1. Why Do Some Solids Dissolve in Water?
The sugar we use to sweeten coffee or tea is a molecular solid, in which the individual molecules are held together by relatively weak intermolecular forces.
When sugar dissolves in water, the weak bonds between the individual sucrose molecules are broken, and these C12H22O11 molecules are released into solution.

2. Why Do Some Solids Dissolve in Water?
It takes energy to break the bonds between the C12H22O11 molecules in sucrose.
It also takes energy to break the hydrogen bonds in water that must be disrupted to insert one of these sucrose molecules into solution.
Sugar dissolves in water because energy is given off when the slightly polar sucrose molecules form intermolecular bonds with the polar water molecules.
The weak bonds that form between the solute and the solvent compensate for the energy needed to disrupt the structure of both the pure solute and the solvent.
In the case of sugar and water, this process works so well that up to 1800 grams of sucrose can dissolve in a liter of water.

4. Saturated Table Sugar Solution
Imagine there is a saturated sugar solution. (Saturated means that the maximum amount is dissolved in the solution, under normal conditions.) There are undissolved sugar crystals at the bottom of the solution. This can be shown by the equation,
The equation describes that sugar crystals (sugar(s)) will dissolve in water (H2O) and produce sugar molecules in solution (sugar(aq)).

5. Saturated Table Sugar Solution
Since the amount of sugar at the bottom does not change once equilibrium is attained, it would seem that the process stops. In other words, it seems that sugar does not go into solution or come out of solution anymore.
However, this is not true. The amounts of undissolved sugar crystals and sugar in solution do not change because the rate at which sugar molecules go into solution is the same rate as sugar molecules coming out of solution (forming crystals).

6. Saturated Table Sugar Solution
The animation represents this process:
the blue "molecules" escape into solution from the ordered crystal. At the same time, molecules are coming out of solution and depositing on the solid. Since this is a continual process and the concentrations do not change, it is called dynamic equilibrium.

7. Dissolving of Ionic Solids
Ionic solids (or salts) contain positive and negative ions, which are held together by the strong force of attraction between particles with opposite charges.
When one of these solids dissolves in water, the ions that form the solid are released into solution, where they become associated with the polar solvent molecules.

8. Dissolving of Ionic Solids

9. Solubility Equilibria

10. Solubility Equilibria

11. Saturated Table Salt Solution
Eventually, the Na+ and Cl- ion concentrations become large enough that the rate at which precipitation occurs exactly balances the rate at which NaCl dissolves.
Once that happens, there is no change in the concentration of these ions with time and the reaction is at equilibrium.
When this system reaches equilibrium it is called a saturated solution, because it contains the maximum concentration of ions that can exist in equilibrium with the solid salt.
The amount of salt that must be added to a given volume of solvent to form a saturated solution is called the solubility of the salt.

12. Saturated NaCl Solution
How can you define solubility?
Solubility is the maximum concentration of the solid in the solution at equilibrium or the maximum amount of the solid that will dissolve in a saturated solution.

13. Oversaturated Solutions
For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated.
These substances have a tendency to form oversaturated solutions.
For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids.
For oversatureated solutions, Qsp is greater than Ksp.
When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring Qsp to approach Ksp.

14. Oversaturated Solutions
Sodium acetate trihydrate, NaCH3COO.3H2O, when heated to 370 K will become a liquid.
The sodium acetate is said to be dissolved in its own water of crystallization.
The substance stays as a liquid when cooled to room temperature or even below 273 K.
As soon as a seed crystal is present, crystallization occur rapidly.
In such a process, heat is released, and the liquid feels warm so used as Hot Pack.

16. Supersaturated Thiosulphate

17. Solubility Rules

20. Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp is the solubility product constant
MgF2 (s) Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO33-]2
Ksp: indicates the solubility of an ionic compound
the smaller value, the less soluble the compound in water
(Ksp: only for salts of low solubility)

21. Precipitation Prediction
Using the ion product (IP or Qsp) to predict whether a precipitate will form
Qsp or IP: has the same expression as Ksp except the conc. of ions are not equilibrium conc. (these are initial conc.)
Dissolution of an ionic solid in aqueous solution:
IP< Ksp
Unsaturated solution
No precipitate
IP = Ksp
Saturated solution
IP > Ksp
Supersaturated solution
Precipitate will form

22. Graphical Representation of [Ag+] [Cl-] = Ksp
The condition
[Ag+] [Cl-] = Ksp, is represented by a curve which divides the plane into two zones.
When [Ag+] [Cl-] < Ksp, no precipitate will be formed.
When [Ag+] [Cl-] > Ksp, a precipitate will be formed.

23. The relationship among Qsp , Ksp and saturation

24. Calculating Solubility from Ksp
Q1. What is the solubility of silver chloride in mg/L and the molar concentrations of ions in the saturated solution?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = 1.69 x 10-10
Initial (M)
Change (M)
Equilibrium (M)
0.00
0.00
Ksp = [Ag+][Cl-] +s +s
Ksp = s2
s = Ksp  s s
s = 1.30 x 10-5 M
[Ag+] = 1.30 x 10-5 M
[Cl-] = 1.30 x 10-5 M
1.30 x 10-5 mol AgCl
1 L soln
g AgCl
1 mol AgCl x
Solubility of AgCl =
= 1.86 x 10-3 g/L
= mg/L

26. Calculating Ksp from Solubility
Q2. A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x 10-4 M. What is the Ksp for Ag2CrO4?
Ag2CrO4 (s)  2 Ag+ (aq) + CrO42- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.00
0.00 2s s
1.3 x 10-4
6.5 x 10-5
Ksp = [Ag+]2[CrO42-]
Ksp = (1.3 x 10-4 )2 (6.5 x 10-5) = 1.1 x 10-12

27. Calculating Ksp from Solubility
Q3. The solubility of silver sulfate is mol/L. This means that mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution. Calculate the value of the equilibrium constant, Ksp for this salt.
Ag2SO4 (s)  2 Ag+ (aq) + SO42- (aq)
+ 2s s
2s s
Ksp = [Ag+]2[SO42-] = (2s)2(s) = (4s2)(s) = 4s3
We know: s = mol/L
Ksp = 4(0.0144)3 = 1.2 x 10-5

28. MgF2 (s)  Mg2+ (aq) + 2 F- (aq)
Q4. Calculate the solubility of MgF2 (s) in a solution of M NaF. Ksp of MgF2 is 7.4 x 10-11
NaF (aq) Na+ (aq) + F- (aq)
0.080 M
0.080 M
0.080 M
MgF2 (s)  Mg2+ (aq) + 2 F- (aq) M
+ s s
sM sM
Ksp = 7.4 x = [Mg2+][F-]2 = (s)( s)2
Since Ksp is so small…assume that 2s << 0.080
7.4 x = (s)(0.080)2
s = 1.2 x 10-8 mol/L

29. Explaining the Common Ion Effect
The presence of a common ion in a solution will lower the solubility of a salt.
Le Chatelier’s Principle:
The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower!

30. Common Ion Effect Movie

31. Precipitation Reactions

32. References http://www.science.uwaterloo.ca/~cchieh/cact/c123/ksp.html