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 between a weak acid (HA) and a weak base (BOH), we can use the relationship between the dissociation constants of the acid and base, as well as the ionization constant of water (Kw).

### Step-by-Step Solution:

1. **Identify the given values:**
- Dissociation constant of the weak acid (HA), \( K_a = 2 \times 10^{-5} \)
- Dissociation constant of the weak base (BOH), \( K_b = 5 \times 10^{-6} \)
- Ionization constant of water, \( K_w = 1 \times 10^{-14} \)

2. **Write the expression for the equilibrium constant (K) for the neutralization reaction:**
The equilibrium constant for the neutralization reaction of a weak acid and a weak base is given by the formula:
\[ K = \frac{K_a \times K_b}{K_w} \]

3. **Substitute the known values into the equation:**
\[ K = \frac{(2 \times 10^{-5}) \times (5 \times 10^{-6})}{1 \times 10^{-14}} \]

4. **Calculate the numerator:**
\[ K_a \times K_b = (2 \times 10^{-5}) \times (5 \times 10^{-6}) = 10 \times 10^{-11} = 1 \times 10^{-10} \]

5. **Now substitute this result back into the equation for K:**
\[ K = \frac{1 \times 10^{-10}}{1 \times 10^{-14}} \]

6. **Perform the division:**
\[ K = 1 \times 10^{-10 + 14} = 1 \times 10^{4} \]

7. **Final result:**
The equilibrium constant for the neutralization reaction is:
\[ K = 1 \times 10^{4} \]

### Conclusion:
The equilibrium constant for the neutralization reaction of the weak acid HA and weak base BOH is \( 1 \times 10^{4} \).

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