question_answer

For a concentrated solution of a weak electrolyte \[{{A}_{x}}{{B}_{y}}\] of concentration \['c',\] the degree of dissociation 'a' is given as

A) \[\alpha =\sqrt{{{K}_{eq}}/c(x+y)}\]

B) \[\alpha =\sqrt{{{K}_{eq}}c/(xy)}\]

C) \[\alpha ={{\left( {{K}_{eq}}/{{c}^{x+y-1}}{{x}^{2}}{{y}^{2}} \right)}^{1/\left( x+y \right)}}\]

D) \[\alpha =\left( {{K}_{eq}}/cxy \right)\]

Correct Answer: C

Solution :

the week electrolyte AxBy dissociates as follows

Where, a =degree of dissociation

c = concentration

\[{{K}_{eq}}=\frac{{{[{{A}^{y+}}]}^{x}}[{{B}^{x-}}]y}{[{{A}_{x}}{{B}_{y}}]}\]

\[=\frac{{{[xc\alpha ]}^{x}}{{[yc\alpha ]}^{y}}}{c(1-\alpha )}\]

\[=\frac{{{x}^{x}}.{{c}^{x}}.{{a}^{x}}.{{y}^{y}}.{{c}^{y}}.{{a}^{y}}}{c}\]               \[[\because 1-\alpha \approx 1]\]

\[={{x}^{x}}.{{y}^{y}}.{{\alpha }^{x+y}}.{{c}^{x+y-1}}\]

\[{{\alpha }^{x+y}}=\frac{{{K}_{eq}}}{{{x}^{x}}.{{y}^{y}}.{{c}^{x+y-1}}}\]

\[\alpha ={{\left( \frac{{{K}_{eq}}}{{{x}^{x}}.{{y}^{y}}.{{c}^{x+y-1}}} \right)}^{\left( \frac{1}{x+y} \right)}}\]