Given is a concentrated solution of a weak electrolyte A_xB_y of...

The dissociation reaction of the weak electrolyte $A_xB_y$ is:

$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$

Initial concentrations:
$c \quad 0 \quad 0$

Equilibrium concentrations:
$c(1-\alpha) \quad cx\alpha \quad cy\alpha$

The dissociation constant $K$ is given by:

$K = \dfrac{[A^{y+}]^x [B^{x-}]^y}{[A_xB_y]}$

$K = \dfrac{(cx\alpha)^x (cy\alpha)^y}{c(1-\alpha)}$

$K = \dfrac{c^{x+y} x^x y^y \alpha^{x+y}}{c(1-\alpha)}$

$K = \dfrac{c^{x+y-1} x^x y^y \alpha^{x+y}}{1-\alpha}$

For a weak electrolyte, the degree of dissociation $\alpha$ is very small, so $1-\alpha \approx 1$.

$K = c^{x+y-1} x^x y^y \alpha^{x+y}$

Solving for $\alpha$:

$\alpha^{x+y} = \dfrac{K}{c^{x+y-1} x^x y^y}$

$\alpha = \left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\dfrac{1}{x+y}}$

Answer: \left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}