If the equilibrium constant of `BOH harr B^(o+) +overset(Theta)OH` at `25^(@)C` is `2.5 xx 10^(-6)`, then equilibrium constant for `BOH +H^(o+) hArr B^(o+)+H_(2)O` at the same temperature is

Options

A. `4.0 xx 10^(-9)`
B. `4.0xx10^(-5)`
C. `2.5xx10^(8)`
D. `2.5xx10^(-6)`

Explanation

To solve the problem, we need to find the equilibrium constant for the reaction:

\[ \text{BOH} + \text{H}^+ \rightleftharpoons \text{B}^+ + \text{H}_2\text{O} \]

Given that the equilibrium constant for the dissociation of BOH is:

\[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]

with \( K_1 = 2.5 \times 10^{-6} \).

### Step-by-Step Solution:

1. **Identify the Given Reaction and its Equilibrium Constant:**
- The first reaction is:
\[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]
- The equilibrium constant for this reaction is given as:
\[ K_1 = 2.5 \times 10^{-6} \]

2. **Write the Second Reaction:**
- The second reaction we need to consider is the dissociation of water:
\[ \text{H}^+ + \text{OH}^- \rightleftharpoons \text{H}_2\text{O} \]
- The equilibrium constant for this reaction, denoted as \( K_2 \), is the inverse of the dissociation constant of water (\( K_w \)):
\[ K_w = 1.0 \times 10^{-14} \quad \text{(at 25°C)} \]
- Therefore, the equilibrium constant for the reverse reaction is:
\[ K_2 = \frac{1}{K_w} = \frac{1}{1.0 \times 10^{-14}} = 1.0 \times 10^{14} \]

3. **Combine the Reactions:**
- To find the equilibrium constant for the desired reaction, we add the two reactions:
\[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \quad (K_1) \]
\[ \text{H}^+ + \text{OH}^- \rightleftharpoons \text{H}_2\text{O} \quad (K_2) \]
- When we add these reactions, the \( \text{OH}^- \) cancels out:
\[ \text{BOH} + \text{H}^+ \rightleftharpoons \text{B}^+ + \text{H}_2\text{O} \]

4. **Calculate the Overall Equilibrium Constant:**
- The equilibrium constant for the overall reaction (\( K_3 \)) is the product of the individual constants:
\[ K_3 = K_1 \times K_2 \]
- Substituting the values:
\[ K_3 = (2.5 \times 10^{-6}) \times (1.0 \times 10^{14}) = 2.5 \times 10^{8} \]

5. **Final Answer:**
- Thus, the equilibrium constant for the reaction \( \text{BOH} + \text{H}^+ \rightleftharpoons \text{B}^+ + \text{H}_2\text{O} \) is:
\[ K_3 = 2.5 \times 10^{8} \]