Ionic Equilibrium Part 2

Let the weak electrolyte $A_x B_y$ dissociate in water as follows:$A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}$Let the initial concentration of $A_x B_y$ be $c$, and let the degree of dissociation be $\alpha$.At equilibrium,Concentration of undissociated $A_x B_y = c(1-\alpha)$Concentration of $A^{y+} = c x \alpha$Concentration of $B^{x-} = c y \alpha$$K = \frac{ [A^{y+1}]^x [B^{x-}]^y }{ [A_x B_y] }$Substituting equilibrium concentrations:$K = \frac{ (c x \alpha)^x (c y \alpha)^y }{ c(1-\alpha) }$Simplify:$K = c^{x+y-1} x^x y^y \frac{ \alpha^{x+y} }{ 1-\alpha }$Since the solution is concentrated, $\alpha$ is small, but generally in such problems the expression is kept exact. However, for simplification, at high concentration the term $(1-\alpha) \approx 1$ is often used if it is a weak electrolyte. So approximately:$K \approx c^{x+y-1} x^x y^y \alpha^{x+y}$$\alpha^{x+y} = \frac{ K }{ c^{x+y-1} x^x y^y }$$\alpha = \left( \frac{ K }{ c^{x+y-1} x^x y^y } \right)^{ \frac{1}{x+y} }$This matches Option B:$\alpha = \left( \frac{ K }{ c^{x+y-1} x^x y^y } \right)^{ \frac{1}{x+y} }$