Question

The ionization constant and degree of dissociation of water at \[25^\circ C\], is
(A) \[1.8 \times {10^{ - 9}},1.8 \times {10^{ - 16}}\]
(B) \[1.8 \times {10^{ - 16}},1.8 \times {10^{ - 9}}\]
(C) \[1.8 \times {10^{ - 12}},1.8 \times {10^{ - 14}}\]
(D) \[1.8 \times {10^{ - 14}},1.8 \times {10^{ - 12}}\]

Answer 1

Hint: In order to find the ionisation constant and degree of dissociation of water, we must know what a degree of dissociation and ionisation constant is. Degree of dissociation is defined as the fraction of a particular molecule which is dissociating at a given time.

Complete Solution :
Let us understand the degree of dissociation. Degree of dissociation can be defined as the fraction of moles of the reactant that has undergone dissociation. It is represented as \[\alpha \].
Ionisation of water is the self-ionisation reaction in which the water will dissociate/ deprotonate to give the hydroxide \[O{H^ - }\] ion. The \[{H^ + }\] ion will immediately combine with another water molecule giving \[{H_3}{O^ + }\] ion. The water does not require ant polar solvents, it will get ionized itself. The activity of solute particles in the dilute solution is almost equal to their concentrations. \[{K_w}\] is the ionisation constant, dissociation constant or ionic product constant.
\[{K_w} = [{H^ + }][O{H^ - }]\]
\[{K_w} = 1 \times {10^{ - 14}}\]

We can calculate the degree of dissociation using the formula,
\[\alpha = \dfrac{{{K_w}}}{C}\]
We know that
\[{K_w} = 1 \times {10^{ - 14}}\]
\[C = 55.5mol{L^{ - 1}}\]

\[\alpha = \dfrac{{{{10}^{ - 14}}}}{{55.5}} = 1.8 \times {10^{ - 16}}\]
Ionisation constant \[ = \dfrac{{1.8 \times {{10}^{ - 16}}}}{{{{10}^{ - 7}}}} = 1.8 \times {10^{ - 9}}\]
So, the correct answer is “Option A”.

Note: Degree of association is totally different from degree of dissociation. We can say that it is opposite to one another. Degree of association is defined as the fraction of total number of molecules that will combine to form a bigger molecule.