The degree of dissociation (alpha) of a weak | vedclass

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Question

(A) For the dissociation of a weak electrolyte $A_xB_y$: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$ At $t=0$,the moles are $1, 0, 0$ respectively.

Vedclass · Accepted Answer

$\alpha = \frac{i - 1}{x + y - 1}$

Vedclass · Answer

(A) For the dissociation of a weak electrolyte $A_xB_y$: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$ At $t=0$,the moles are $1, 0, 0$ respectively. At equilibrium,the moles are $(1 - \alpha), x\alpha, y\alpha$. The total number of moles at equilibrium is: $n_{total} = 1 - \alpha + x\alpha + y\alpha = 1 + \alpha(x + y - 1)$ The van't Hoff factor $(i)$ is defined as the ratio of observed moles to initial moles: $i = \frac{1 + \alpha(x + y - 1)}{1} = 1 + \alpha(x + y - 1)$ Rearranging for $\alpha$: $i - 1 = \alpha(x + y - 1)$ $\alpha = \frac{i - 1}{x + y - 1}$

The degree of dissociation $(\alpha)$ of a weak electrolyte,$A_xB_y$ is related to van't Hoff factor $(i)$ by the expression

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