Question

The degree of dissociation $\left(\alpha \right)$ 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. $\alpha =\frac{x+y+1}{i-1}$

Answer 1

The correct option is A $\alpha =\frac{i-1}{(x+y-1)}$

The dissociation of the weak electrolyte is as shown below.

$\underset{1-\alpha }{A_x{B}_y}\to \underset{x\alpha }{x{A}^{y+}}+\underset{y\alpha }{y{B}^{x-}}$

The vant hoff factor is $i=1-\alpha +x\alpha +y\alpha$

Hence, $\alpha =\frac{i-1}{(x+y-1)}$

Option A is correct. .