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

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

If we start with one mole of the compound,

$A_x{B}_y\rightleftharpoons x{A}^{y+}+y{B}^{x-}$

After dissociation

$(1-\alpha )x\alpha y\alpha$

$i=n\left(A_x{B}_y\right)+n\left(A^{y+}\right)+n\left(B^{x-}\right)$

= $1-\alpha +x\alpha +y\alpha$

= $1+\alpha (x+y-1)$

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