Which one of the following is the correct quadratic form of the Ostwald's dilution law equation

Options

A. `alpha^(2)C + alpha K - K = 0`
B. `alpha^(2)C - alpha K - K = 0`
C. `alpha^(2)C - alpha K + K = 0`
D. `alpha^(2)C+alpha K + K = 0`

Explanation

To derive the quadratic form of Ostwald's dilution law equation, we start with the dissociation of a weak acid. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Identify the Weak Acid**:
We consider a mono-basic weak acid (HA) that dissociates in water:
\[ HA \rightleftharpoons H^+ + A^- \]

2. **Define Initial Concentration**:
Let the initial concentration of the weak acid be \( C \). When the acid dissociates, let \( \alpha \) be the degree of dissociation.

3. **Concentration After Dissociation**:
After dissociation, the concentrations at equilibrium will be:
- Concentration of \( HA \): \( C(1 - \alpha) \)
- Concentration of \( H^+ \): \( C\alpha \)
- Concentration of \( A^- \): \( C\alpha \)

4. **Write the Expression for Equilibrium Constant (K)**:
The equilibrium constant \( K \) for the dissociation of the weak acid can be expressed as:
\[ K = \frac{[H^+][A^-]}{[HA]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

5. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ K(1 - \alpha) = C\alpha^2 \]
Expanding this, we have:
\[ K - K\alpha = C\alpha^2 \]

6. **Forming the Quadratic Equation**:
Rearranging the terms leads to:
\[ C\alpha^2 + K\alpha - K = 0 \]
This is the quadratic form of Ostwald's dilution law equation.

7. **Conclusion**:
Thus, the correct quadratic form of the Ostwald's dilution law equation is:
\[ C\alpha^2 + K\alpha - K = 0 \]

### Final Answer:
The correct quadratic form of the Ostwald's dilution law equation is:
\[ C\alpha^2 + K\alpha - K = 0 \]