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:

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

with the provided base dissociation constant \( K_b = 1 \times 10^{-5} \), we can follow these steps:

### Step 1: Understand the relationship between K_h and K_b
The hydrolysis constant \( K_h \) can be calculated using the relationship:

\[ K_h = \frac{K_w}{K_b} \]

where \( K_w \) is the ion product of water, which is typically \( 1 \times 10^{-14} \) at 25°C.

### Step 2: Substitute the known values into the equation
We know:
- \( K_w = 1 \times 10^{-14} \)
- \( K_b = 1 \times 10^{-5} \)

Substituting these values into the equation gives:

\[ K_h = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} \]

### Step 3: Perform the calculation
Now, we simplify the expression:

\[ K_h = 1 \times 10^{-14} \div 1 \times 10^{-5} = 1 \times 10^{-14 + 5} = 1 \times 10^{-9} \]

### Step 4: State the final answer
Thus, the hydrolysis constant \( K_h \) is:

\[ K_h = 1 \times 10^{-9} \]

### Final Answer
The hydrolysis constant is \( 1 \times 10^{-9} \).

---