The degree of dissociation of acetic acid in a 0.1 M solution is `1.32xx10^(-2)`, find out the pKa :-

Options

A. 5.75
B. 3.75
C. 4. 00
D. 4. 76

Explanation

To find the pKa of acetic acid given its degree of dissociation and concentration, we can follow these steps:

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

### Step 2: Define the variables
Let:
- \( C \) = initial concentration of acetic acid = 0.1 M
- \( \alpha \) = degree of dissociation = \( 1.32 \times 10^{-2} \)

### Step 3: Calculate the equilibrium concentrations
At equilibrium:
- Concentration of CH₃COOH = \( C(1 - \alpha) = 0.1(1 - 1.32 \times 10^{-2}) \)
- Concentration of CH₃COO⁻ = \( C\alpha = 0.1 \times 1.32 \times 10^{-2} \)
- Concentration of H⁺ = \( C\alpha = 0.1 \times 1.32 \times 10^{-2} \)

Calculating these:
- Concentration of CH₃COOH = \( 0.1(1 - 0.0132) = 0.1 \times 0.9868 = 0.09868 \) M
- Concentration of CH₃COO⁻ = \( 0.1 \times 1.32 \times 10^{-2} = 0.00132 \) M
- Concentration of H⁺ = \( 0.00132 \) M

### Step 4: Write the expression for Ka
The dissociation constant \( K_a \) 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:
\[ K_a = \frac{(0.00132)(0.00132)}{0.09868} \]

### Step 5: Calculate Ka
Calculating \( K_a \):
\[ K_a = \frac{(0.00132)^2}{0.09868} = \frac{1.7424 \times 10^{-6}}{0.09868} \approx 1.76 \times 10^{-5} \]

### Step 6: Calculate pKa
The pKa is calculated using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1.76 \times 10^{-5}) \]

### Step 7: Simplify the logarithm
Using properties of logarithms:
\[ pK_a = -\log(1.76) - \log(10^{-5}) \]
\[ pK_a = -0.246 + 5 \]
\[ pK_a \approx 4.754 \]

### Final Answer
Thus, the pKa of acetic acid is approximately **4.76**.

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