A certain weak acid has a dissocation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base is

Options

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

Answer

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

### Step 1: Identify the weak acid and its dissociation
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 dissociation constant \( K_a \) for this reaction is given as:
\[ K_a = 1.0 \times 10^{-4} \]

### Step 2: Write the reaction with a strong base
When the weak acid reacts with a strong base (let's denote the strong base as BOH), the neutralization reaction can be represented as:
\[ HA + BOH \rightarrow AB + H_2O \]
where AB is the salt formed from the reaction.

### Step 3: Consider the hydrolysis of the anion
The anion \( A^- \) from the weak acid can undergo hydrolysis in water:
\[ A^- + H_2O \rightleftharpoons HA + OH^- \]
The equilibrium constant for this hydrolysis reaction is known as the hydrolysis constant \( K_h \).

### Step 4: Relate \( K_h \) to \( K_a \) and \( K_w \)
The hydrolysis constant \( K_h \) can be calculated using the relationship:
\[ K_h = \frac{K_w}{K_a} \]
where \( K_w \) is the ion product of water, which at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

### Step 5: Substitute the values into the equation
Now, substituting the known values into the equation:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} \]

### Step 6: Calculate \( K_h \)
Performing the calculation:
\[ K_h = 1.0 \times 10^{-14} \div 1.0 \times 10^{-4} = 1.0 \times 10^{-10} \]

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

### Final Answer
The equilibrium constant for the reaction of the weak acid with the strong base is \( 1.0 \times 10^{-10} \).

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