The dissociation constant of anilinium hydroxide is `4.6 xx 10^(-10)`. What is hydrolysis constant?

Options

A. `2.18 xx 10^(-5)`
B. `2.18 xx 10^(-4)`
C. `1 xx 10^(-5)`
D. `1 xx 10^(-4)`

Answer

To find the hydrolysis constant (K_h) of anilinium hydroxide, we can use the relationship between the ionization constant of water (K_w) and the dissociation constant of the base (K_b). Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The hydrolysis constant (K_h) can be calculated using the formula:
\[ K_h = \frac{K_w}{K_b} \]
where:
- \( K_h \) = hydrolysis constant
- \( K_w \) = ionization constant of water
- \( K_b \) = dissociation constant of the base

2. **Identify Given Values**:
- The dissociation constant of anilinium hydroxide (K_b) is given as:
\[ K_b = 4.6 \times 10^{-10} \]
- The ionization constant of water (K_w) at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

3. **Substitute Values into the Formula**:
Now, substitute the values of \( K_w \) and \( K_b \) into the formula:
\[ K_h = \frac{1.0 \times 10^{-14}}{4.6 \times 10^{-10}} \]

4. **Perform the Calculation**:
Calculate \( K_h \):
\[ K_h = \frac{1.0 \times 10^{-14}}{4.6 \times 10^{-10}} \approx 2.17 \times 10^{-5} \]

5. **Final Answer**:
The hydrolysis constant (K_h) of anilinium hydroxide is:
\[ K_h \approx 2.17 \times 10^{-5} \]

### Conclusion:
The correct option that represents this hydrolysis constant is option 1.