Question

For a concentrated solution of a weak electrolyte $A_x{B}_y$, the degree of dissociation is given as:

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

Answer 1

Answer: (C)

$\alpha ={\left(K_{eq}/c^{x+y-1}{x}^x{y}^y\right)}^{1/(x+y)}$

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

c 0 0

$c(1-\alpha )$ $xc\alpha$ $yc\alpha$

$\alpha <<1$ $\left\{weakelectrolyte\right\}$

$\therefore c(1-\alpha )\approx c$

$K_{eq}=\frac{\left(nc\alpha {)}^x\right(yc\alpha {)}^y}{c(1-\alpha )}$

$K_{eq}=\frac{c^{x+y-1}{\alpha }^{x+y}\times x^x\times y^y}{(1-\alpha )}$

$K_{eq}=C^{x+y-1}{\alpha }^{x+y}{x}^x{y}^y$

${\alpha }^{x+y}=\frac{K_{eq}}{C^{x+y-1}{x}^x{y}^y}$

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