Question

The Van’t Hoff factor ‘i’ for an electrolyte which undergoes dissociation and association in solvents respectively are:
[A] Greater than one and less than one.
[B] Less than one and greater than one.
[C] Less than one and less than one.
[D] Greater than one and greater than one.

Answer 1

Hint: Van’t Hoff factor is the number of splitting or forming of ions of the solute in a solution. To solve this you can use the relation of the Van’t Hoff factor with the dissociation constant and find out the correct answer.
For dissociation, $$i=1+\alpha (n-1)$$ and for association, $$i=1-(1-{\displaystyle \frac{1}{n}})\alpha$$.

Complete step by step answer:
We represent the van’t Hoff factor as ‘i’. It is the measure of the effect of colligative properties upon a solute. Colligative properties are osmotic pressure, depression of freezing point, elevation of boiling point and relative lowering of vapour pressure.
‘i’ is the Van’t Hoff factor which is basically the number of splitting or forming of the particles of the solute.
It is the ratio of the actual concentration of the particles when the substance is dissolved and the actual calculated concentration of the substance.
Now, let us see how it varies with association and dissociation of an electrolyte-
We see a relation between the degree of dissociation and the Van’t Hoff factor. If fraction of the solute is dissociated into ions then-
$$i=\alpha n+(1-\alpha )=1+\alpha (n-1)$$
Therefore, if we take KCl and write its dissociation reaction
$$KCl\rightleftharpoons K^{+}+C{l}^{-}$$

As we can see it yields 2 ions so, n = 2 ions and $i=1+\alpha$
Therefore, for a dissociation reaction i is greater than 1.

Now, for an association of solutes, the relation becomes
$$i=1-(1-{\displaystyle \frac{1}{n}})\alpha$$

For example if we take dimerization of acetic acid in benzene,
$$2C{H}_{3}COOH\rightleftharpoons {\left(C{H}_{3}COOH\right)}_2$$
Here, 2 moles of acetic acid associated into 1 mole of dimer so,
$$i=1-(1-{\displaystyle \frac{1}{2}})\alpha =1-{\displaystyle \frac{\alpha }{2}}$$

Therefore, for association, the Van't Hoff factor is less than 1.
It’s clear from the above discussion that the Van’t Hoff factor ‘i’ for an electrolyte which undergoes dissociation and association in solvents respectively are greater than one and less than one.
So, the correct answer is “Option A”.

Note: Non-electrolytes dissolved in water have the value of i equal to 1.
- Ionic compounds dissolve in water and thus the Van’t Hoff factor is the number of ions in a formula unit of the substance. (This is true only for ideal solutions)
We know that ion pairing occurs in solutions and at a given instant only a small percentage of ions are paired and thus counted as a single particle. This pairing occurs in all electrolytic solutions. For this reason the Van't Hoff factor is less than that predicted for any ideal solution. The deviation is greatest when ions have multiple charges.