The acid dissociation constant `K_(a)` of acetic acid is `1.74 xx 10^(-5) ` at 298 K. The pH of a solution of 0.1 M acetic acid is

Options

A. 2.88
B. 3.6
C. `4.0`
D. `1.0`

Explanation

To find the pH of a 0.1 M acetic acid solution, we will follow these steps:

### Step 1: Write the dissociation equation and expression for \( K_a \)
Acetic acid (\( CH_3COOH \)) dissociates in water as follows:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]

### Step 2: Set up the initial concentrations
Let the initial concentration of acetic acid be \( C = 0.1 \, M \). At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- The concentration of \( CH_3COOH \) at equilibrium will be \( C(1 - \alpha) \)
- The concentrations of \( CH_3COO^- \) and \( H^+ \) will both be \( C\alpha \)

### Step 3: Substitute into the \( K_a \) expression
Substituting the equilibrium concentrations into the \( K_a \) expression gives:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C^2\alpha^2}{C(1 - \alpha)} \]
Since \( \alpha \) is very small (the dissociation of acetic acid is weak), we can approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx C\alpha^2 \]

### Step 4: Solve for \( \alpha \)
Rearranging the equation gives:
\[ \alpha^2 = \frac{K_a}{C} \]
Substituting the values:
\[ K_a = 1.74 \times 10^{-5}, \quad C = 0.1 \]
\[ \alpha^2 = \frac{1.74 \times 10^{-5}}{0.1} = 1.74 \times 10^{-4} \]
Taking the square root:
\[ \alpha = \sqrt{1.74 \times 10^{-4}} \approx 0.0132 \]

### Step 5: Calculate the concentration of \( H^+ \)
The concentration of \( H^+ \) ions at equilibrium is given by:
\[ [H^+] = C\alpha = 0.1 \times 0.0132 = 0.00132 \, M \]

### Step 6: Calculate the pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ pH = -\log(0.00132) \]
Calculating this gives:
\[ pH \approx 2.88 \]

### Final Answer
The pH of a 0.1 M acetic acid solution is approximately **2.88**.

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