Acetic acid is a weak acid. The molar conductances of 0.05 M acetic acid at `25^@C` is `7.36 S " cm"^(2) "mol"^(-1)` . If its `^^^_m^(oo)` is 390.72 S `cm^(2)"mol"^(-1)` . The value of equilibrium constant is

Options

A. `1.8 xx10^(-5)`
B. `1.8 xx10^(-7)`
C. `3.6 xx10^(-5)`
D. `5.4 xx10^(-5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:

\[ \alpha = \frac{\lambda_m}{\lambda_{\infty}} \]

Where:
- \(\lambda_m = 7.36 \, S \, cm^{-2} \, mol^{-1}\) (molar conductance of the weak acid solution)
- \(\lambda_{\infty} = 390.72 \, S \, cm^{-2} \, mol^{-1}\) (molar conductance at infinite dilution)

Substituting the values:

\[ \alpha = \frac{7.36}{390.72} \approx 0.0188 \]

### Step 2: Determine the concentration (C)
The concentration of the acetic acid solution is given as:

\[ C = 0.05 \, mol \, L^{-1} \]

### Step 3: Calculate the concentration of dissociated acetic acid
The concentration of the dissociated acetic acid can be calculated using:

\[ C \cdot \alpha \]

Substituting the values:

\[ \text{Concentration of dissociated acetic acid} = 0.05 \cdot 0.0188 \approx 0.00094 \, mol \, L^{-1} \]

### Step 4: Calculate the concentration of undissociated acetic acid
The concentration of undissociated acetic acid can be calculated as:

\[ C - C \cdot \alpha = 0.05 - 0.00094 \approx 0.04906 \, mol \, L^{-1} \]

### Step 5: Set up the equilibrium expression
For the dissociation of acetic acid (\(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\)), the equilibrium constant \(K_a\) can be expressed as:

\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]

At equilibrium:
- \([CH_3COO^-] = C \cdot \alpha = 0.00094 \, mol \, L^{-1}\)
- \([H^+] = C \cdot \alpha = 0.00094 \, mol \, L^{-1}\)
- \([CH_3COOH] = 0.04906 \, mol \, L^{-1}\)

### Step 6: Substitute values into the equilibrium expression
Substituting the values into the expression for \(K_a\):

\[ K_a = \frac{(0.00094)(0.00094)}{0.04906} \]

Calculating the numerator:

\[ 0.00094 \times 0.00094 = 0.0000008836 \]

Now substituting into the \(K_a\) expression:

\[ K_a = \frac{0.0000008836}{0.04906} \approx 1.80 \times 10^{-5} \]

### Final Result
Thus, the equilibrium constant \(K_a\) for acetic acid is approximately:

\[ K_a \approx 1.8 \times 10^{-5} \]

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