If the `K_b` value in the hydrolysis reaction
`B^(+)+H_2O hArr BOH +H^+`
is `1.0 xx 10^(-6)`, then the hydrolysis constant of the salt would be

Options

A. `1 xx 10^(-6)`
B. `1 xx 10^(-7)`
C. `1 xx 10^(-8)`
D. `1xx 10^(-9)`

Answer

To find the hydrolysis constant of the salt (K_H) given the base dissociation constant (K_B) for the reaction:

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

where \( K_B = 1.0 \times 10^{-6} \), 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 values needed**:
- We know that \( K_B = 1.0 \times 10^{-6} \).
- The ionic product of water \( K_W \) at 25°C is \( 1.0 \times 10^{-14} \).

2. **Use the formula for hydrolysis constant**:
The hydrolysis constant \( K_H \) is given by the formula:
\[ K_H = \frac{K_W}{K_B} \]

3. **Substitute the known values**:
Now, substituting the values we have:
\[ K_H = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-6}} \]

4. **Perform the calculation**:
Dividing the two values:
\[ K_H = 1.0 \times 10^{-14} \div 1.0 \times 10^{-6} = 1.0 \times 10^{-8} \]

5. **Conclusion**:
Therefore, the hydrolysis constant of the salt is:
\[ K_H = 1.0 \times 10^{-8} \]

### Final Answer:
The hydrolysis constant of the salt is \( 1.0 \times 10^{-8} \).

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