The percentage of pyridine `(C_(5)H_(5)N)` that forms pyridinium ion `(C_(5)H_(5)N^(+)H)` in a 0.10 M aqueous pyridine solution (`K_(b)` for `C_(5)H_(5)N=1.7xx10^(-9)`) is

Options

A. `0.0060%`
B. `0.013%`
C. `0.77%`
D. `1.6%`

Explanation

To solve the problem of determining the percentage of pyridine that forms pyridinium ion in a 0.10 M aqueous solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Concentration of pyridine, \( C = 0.10 \, \text{M} \)
- Base dissociation constant, \( K_b = 1.7 \times 10^{-9} \)

2. **Use the formula for the degree of ionization (\( \alpha \))**:
The relationship between \( K_b \), concentration \( C \), and degree of ionization \( \alpha \) is given by:
\[ K_b = C \cdot \alpha^2 \]
Rearranging this gives:
\[ \alpha = \sqrt{\frac{K_b}{C}} \]

3. **Substitute the values into the equation**:
\[ \alpha = \sqrt{\frac{1.7 \times 10^{-9}}{0.10}} \]

4. **Calculate \( \alpha \)**:
\[ \alpha = \sqrt{1.7 \times 10^{-8}} \approx 1.30 \times 10^{-4} \]

5. **Calculate the percentage of pyridine that forms pyridinium ion**:
The percentage of ionization is given by:
\[ \text{Percentage} = \alpha \times 100 \]
Substituting the value of \( \alpha \):
\[ \text{Percentage} = 1.30 \times 10^{-4} \times 100 = 0.013\% \]

6. **Final Answer**:
The percentage of pyridine that forms pyridinium ion in a 0.10 M aqueous pyridine solution is **0.013%**.