The measured freezing point depression for a 0.1 m aqueous `CH_(3)COOH` solution is `0.19^(@)C`. The acid dissociation constant `K_(a)` at this concentration will be (Given `K_(f)`, the moala cryoscopic constant = 1.86 K Kg `mol^(-1))`

Options

A. `4.76xx10^(-5)`
B. `4xx10^(-5)`
C. `8xx10^(-5)`
D. `2xx10^(-5)`

Explanation

To find the acid dissociation constant \( K_a \) for the acetic acid solution, we will follow these steps:

### Step 1: Determine the Freezing Point Depression
The freezing point depression (\( \Delta T_f \)) is given as \( 0.19^\circ C \).

### Step 2: Use the Freezing Point Depression Formula
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = freezing point depression
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_f \) = molal cryoscopic constant (given as \( 1.86 \, K \, kg \, mol^{-1} \))
- \( m \) = molality of the solution (given as \( 0.1 \, m \))

### Step 3: Substitute Known Values into the Formula
Substituting the known values into the formula:
\[ 0.19 = i \cdot 1.86 \cdot 0.1 \]

### Step 4: Solve for \( i \)
Rearranging the equation to solve for \( i \):
\[ i = \frac{0.19}{1.86 \cdot 0.1} \]
Calculating this gives:
\[ i = \frac{0.19}{0.186} \approx 1.02 \]

### Step 5: Determine the Degree of Dissociation
For acetic acid (\( CH_3COOH \)), it dissociates as follows:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
Let \( C \) be the initial concentration (0.1 m), and \( \alpha \) be the degree of dissociation. At equilibrium:
- Concentration of \( CH_3COOH \) = \( C(1 - \alpha) \)
- Concentration of \( CH_3COO^- \) = \( C\alpha \)
- Concentration of \( H^+ \) = \( C\alpha \)

The van 't Hoff factor \( i \) can also be expressed as:
\[ i = 1 + \alpha \]
From our calculation, we have:
\[ 1.02 = 1 + \alpha \implies \alpha = 0.02 \]

### Step 6: Calculate the Acid Dissociation Constant \( K_a \)
The expression for \( K_a \) is:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]
Substituting \( C = 0.1 \, m \) and \( \alpha = 0.02 \):
\[ K_a = \frac{(0.1 \cdot 0.02)(0.1 \cdot 0.02)}{0.1(1 - 0.02)} = \frac{(0.002)(0.002)}{0.1 \cdot 0.98} \]
Calculating this gives:
\[ K_a = \frac{0.000004}{0.098} \approx 4.08 \times 10^{-5} \]

### Final Answer
The acid dissociation constant \( K_a \) at this concentration is approximately \( 4 \times 10^{-5} \).

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