Degree of dissociation of 0.1 molar acetic acid at `25^(@)C`
`(K_(a) = 1.0 xx 10^(-5))` is

Options

A. 0.01
B. 0.1
C. 1
D. 0.02

Answer

To find the degree of dissociation (α) of 0.1 M acetic acid at 25°C, we can follow these steps:

### Step 1: Write the dissociation equation
Acetic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the initial concentrations
Let the initial concentration of acetic acid (C) be 0.1 M. At the start (t = 0):
- [CH₃COOH] = 0.1 M
- [CH₃COO⁻] = 0 M
- [H⁺] = 0 M

### Step 3: Define the change in concentration
Let α be the degree of dissociation. At equilibrium, the concentrations will be:
- [CH₃COOH] = C(1 - α) = 0.1(1 - α) M
- [CH₃COO⁻] = Cα = 0.1α M
- [H⁺] = Cα = 0.1α M

### Step 4: Write the expression for Ka
The dissociation constant (Kₐ) for acetic acid is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.1α)(0.1α)}{0.1(1 - α)} \]

### Step 5: Simplify the expression
This simplifies to:
\[ K_a = \frac{0.01α^2}{0.1(1 - α)} \]
\[ K_a = \frac{0.1α^2}{1 - α} \]

### Step 6: Substitute the value of Ka
Given that \( K_a = 1.0 \times 10^{-5} \):
\[ 1.0 \times 10^{-5} = \frac{0.1α^2}{1 - α} \]

### Step 7: Assume α is small
Since acetic acid is a weak acid, we can assume that α is small, so \( 1 - α \approx 1 \):
\[ 1.0 \times 10^{-5} = 0.1α^2 \]

### Step 8: Solve for α
Rearranging gives:
\[ α^2 = \frac{1.0 \times 10^{-5}}{0.1} \]
\[ α^2 = 1.0 \times 10^{-4} \]
Taking the square root:
\[ α = \sqrt{1.0 \times 10^{-4}} \]
\[ α = 0.01 \]

### Conclusion
The degree of dissociation (α) of 0.1 M acetic acid at 25°C is:
\[ α = 0.01 \]