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Questions
Derive the relationship between the degree of dissociation of an electrolyte and van’t Hoff factor.
A substance dissociates in solution to give n ions. Derive an expression for degree of dissociation in terms of van’t Hoff factor i and n.
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Solution
The weak electrolytes involve the concept of degree of dissociation (α) that changes the van’t Hoff factor.
Consider an electrolyte AxBy that dissociates in aqueous solution as
| \[\ce{A_xB_y \phantom{...}<=> \phantom{..} xA^{y+} + \phantom{..} yB^{x−}}\] | |
| Initially | 1 mol 0 0 |
| At equilibrium | (1 − α) mol (xα mol) (yα mol) |
If α is the degree of dissociation of the 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.
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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