Question

For a concentrated solution of a weak electrolyte $ {{A}_{x}}{{B}_{y}} $ By of concentration $ C $ , the degree of dissociation $ \alpha $ is given as

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

Answer 1

Answer: (C)

The weak electrolyte $ {{A}_{x}}{{B}_{y}} $ dissociates as follows
$ {{A}_{x}}{{B}_{y}}\rightleftharpoons x{{A}^{y+}}+y{{B}^{x-}} $ $ \begin{matrix} C & \,\,\,\,\,\,\,\,0 & 0 & Initially \\ C(1-\alpha ) & \,\,\,\,\,\,\,\,\,xC\alpha & yC\alpha & At\,equilibrium \\ \end{matrix} $
where, $ \alpha $ degree of dissociation
$ C= $ concentration
$ {{K}_{eq}}=\frac{{{[{{A}^{y+}}]}^{x}}{{[{{B}^{x-}}]}^{y}}}{[{{A}_{x}}{{B}_{y}}]} $
$ =\frac{{{[xC\alpha ]}^{x}}{{[yC\alpha ]}^{y}}}{C(1-\alpha )} $
$ =\frac{{{x}^{x}}.{{C}^{x}}.{{\alpha }^{x}}.{{y}^{y}}.{{C}^{y}}.{{\alpha }^{y}}}{C} $ $ [\because 1-\alpha \approx 1] $
$ ={{x}^{x}}.{{y}^{y}}.{{\alpha }^{x+y}}.{{C}^{x+y-1}} $
$ {{a}^{x+y}}=\frac{{{K}_{eq}}}{{{x}^{x}}.{{y}^{y}}.{{C}^{x+y-1}}} $
$ \alpha ={{\left( \frac{{{K}_{eq}}}{{{x}^{x}}.{{y}^{y}}.{{C}^{x+t-1}}} \right)}^{\left( \frac{1}{x+y} \right)}} $