Question

Ionization constant of a weak base $MOH$, is given by the expression $K_b=\frac{\left[M^{+}\right]\left[O{H}^{-}\right]}{\left[MOH\right]}$ Values of ionization constant of some weak bases at a particular temperature are given below:

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

Arrange the bases in decreasing order of the extent of their ionization at equilibrium. Which of the above base is the strongest?

Answer 1

$K_b$ value for a base indicates the extent of dissociation of base to produce $O{H}^{-}$ ions and cation.

Therefore, higher value of $K_b$ indicates that more $O{H}^{-}$ and $M^{+}$ions are produced.

$MOH\rightleftharpoons M^{+}+O{H}^{-}$

As more the value of $K_b$ , more will be the concentration of $O{H}^{-}$ ions and more will be $pH$ or more basicity.

Hence, the most basic is Dimethylamine.

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

Therefore, the decreasing order of basicity or extent of dissociation is

Diethylamine > ammonia > pyridine > urea