i. Consider 1 mol of a weak base BOH dissolved in V dm³ of a solution. The base dissociates partially as

\[\ce{BOH_{(aq)} ⇌ B^+_{(aq)} + OH^-_{(aq)}}\]

ii. The base dissociation constant is given as:

`K_b = ([B^+][OH^-])/[BOH]`  ....(1)

iii. Let the fraction dissociated at equilibrium be \(\alpha\) and that remains undissociated be \((1 - \alpha)\).

\[\ce{BOH_{(aq)}⇌ B^+_{(aq)} + OH^-_{(aq)}}\]
Amount present at equilibrium (mol)	(1 - \(\alpha\))	\(\alpha\)	\(\alpha\)
Concentration at equilibrium (mol dm−3)	\((1 - \(\alpha\))/V\)	\(\alpha/V\)	\(\alpha/V\)

iv. Thus, at equilibrium,

\([BOH] = (1 - \(\alpha\))/V\) mol dm^-3,

\([B^+] = [OH^-] = \(\alpha/V\) mol dm^-3

v. Substituting these concentrations into equation (1),

\(K_b = (\(\alpha/V\)\(\alpha/V\))/((1 - \(\alpha\))/V) = \(\alpha^2/(1 - \(\alpha\)/V)\) ....(2)

vi. If c is the initial concentration of the base in mol dm^–3 and V is the volume in dm³ mol^–1, then \(c = 1/V\). Replacing \(1/V\) in equation (2) by c,

we get

\(K_b = (\(\alpha^2c\)/(1 - \(\alpha\)))\) ....(3)

vii. For the weak base BOH, \(\alpha\) is very small, or \((1 - \(\alpha\)) \(\approx\) 1. With this, equation (2) and (3) become:

\(K_b = \(\alpha^2/V\)\) and \(K_b = \(\alpha^2c\)\) ....(4)

\(\alpha = \(\sqrt{K_b/c}\)\) or \(\alpha = \(\sqrt{K_bV}\)\) ....(5)

The equation (5) implies that the degree of dissociation of a weak base is inversely proportional to the square root of its concentration and is directly proportional to the square root of the volume of the solution containing 1 mol of the weak base.