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

### Step 1: Write the equilibrium reaction
The equilibrium reaction for the formation of pyridinium ion from pyridine can be expressed as:
\[ \text{C}_5\text{H}_5\text{N} + \text{H}_2\text{O} \rightleftharpoons \text{C}_5\text{H}_5\text{N}^+ + \text{OH}^- \]

### Step 2: Set up the equilibrium expression
The base dissociation constant \( K_b \) for this reaction is given by:
\[ K_b = \frac{[\text{C}_5\text{H}_5\text{N}^+][\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}]} \]

### Step 3: Define the initial concentrations and changes
Let:
- Initial concentration of pyridine, \([C_5H_5N] = 0.10 \, M\)
- Change in concentration of pyridine that ionizes to form pyridinium ion = \(x\)

At equilibrium:
- \([C_5H_5N] = 0.10 - x\)
- \([C_5H_5N^+] = x\)
- \([\text{OH}^-] = x\)

### Step 4: Substitute into the equilibrium expression
Substituting these values into the \( K_b \) expression gives:
\[ K_b = \frac{x \cdot x}{0.10 - x} = \frac{x^2}{0.10 - x} \]

### Step 5: Substitute the value of \( K_b \)
We know \( K_b = 1.7 \times 10^{-9} \), so we can write:
\[ 1.7 \times 10^{-9} = \frac{x^2}{0.10 - x} \]

### Step 6: Make an assumption
Since \( K_b \) is very small, we can assume that \( x \) is much smaller than 0.10 M, thus \( 0.10 - x \approx 0.10 \). This simplifies our equation to:
\[ 1.7 \times 10^{-9} = \frac{x^2}{0.10} \]

### Step 7: Solve for \( x \)
Rearranging gives:
\[ x^2 = 1.7 \times 10^{-9} \times 0.10 \]
\[ x^2 = 1.7 \times 10^{-10} \]
\[ x = \sqrt{1.7 \times 10^{-10}} \]
\[ x \approx 1.303 \times 10^{-5} \, M \]

### Step 8: Calculate the percentage of pyridine that ionizes
The percentage of pyridine that forms pyridinium ion is given by:
\[ \text{Percentage} = \left( \frac{x}{0.10} \right) \times 100 \]
\[ \text{Percentage} = \left( \frac{1.303 \times 10^{-5}}{0.10} \right) \times 100 \]
\[ \text{Percentage} \approx 0.01303\% \]

### Final Answer
The percentage of pyridine that forms pyridinium ion in a 0.10 M aqueous solution is approximately **0.01303%**.