Solubility Curves (g/100 g)
Key Concepts
 Solubility of a solute in a particular solvent is the concentration of its saturated solution at the temperature specified.
 Solubility data for a solubility curve is usually expressed in units of grams of solute per 100 g of solvent (g/100 g) which is a weight ratio concentration (or mass ratio concentration)
 The solubility of a substance depends on:
(i) nature of the solute (intermolecular forces)
(ii) nature of the solvent (intermolecular forces)
(iii) temperature (Solubility & Le Chatelier's Principle)
 Solubility Rules (charts) and solubility tables (tables of solubility) are usually given for the solubility of substances in water at a specified temperature (often 25°C).
 Solubility curves show how the solubility of a solute in a given solvent changes as the temperature changes.
 Points on the solubility curve give the concentration, in g/100 g, of a saturated solution at that temperature.
Points below the solubility curve give concentrations for unsaturated solutions.
Points above the solubility curve give concentrations for supersaturated solutions.
 Solubility curves can be used to
(i) determine how much solute will dissolve in the solvent at a given temperature
(ii) compare the solubilities of different solutes in the same solvent
(iii) determine the mass of solid precipitated, or crystallised, out of solution when the temperature of the solution changes
Units of Solubility
Why use grams of solute per 100 g of water as the unit of choice for describing solubility?
Mass is chosen primarily because the mass of a substance does not change with temperature.
The density of water changes with temperature:
temperature °C 
density g cm^{3} 
Trend 
Comment 
4 
0.9999 
most dense 
The density of liquid water is a maximum value of 0.9999 g cm^{3} at 4°C . As water is heated, its density decreases. 
10 
0.9997 
↓ 
20 
0.9982 
↓ 
30 
0.9957 
↓ 
40 
0.9922 
least dense 
This means that if you measure out 100 mL (100 cm^{3}) of water at 20°C in a 100 mL volumetric flask, it will have a mass of:
Write the equation for density: 
density (g cm^{3}) 
= 
mass (g) volume (cm^{3}) 
Multiply both sides of equation by volume: 
density (g cm^{3}) × volume (cm^{3}) 
= 
mass (g) × volume (cm^{3})
volume (cm^{3}) 
and simplify: 
density × volume 
= 
mass (g) 
substitute in values: 
0.9982 × 100 
= 
mass (g) 
and solve: 
99.82 g 
= 
mass (g) 
100 mL of water at 20°C has a mass of 99.82 g
If we then heat this 99.82 g of water in the 100 mL volumetric flask to 40°C we could see the water level rise above the mark as the volume increases (as a result of the decrease in density at higher temperature), we can calculate the volume of this mass of water at 40°C:
Write the equation for mass given density: 
mass (g) 
= 
density × volume 
divide both sides of equation by density: 
mass (g) density (g cm^{3}) 
= 
density (g cm^{3}) × volume (cm^{3})
density (g cm^{3}) 
simplify: 
mass (g) density (g cm^{3}) 
= 
volume (cm^{3}) 
substitute values for 99.82 g water at 40°C: 
99.82 (g) 0.9922 (g cm^{3}) 
= 
volume (cm^{3}) 
solve for 99.82 g water at 40°C: 
100.6 cm^{3} 
= 
volume (cm^{3}) 
At 20°C, 99.82 g of water has a volume of 100 cm^{3}, but if you increase the temperature to 40°C, the same mass of water has a greater volume, that is, 100.6 cm^{3}.
This means that you can't use the volume of water to determine concentration if you are going to the change the temperature because the volume of the water also changes with temperature!
Solubility data are therefore given as a ratio of the mass of solute to a fixed mass of solvent.
Typically, the units are grams of solute per 100 g of solvent (g/100 g).
For aqueous solutions, the units are grams of solute per 100 g of water (g/100 g).
Solubility Curves
Many salts, such as sodium chloride (NaCl_{(s)}), are soluble in water at 25°C.
But what happens if we change the temperature of the solution?
The table below shows experimentally determined values for the solubility of sodium chloride in water at various temperatures.
Note the units are grams of solute dissolved in 100 g of water (g/100 g).
The points have been plotted on a graph to produce a solubility curve.
Solubility of NaCl in water 
Temperature °C 
Solubility g/100 g 
Solubility Curve 
0 
35.65 
mass solute (g) per 100 g water

temperature (°C)


