The dissociation constant of a weak acid HA and weak base BOH are `2 xx 10^(-5) and 5 xx 10^(-6)` respectively.
The equilibrium constant for the neutralization reaction of the two is (ignnore hydrolysis of resulting salt )

Options

A. `1.0 xx 10^(-4)`
B. `1.0 xx 10^(-10)`
C. `2.5 xx 10^(-1)`
D. `1.0 xx 10^4`

Answer

To find the equilibrium constant for the neutralization reaction of a weak acid (HA) and a weak base (BOH), we can follow these steps:

### Step 1: Write the neutralization reaction
The neutralization reaction between the weak acid HA and the weak base BOH can be represented as:
\[ \text{HA} + \text{BOH} \rightleftharpoons \text{A}^- + \text{B}^+ + \text{H}_2\text{O} \]

### Step 2: Define the equilibrium constant expression
The equilibrium constant \( K \) for this reaction can be expressed as:
\[ K = \frac{[\text{A}^-][\text{B}^+]}{[\text{HA}][\text{BOH}]} \]

### Step 3: Relate \( K \) to \( K_a \) and \( K_b \)
We know the dissociation constants for the weak acid and weak base:
- \( K_a \) for HA:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
- \( K_b \) for BOH:
\[ K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]} \]

### Step 4: Use the ion product of water
The ion product of water \( K_w \) is given by:
\[ K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \]

### Step 5: Relate \( K \) to \( K_a \) and \( K_b \)
From the above expressions, we can derive the relationship:
\[ K = \frac{K_a \cdot K_b}{K_w} \]

### Step 6: Substitute the values
Given:
- \( K_a = 2 \times 10^{-5} \)
- \( K_b = 5 \times 10^{-6} \)
- \( K_w = 1 \times 10^{-14} \)

Substituting these values into the equation:
\[ K = \frac{(2 \times 10^{-5}) \cdot (5 \times 10^{-6})}{1 \times 10^{-14}} \]

### Step 7: Calculate \( K \)
Calculating the numerator:
\[ (2 \times 5) = 10 \]
And the powers of ten:
\[ 10^{-5} \cdot 10^{-6} = 10^{-11} \]
So,
\[ K = \frac{10 \times 10^{-11}}{10^{-14}} = 10 \times 10^{3} = 1 \times 10^{4} \]

### Final Answer
Thus, the equilibrium constant for the neutralization reaction is:
\[ K = 1 \times 10^{4} \]