Aniline, `C_(6)H_(5)NH_(2)` react with water according to the equation
`C_(6)H_(5)NH_(2)(aq)+H_(2)O(I)hArrC_(6)H_(5)NH_(3)^(+)(aq)+OH^(-)(aq)`
In a 0.180 M aqueous aniline solution the `[OH^(-)]=8.80xx10^(-6)M`
The value of the base ionization constant `K_(b)` for `C_(6)H_(5)NH_(2)(aq)` and the percent ionization of `C_(6)H_(5)NH_(2)` in this solution are,

Options

A. `4.3xx10^(-10)`
B. `3.1xx10^(-10)`
C. `4.9xx10^(-3)%`
D. `2.4xx10^(-3)%`

Explanation

To solve the problem, we need to find the base ionization constant \( K_b \) for aniline and the percent ionization of aniline in the given solution. Let's break down the steps:

### Step 1: Write the ionization equation
The ionization of aniline in water can be represented as:
\[ C_6H_5NH_2(aq) + H_2O(l) \rightleftharpoons C_6H_5NH_3^+(aq) + OH^-(aq) \]

### Step 2: Set up the initial concentrations
Let the initial concentration of aniline \( C_0 = 0.180 \, M \). Initially, the concentrations of \( C_6H_5NH_3^+ \) and \( OH^- \) are both 0.

### Step 3: Define the change in concentration
Let \( \alpha \) be the degree of ionization. At equilibrium, the concentrations will be:
- For aniline: \( [C_6H_5NH_2] = C_0(1 - \alpha) \)
- For aniline cation: \( [C_6H_5NH_3^+] = C_0 \alpha \)
- For hydroxide ions: \( [OH^-] = C_0 \alpha \)

### Step 4: Use the given concentration of hydroxide ions
We are given that \( [OH^-] = 8.80 \times 10^{-6} \, M \). Therefore, we can write:
\[ C_0 \alpha = 8.80 \times 10^{-6} \]
Substituting \( C_0 = 0.180 \, M \):
\[ 0.180 \alpha = 8.80 \times 10^{-6} \]
Solving for \( \alpha \):
\[ \alpha = \frac{8.80 \times 10^{-6}}{0.180} = 4.89 \times 10^{-5} \]

### Step 5: Calculate \( K_b \)
The base ionization constant \( K_b \) is given by:
\[ K_b = \frac{[OH^-][C_6H_5NH_3^+]}{[C_6H_5NH_2]} = \frac{(C_0 \alpha)(C_0 \alpha)}{C_0(1 - \alpha)} \]
Since \( \alpha \) is very small compared to 1, we can approximate \( 1 - \alpha \approx 1 \):
\[ K_b \approx \frac{(C_0 \alpha)^2}{C_0} \]
Substituting the values:
\[ K_b \approx C_0 \alpha^2 = 0.180 \times (4.89 \times 10^{-5})^2 \]
Calculating \( K_b \):
\[ K_b \approx 0.180 \times 2.40 \times 10^{-9} \approx 4.32 \times 10^{-10} \]

### Step 6: Calculate percent ionization
Percent ionization is given by:
\[ \text{Percent Ionization} = \left( \frac{\alpha}{C_0} \right) \times 100 \]
Substituting the values:
\[ \text{Percent Ionization} = \left( \frac{4.89 \times 10^{-5}}{0.180} \right) \times 100 \approx 0.0272\% \]

### Final Results
- The value of the base ionization constant \( K_b \) for aniline is approximately \( 4.32 \times 10^{-10} \).
- The percent ionization of aniline in this solution is approximately \( 0.0272\% \).