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Solubility and Le Chatelier′s Principle

Key Concepts

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Saturated Solutions at Equilibrium

Water is the most common solvent you will use.
When a solute dissolves in water we refer to the solution as an aqueous solution.
The solute can be a solid, liquid or gas.

We can prepare a saturated aqueous solution by using the known solubility of a solute in a specified amount of water at a specified temperature. The solubility of some solutes in water is shown in the table below1.

To prepare a saturated aqueous solution of copper(II) sulfate at 25oC we only need to add a minimum of 14 grams of solid copper(II) sulfate to 100 grams of liquid water (see table).
If we add more than 14 g of CuSO4(s) to 100 g of water at 25o, only 14 g will dissolve and the rest will not dissolve. The undissolved CuSO4(s) will appear as a blue solid, and, if left alone, will settle to the bottom of the vessel containing the solution.

When can represent this saturated solution as shown below:

O = dissolved copper sulfate particles
Black box represents the volume of 100 grams of water at 25oC
  = undissolved copper sulfate particles
O O
O O
O O
O
O O O
O O O O O
O O O O O O

Because all the particles are in constant motion, the lattice of copper sulfate particles in the undissolved solute is constantly being broken up and these freed particles being surrounded by water molecules which then disperse through the solution.
At the same time, dissolved copper sulfate particles in the solution, which are in constant motion, come into contact with the undissolved copper particles and "stick" to it, becoming part of the lattice.
At equilibrium, the rate at which solute particles leave the lattice and enter solution is the same as the rate at which dissolved solute particles leave solution and become part of the undissolved solute′s lattice.
So, at equilibrium, the apparent mass of undissolved solute particles remains constant, and the apparent mass of solute particles forming part of the solution also remains constant.
Therefore we can write a general chemical equation for the system in which a saturated solution is in equilibrium with its solute:

solute + solvent solution
Solubilities of some common substances in water at 25oC
Solute
Name
Solute
Formula
Solubility
g/100 g water
silver nitrate AgNO3(s) 245
sodium hydroxide NaOH(s) 80
sucrose C12H22O11(s) 70
lead(II) nitrate Pb(NO3)2(s) 60
ammonia NH3(g) 48
glucose C6H12O6(l) 45
ammonium chloride NH4Cl(s) 39
sodium chloride NaCl(s) 36
potassium chloride KCl(s) 35
magnesium sulfate MgSO4(s) 18
copper(II) sulfate CuSO4(s) 14
chlorine Cl2(g) 0.6
carbon dioxide CO2(g) 0.15
calcium hydroxide Ca(OH)2(s) 0.12
iodine I2(s) 0.03
oxygen O2(g) 0.004
nitrogen N2(g) 0.002
silver chloride AgCl(s) 0.0002

For a saturated aqueous solution in equilibrium with gaseous solute, gas particles dissolved in water are in equilibrium with undissolved gas particles above the solution and we can write:

general equations: gaseous solute + liquid water aqueous solution
solute(g) + H2O(l) solution(aq)
word equation example: ammonia gas + liquid water aqueous solution of ammonia
chemical equation example: NH3(g) + H2O(l) NH3(aq)

For a saturated aqueous solution in equilibrium with a molecular (covalent) solute which is a solid, molecules of solute dissolved in water are in equilibrium with undissolved molecules of solute and we can write:

general equations: solid solute + liquid water aqueous solution
solute(s) + H2O(l) solution(aq)
word equation example: sucrose + liquid water aqueous solution of sucrose
chemical equation example: C12H22O11(s) + H2O(l) C12H22O11(aq)

For a saturated aqueous solution in equilibrium with an ionic solid solute, undissolved ions making up the lattice are in equilibrium with dissolved ions in solution and we can write:

general equations: solid solute + liquid water aqueous solution
solute(s) + H2O(l) solution(aq)
word equation example: copper(II) sulfate solid + liquid water aqueous solution of copper(II) sulfate
molecular equation example: CuSO4(s) + H2O(l) CuSO4(aq)
ionic equation example: CuSO4(s) + H2O(l) Cu2+(aq) + SO42-(aq)

Note that, at constant temperature and pressure, we can NOT increase the concentration of a saturated solution by adding more solute!

