Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor. - Chemistry

Exercise | Q 4 | Page 46

Derive the relationship between the degree of dissociation of an electrolyte and van’t Hoff factor.

Solution:

1. The weak electrolytes involve the concept of degree of dissociation (α) that changes the van’t Hoff factor.

2. Consider an electrolyte A_xB_y that dissociates in aqueous solution as

	AxBy        ⇌      xAy+    +    yBx-
Initially	1 mol                 0               0
At equilibrium	(1 - α) mol      (x α mol)    (yα mol)

3. If α is the degree of dissociation of electrolyte, then the moles of cations are xα and those of anions are yα equilibrium. We have dissolved just 1 mol of electrolyte initially. α mol of electrolyte dissociates and (1 – α) mol remains undissociated at equilibrium.

Total moles after dissociation = (1 – α) + (xα) + (yα)

= 1 + α (x + y - 1)

= 1 + α (n - 1)

where, n = x + y = moles of ions obtained from dissociation of 1 mole of electrolyte.

4. The van’t Hoff factor given as

i = ${\displaystyle \frac{\text{actual moles of particles insolution after dissociation}}{\text{moles of formula units dissoved in solution}}}$

${\displaystyle =\frac{1+\alpha \left(\text{n - 1}\right)}{1}}$

Hence, i = 1 + α(n - 1) or α = ${\displaystyle \frac{\text{i}+1}{\text{n}-1}}$