Question

If $\alpha $ is the degree of ionization, $C$ the concentration of a weak electrolyte and $Ka$, the acid ionization constant, then the correct relationship between $\alpha ,C$ and $Ka$​ is:
A. ${\alpha ^2} = \sqrt {\dfrac{{{K_a}}}{C}} $
B. ${\alpha ^2} = \sqrt {\dfrac{C}{{{K_a}}}} $
C. $\alpha = \sqrt {\dfrac{{{K_a}}}{C}} $
D. $\alpha = \sqrt {\dfrac{C}{{{K_a}}}} $

Answer 1

Hint: A weak electrolyte's degree of an ionisation that may be influenced by the variety of variables which is sometimes combined with other chemical changes. Weak electrolytes can dissolve in small amounts per liter of solution.

Complete step-by-step answer:In order to know that the weak electrolyte's dissociation constant is described by Ostwald's dilution law along with the weak electrolyte's concentration and degree of dissociation $(\alpha )$,
Only a small portion of the weak electrolytes are soluble in solution. As a result the electrolyte do not fully ionise in solution. However, even weak acids begin to ionise at high dilutions. As a result, dilution increases along with the degree of ionisation.
The binary electrolyte $HA$ separates into the ions ${H^ + }$ and ${A^ - }$.
$HA{\text{ }} \rightleftharpoons {\text{ }}{H^ + }\; + {\text{ }}{A^-}$
For very weak electrolytes, since $\alpha {\text{ }} < < < {\text{ }}1,{\text{ }}\left( {1{\text{ }}-{\text{ }}\alpha } \right){\text{ }} = {\text{ }}1$, then we have:
$\;{K_a} = C{\alpha ^2} \\$
 $\Rightarrow \alpha = \sqrt {\dfrac{{{K_a}}}{C}} \\$
Here, Degree of ionization $\alpha = \dfrac{{Number\,of\,molecules\,get\,ionized\,in\,solution}}{{Total\,number\,of\,molecules\,taken}} \times 100$
$\alpha = $degree of ionization.
From the above equation, it is clear that when concentration decreases then dilution increases and as a result, so does the degree of ionisation.
Thus, $\alpha = \sqrt {\dfrac{{{K_a}}}{C}} $.

Option ‘C’ is correct

Note:Since ionization does, to a limited extent, occur in weak electrolytes. As a result, the unionised electrolyte and the ions created in the solution will be in balance. The degree of ionisation is affected by a variety of variables, such as dilution, solvent type, temperature, the presence of additional ions, etc.