For a concentrated solution of a weak electrolyte (`K_(eq)` = equilibrium constant) `A_2B_3` of concentration ‘c’, the degree of dissociation `alpha` is

Options

A. `((K_(eq))/(5c)^4)^(1/5)`
B. `((K_(eq))/(108c)^4)^(1/5)`
C. `((K_(eq))/(25c)^2)^(1/5)`
D. `((K_(eq))/(6c)^5)^(1/5)`

Answer

To find the degree of dissociation (α) for a concentrated solution of a weak electrolyte \( A_2B_3 \) with an equilibrium constant \( K_{eq} \) and initial concentration \( c \), we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the weak electrolyte \( A_2B_3 \) can be represented as:
\[ A_2B_3 \rightleftharpoons 2A^+ + 3B^- \]

### Step 2: Set up the initial concentration and changes
Let the initial concentration of \( A_2B_3 \) be \( c \). When it dissociates, let \( \alpha \) be the degree of dissociation. The changes in concentration at equilibrium will be:
- For \( A_2B_3 \): \( c - c\alpha \) (which can be approximated as \( c \) since \( \alpha \) is small)
- For \( A^+ \): \( 2c\alpha \)
- For \( B^- \): \( 3c\alpha \)

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_{eq} \) for the reaction can be expressed as:
\[ K_{eq} = \frac{[A^+]^2[B^-]^3}{[A_2B_3]} \]
Substituting the equilibrium concentrations:
\[ K_{eq} = \frac{(2c\alpha)^2(3c\alpha)^3}{c - c\alpha} \]

### Step 4: Simplify the expression
Substituting the values into the equation:
\[ K_{eq} = \frac{(4c^2\alpha^2)(27c^3\alpha^3)}{c(1 - \alpha)} \approx \frac{108c^5\alpha^5}{c} \]
Since \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \).

### Step 5: Further simplification
This simplifies to:
\[ K_{eq} \approx 108c^4\alpha^5 \]

### Step 6: Solve for α
Rearranging gives:
\[ \alpha^5 = \frac{K_{eq}}{108c^4} \]
Taking the fifth root:
\[ \alpha = \left( \frac{K_{eq}}{108c^4} \right)^{\frac{1}{5}} \]

### Final Answer
Thus, the degree of dissociation \( \alpha \) is given by:
\[ \alpha = \left( \frac{K_{eq}}{108c^4} \right)^{\frac{1}{5}} \]