Question

For a concentrated solution of a weak electrolyte $A_x{B}_y$ of concentration '$C$', the degree of dissociation '$\alpha$' is given as

• A. $\alpha =\sqrt{K_{eq.}/C(x+y)}$
• B. $\alpha =\sqrt{\frac{C{K}_{eq.}}{xy}}$
• C. $\alpha ={\left[\frac{K_{eq.}}{\left(C^{x+y-1}{x}^x{y}^y\right)}\right]}^{1/x+y}$
• D. $\alpha =\frac{K_{eq.}}{Cxy}$
• E. $\alpha =\frac{K_{eq.}}{C^{xy}}$

Answer 1

Answer: (C)

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

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

$C$

$C-C(x\alpha +y\alpha )$ $Cx\alpha$ $Cy\alpha$

$K_{eq}=\frac{{\left(C\alpha x\right)}^x{\left(C\alpha y\right)}^y}{C-C(\alpha x+\alpha y)}=\frac{C^{(x+y-1)}{x}^x{y}^y{\alpha }^{x+y}}{1-\alpha (x+y)}\alpha <<11-\alpha (x+y)\approx 1$

$K_{eq}=\frac{C^{(x+y-1)}{x}^x{y}^y{\alpha }^{x+y}}{1}\alpha ={\left[\frac{K_{eq}}{C^{(x+y-1)}{x}^x{y}^y}\right]}^{\frac{1}{x+y}}$
Option C is correct.