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) \[\alpha =({{K}_{eq}}/{{C}^{xy}})\]
Correct Answer: C
Solution :
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 \[{{K}_{eq}}=\frac{{{[{{A}^{y+}}]}^{x}}{{[{{B}^{x-}}]}^{y}}}{[{{A}_{x}}{{B}_{y}}]}\] \[=\frac{{{[xC\alpha ]}^{x}}{{[yC\alpha ]}^{y}}}{C(1-\alpha )}\] \[=\frac{{{x}^{x}}.{{C}^{x}}.{{\alpha }^{x}}.{{y}^{y}}.{{C}^{y}}.{{\alpha }^{y}}}{C}\] \[[\because 1-\alpha \approx 1]\] \[={{x}^{x}}.{{y}^{y}}.{{\alpha }^{x+y}}.{{C}^{x+y-1}}\] \[{{a}^{x+y}}=\frac{{{K}_{eq}}}{{{x}^{x}}.{{y}^{y}}.{{C}^{x+y-1}}}\] \[\alpha ={{\left( \frac{{{K}_{eq}}}{{{x}^{x}}.{{y}^{y}}.{{C}^{x+t-1}}} \right)}^{\left( \frac{1}{x+y} \right)}}\]
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