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 (BOH), we can use the relationship between the dissociation constant (K_b) and the hydrolysis constant. The formula we will use is:

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

Where:
- \( K_w \) is the ion product of water, which is \( 1.0 \times 10^{-14} \) at 25°C.
- \( K_b \) is the dissociation constant of the base, which is given as \( 4.6 \times 10^{-10} \).

### Step-by-Step Solution:

1. **Identify the given values:**
- \( K_b = 4.6 \times 10^{-10} \)
- \( K_w = 1.0 \times 10^{-14} \)

2. **Substitute the values into the formula:**
\[ K_h = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{4.6 \times 10^{-10}} \]

3. **Perform the division:**
\[ K_h = \frac{1.0}{4.6} \times 10^{-14 + 10} = \frac{1.0}{4.6} \times 10^{-4} \]

4. **Calculate \( \frac{1.0}{4.6} \):**
\[ \frac{1.0}{4.6} \approx 0.217 \]

5. **Combine the results:**
\[ K_h \approx 0.217 \times 10^{-4} = 2.17 \times 10^{-5} \]

6. **Final Answer:**
The hydrolysis constant \( K_h \) is approximately \( 2.17 \times 10^{-5} \).