Adding more CuSO4(s) to saturated CuSO4(aq) at constant temperature will NOT increase the concentration of CuSO4(aq) because the water has already dissolved as much CuSO4 as it can.
Adding more CuSO4(s) to saturated CuSO4(aq) at constant temperature will only deposit more CuSO4(s) on the bottom of the vessel.

Adding more NH3(g) to saturated NH3(aq) at constant temperature and pressure will NOT increase the concentration of NH3(aq) because the water has already dissolved as much of the NH3(g) as it can.
Adding more NH3(g) to saturated NH3(aq) at constant temperature and pressure will result in more NH3(g) being suspended above the solution, and, in order to maintain the same pressure, this will result in the gas occupying a greater volume (see Ideal Gas Law).

It is possible to force more solute to dissolve in a given amount of solvent, if you change the conditions, that is, if you disturb the equilibrium .....

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Applying Le Chatelier′s Principle to Saturated Solutions

Solubility tables like the one above, tell us the maximum mass of solute that will dissolve in a given mass of solvent under certain conditions. Typically, the conditions are a temperature of 25oC and a pressure of 101.3 kPa (1 atm).
Let′s see what happens when we change either of these conditions.

Changing Pressure

For a gaseous solute in equilibrium with its saturated aqueous solution, we can write
solute(g) + H2O(l) solute(aq)

By Le Chatelier′s Principle, increasing the partial pressure of a gaseous solute will favour the side of the equation with the fewest gas molecules in order to minimise the effect of the increased pressure, that is, the equilibrium position will shift to the right and the concentration of solution will increase according to Henry′s Law.
Similarly, decreasing the partial pressure of the gaseous solute will force the equilibrium position to shift to the left to increase the number of gas molecules and hence the pressure, so solute molecules escape from the solution phase and enter the gaseous solute phase and the concentration of the solution decreases.

In the school laboratory, increasing the pressure on a solid or liquid solute in equilbrium with its aqueous solution will have no noticeable effect.

Changing Temperature

Changing the temperature of a saturated solution will result in more, or less, of the solute being dissolved depending on whether the process is endothermic or exothermic.

When a solute dissolves in a solvent to form a solution, energy is either absorbed or released.

If heat is released when the solute dissolves:


By Le Chatelier′s Principle, if we add more heat to this system, the equilibrium position will shift to the left to minimise the effect of the change by consuming more heat, which will result in the concentration of the solution decreasing as more solute is formed on the reactant side of the equation.
Similarly, if the system is cooled, that is, heat is removed from the system, the equilibrium position will shift to the right to minimise the effect of the change by producing more heat, so the concentration of the solution increases as more solute on the left hand side of the equation dissolves in the solvent.

If heat is absorbed when the solute dissolves:


By Le Chatelier′s Principle, if we add more heat to this system, the equilibrium position will shift to the right to minimise the effect of the change by consuming more heat, which will result in the concentration of the solution increasing as more solute is dissolved in solvent.
Similarly, if the system is cooled, that is, heat is removed from the system, the equilibrium position will shift to the left to minimise the effect of the change by producing more heat, so the concentration of the solution decreases as more solute particles leave the solution phase.

This means that in order to understand the impact of a change of temperature on the concentration of a specified saturated solution, we need to know whether the dissolving process is endothermic or exothermic.

The energy absorbed or released per mole of solute is referred to as the molar heat of solution, or the molar enthalpy of solution, ΔHsoln.

If water is the solvent, then the energy absorbed or released per mole of solute is referred to as the molar heat of hydration, or the molar enthalpy of hydration, ΔHhyd.

