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Question
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
| AxBy ⇌ xAy+ + yBx- | |
| Initially | 1 mol 0 0 |
| At equilibrium | (1 - α) mol (xα 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 the dissociation of 1 mole of electrolyte.
4. The van’t Hoff factor given as
i = `"actual moles of particles insolution after dissociation"/"moles of formula units dissolved in solution"`
`= (1 + alpha ("n - 1"))/1`
Hence, i = 1 + α(n - 1) or α = `("i" - 1)/("n" - 1)`
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