If the `K_(a)` value in the hydrolysis reaction, `B^(+) + H_(2)O to BOH + H^(+)` is `1.0xx10^(-6)`,then the hydrolysis constant of the salt would be :

Options

A. `1.0xx10^(-6)`
B. `1.0xx10^(-7)`
C. `1.0xx10^(-8)`
D. `1.0xx10^(-9)`

Explanation

To find the hydrolysis constant (K_h) of the salt formed from a weak base (B) and a strong acid, we can use the relationship between the hydrolysis constant, the dissociation constant of water (K_w), and the base dissociation constant (K_b). The steps to solve the problem are as follows:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The acid dissociation constant (K_a) is given as \(1.0 \times 10^{-6}\).
- We need to find the base dissociation constant (K_b) using the relationship \(K_w = K_a \times K_b\), where \(K_w\) is the dissociation constant of water at 25°C, which is \(1.0 \times 10^{-14}\).

2. **Calculate K_b:**
- Rearranging the equation gives us:
\[ K_b = \frac{K_w}{K_a} \]
- Substituting the known values:
\[ K_b = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-6}} = 1.0 \times 10^{-8} \]

3. **Use the Relationship Between K_h, K_w, and K_b:**
- The hydrolysis constant (K_h) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_b} \]
- Substitute the values we have:
\[ K_h = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-8}} = 1.0 \times 10^{-6} \]

4. **Final Result:**
- Therefore, the hydrolysis constant of the salt is:
\[ K_h = 1.0 \times 10^{-6} \]