Q.Ionisation constant of a weak base MOH, is given by the expression
Kb = [M^+][OH^-] / [MOH]
Values of ionisation constant of some weak bases at a particular temperature are given below:
Base: Dimethylamine, Urea, Pyridine, Ammonia
Kb: 5.4 × 10^-4, 1.3 × 10^-14, 1.77 × 10^-9, 1.77 × 10^-5
Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?

Concept: Weak Base Ionization – For a weak base MOH, a larger $K_b$ means greater ionization at equilibrium, because the equilibrium lies further to the right: $\text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^-$.

Step 1 – Compare $K_b$ values directly.

The extent of ionization is proportional to $K_b$: higher $K_b$ → more ions formed → stronger base.

Step 2 – Arrange in decreasing order of $K_b$.

$5.4 \times 10^{-4} > 1.77 \times 10^{-5} > 1.77 \times 10^{-9} > 1.3 \times 10^{-14}$ …

The key idea here is that the ionisation constant $K_b$ is a direct measure of how far the equilibrium $\text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^-$ lies to the right. A larger $K_b$ means a greater fraction of the base molecules have dissociated into ions at equilibrium — that is, a greater extent of ionisation. The strength of a base is also judged by the same constant: the larger the $K_b$, the stronger the base.

Let’s work through the reasoning step by step.

1. Understand what $K_b$ tells us. For a weak base MOH, the equilibrium constant is

If $K_b$ is large, the numerator (product of ion concentrations) is large relative to the denominator (concentration of unionised base). This directly implies that a larger proportion of the base has ionised. So, higher $K_b$ → greater extent of ionisation → stronger base.

2. List the given $K_b$ values in order.

   From the data:

   • Dimethylamine: $5.4 \times 10^{-4}$
   • Ammonia: $1.77 \times 10^{-5}$
   • Pyridine: $1.77 \times 10^{-9}$
   • Urea: $1.3 \times 10^{-14}$

   Notice that $5.4 \times 10^{-4}$ is the largest, followed by $1.77 \times 10^{-5}$, then $1.77 \times 10^{-9}$, and finally $1.3 \times 10^{-14}$ is the smallest.

3. Arrange in decreasing order of ionisation.

   Since ionisation extent follows $K_b$, we simply write the bases from the largest $K_b$ to the smallest:

   $\text{Dimethylamine} > \text{Ammonia} > \text{Pyridine} > \text{Urea}$ …