Solubility refers to the maximum amount of a substance that can dissolve in a solvent to form a homogeneous solution at a given temperature and pressure. In comparison, solubility product (Ksp) is a type of equilibrium constant related to the solubility of a sparingly soluble ionic compound. Both solubility (S) and solubility product (Ksp) can be determined for given salt from each other.

Considering the relation between solubility and solubility product one can study the solubility of ionic compounds.

For example, the lead sulfate dissociates into its ions as given in the equation below:

PbSO4 ⇌ Pb2+ + SO42-

The equilibrium constant for this reaction is given by:

Kc = [Pb2+] [SO42-] / [PbSO4]

Note that, the concentration of lead sulfate almost remains constant. This is because of the low solubility. So, When this solid combines with the Kc it gives a new constant known as the solubility product.

Kc [PbSO4] = [Pb2+] [SO42-]

  • Kc [PbSO4] = Ksp

Ksp = [Pb2+] [SO42-]

= [S][S]

Thus for lead sulfate the solubility and solubility product can be related as follows:

Ksp = S2

  • Ksp is known as the solubility product of sulfate
  • S is the solubility

It is to be noted the solubility of salt is different at different temperatures.

Determination of solubility Ksp from solubility

Solubility can be calculated from solubility product and vice versa. This can be done by using a balanced chemical equation. First, we find the moles of each ion and then the solubility product.

Example # 01

The solubility of PbF2 is 0.64 gdm-3. Calculate Ksp of PbF2.

Solution:

First, find out moles from concentration.

Mass of PbF2 dissolved  = 0.64gdm-3

Molecular mass of PbF2 = 245.2 gmol-1

No of moles of PbF2 = 0.64 gdm-3/ 245.2 gmol-1 = 2.6 × 10-3

The balanced chemical equation for PbF2 is the following:

PbF2                              ⇌                      Pb2+   +         2F– 

initially,                                      2.6 × 10-3                                              0       +            0

At equilibrium,                           zero moles                                     2.6 × 10-3 + 2(2.6 × 10-3)

The expression of Ksp is:

Ksp = [Pb2+][F]2

Putting the values

Ksp =  (2.6 × 10-3 ) x {2(2.6 × 10-3)}2 = 7 × 10-8

Ksp= 7 × 10-8 units

Example # 02

The solubility of the strontium oxalate is 0.00054M. Calculate the solubility product.

The balanced chemical equation for strontium oxalate is the following:

SrOx ⇌ Sr2+ + Ox2-

Therefore,

[Sr2+] = [Ox2-] = 0.00054M = 5.4 × 10-4

The expression of Ksp is the following:

Ksp = [Sr2+] [Ox2-]

Putting the values

Ksp = (5.4 × 10-4)(5.4 × 10-4) = 2.92 × 10-7mol(dm-3)2

Hence,

Ksp= 2.92 × 10-7mol(dm-3)2

Determination of solubility from Ksp

Example # 01 

Ca(OH)2 is a sparingly soluble substance. Its solubility product is 6.5 × 10-6. Calculate its solubility.

Solution 

Solubility is represented by the ‘S’. The balanced chemical equation is given below:

Ca(OH)2 ⇌ Ca2+ + 2OH

At initial stage                                                    Ca(OH)2   0    +     0

At equilibrium stage                                        Ca(OH)2 ⇌     S    +    2S

The expression of Ksp is the following:

Ksp = [Ca2+][OH]2

Ksp= S × 2S

So,

4S3 = 6.5 × 10-6

Therefore,

S = ( 6.5 × 10-6 / 4)1/3

S = 1.175 × 10-2

Example # 02

What is the solubility of Ag2(CrO4) in water? If the value of the solubility product is 1.3 × 10-11.

Solution 

The balanced chemical equation is given below:

Ag2CrO4 ⇌ 2Ag+ + CrO42-

The solubility expression is given below:

Ksp = [Ag+]2[CrO42-]

So,

Ksp = (2S)2(S)

Therefore,

4S3 = 1.3 × 10-11.

S = (3.25 × 10-12)1/3 = 1.48 × 10-4M

Hence,

S= 1.48 × 10-4M

Related Resources

Concepts Berg

What Factors affect Ksp?

There are different factors that affect the Ksp such as:

How Do You Calculate Ksp?

We can calculate Ksp values from solubility. However, this can be done by using a balanced chemical equation.

How do you calculate the solubility?

We can calculate the solubility by using the Ksp values.

References 

  • A textbook of physical chemistry by K.L.Kapoor