In the hydrolysis equilibrium
`B^(+)H_(2)OhArrBOH+H^(+),K_(b)=1xx10^(-5)`
The hydrolysis constant is

Options

A. `10^(-5)`
B. `10^(-19)`
C. `10^(-10)`
D. `10^(-9)`

Explanation

To find the hydrolysis constant \( K_H \) for the given equilibrium reaction:

\[ B^+ + H_2O \rightleftharpoons BOH + H^+ \]

we can use the relationship between the hydrolysis constant \( K_H \), the ion product of water \( K_W \), and the base dissociation constant \( K_B \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The base dissociation constant \( K_B \) is given as:
\[ K_B = 1 \times 10^{-5} \]
- The ion product of water \( K_W \) at 25°C is:
\[ K_W = 1 \times 10^{-14} \]

2. **Write the Relationship Between Constants:**
The relationship between the hydrolysis constant \( K_H \), the ion product of water \( K_W \), and the base dissociation constant \( K_B \) is given by:
\[ K_H = \frac{K_W}{K_B} \]

3. **Substitute the Values:**
Now, substitute the known values into the equation:
\[ K_H = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} \]

4. **Perform the Calculation:**
\[ K_H = 1 \times 10^{-14} \div 1 \times 10^{-5} = 1 \times 10^{-9} \]

5. **Final Result:**
Therefore, the hydrolysis constant \( K_H \) is:
\[ K_H = 1 \times 10^{-9} \]

### Conclusion:
The hydrolysis constant \( K_H \) for the given equilibrium is \( 1 \times 10^{-9} \).