The molar heat of hydration (molar enthalpy of hydration) is given for a number of different solutes in the table on the right2.

In general, the molar heat of hydration for gases is negative, ΔHhyd = - kJ mol-1, that is, dissolving a gas in water is an exothermic process, energy is released.

solute(g) + H2O(l) solute(aq) + ΔHhyd

solute(g) + H2O(l) solute(aq)     ΔHhyd = - kJ mol-1

In general, if a gaseous solute is in equilibrium with its saturated aqueous solution and we increase the temperature of the system, gas will escape from the solution which will decrease the concentration of the solution.
Similarly, if a gaseous solute is in equilibrium with its saturated aqueous solution and we decrease the temperature of the system, more gaseous solute will dissolve in the solvent to increase the concentration of the solution.

We can NOT make a similar generalisation about dissolving solids however, since we can see from the data in the table that some solids, like sodium hydroxide (NaOH), dissolve in water to release heat, while others, such as sodium nitrate (NaNO3), absorb heat when they dissolve in water.

Many salts, like potassium chloride (KCl), dissolve by absorbing energy from the surroundings (ΔHhyd = + kJ mol-1, endothermic) but there are also salts that release energy when they dissolve in water (ΔHhyd = - kJ mol-1, exothermic), such as lithium chloride (LiCl).

Molar Heat of Hydration at 25oC
Type of
Solute
Solute name
(formula)
ΔHhyd
kJ mol-1
Gaseous
solutes
hydrogen fluoride
HF(g)
-62
hydrogen chloride
HCl(g)
-75
ammonia
NH3(g)
-31
chlorine
Cl2(g)
-25
oxygen
O2(g)
-13
Solid
solutes
potassium hydroxide
KOH(s)
-58
sodium hydroxide
NaOH(s)
-45
lithium chloride
LiCl(s)
-37
lithium hydroxide
LiOH(s)
-24
potassium fluoride
KF(s)
-18
lithium nitrate
LiNO3(s)
-3
sodium fluoride
NaF(s)
+1
sodium chloride
NaCl(s)
+4
lithium fluoride
LiF(s)
+5
potassium chloride
KCl(s)
+17
sodium nitrate
NaNO3(s)
+21
amonium nitrate
NH4NO3(s)
+26
potassium nitrate
KNO3(s)
+35

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Temperature and Solubility

We saw above that changing the temperature of a system in which a solute is in equilibrium with its saturated aqueous solution will change the concentration of the solution.
Solubility of a solute refers to the maximum amount of solute that can be dissolved in a solvent under specified conditions.
Applying Le Chatelier′s Principle, we can see that

Exothermic process: ΔHhyd is negative

If a solute dissolves in a solvent to release energy, the process is exothermic, energy is a product, and the heat (enthalpy) of solution is negative.
For a saturated aqueous solution in equilibrium with its solute, we can write:

Gaseous
solute
solute(g) + H2O(l) solution(aq)   ΔHhyd = - kJ mol-1
solute(g) + H2O(l) solution(aq) +ΔHhyd kJ mol-1  
Solid
solute
solute(s) + H2O(l) solution(aq)   ΔHhyd = - kJ mol-1
solute(s) + H2O(l) solution(aq) +ΔHhyd kJ mol-1  

If we increase the temperature of the system, the equilibrium position will shift to the left according to Le Chatelier′s Principle in order to minimise the effect of the change by consuming some of this energy.
This means that the concentration of the solution will decrease and the amount of undissolved solute will increase.
Another way to say this is to say that the solubility of the solute decreases with increasing temperature.

Similarly, if we were to decrease the temperature of the system, the equilibrium position will shift to the right according to Le Chatelier′s Principle in order to minimise the effect of the change by producing more heat.
This means that the concentration of the solution will increase and the amount of undissolved solute will decrease.
Another way to say this is to say that the solubility of the solute increases with decreasing temperature.

