Question

The degree of dissociation$\left( \alpha \right)$ of a weak electrolyte, ${{A}_{x}}{{B}_{y}}$ is related to van’t hoff factor (i) by the expression:
A. \[\alpha =\dfrac{i-1}{\left( x+y-1 \right)}\]
B. \[\alpha =\dfrac{i-1}{\left( x+y+1 \right)}\]
C. \[\alpha =\dfrac{\left( x+y-1 \right)}{i-1}\]
D. \[\alpha =\dfrac{\left( x+y+1 \right)}{i-1}\]

Answer 1

Hint: For us to be able to solve the question, we first need to know what are weak electrolytes and how do they differ with the strong electrolytes. Weak electrolytes do not completely dissociate themselves into their ions. Only a part of them dissociates and this is important to be studied to find various quantities for them like the moles dissociated, reaction rate, etc.

Complete step by step solution:
-Let us first talk about what an electrolyte is. Electrolyte is a compound which dissociates into its constituent ions, cation and anion in presence of DC current under the process of electrolysis.
-Many types of compound can be used as electrolytes but the most preferred compounds are acids, bases and salts. Electrolytes dissolve in water to form a solution and that solution conducts electricity by dissociating into ions.
Eg. $NaCl\left( s \right)\to N{{a}^{+}}\left( aq \right)+C{{l}^{-}}\left( aq \right)$
Water is added as solvent so that the ions can become mobile. Only then they can conduct electricity.
-Now coming to the strength of the electrolytes. There are two types of electrolytes, strong and weak. Strong electrolytes are those which can dissociate to a very large extent to give the respective cations and anions. Weak electrolytes are those which dissociate very less and so the ions formed are not much which reduces the conductivity of the solution.
-Weak electrolytes do not dissociate completely into their ions and the difference in their dissociation can be shown in the form of a quantity called the dissociation degree$\left( \alpha \right)$
-Dissociation of strong electrolytes is shown by a forward arrow and that of weak electrolytes is shown by double arrows.
$NaCl\left( s \right)\to N{{a}^{+}}\left( aq \right)+C{{l}^{-}}\left( aq \right)$
It signifies that this electrolyte is a strong electrolyte.
\[{{H}_{2}}O\rightleftarrows {{H}^{+}}+O{{H}^{-}}\]
It signifies that water is a weak electrolyte.
-Van't hoff factor is the measure of the degree of dissociation for weak electrolytes. It is directly proportional to the degree of dissociation. It means that if the value of the van't hoff factor is more, more electrolyte will dissociate itself to form ions. If the value is less, then a lesser amount of electrolyte will dissociate.
-There is a relationship between the van't hoff factor I and degree of dissociation $\left( \alpha \right)$
\[\alpha =\dfrac{i-1}{n-1}\]
Here n is the number of ions that can be produced by 1 mole of the electrolyte on dissociation.
-Van't hoff factor can be studied by the following expression. Suppose a weak electrolyte dissociates itself into its ions as
${{A}_{x}}{{B}_{y}}\rightleftharpoons x{{A}^{y+}}+y{{B}^{x-}}$
If the degree of dissociation is $\left( \alpha \right)$, then equal moles will be formed as ions and the moles of the electrolyte left will be given as $\left( 1-\alpha \right)$ .
-So the total number of species will now become equal to 1+(x+y-1) $\left( \alpha \right)$
Now the van't hoff factor will be i=1+(x+y-1) $\left( \alpha \right)$
From this we get the relation between i and $\left( \alpha \right)$ as \[\alpha =\dfrac{i-1}{\left( x+y-1 \right)}\]

Therefore the correct option is A.

Note: The compounds which do not dissociate at all into its cation and anion are termed as non-electrolytes. Do not confuse it with weak electrolytes. Weak electrolytes do dissociate themselves, though very less. Eg include glucose, fructose, sucrose, galactose, etc.