Explain the Dissociation of bases in water.

The ionization constant or dissociation of a base BOH in water may be represented as:
BOH (aq) <---------> B⁺ (aq) + OH^– (aq)
Therefore the ionization constant of the base, K_b,
Therefore, K_b = [B⁺] [OH^–] / [BOH]
(Concentration of water remains constant)
If c is the number of moles of the base in one litre of the solution and α is the degree of ionization, then the concentrations of each species at equilibrium are:
[B⁺] = cα
[OH^–] = cα
[BOH] = c (1 – α)
Therefore, K_b = (cα) X (cα) / c (1 – α)
But for weak bases, α is very small so that 1 – α ≈ 1
Therefore,
K_b = (cα) ²
α= (K_b / c) ^1/2
And
[OH^–] = c x (K_b / c) ^1/2
[OH^–] = (K_b. c) ^1/2