If we were to graph the changes in solubility with temperature the graph would like the one shown on the right3.

  • As temperature increases, the amount of solute that can dissolve in 100 g of water decreases, that is, the solubility of the solute in water decreases.

  • As temperature decreases, the amount of solute that can be dissolved in 100 g of water increases, that is, the solubility of the solute in water increases.

solubility
( g/100 g water )
Solubility Changes with Temperature when ΔH is negative

temperature →

For example, consider a saturated aqueous solution of ammonia gas in equilibrium with gaseous ammonia at 25oC:

NH3(g) + H2O(l) NH3(aq)   ΔHhyd = -31 kJ mol-1
NH3(g) + H2O(l) NH3(aq) + 31 kJ mol-1  

If we increase the temperature of the system while maintaining constant pressure, the equilibrium position will shift to the left by Le Chatelier′s Principle so that the concentration of the aqueous ammonia solution will decrease and the amount of ammonia gas above the solution will increase.
The solubility of ammonia gas in water is said to decrease with increasing temperature.

Endothermic process: ΔHhyd is positive

If a solute absorbs energy from the surroundings when it dissolves in a solvent, the process is endothermic, energy is a reactant, and the heat (enthalpy) of solution is positive.
For a saturated aqueous solution in equilibrium with its solute, we can write:

Solid
solute
solute(s) + H2O(l) solution(aq)   ΔHhyd = + kJ mol-1
solute(s) + H2O(l) + ΔHhyd kJ mol-1 solution(aq)    

If we increase the temperature of the system, the equilibrium position will shift to the right according to Le Chatelier′s Principle in order to minimise the effect of the change by consuming more energy.
This means that the concentration of the solution will increase and the amount of undissolved solute will decrease.
Another way to say this is to say that the solubility of the solute increases with increasing temperature.

Similarly, if we were to decrease the temperature of the system, the equilibrium position will shift to the left according to Le Chatelier′s Principle in order to minimise the effect of the change by producing more heat.
This means that the concentration of the solution will decrease and the amount of undissolved solute will increase.
Another way to say this is to say that the solubility of the solute decreases with decreasing temperature.

If we were to graph the changes in solubility with temperature the graph would like the one shown on the right3.

  • As temperature increases, the amount of solute that can dissolve in 100 g of water increases, that is, the solubility of the solute in water increases.

  • As temperature decreases, the amount of solute that can be dissolved in 100 g of water decreases, that is, the solubility of the solute in water decreases.

solubility
( g/100 g water )
Solubility Changes with Temperature when ΔH is positive

temperature →

For example, consider a saturated aqueous solution of ammonium nitrate in equilibrium with solid ammonium nitrate at 25oC:

molecular equations NH4NO3(s) + H2O(l) NH4NO3(aq)   ΔHhyd = +26 kJ mol-1
NH4NO3(s) + H2O(l) + 26 kJ mol-1 NH4NO3(aq)    
ionic equations NH4NO3(s) + H2O(l) NH4+(aq) + NO3-(aq)   ΔHhyd = +26 kJ mol-1
NH4NO3(s) + H2O(l) + 26 kJ mol-1 NH4+(aq) + NO3-(aq)    

If we increase the temperature of the system, the equilibrium position will shift to the right by Le Chatelier′s Principle so that the concentration of the aqueous ammonium nitrate solution will increase and the amount of undissolved solid ammonium chloride will decrease.
The solubility of ammonium nitrate in water is said to increase with increasing temperature.

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1It is assumed that gaseous solutes are acting as Ideal Gases, that is, there is no chemical reaction between the gaseous solute and the solvent.

2The values given are for solutions at infinite dilution which is the enthalpy change when 1 mole of solute in its standard state is dissolved in an infinite amount of water.

3These graphs are only descriptive in order to show general trends. The solubility curve for an given solute is much more likely to be a curve than a straight line.