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 AxBy that dissociates in aqueous solution as

  AxB       ⇌      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 = actual moles of particles insolution after dissociationmoles of formula units dissoved in solution

=1+α(n - 1)1

Hence, i = 1 + α(n - 1) or α = i+1n-1