Discuss the ionization of water.

Ionization of water – ionic product of water
Water is a weak electrolyte and undergoes self dissociation to a small extent as:
H₂O (l) + H₂O (l) < --------------> H₃O⁺ (aq) + OH^–
The dissociation constant foe the ionization of water is:
K = [H₃O⁺] [OH^–] / H₃O⁺]
Since the degree of dissociation of water is very small, the concentration of undissociated water may be practically taken as constant. Thus, we have
K x H₃O⁺] = [H₃O⁺] [OH^–]
K x a constant = [H₃O⁺] [OH^–]
K_w = [H₃O⁺] [OH^–]
Where K_w is a constant and is known as ionic product of water. Its value is constant at a particular temperature and varies with the change in temperature. At 298 K, the value of K_w is 1.008 x 10^-14 mol² L^-2. That is,
K_w = [H₃O⁺] [OH^–]
= 1.008 x 10^-14 = 1 x 10^-14 at 298 K
It is quite evident that the concentrations of [H₃O⁺] and [OH^–] ions are equal in pure water so that
[H₃O⁺] = [OH^–]
Therefore, K_w = [H₃O⁺] [OH^–]
= 1.008 x 10^-14 mol² L^-2
[H₃O⁺] ² = 1.008 x 10^-14
[H₃O⁺] = (1.008 x 10^-14) ^1/2
= 1.0 x 10^-7 mol L^-1
Thus, in pure water,
[H₃O⁺] = [OH^–] = 1.0 x 10^-7 mol L^-1 at 298 K