If C the concentration of a weak electrolyte , `alpha` is the degree of ionization and `K_a` is the acid ionization constant , then the correct relationship between `alpha` , C and `K_a` is

Options

A. `a^2=sqrt((K_a)/C)`
B. `a^2=sqrt((C)/(K_a))`
C. `alpha = sqrt((C)/(K_a))`
D. `alpha = sqrt ((K_a)/(C))`

Answer

To derive the relationship between the concentration (C), degree of ionization (α), and the acid ionization constant (K_a) for a weak electrolyte, we can follow these steps:

### Step 1: Write the dissociation equation
For a weak acid (HA), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Define the concentration and degree of ionization
Let:
- \( C \) = initial concentration of the weak acid (HA)
- \( \alpha \) = degree of ionization (the fraction of the acid that dissociates)

### Step 3: Express the concentrations at equilibrium
At equilibrium, the concentrations can be expressed as:
- Concentration of \( HA \) = \( C(1 - \alpha) \)
- Concentration of \( H^+ \) = \( C\alpha \)
- Concentration of \( A^- \) = \( C\alpha \)

### Step 4: Write the expression for the acid dissociation constant (K_a)
The acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Substituting the equilibrium concentrations into this expression:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 5: Simplify the equation
This simplifies to:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

### Step 6: Make an approximation for weak acids
Since \( \alpha \) is small for weak acids, we can approximate \( 1 - \alpha \approx 1 \). Thus, the equation simplifies to:
\[ K_a \approx C\alpha^2 \]

### Step 7: Solve for α
Rearranging the equation gives us:
\[ \alpha^2 = \frac{K_a}{C} \]

Taking the square root of both sides:
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

### Conclusion
The correct relationship between the degree of ionization (α), concentration (C), and the acid ionization constant (K_a) is:
\[ \alpha = \sqrt{\frac{K_a}{C}} \]