Question

For `C' molar solution of weak electrolyte \( A_xB_y \), dissociation constant is \( K \), then degree of dissociation of \( A_xB_y \) at equilibrium is

• A. \(\left(\dfrac{K}{C^{x+y-1}x^x y^y}\right)^{\frac{1}{x+y}}\)
• B. \(\left(\dfrac{K \times C^{x+y-1}}{x^x - y^y}\right)^{\frac{1}{x+y}}\)
• C. \(\left(\dfrac{K \cdot x^x \cdot y^y}{C^{x+y-1}}\right)^{\frac{1}{x+y}}\)
• D. \(\left(\dfrac{K \times C^{x+y-1}}{x^x \times y^y}\right)^{x+y}\)

Hint:

For a weak electrolyte, always use the approximation \(1-\alpha \approx 1\) because the degree of dissociation is very small. Then substitute equilibrium concentrations carefully in the dissociation constant expression.

Answer 1

The Correct Option is A

Step 1: Write the dissociation of the weak electrolyte.
The weak electrolyte \( A_xB_y \) dissociates as:
\[ A_xB_y \rightleftharpoons xA^{\cdots} + yB^{\cdots} \] If the initial concentration of the electrolyte is \( C \) molar and the degree of dissociation is \( \alpha \), then at equilibrium:
\[ [A_xB_y] = C(1-\alpha) \] \[ [A] = xC\alpha \] \[ [B] = yC\alpha \]
Step 2: Write the expression for dissociation constant.
The dissociation constant is:
\[ K = \frac{[A]^x[B]^y}{[A_xB_y]} \] Substituting the equilibrium concentrations:
\[ K = \frac{(xC\alpha)^x (yC\alpha)^y}{C(1-\alpha)} \]
Step 3: Simplify the expression.
Expanding the numerator:
\[ K = \frac{x^x y^y C^{x+y}\alpha^{x+y}}{C(1-\alpha)} \] \[ K = \frac{x^x y^y C^{x+y-1}\alpha^{x+y}}{(1-\alpha)} \] Since \( A_xB_y \) is a weak electrolyte, its degree of dissociation is very small, so:
\[ 1-\alpha \approx 1 \] Therefore:
\[ K = x^x y^y C^{x+y-1}\alpha^{x+y} \]
Step 4: Solve for degree of dissociation \( \alpha \).
Rearranging the equation:
\[ \alpha^{x+y} = \frac{K}{C^{x+y-1}x^x y^y} \] Taking power \( \dfrac{1}{x+y} \) on both sides:
\[ \alpha = \left(\frac{K}{C^{x+y-1}x^x y^y}\right)^{\frac{1}{x+y}} \]
Step 5: Match with the given options.
This expression matches option (A). Hence, the required degree of dissociation at equilibrium is:
\[ \left(\frac{K}{C^{x+y-1}x^x y^y}\right)^{\frac{1}{x+y}} \] Final Answer: (A) \(\left(\dfrac{K}{C^{x+y-1}x^x y^y}\right)^{\frac{1}{x+y}}\)