Question

The degree of dissociation $\left(\alpha \right)$ of a weak electrolyte, $A_x{B}_y$ is related to the 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 $\alpha =\frac{i-1}{(x+y-1)}$

For the weak electrolyte $A_x{B}_y$

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

Number of species$=1+(x+y-\alpha )$

So van 't Hoff factor $i=\frac{1+(x+y-\alpha )}{1}$
Hence $\alpha =\frac{i-1}{(x+y-1)}$