10 
35.72 
20 
35.89 
30 
36.09 
40 
36.37 
50 
36.69 
60 
37.04 
70 
37.46 
80 
37.93 
90 
38.47 
100 
38.99 
Remember that solubility refers to the maximum mass of solute that can be dissolved in a given mass of solvent at a specified temperature.
In this case the solute is sodium chloride (NaCl_{(s)}) and the solvent is 100 g of water.
Each point on the curve in the graph above tells how much solute we can add to 100 g of water at that temperature in order to form a saturated solution.
That is, the curve in the graph represents the concentration, in grams of NaCl_{(s)} per 100 g of water, of a saturated aqueous solution of sodium chloride (NaCl_{(aq)}) at different temperatures.
The first thing we notice is that more NaCl_{(s)} can be dissolved in 100 g of water if we increase the temperature.
At 20°C (about room temperature), dissolving 35.89 g of NaCl_{(s)} in 100 g of water produces a saturated solution.
If the temperature is increased to 60°C (temperature of hot water from the tap), 1.14 g more of NaCl_{(s)} needs to be dissolved in the same 100 g of water to make a saturated solution.
What does this tell us about the region in the graph below the solubility curve?
It tells us that at any point below the solubility curve the solution is unsaturated, that is, more solute could be added to the water at that temperature until you reach the line at which point the solution will be saturated.
mass NaCl (g) per 100 g water

temperature (°C)

Consider point A, 37 g of NaCl_{(s)} will dissolve in 100 g of water at 80°C.
This solution will be unsaturated because it is possible to dissolve more than 37 g of NaCl_{(s)} in 100 g of water at this temperature.

But what happens if we could instantaneously cool the solution made up of 37 g of NaCl_{(s)} dissolved in 100 g of water from 80°C down to 20°C?
mass NaCl (g) per 100 g water

temperature (°C)

Solution containing 37 g of NaCl_{(s)} dissolved in 100 g of water at 80°C, point A, is instantaneously cooled to 20°C, point B.
Temporarily the water has more NaCl dissolved in it than is possible based on the solubility curve data. This is referred to as a supersaturated solution.

mass NaCl (g) per 100 g water

temperature (°C)

This supersaturated solution at point B is not stable.
NaCl_{(s)} will begin to precipitate out until the amount of dissolved NaCl in the 100 g of water now at 20°C is matched by that shown by the solubility curve at C.

Different substances will have different solubility curves.
For lots of salts, like sodium chloride, the solubility of the salt in water increases with increasing temperature.
But for gases and some other salts, like lithium sulfate, the solubility decreases as the temperature increases.
Solubility of Li_{2}SO_{4} in water 
Temperature °C 
Solubility g/100 g 
Solubility Curve 
0 
36.1 
mass Li_{2}SO_{4} (g) per 100 g water

temperature (°C)


10 
35.5 
20 
34.8 
30 
34.2 
40 
33.7 
60 
32.6 
80 
31.4 
90 
30.9 
As shown above, the mass of lithium sulfate that can be dissolved in 100 g of water actually decreases as you increase the temperature of the solution.
Comparing Solubility
The solubility of different solutes in 100 g of water at different temperatures is usually plotted on the same set of axes.
The solubility curves for sodium chloride (NaCl) and lithium sulfate (Li_{2}SO_{4}) are plotted in the graph below:
mass solute (g) per 100 g water

temperature (°C)


The graph shows that as temperature increases
 solubility of NaCl increases
 solubility of Li_{2}SO_{4} decreases
From the graph we see that:
 At 0°C, lithium sulfate (36.10 g/100 g) is a bit more soluble than sodium chloride (35.65 g/100 g)
 At 20°C, lithium sulfate (34.80 g/100 g) is less soluble than sodium chloride (35.89 g/100 g)
From the graph we can see that if we have a solution containing 35 g of solute in 100 g of water at 40°C, then
 if the solute is Li_{2}SO_{4}, the solution will be supersaturated
 if the solute is NaCl, the solution will be unsaturated
Crystallisation (precipitation)
A supersaturated solution is one in which the mass of solute dissolved is more than that required for the solution to be saturated.
For a supersaturated solution of a solid dissolved in water, the solid (crystals or precipitate) will spontaneously form to reduce the mass of solute dissolved in the solution until the solution becomes saturated.
We can use the solubility curve to predict the mass of solid that should precipitate (or crystallise) out of solution.
Let's assume you have a supersaturated solution that contains 35.00 g of litium sulfate dissolved in 100 g of water at a temperature of 30°C as shown by point A (×) on the graph below:
mass solute (g) per 100 g water

temperature (°C)


