Solubility and Le Chatelier's Principle Chemistry Tutorial

Key Concepts

• A solution is formed when a solute dissolves in a solvent.
• Solubility refers to the maximum amount of solute that can be dissolved in the solvent at a specified temperature and pressure.
• A saturated solution is one in which no more solute can be dissolved in a given amount of solvent at a specified temperature and pressure.
• If excess solute is present, then the undissolved solute is in equilibrium with dissolved solute in the saturated solution.
• When a solute dissolves in a solvent heat is either :

  (i) absorbed (endothermic, ΔH is a positive value)
  or
  (ii) released (endothermic, ΔH is a negative value)

• The amount of heat absorbed or released when the solute dissolves in the solvent is known as its heat of solution (or enthalpy of solution), symbol is ΔH_soln

  If water is the solvent, the term heat of hydration, or enthalpy of hydration, is used and the symbol is ΔH_hyd

• Le Chatelier's Principle can be used to decide the effect of changing the temperature or the pressure of a saturated solution at equilibrium.
• Temperature changes:

  (i) Solubility of a solute decreases with increasing temperature if ΔH_soln is negative (exothermic)

  (ii) Solubility of a solute increases with increasing temperature if ΔH_soln is positive (endothermic)

• Pressure changes:

  (i) Solubility of a gaseous solute increases if its partial pressure increases.

  (ii) Solubility of a gaseous solute decreases if its partial pressure decreases.

Saturated Solutions at Equilibrium

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 .....

Applying Le Chatelier's Principle to Saturated Solutions

Changing Pressure

solute(g) + H₂O(l) solute(aq)

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

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:

• the process is said to be exothermic

• heat is a product of the reaction
  for a saturated solution in equilibrium with its solute:
  solute + solvent solution + ΔH

• the sign for the enthalpy change is negative
  for a saturated solution in equilibrium with its solute:
  solute + solvent solution     ΔH = - kJ mol^-1

If heat is absorbed when the solute dissolves:

• the process is said to be endothermic

• heat is a reactant
  for a saturated solution in equilibrium with its solute:
  solute + solvent + ΔH solution

• the sign for the enthalpy change is positive
  for a saturated solution in equilibrium with its solute:
  solute + solvent solution     ΔH = + kJ mol^-1

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

Temperature and Solubility

• if ΔH_hyd is negative: solute + H₂O(l) solute(aq) + ΔH_hyd
  (i) decreasing temperature increases concentration of solution (more solute dissolves),
  therefore more solute can be dissolved in the water and solubility of solute increases
  (ii) increasing temperature decreases concentration of solution (less solute dissovles),
  therefore less solute can be dissolved in the water and solubility of solute decreases

• if ΔH_hyd is positive: solute + H₂O(l) + ΔH_hyd solute(aq)
  (i) decreasing temperature decreases concentration of solution (less solute dissolves),
  therefore less solute can be dissolved in the water and solubility of solute decreases
  (ii) increasing temperature increases concentration of solution (more solute dissovles),
  therefore more solute can be dissolved in the water and solubility of solute increases

Exothermic process: ΔH_hyd is negative

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.

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

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: ΔH_hyd is positive

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.

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

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.

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.