Dissociation constant of a weak acid is `10^(-6)` . What is the value of equilibrium constant for its reaction with strong base

Options

A. `10^(-5)`
B. 6.9
C. None
D. `10^(9)`

Explanation

To solve the problem of finding the equilibrium constant for the reaction of a weak acid with a strong base given that the dissociation constant (Ka) of the weak acid is \(10^{-6}\), we can follow these steps:

### Step 1: Write the dissociation reaction of the weak acid.
Let’s denote the weak acid as HA. The dissociation of the weak acid in water can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The equilibrium constant for this dissociation is given by:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Given that \(K_a = 10^{-6}\).

### Step 2: Write the neutralization reaction with a strong base.
When the weak acid reacts with a strong base (let's denote the strong base as BOH), the reaction can be represented as:
\[ HA + BOH \rightarrow A^- + H_2O \]
This reaction essentially shows that the weak acid (HA) reacts with the strong base (BOH) to form the conjugate base (A^-) and water.

### Step 3: Determine the equilibrium constant for the reaction.
The equilibrium constant for the reaction of the weak acid with the strong base (K) can be derived from the dissociation constant of the weak acid and the dissociation constant of water (Kw). The relationship is given by:
\[ K = \frac{K_w}{K_a} \]
Where \(K_w\) is the ion product of water, which at 25°C is \(1.0 \times 10^{-14}\).

### Step 4: Substitute the values into the equation.
Now substituting the known values:
\[ K_w = 1.0 \times 10^{-14} \]
\[ K_a = 10^{-6} \]
Thus,
\[ K = \frac{1.0 \times 10^{-14}}{10^{-6}} = 1.0 \times 10^{-8} \]

### Conclusion
The equilibrium constant for the reaction of the weak acid with the strong base is:
\[ K = 1.0 \times 10^{-8} \]