Dissociation constant of a weak acid is `1 xx 10^(-4)` . Equilibrium constant of its reaction with strong base is

Options

A. `1 xx 10^(-4)`
B. `1 xx 10^(10)`
C. `1 xx 10^(-10)`
D. `1 xx 10^(14)`

Answer

To find the equilibrium constant for the reaction of a weak acid with a strong base, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Weak Acid and its Dissociation Constant**:
- Let the weak acid be represented as \( HA \).
- The dissociation of the weak acid in water can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
- The dissociation constant (\( K_a \)) for this reaction is given as:
\[ K_a = 1 \times 10^{-4} \]

2. **Write the Reaction with a Strong Base**:
- When the weak acid \( HA \) reacts with a strong base (e.g., \( NaOH \)), the reaction can be represented as:
\[ HA + OH^- \rightarrow A^- + H_2O \]

3. **Define the Equilibrium Constant for the Reaction**:
- The equilibrium constant (\( K_{eq} \)) for the reaction can be expressed as:
\[ K_{eq} = \frac{[A^-][H_2O]}{[HA][OH^-]} \]
- Since water is a liquid and its concentration remains constant, we can simplify this to:
\[ K_{eq} = \frac{[A^-]}{[HA][OH^-]} \]

4. **Relate \( K_{eq} \) to \( K_a \) and \( K_w \)**:
- The dissociation constant of water (\( K_w \)) at 25°C is:
\[ K_w = [H^+][OH^-] = 1 \times 10^{-14} \]
- From the dissociation of the weak acid, we know:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
- Rearranging this gives:
\[ [H^+] = \frac{K_a \cdot [HA]}{[A^-]} \]

5. **Substituting \( [H^+] \) into the \( K_{eq} \) Expression**:
- We can express \( K_{eq} \) in terms of \( K_a \) and \( K_w \):
\[ K_{eq} = \frac{K_w}{K_a} \]

6. **Calculate \( K_{eq} \)**:
- Substitute the known values of \( K_w \) and \( K_a \):
\[ K_{eq} = \frac{1 \times 10^{-14}}{1 \times 10^{-4}} = 1 \times 10^{-10} \]

### Final Answer:
The equilibrium constant (\( K_{eq} \)) for the reaction of the weak acid with a strong base is:
\[ K_{eq} = 1 \times 10^{-10} \]

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