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_{x}B_{y} that dissociates in aqueous solution as
A_{x}B_{y } ⇌ xA^{y+} + yB^{x-} | |
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 =
Hence, i = 1 + α(n - 1) or α =