The degree of dissociation of `0.1N CH_(3)COOH` is `(K_(a)= 1xx10^(-5))`

Options

A. `10^(-5)`
B. `10^(-2)`
C. `10^(-3)`
D. `10^(-4)`

Answer

To find the degree of dissociation (α) of 0.1N acetic acid (CH₃COOH) given that the dissociation constant (Kₐ) is \(1 \times 10^{-5}\), we can follow these steps:

### Step 1: Understand the relationship between Normality and Molarity
For acetic acid, which is a monoprotic acid, the normality (N) is equal to the molarity (M). Therefore, for 0.1N acetic acid:
\[ \text{Molarity (C)} = 0.1 \, \text{M} \]

### Step 2: Write the dissociation equation
The dissociation of acetic acid can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 3: Set up the expression for the equilibrium concentrations
Let the degree of dissociation be α. Initially, we have:
- Concentration of CH₃COOH = C = 0.1 M
- Concentration of CH₃COO⁻ = 0
- Concentration of H⁺ = 0

At equilibrium, the concentrations will be:
- CH₃COOH = C(1 - α) = 0.1(1 - α)
- CH₃COO⁻ = Cα = 0.1α
- H⁺ = Cα = 0.1α

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

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

### Step 5: Simplify the equation
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 \( K_a = 1 \times 10^{-5} \):
\[ 1 \times 10^{-5} = \frac{0.1α^2}{1 - α} \]

### Step 7: Assume α is small
Since α is very small compared to 1, we can approximate \(1 - α \approx 1\):
\[ 1 \times 10^{-5} = 0.1α^2 \]

### Step 8: Solve for α
Rearranging gives:
\[ α^2 = \frac{1 \times 10^{-5}}{0.1} \]
\[ α^2 = 1 \times 10^{-4} \]
\[ α = \sqrt{1 \times 10^{-4}} \]
\[ α = 10^{-2} \]
\[ α = 0.01 \]

### Conclusion
The degree of dissociation (α) of 0.1N CH₃COOH is 0.01.