Question

For a concentrated solution of a weak electrolyte $A_x$ , $B_y$ , the degree of dissociation is given as

• A. $\alpha =\sqrt{\frac{K_{eq}}{C\left(x+y\right)}}$
• B. $\alpha =\sqrt{\frac{K_{eq}C}{\left(xy\right)}}$
• C. $\alpha ={\left(\frac{K_{eq}}{\left(C^{x+y-1}\right)\cdot x^x{y}^y}\right)}^{\frac{1}{\left(x+y\right)}}$
• D. $\alpha =\sqrt{\frac{K_{eq}}{xyC}}$

Answer 1

Answer: (C)

Dissociation of weak electrolyte $A_x{B}_y$ as;

Applying law of mass action
$K_{eq}=\frac{[xC\alpha {]}^x[yC\alpha {]}^y}{C(1-\alpha )}$
For weak electrolytes $a<<<1$,
$\therefore 1-\alpha =1$
$\therefore K_{eq}C=x^x{y}^y[C\alpha {]}^{x+y}$
or $\frac{K_{eq}}{C^{x+y-1}}=x^x{y}^y(\alpha {)}^{x+y}$
or $\alpha ={\left(\frac{K_{eq}}{\left(C^{x+y-1}\right)x^x{y}^y}\right)}^{\frac{1}{x+y}}$