A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :

Options

A. `10^(-14)`
B. `10^(-9)`
C. `10^(-10)`
D. `10^(10)`

Explanation

To solve the problem, we need to find the equilibrium constant for the reaction of a weak acid with a strong base. Here are the steps to arrive at the solution:

### Step 1: Write the dissociation reaction of the weak acid
Let’s assume the weak acid is represented as HX. The dissociation of the weak acid can be written as:
\[ HX \rightleftharpoons H^+ + X^- \]
The dissociation constant \( K_a \) for this reaction is given as:
\[ K_a = 1.0 \times 10^{-4} \]

### Step 2: Write the reaction of the weak acid with the strong base
When the weak acid reacts with a strong base (like NaOH), the following reaction occurs:
\[ H^+ + OH^- \rightarrow H_2O \]
This reaction can also be expressed in terms of the weak acid:
\[ HX + OH^- \rightarrow H_2O + X^- \]

### Step 3: Relate the equilibrium constant for the reaction with the dissociation constant
The equilibrium constant \( K \) for the reaction of the weak acid with the strong base can be expressed in terms of \( K_a \) and the ion product of water \( K_w \). The relationship is given by:
\[ K = \frac{K_w}{K_a} \]

### Step 4: Substitute the values of \( K_a \) and \( K_w \)
We know that:
- \( K_a = 1.0 \times 10^{-4} \)
- The ion product of water \( K_w = 1.0 \times 10^{-14} \)

Now, substituting these values into the equation:
\[ K = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} \]

### Step 5: Calculate the equilibrium constant
Calculating the above expression:
\[ K = 1.0 \times 10^{-14} \times 10^{4} = 1.0 \times 10^{-10} \]

### Final Answer
Thus, the equilibrium constant for the reaction of the weak acid with the strong base is:
\[ K = 1.0 \times 10^{10} \]