For a 'C'`M` concentrated solution of a weak electrolyte 'A_(x)B_(y) 'alpha' degree of dissociation) is

Options

A. `sqrt(K_(eq)//xyC)`
B. `[K_(eq).C//(xy)]^(1//2)`
C. `(K_(eq)//C^(x+y-1).x^(x)x^(y))^((1)/(x+y))`
D. `[K_(eq)//C(x+y)]^(1//2)`

Explanation

To solve the problem regarding the degree of dissociation (α) of a weak electrolyte \( A_xB_y \) in a \( C \) molar solution, we will follow these steps:

### Step 1: Write the dissociation equation
The weak electrolyte \( A_xB_y \) dissociates into its constituent ions. The dissociation can be represented as:
\[ A_xB_y \rightleftharpoons xA^+ + yB^- \]

### Step 2: Set up initial concentrations
Initially, the concentration of \( A_xB_y \) is \( C \) mol/L. The initial concentrations of the ions \( A^+ \) and \( B^- \) are both 0.

### Step 3: Establish equilibrium concentrations
At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- For \( A^+ \): \( [A^+] = xC\alpha \)
- For \( B^- \): \( [B^-] = yC\alpha \)
- For \( A_xB_y \): \( [A_xB_y] = C(1 - \alpha) \)

### Step 4: Write the expression for the equilibrium constant (K)
The equilibrium constant \( K \) for the dissociation can be expressed as:
\[ K = \frac{[A^+]^x [B^-]^y}{[A_xB_y]} = \frac{(xC\alpha)^x (yC\alpha)^y}{C(1 - \alpha)} \]

### Step 5: Simplify the expression
Substituting the equilibrium concentrations into the expression for \( K \):
\[ K = \frac{(xC\alpha)^x (yC\alpha)^y}{C(1 - \alpha)} = \frac{x^x y^y C^{x+y} \alpha^{x+y}}{C(1 - \alpha)} \]
This simplifies to:
\[ K = \frac{x^x y^y C^{x+y-1} \alpha^{x+y}}{1 - \alpha} \]

### Step 6: Assume \( \alpha \) is small
Since \( A_xB_y \) is a weak electrolyte, we can assume that \( \alpha \) is very small compared to 1, which allows us to neglect \( \alpha \) in the denominator:
\[ K \approx x^x y^y C^{x+y-1} \alpha^{x+y} \]

### Step 7: Solve for α
Rearranging the equation to solve for \( \alpha \):
\[ \alpha^{x+y} = \frac{K(1 - \alpha)}{x^x y^y C^{x+y-1}} \]
Since \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \):
\[ \alpha^{x+y} \approx \frac{K}{x^x y^y C^{x+y-1}} \]
Taking the \( (x+y) \)th root gives:
\[ \alpha \approx \left( \frac{K}{x^x y^y C^{x+y-1}} \right)^{\frac{1}{x+y}} \]

### Final Answer
Thus, the degree of dissociation \( \alpha \) for a \( C \) molar concentrated solution of the weak electrolyte \( A_xB_y \) is given by:
\[ \alpha \approx \left( \frac{K}{x^x y^y C^{x+y-1}} \right)^{\frac{1}{x+y}} \]