For a weak electrolyte A_xB_y,the degree of d | vedclass

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(C) For a weak electrolyte $A_xB_y$,the dissociation reaction is: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$.Initially,concentration is $c$ for $A_

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a weak electrolyte $A_xB_y$,the dissociation reaction is: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$.Initially,concentration is $c$ for $A_xB_y$ and $0$ for products.At equilibrium,concentrations are: $[A_xB_y] = c(1-\alpha)$,$[A^{y+}] = xc\alpha$,and $[B^{x-}] = yc\alpha$.The equilibrium constant $K_{eq}$ is given by: $K_{eq} = \frac{[A^{y+}]^x [B^{x-}]^y}{[A_xB_y]} = \frac{(xc\alpha)^x (yc\alpha)^y}{c(1-\alpha)}$.Since $\alpha$ is very small for a weak electrolyte,$(1-\alpha) \approx 1$.Thus,$K_{eq} = \frac{x^x c^x \alpha^x y^y c^y \alpha^y}{c} = x^x y^y c^{x+y-1} \alpha^{x+y}$.Solving for $\alpha$: $\alpha^{x+y} = \frac{K_{eq}}{c^{x+y-1} x^x y^y}$.Therefore,$\alpha = \left(\frac{K_{eq}}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$.

For a weak electrolyte $A_xB_y$,the degree of dissociation $\alpha$ in terms of concentration $c$ is given by:

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