For a weak acid HA, Ostwald's dilution law is representented by the equation

Options

A. `K_(a) = (alphaC)/(1 - alpha^(2)) `
B. `K_(a) = (alpha^(2)C)/(1 - alpha ) `
C. `alpha = (K_(a)C)/(1 - C) `
D. `K_(a) = (alpha^(2)C)/(1 - alpha^(2))`

Answer

To derive Ostwald's dilution law for a weak acid HA, we can follow these steps:

### Step 1: Understand the dissociation of the weak acid
The weak acid HA dissociates in water as follows:
\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

### Step 2: Set up the initial and equilibrium concentrations
Let the initial concentration of the weak acid HA be \( C \). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- Concentration of HA at equilibrium: \( C(1 - \alpha) \)
- Concentration of \( \text{H}^+ \) at equilibrium: \( C\alpha \)
- Concentration of \( \text{A}^- \) at equilibrium: \( C\alpha \)

### Step 3: Write the expression for the acid dissociation constant \( K_a \)
The acid dissociation constant \( K_a \) for the reaction is given by:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Substituting the equilibrium concentrations into this expression, we get:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 4: Simplify the expression
This can be simplified to:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]

### Step 5: Consider the case of weak electrolytes
For weak electrolytes, the degree of dissociation \( \alpha \) is very small (i.e., \( \alpha << 1 \)). Therefore, we can approximate \( 1 - \alpha \approx 1 \). This allows us to simplify our expression further:
\[ K_a \approx C\alpha^2 \]

### Final Expression
Thus, the Ostwald's dilution law for a weak acid can be represented as:
\[ K_a = \frac{C\alpha^2}{1 - \alpha} \]