When acetic acid is dissolved in water, it dissociates partly into H^{+} and H_{3}O^{+} and CH_{3}COO‾ ions as:

CH_{3}COOH + H_{2}O CH_{3}COO‾ + H_{3}O^{+}

In dilute solution , concentration of water is constant. The product of K and constant _{}

The product of K and _{ }is denoted by **K _{a}**, the

**ionization constant or dissociation constant of the acid**.

If C represents the initial concentration of the acid in moles L^{-1}

and α , the **degree of dissociation** , then equilibrium concentration of the ions ( H_{3}O^{+} and CH_{3}COO‾ ) is equal to Cα and that of the undissociated acetic acid = C ( 1- α ) i.e. we have

In case of weak electrolyte, The value of α is very small and can be neglected in comparison to 1 i.e. 1-α =1.Hence we get

α = √ Ka / C

If V is the volume of the solution in litres containing 1 mole of the electrolyte , C = 1/ V.Hence, we have,

α = √ Ka × V

For a weak base like NH_{4}OH we have

α = √ K_{b} / C

α = √ K_{b} × V

For a weak electrolyte , the degree of ionisation is inversely proportional to the square root of molar concentration or directly proportional to the square root of volume containing one mole of solute.This is called **Ostwald’s dilution law**.

Man Singh says

Congrats Mrs Shilpi Nagpal, for preparing a very simple text to understand the ionization of weak electrolytes. Many topics also need your methodology to understand them. I do not know whether you could prefer to write simple text to understand the friccohesity.