At `298 K` a`0.01 M CH_(3)COOH` solution is `1.34%` ionized. The ionization constant `K_(a)` for acetic acid will be

Options

A. `1.82xx10^(-5)`
B. `18.2xx10^(-5)`
C. `0.182xx10^(-5)`
D. None of these

Explanation

To find the ionization constant \( K_a \) for acetic acid (\( CH_3COOH \)), we can follow these steps:

### Step 1: Understand the given data
We are given:
- Concentration of acetic acid, \( C = 0.01 \, M \)
- Degree of ionization, \( \alpha = 1.34\% = \frac{1.34}{100} = 0.0134 \)

### Step 2: Calculate the concentration of ionized species
The degree of ionization \( \alpha \) tells us the fraction of the acetic acid that has ionized. Therefore, the concentration of ionized acetic acid can be calculated as:
\[ \text{Concentration of ionized } CH_3COOH = C \times \alpha = 0.01 \, M \times 0.0134 = 1.34 \times 10^{-4} \, M \]

### Step 3: Determine the concentrations of the products at equilibrium
When acetic acid ionizes, it dissociates as follows:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
At equilibrium, the concentrations of the products \( [CH_3COO^-] \) and \( [H^+] \) will both be equal to the concentration of ionized acetic acid:
\[ [CH_3COO^-] = [H^+] = 1.34 \times 10^{-4} \, M \]

The concentration of the undissociated acetic acid at equilibrium will be:
\[ [CH_3COOH] = C - [CH_3COO^-] = 0.01 - 1.34 \times 10^{-4} \approx 0.01 \, M \]
(Since \( 1.34 \times 10^{-4} \) is very small compared to \( 0.01 \), we can approximate it as \( 0.01 \, M \).)

### Step 4: Write the expression for the ionization constant \( K_a \)
The ionization constant \( K_a \) is given by the expression:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]

Substituting the values we have:
\[ K_a = \frac{(1.34 \times 10^{-4})(1.34 \times 10^{-4})}{0.01} \]

### Step 5: Calculate \( K_a \)
Now, we perform the calculation:
\[ K_a = \frac{(1.34 \times 10^{-4})^2}{0.01} = \frac{1.7956 \times 10^{-8}}{0.01} = 1.7956 \times 10^{-6} \]

### Final Answer
Thus, the ionization constant \( K_a \) for acetic acid at \( 298 \, K \) is approximately:
\[ K_a \approx 1.80 \times 10^{-5} \]