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 step by step, we will use the information provided about acetic acid, its molar conductance, and the concept of equilibrium constant.

### Step 1: Identify the given values
- Molar conductance of 0.05 M acetic acid, \( \Lambda_m = 7.36 \, S \, cm^{-2} \, mol^{-1} \)
- Limiting molar conductance, \( \Lambda_m^{\infty} = 390.72 \, S \, cm^{-2} \, mol^{-1} \)
- Concentration of acetic acid, \( C = 0.05 \, mol/L \)

### Step 2: Calculate the degree of dissociation (α)
The degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}} \]
Substituting the values:
\[ \alpha = \frac{7.36}{390.72} \approx 0.0188 \]

### Step 3: Write the expression for the equilibrium constant (K)
For the dissociation of acetic acid:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
At equilibrium, the concentrations are:
- Concentration of \( CH_3COO^- \) = \( C \alpha \)
- Concentration of \( H^+ \) = \( C \alpha \)
- Concentration of \( CH_3COOH \) = \( C(1 - \alpha) \)

The equilibrium constant \( K \) is given by:
\[ K = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{C \alpha \cdot C \alpha}{C(1 - \alpha)} = \frac{C \alpha^2}{1 - \alpha} \]

### Step 4: Substitute the values into the equilibrium constant expression
Substituting \( C = 0.05 \, mol/L \) and \( \alpha = 0.0188 \):
\[ K = \frac{0.05 \cdot (0.0188)^2}{1 - 0.0188} \]
Calculating \( (0.0188)^2 \):
\[ (0.0188)^2 \approx 0.00035344 \]
Now substituting this value:
\[ K = \frac{0.05 \cdot 0.00035344}{0.9812} \approx \frac{0.000017672}{0.9812} \approx 0.000018 \]
This can be expressed in scientific notation:
\[ K \approx 1.8 \times 10^{-5} \]

### Step 5: Final answer
Thus, the value of the equilibrium constant \( K \) for acetic acid is:
\[ K \approx 1.8 \times 10^{-5} \]