Ksp Chemistry (What is Solubility Product Constant)

Ksp Chemistry or Ksp in Chemistry is a term commonly used in Solubility and solubility-related tasks discussion in Chemistry.

However before I go deep into Ksp chemistry and its calculations, I need to recap solubility in chemistry. Solubility is the amount of solute in grams or moles that will dissolve in 1 dm³  of solvent.

There are about four factors that determine the solubility of a solute in a solvent; temperature, pressure for gases, nature of solute, and nature of the solvent.

To determine the solubility of a solute at a particular temperature, a saturated solution of that solute must be prepared at that temperature.

A standard solution is a solution of known concentration.

Please it is important to note that Ksp is commonly attributed to sparingly soluble salts to measure the degree of solubility of the salts.

Solubility Constant Product Ksp

Solubility Product Constant is the equilibrium established between sparingly soluble salt and its ions.

For emphasis, it is pertinent to note that solubility product relates to every salt but most times to estimate the degree of solubility of salts that are not readily soluble.  You can determine if salt will be soluble or not from the solubility rule.

When is a salt said to be very soluble or sparingly soluble?

A salt is said to be readily soluble or soluble if 1 gram of salt can dissolve in 100cm³ of solvent.

A salt is considered sparingly soluble if less than 1 gram dissolves in 100 cm³ of the solvent.

A salt is insoluble when nothing or a very negligible amount dissolves in the solvent.

Ksp increases as the solubility of the solute increases and it does not have a unit due to the changes in the concentrations of reactants and products.

Most ionic compounds are very soluble in water and as such have high values for the solubility product constant.

Write the expression of Ksp in Chemistry

I want to show you the model expression for solubility product constant calculations

         aA + bB à cC + dD

         Ksp = [C]^c[D]^d

Factors that affect the solubility of salts

1. Temperature
2. Pressure
3. Nature of solute
4. Nature of solvent

10 Solubility calculations in Ksp Chemistry

Example 1

Calculate the solubility product constant of Barium Sulphate at 280K if the solubility of Barium sulfate is 1.0 x 10^-5.

Solution

Express the equilibrium constant

 BaSO₄ à [Ba²⁺] [SO₄^2-]

Since the ions are equimolar

         [Ba²⁺] = [SO₄^2-]

  K_sp of BaSO₄   –> [1.0 x 10^-5] [1.0 x 10^-5]

  K_sp = 1.0 x 10^-10 mol²/dm⁶

Example 2

Calculate the solubility product constant of Lead II Sulphate at 25⁰C if the solubility of Lead II sulfate is 1.25 x 10^-4?

Solution

First express equilibrium constant

            PbSO₄ –> [Pb²⁺] [SO₄^2-]

Since the ions are equimolar

         [Pb²⁺] = [SO₄^2-]

  K_sp of PbSO₄   = [1.25 x 10^-4] [1.25 x 10^-4]

  K_sp = 1.6 x 10^-8 mol²/dm⁶

Example 3

Calculate the solubility of Lead II Sulphide at 25⁰C if the solubility product constant (K_sp) of Lead II sulfide is 1.0 x 10^-30?

Solution

            PbS à [Pb²⁺] [S^2-]

        Since the ions are equimolar

         [Pb²⁺] = [S^2-]

        K_sp of PbS   à [x] [x]

         K_sp =  x²

              X = /1.0 x 10^-30

               x =  1.0 x 10^-15

Example 4

Calculate the solubility product constant of Calcium Fluoride at 250C if the solubility of  OF calcium Flouride is 2.14 x 10^-2?

Solution

First express equilibrium constant

            CaF₂ –> [Ca²⁺] [F^–]²

Be careful with this because, for every 1 mol/dm3 of CaF2, there are I mol/dm³ Ca²⁺ and 2 mol/dm³.

     K_sp = K_sp = [Ca²⁺] [F^–]²

Since the ions are  not equimolar

         [Ca²⁺] then 2[F^–]

         [Ca²⁺] = 0.214 X 10^-2

         [F^–] = 0.428 X 10^-2

  K_sp of CaF₂   à [0.214 x 10^-2] [0.428 x 10^-3]²

  K_sp = 1.92 x 10^-8 mol²/dm⁶

The solubility product constant is always used to measure the solubility of sparingly soluble salts in water.