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

Options

A. `1.0xx10^(-4)`
B. `1.0xx10^(-10)`
C. `1xx10^(-10)`
D. `1.0xx10^(-14)`

Explanation

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

### Step 1: Write the dissociation equation for the weak acid (HA)
The weak acid (HA) dissociates in water as follows:
\[ HA \rightleftharpoons H^+ + A^- \]
The equilibrium constant for this dissociation is given as:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
Given \( K_a = 1.0 \times 10^{-4} \).

### Step 2: Write the reaction equation for the weak acid with the strong base
When the weak acid reacts with a strong base (BOH), the reaction can be represented as:
\[ HA + BOH \rightarrow A^- + H_2O + B^+ \]
In this case, the \( H^+ \) from the weak acid reacts with \( OH^- \) from the strong base to form water.

### Step 3: Write the equilibrium expression for the reaction
The equilibrium constant (K) for the reaction can be expressed as:
\[ K = \frac{[A^-][H_2O]}{[HA][OH^-]} \]

### Step 4: Relate K to \( K_a \) and \( K_w \)
We know that:
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
From the equilibrium expressions, we can relate K to \( K_a \) and \( K_w \):
\[ K = \frac{K_w}{K_a} \]

### Step 5: Substitute the values of \( K_a \) and \( K_w \)
Substituting the known values:
\[ K = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} \]

### Step 6: Calculate K
Now, we can perform the calculation:
\[ K = 1.0 \times 10^{-14} \div 1.0 \times 10^{-4} = 1.0 \times 10^{-10} \]

### Step 7: Final answer
Thus, the equilibrium constant for the reaction of the weak acid with the strong base is:
\[ K = 1.0 \times 10^{-10} \]

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