
In weak electrolyte solution, degree of ionization
A. Will be proportional to dilute
B. Will be proportional to concentration of electrolyte
C. Will be proportional to the square root of dilution
D. Will be reciprocal to the dilution
Answer
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Hint: Ionization, which is sometimes combined with other chemical changes, is the process by which an atom or molecule obtains a negative or positive charge by gaining or losing electrons. The degree of ionisation depends on how far along this process is.
Complete step-by-step answer:The Ostwald dilution law must first be understood. The link between the weak electrolyte's dissociation constant, degree of dissociation, and concentration is provided by Ostwald's dilution law. In other terms, it can be said that the weak electrolyte will completely dilute with infinite dilution. Weak electrolytes are those that will only partially dissociate when dissolved in water. When an electrolyte is dissolved in water, it totally dissociates, making it a strong electrolyte.
Take into account a binary electrolyte solution \[AB\] that will irreversibly split into \[{A^ + }\] and \[{B^ - }\] ions. Below is a breakdown of the binary electrolyte solution AB's reactions.
$AB\,\, \rightleftharpoons \,\,{A^ + }\, + \,{B^ - }$
The dissociation constant will then be
$K = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}}$
Now if $\alpha $ is the degree of dissociation and $C$ is the concentration of the solution, then dissociation constant becomes
$K = \dfrac{{{\alpha ^2}{C^2}}}{{\left( {1 - \alpha } \right)C}}$
Since, $\alpha $ is very less than 1, so we get,$\left( {1 - \alpha } \right) = 1$
This implies that $K = {\alpha ^2}C$
Hence, we get $\alpha = \sqrt {\dfrac{K}{C}} $ .
Any weak electrolyte's degree of ionisation (or dissociation) is inversely proportional to its concentration squared and directly proportional to its dilution squared.
Option ‘C’ is correct
Note: We must keep in mind that the Ostwald's Dilution Law has some restrictions, namely that it only applies to weak electrolytes. For saturated solutions with strong electrolytes, this law does not apply. At a typical dilution, only a small part of the electrolyte will be separated. But at infinite dilution, the entire electrolyte will disintegrate.
Complete step-by-step answer:The Ostwald dilution law must first be understood. The link between the weak electrolyte's dissociation constant, degree of dissociation, and concentration is provided by Ostwald's dilution law. In other terms, it can be said that the weak electrolyte will completely dilute with infinite dilution. Weak electrolytes are those that will only partially dissociate when dissolved in water. When an electrolyte is dissolved in water, it totally dissociates, making it a strong electrolyte.
Take into account a binary electrolyte solution \[AB\] that will irreversibly split into \[{A^ + }\] and \[{B^ - }\] ions. Below is a breakdown of the binary electrolyte solution AB's reactions.
$AB\,\, \rightleftharpoons \,\,{A^ + }\, + \,{B^ - }$
The dissociation constant will then be
$K = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}}$
Now if $\alpha $ is the degree of dissociation and $C$ is the concentration of the solution, then dissociation constant becomes
$K = \dfrac{{{\alpha ^2}{C^2}}}{{\left( {1 - \alpha } \right)C}}$
Since, $\alpha $ is very less than 1, so we get,$\left( {1 - \alpha } \right) = 1$
This implies that $K = {\alpha ^2}C$
Hence, we get $\alpha = \sqrt {\dfrac{K}{C}} $ .
Any weak electrolyte's degree of ionisation (or dissociation) is inversely proportional to its concentration squared and directly proportional to its dilution squared.
Option ‘C’ is correct
Note: We must keep in mind that the Ostwald's Dilution Law has some restrictions, namely that it only applies to weak electrolytes. For saturated solutions with strong electrolytes, this law does not apply. At a typical dilution, only a small part of the electrolyte will be separated. But at infinite dilution, the entire electrolyte will disintegrate.
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