What is the percentage dissociation of 0.1 M solution of acetic acid ? `(K_(a)=10^(-5))`

Options

A. `10%`
B. `100%`
C. `1%`
D. `0.01%`

Explanation

To find the percentage dissociation of a 0.1 M solution of acetic acid (with \( K_a = 10^{-5} \)), we can follow these steps:

### Step 1: Understand the dissociation of acetic acid
Acetic acid (\( CH_3COOH \)) dissociates in water according to the following equilibrium reaction:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
Let \( \alpha \) be the degree of dissociation, which represents the fraction of acetic acid that dissociates.

### Step 2: Set up the expression for \( K_a \)
The equilibrium constant \( K_a \) for the dissociation of acetic acid can be expressed as:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
At the start, the concentration of acetic acid is 0.1 M. After dissociation, the concentrations at equilibrium will be:
- \( [CH_3COOH] = 0.1 - \alpha \)
- \( [CH_3COO^-] = \alpha \)
- \( [H^+] = \alpha \)

Substituting these into the \( K_a \) expression gives:
\[ K_a = \frac{\alpha \cdot \alpha}{0.1 - \alpha} = \frac{\alpha^2}{0.1 - \alpha} \]

### Step 3: Simplify the equation
Assuming \( \alpha \) is small compared to 0.1, we can approximate \( 0.1 - \alpha \approx 0.1 \):
\[ K_a \approx \frac{\alpha^2}{0.1} \]

### Step 4: Substitute the known values
Substituting \( K_a = 10^{-5} \):
\[ 10^{-5} = \frac{\alpha^2}{0.1} \]

### Step 5: Solve for \( \alpha^2 \)
Rearranging the equation gives:
\[ \alpha^2 = 10^{-5} \times 0.1 = 10^{-6} \]

### Step 6: Calculate \( \alpha \)
Taking the square root of both sides:
\[ \alpha = \sqrt{10^{-6}} = 10^{-3} \]

### Step 7: Calculate the percentage dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = 10^{-3} \times 100 = 0.1\% \]

### Final Answer
Thus, the percentage dissociation of a 0.1 M solution of acetic acid is **0.1%**.

---