Q. The degree of dissociation ($\alpha)$ of a weak electrolyte , $A_{x}B_{y}$ is related to van’t Hoff factor (i) by the expression
A) $\alpha = \frac{i - 1}{(x + y - 1)}$
B) $\alpha = \frac{i - 1}{x + y + 1}$
C) $\alpha = \frac{x + y - 1}{i - 1}$
D) $\alpha = \frac{x + y + 1}{i - 1}$
Explanation
(a)
$\ \ \ \ A_{x}B_{y}\ \ \ \ \ \ \ \rightleftharpoons \ \ \ xA^{y +}\ \ \ \ \ + \ \ \ \ \ yB^{x -}$
After dissociation (1-$\alpha)$ $x\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ y\alpha$
i=n($A_{x}B_{y})$+n($A^{y +}) + n$($B^{x - \ })$
=2-$\alpha$+$x\ \alpha + y\alpha = 1 + \alpha(x + y - 1)\ \$
$\ \ \therefore$ $\alpha = \frac{i - 1}{(x + y - 1)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.