Degree of Dissociation for Weak Electrolyte

To find the degree of dissociation ($\alpha$) for the weak electrolyte $\text{A}_x\text{B}_y$, consider its dissociation equilibrium in an aqueous solution:

$\text{A}_x\text{B}_y(aq) \rightleftharpoons x \text{A}^{y+}(aq) + y \text{B}^{x-}(aq)$

Let $c$ be the initial molar concentration of the electrolyte and $\alpha$ be its degree of dissociation.

The initial and equilibrium concentrations of the species are as follows:

$$\begin{array}{lcccc} & \text{A}_x\text{B}_y & \rightleftharpoons & x \text{A}^{y+} & + & y \text{B}^{x-} \\ \text{Initial Concentration:} & c & & 0 & & 0 \\ \text{Equilibrium Concentration:} & c(1 - \alpha) & & x c \alpha & & y c \alpha \end{array}$$

The expression for the dissociation constant $K$ of the weak electrolyte is given by:

$K = \frac{[\text{A}^{y+}]^x [\text{B}^{x-}]^y}{[\text{A}_x\text{B}_y]}$

Substitute the equilibrium concentrations into the expression for $K$:

$K = \frac{(xc\alpha)^x \cdot (yc\alpha)^y}{c(1 - \alpha)}$

For a weak electrolyte, the degree of dissociation is very small ($\alpha \ll 1$), so we can approximate $1 - \alpha \approx 1$. Thus, the expression simplifies to:

$K = \frac{x^x c^x \alpha^x \cdot y^y c^y \alpha^y}{c}$

Combining the terms with like bases:

$K = \frac{x^x y^y \cdot c^{x+y} \cdot \alpha^{x+y}}{c}$

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

Solving for $\alpha^{x+y}$:

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

Taking the $(x+y)^{\text{th}}$ root on both sides yields the degree of dissociation:

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

This corresponds to Option B.