Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium

For a concentrated solution of a weak electrolytic $A_xB_y$ of concentration C, the degree of dissociation $\alpha$ is given by

(a) $\alpha = \sqrt{\frac{K_{eq}}{(x+y)}}$

(b) $\alpha = (\frac{K_{eq}}{C^{x+y-1} X^x Y^y})^{\frac{1}{x + y}}$

(c) $\alpha = \frac{k_{eq}}{C^{xy}}$

(d) $\alpha = \sqrt{\frac{K_{eq} C}{x \times y}}$

1 Answer

Answer: $\alpha = [\frac{K_{eq}}{x^x y^y c^{x+y-1}}]^{\frac{1}{x+y}}$
$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
Initially, $[A_xB_y] = c, [xA^{y+}] = 0, [yB^{x-}] = 0$
At equilibrium, $[A_xB_y] = c(1-\alpha), [xA^{y+}]= xc\alpha, [yB^{x-}] = yc\alpha$
Equilibrium Constant, $K_{eq} = \frac{(xc\alpha)(yc\alpha)}{c(1-\alpha)} = \frac{(xc\alpha)^x (yc\alpha)^y}{c}$
Taking $1-\alpha \simeq1$
or $K_{eq} = \frac{x^x y^y c^{x+y} \alpha^{x+y}}{c} = x^x y^y c^{x+y-1} \alpha^{x+y}$
$\therefore \alpha = [\frac{K_{eq}}{x^x y^y c^{x+y-1}}]^{\frac{1}{x+y}}$


answered Nov 29, 2013 by mosymeow_1

Related questions

Download clay6 mobile appDownload clay6 mobile app