Question

1. What is Van’t Hoff factor in ${K_4}[Fe{(CN)_6}]$ and $BaC{l_2}$ ?
2. How is it related to the degree of dissociation?

Answer 1

Van’t Hoff factor is actually the ratio of actual number of particles after dissociation in solution to the number of particles before dissociation for strong electrolytes.
While in case of weak electrolytes, Van’t Hoff factor is the ratio of concentration of all substances to the initial concentration.

Complete step by step answer:
For coordination compound, we know that only the ions which are out of square bracket will dissociate in solution, thus we can write dissociation of ${K_4}[Fe{(CN)_6}]$ as given below:
${K_4}[Fe{(CN)_6}] \to 4{K^ + } + {[Fe{(CN)_6}]^{4 - }}$
Thus, we know Van’t Hoff factor, ‘i’ can be written as:
$i = \dfrac{{{\text{total ions after dissociation}}}}{{{\text{total ions before dissociation}}}}$
Here, we see that 5 ions are formed after dissociation, while there was only 1 before dissociation. Thus, we can write:
$i = \dfrac{5}{1} = 5$
We know for Strong electrolyte salt, it will dissociate completely into ions, so for $BaC{l_2}$ we can write dissociation as below:
$BaC{l_2} \to B{a^{2 + }} + 2C{l^ - }$
Here, the salt is dissociated into 3 ions, so we can write Vant Hoff factor for this salt as:
$i = \dfrac{3}{1} = 3$
Thus for strong electrolytes, Van’t Hoff factor is the number of ions into which it dissociates.
Thus Van’t Hoff factor in ${K_4}[Fe{(CN)_6}]$ and $BaC{l_2}$ are 5 and 3 respectively.
For relation of degree of dissociation and Van’t Hoff factor, we need to take an example reaction.
${A_n} \rightleftharpoons nA$
Let us consider we have 1 mole initially of weak electrolyte, and degree of dissociation is represented by $\alpha$
At initial and equilibrium condition, the concentration is written below,
${A_n} \rightleftharpoons nA$
$\text{ 1 0}$
${\text{1}} - \alpha { \text{ n}}\alpha$
Total concentration at equilibrium = $n\alpha + 1 - \alpha$
Total Concentration initially= 1
Now, Van’t Hoff factor, $i = \dfrac{{{\text{total concentration at equilibrium}}}}{{{\text{total concentration initially}}}}$
On substituting values, we get:
$i = \dfrac{{n\alpha + 1 - \alpha }}{1}$
Rearranging the above, to get degree of dissociation on one side, we can write:

$$i = \alpha (n - 1) + 1 \\\ i - 1 = \alpha (n - 1) \\\$$

Now, we keep degree of dissociation on one side, and rewrite equation as:
$\alpha = \dfrac{{i - 1}}{{n - 1}}$
Thus, the above is the relation between degree of dissociation and Van’t Hoff factor for weak electrolyte. Here ‘n’ is the number of particles in which the electrolyte dissociates, ‘i’ is the Van't Hoff factor and ‘$\alpha$’ is the degree of dissociation.

Note: For non-electrolytes, the Van’t Hoff factor is unity (1).
Be careful is dissociation of coordination compounds, as the coordinate bond does not break or dissociate in solution. The Van’t Hoff factor for strong electrolytes and weak electrolytes are also different. But, the basic definition of the Van't Hoff factor is the same.