The degree of dissociation ( α ) of a weak electrolyte A_xB_y is rela…

The correct answer is Option (1) → \(α = (i-1)/(x + y -1)\)

The van’t Hoff factor i is the ratio of the number of particles actually present in solution to the number of formula units dissolved. For a weak electrolyte it depends on the degree of dissociation α, the fraction of formula units that split into ions.

One formula unit of \( \mathrm{A_xB_y} \) gives x cations and y anions:

    \( \displaystyle \mathrm{A_xB_y} \ce{<=>} x\,\mathrm{A^{y+}} + y\,\mathrm{B^{x-}} \)

Starting with 1 mol, at equilibrium:

    \( \displaystyle \text{undissociated: } 1 - \alpha;\quad \text{cations: } x\alpha;\quad \text{anions: } y\alpha \)

    \( \displaystyle \text{total particles} = 1 - \alpha + x\alpha + y\alpha = 1 + (x + y - 1)\alpha \)

    \( \displaystyle i = \frac{1 + (x + y - 1)\alpha}{1} \)

    \( \displaystyle \alpha = \frac{i - 1}{x + y - 1} \)

So the relation is \( \alpha = (i - 1)/(x + y - 1) \).

Shortcut: Test the options on a 1 : 1 electrolyte such as acetic acid (x = y = 1), for which \( i = 1 + \alpha \), so \( \alpha = i - 1 \). Only Option (1) reduces to this.