If degree of ionization `(alpha)` of a weak electrolyte AB is very less then `alpha` is :

Options

A. Directly proportional to the square root of volume of solution
B. inversely proportional to the dilution
C. inversely proportional to the square root of concentration
D. directly proportional to concentration

Answer

To solve the question regarding the degree of ionization (α) of a weak electrolyte AB, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Weak Electrolyte:**
- A weak electrolyte AB dissociates in water to form ions:
\[ AB \rightleftharpoons A^+ + B^- \]

2. **Initial Concentration:**
- Let the initial concentration of AB be \( C \). Initially, there are no ions present, so:
- Concentration of \( A^+ = 0 \)
- Concentration of \( B^- = 0 \)

3. **Degree of Ionization (α):**
- The degree of ionization \( \alpha \) is defined as the fraction of the original solute that has dissociated into ions. If \( \alpha \) is very small, we can express the concentrations at equilibrium as:
- Concentration of \( A^+ = C\alpha \)
- Concentration of \( B^- = C\alpha \)
- Concentration of undissociated \( AB = C - C\alpha = C(1 - \alpha) \)

4. **Applying Ostwald's Dilution Law:**
- According to Ostwald's dilution law, the dissociation constant \( K_a \) for the weak electrolyte can be expressed as:
\[ K_a = \frac{[A^+][B^-]}{[AB]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{C(1 - \alpha)} \]
- Simplifying this gives:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

5. **Assuming α is Very Small:**
- Since \( \alpha \) is very small, we can approximate \( 1 - \alpha \approx 1 \). Thus, the equation simplifies to:
\[ K_a \approx C\alpha^2 \]

6. **Rearranging for α:**
- From the above equation, we can solve for \( \alpha \):
\[ \alpha^2 \approx \frac{K_a}{C} \]
\[ \alpha \approx \sqrt{\frac{K_a}{C}} \]

7. **Conclusion:**
- Therefore, the degree of ionization \( \alpha \) of the weak electrolyte AB is given by:
\[ \alpha \approx \sqrt{\frac{K_a}{C}} \]