) solution contains _____ mole solute dissolved in each one
_______ of solution: 1 mole per litre, 1 mol L-1, 1mol/L or M.
The amount of solute is linked to the concentration (molarity) and volume of the
solution:
amount = concentration x
Volume
n = c x V
mol
mol L-1
L
Unit Conversion
gram
mole
s
s
Examples:
1. Calculate the molarLitr
concentration of a solutionLitr
that contains
e
potassium nitrate dissolved
in 200mL of solution: e
0.105 mol of
n = 0.105 mol
V = 200ml = _____ L
c = ?
n .
C(KNO3) = V
2. Calculate the amount, in moles, of ammonia (NH3) in 25.0mL of a 0.3277M
ammonia solution.
n = ?
V = 25.0mL = _______ L
c = 0.3277M
3. Calculate the concentration, in mol L-1, of a solution that contains 16.8mg of
silver nitrate (AgNO3) dissolved in 150mL of solution.
Step 1 Convert:
n = ?
Two unknowns though we do have
V = 150mL = 0.150 L
mass so we can work out mol (n).
c = ?
Step 2: Calculate n using mass equation
n = ?
m = 16.8mg = 0.0168g
M = 169.9 g mol-1
M(AgNO3)= 107.87 + 14.01 + 16.00x3
= 169.9 g mol-1
Step 3: Calculate c using concentration equation
Dilution
The process of adding more solvent to a solution is known as dilution. When a solution is
diluted, the solute particles are more widely spread.
Looking at figure 11.11 The concentration of the solutions has decreased by a factor of
5, however the amount of solute has not changed though they are just spaced further
apart.
Amount doesn’t change therefore n does not change.
Dilution Formula:~ c1V1 =c2V2
c1
V1
c2
V1
=
=
=
=
initial conc
intial volume
final conc
final volume
Can be measured
in mL or L as
long as they’re
the same.
Example:
1. The concentration of a seaweed extract in a bottle of seaweed fertiliser
solution is 9.0 g L-1. When used to fertilise plants the seaweed extract must
be diluted. If 10mL of seaweed fertiliser is diluted with water to fill a 2.0L
container, what is the new concentration?
c1V1 =c2V2
c1 = 9.0 g/L
V1 = 10mL = 0.010L
C2 = ?
V2 = 2.0L
2. How much water must be added to 30mL of 2.5M solution of a sodium
hydroxide solution to dilute it to 1.0M.
QUESTIONS:
 10, 11, 12ac, 13bd, E12345, 17, 19, 21, 22ac, 23ab, 24 & 26.
 27, 28, 29, 30, 31, 32, 33 & 36.