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 molar 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 be \( C = 0.1 \, \text{M} \). At the start:
- \([ \text{CH}_3\text{COOH} ] = C = 0.1 \, \text{M}\)
- \([ \text{CH}_3\text{COO}^- ] = 0 \)
- \([ \text{H}^+ ] = 0 \)

### Step 3: Define the changes at equilibrium
Let α be the degree of dissociation. At equilibrium:
- \([ \text{CH}_3\text{COOH} ] = C(1 - \alpha) = 0.1(1 - \alpha)\)
- \([ \text{CH}_3\text{COO}^- ] = \alpha C = 0.1\alpha\)
- \([ \text{H}^+ ] = \alpha C = 0.1\alpha\)

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

Substituting the equilibrium concentrations:
\[ K_a = \frac{(0.1\alpha)(0.1\alpha)}{0.1(1 - \alpha)} \]

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

### Step 6: Substitute the known values
Given \( K_a = 1.0 \times 10^{-5} \) and \( C = 0.1 \):
\[ 1.0 \times 10^{-5} = \frac{0.1\alpha^2}{1 - \alpha} \]

### Step 7: Rearranging the equation
Multiplying both sides by \( (1 - \alpha) \):
\[ 1.0 \times 10^{-5} (1 - \alpha) = 0.1\alpha^2 \]
\[ 1.0 \times 10^{-5} - 1.0 \times 10^{-5}\alpha = 0.1\alpha^2 \]

### Step 8: Rearranging to form a quadratic equation
Rearranging gives:
\[ 0.1\alpha^2 + 1.0 \times 10^{-5}\alpha - 1.0 \times 10^{-5} = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 0.1 \), \( b = 1.0 \times 10^{-5} \), and \( c = -1.0 \times 10^{-5} \).

Calculating the discriminant:
\[ b^2 - 4ac = (1.0 \times 10^{-5})^2 - 4(0.1)(-1.0 \times 10^{-5}) \]
\[ = 1.0 \times 10^{-10} + 4.0 \times 10^{-6} \]
\[ = 4.0 \times 10^{-6} + 1.0 \times 10^{-10} \approx 4.0 \times 10^{-6} \]

Now, substituting back into the quadratic formula:
\[ \alpha = \frac{-1.0 \times 10^{-5} \pm \sqrt{4.0 \times 10^{-6}}}{2(0.1)} \]
\[ = \frac{-1.0 \times 10^{-5} \pm 2.0 \times 10^{-3}}{0.2} \]

Taking the positive root:
\[ \alpha = \frac{1.99 \times 10^{-3}}{0.2} \approx 0.01 \]

### Final Answer
Thus, the degree of dissociation (α) of 0.1 molar acetic acid at 25°C is approximately:
\[ \alpha \approx 0.01 \]