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 $A_xB_y$ 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 $\alpha = \left(\frac{K_{eq}}{\left(C^{x+y-1}\right)x^{x}y^{y}}\right)^{\frac{1}{x+y}}$