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 (K_{sp}) 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:

PbSO_{4} ⇌ Pb^{2+} + SO_{4}^{2-}

The equilibrium constant for this reaction is given by:

K_{c} = [Pb^{2+}] [SO_{4}^{2-}] / [PbSO_{4}]

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

K_{c} [PbSO4] = [Pb^{2+}] [SO_{4}^{2-}]

- K
_{c}[PbSO4] = K_{sp}

K_{sp} = [Pb^{2+}] [SO_{4}^{2-}]

= [S][S]

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

Ksp = S^{2}

- K
_{sp}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 K_{sp} 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 PbF _{2} is 0.64 gdm^{-3}. Calculate K_{sp} of PbF_{2}.**

**Solution:**

First, find out moles from concentration.

Mass of PbF_{2} dissolved = 0.64gdm^{-3}

Molecular mass of PbF_{2} = 245.2 gmol^{-1}

No of moles of PbF_{2} = 0.64 gdm^{-3}/ 245.2 gmol^{-1} = 2.6 × 10^{-3}

The balanced chemical equation for PbF_{2} is the following:

PbF_{2 }⇌ Pb^{2+ }+ 2F^{– }

initially, 2.6 × 10^{-3} 0 + 0

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

The expression of K_{sp} is:

K_{sp} = [Pb^{2+}][F^{–}]^{2}

Putting the values

K_{sp} = (2.6 × 10^{-3 }) x {2(2.6 × 10^{-3})}^{2} = 7 × 10^{-8}

K_{sp}= 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:

SrO_{x} ⇌ Sr^{2+} + O_{x}^{2-}

Therefore,

[Sr^{2+}] = [O_{x}^{2-}] = 0.00054M = 5.4 × 10^{-4}

The expression of K_{sp} is the following:

K_{sp} = [Sr^{2+}] [O_{x}^{2-}]

Putting the values

K_{sp} = (5.4 × 10^{-4})(5.4 × 10^{-4}) = 2.92 × 10^{-7}mol^{2 }(dm^{-3})^{2}

Hence,

K_{sp}= 2.92 × 10^{-7}mol^{2 }(dm^{-3})^{2}

## Determination of solubility from K_{sp}

**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} ⇌ Ca^{2+} + 2OH^{–}

At initial stage Ca(OH)_{2 }⇌ 0 + 0

At equilibrium stage Ca(OH)_{2} ⇌ S + 2S

The expression of K_{sp} is the following:

K_{sp} = [Ca^{2+}][OH^{–}]^{2}

K_{sp}= S × 2S

So,

4S^{3} = 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 Ag _{2}(CrO_{4}) in water? If the value of the solubility product is 1.3 × 10^{-11}.**

**Solution **

The balanced chemical equation is given below:

Ag_{2}CrO_{4} ⇌ 2Ag^{+} + CrO_{4}^{2-}

The solubility expression is given below:

K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2-}]

So,

K_{sp} = (2S)^{2}(S)

Therefore,

4S^{3} = 1.3 × 10^{-11}.

S = (3.25 × 10^{-12})^{1/3} = 1.48 × 10^{-4}M

Hence,

S= 1.48 × 10^{-4}M

**Related Resources**

## Concepts Berg

**What Factors affect Ksp?**

There are different factors that affect the K_{sp} such as:

- Common ion effect
- Temperature
- presence of ion pairs

**How Do You Calculate Ksp?**

We can calculate K_{sp} 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 K_{sp} values.

**References **

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