Reading off the light blue solubility curve for Li_{2}SO_{4} at a temperature of 30°C, we see that a saturated solution will contain 34.20 g of Li_{2}SO_{4} dissolved in 100 g of water.
We predict that the "excess" mass of Li_{2}SO_{4} in the supersaturated solution will precipitate out of solution (crystallise) until the solution becomes saturated:
Solution 
Concentration (g/100 g) 
Mass solute (g) (in 100 g water) 
supersaturated 
35.00 g/100 g 
35.00 g 
saturated 
34.20 g/100 g 
34.20 g 

"excess" mass 

35.00  34.20 = 0.80 g 
We predict that 0.80 g of Li_{2}SO_{(s)} will precipitate, or crystallise, out of this solution.
Worked Example (using the StoPGoPS approach to problem solving)
Chris the Chemist has dissolved 19.50 g of copper(II) sulfate in 50.00 g of water in a 250 mL conical (erlenmeyer) flask at a temperature of 60°C.
Chris cooled the solution to 20°C and observed the formation of blue copper(II) sulfate crystals.
Predict the mass of copper(II) sulfate crystals in the conical (erlenmyer) flask at 20°C.
The solubility curve for copper(II) sulfate in water is shown below:
mass solute (g) per 100 g water

temperature (°C)


What is the question asking you to do?
Calculate the mass of copper(II) sulfate crystals.
m(CuSO_{4(s)}) = ? g
What information have you been given in the question?
mass copper(II) sulfate dissolved at 60°C = m(CuSO_{4(dissolved)}) = 19.50 g
mass of water = m(H_{2}O) = 50.00
From the solubility curve: concentration of saturated solution at 20°C = 32.00 g/100 g
What is the relationship between what you have been given and what you need to find out?
mass CuSO_{4(s)} = mass CuSO_{4(dissolved)}  mass CuSO_{4(saturated)}
Note: Chris the Chemist has a solution containing 50 g of water, NOT 100 g of water.
(i) Calculate mass CuSO_{4} in saturated solution containing 50 g of water.
(ii) Subtract this mass of CuSO_{4} from the original mass of CuSO_{4} present in the "hot" solution.
Perform the calculations
(i) Calculate mass of CuSO_{4} present in a saturated solution of 50 g water at 20°C:
m(CuSO_{4(saturated)}) g 50 g water 
= 
32.00 g 100 g water 
Multiply both sides by 50 g water 
m(CuSO_{4(saturated)}) g × 50 g water
50 g water 
= 
32.00 g × 50 g water 100 g water 
Simplify the equation 
m(CuSO_{4(saturated)}) g 
= 
32.00 g × 50 g water 100 g water 
Solve the equation 
m(CuSO_{4(saturated)}) g 
= 
16.00 g 
(ii) Calculate mass of CuSO_{4} that precipitates (crystallises) out of solution when cooled.
Write the equation: 
mass CuSO_{4(s)} 
= 
mass CuSO_{4(dissolved)} 
 
mass CuSO_{4(saturated)} 
Substitute the values into the equation: 
mass CuSO_{4(s)} g 
= 
19.50 g 
 
16.00 g 
Solve the equation: 
mass CuSO_{4(s)} g 
= 
3.50 g 


Is your answer plausible?
Mass of crystals formed must be less than 19.50 g because we can't crystallise out more CuSO_{4} than was present in the original solution.
3.5 g < 19.50 g so we are confident our answer is plausible.
The mass of CuSO_{4(s)} must be a positive value, that is, mass CuSO_{4} dissolved in "hot" solution must be greater than the mass of CuSO_{4} dissolved at lower temperature in order for some solid to precipitate (crystallise) out of solution.
19.50 g > 16.00 g so we are confident our calculation is plausible.
Our value for mass of CuSO_{4(s)} is positive so we are confident our answer is plausible.
State your solution to the problem.
3.50 g of CuSO_{4(s)} precipitates (crystallises) out of solution at 20°C.