The molar conductivity of 0.007 M acetic acid is `20S cm^(2) mol^(-1)`. What is the dissociation constant of acetic acid ? Choose the correct option `[(Lamda_(H^(+))^(@)= 350S cm^(2) mol^(-1)),(Lamda_(CH_(3)COO^(-))^(@)= 50S cm^(2) mol^(-1))]`

Options

A. `2.50 xx 10^(-4) mol L^(-1)`
B. `1.75 xx 10^(-5) mol L^(-1)`
C. `2.50 xx 10^(-5) mol L^(-1)`
D. `1.75 xx 10^(-4) mol L^(-1)`

Explanation

To find the dissociation constant (Ka) of acetic acid from the given molar conductivity, we can follow these steps:

### Step 1: Understand the relationship between molar conductivity and dissociation
The molar conductivity (Λ) of a weak electrolyte like acetic acid can be expressed as:
\[ \Lambda = \Lambda^0_{H^+} + \Lambda^0_{CH_3COO^-} - \Lambda^0_{CH_3COOH} \]
where:
- \(\Lambda^0_{H^+}\) is the molar conductivity of the hydrogen ion,
- \(\Lambda^0_{CH_3COO^-}\) is the molar conductivity of the acetate ion,
- \(\Lambda^0_{CH_3COOH}\) is the molar conductivity of acetic acid (which is negligible for very dilute solutions).

### Step 2: Calculate the degree of dissociation (α)
The degree of dissociation (α) of acetic acid can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda^0} \]
where \(\Lambda^0\) is the molar conductivity at infinite dilution. For acetic acid, the total molar conductivity at infinite dilution can be calculated as:
\[ \Lambda^0 = \Lambda^0_{H^+} + \Lambda^0_{CH_3COO^-} \]
Substituting the values given:
\[ \Lambda^0 = 350 \, S \, cm^2 \, mol^{-1} + 50 \, S \, cm^2 \, mol^{-1} = 400 \, S \, cm^2 \, mol^{-1} \]
Now substituting the values into the degree of dissociation formula:
\[ \alpha = \frac{20 \, S \, cm^2 \, mol^{-1}}{400 \, S \, cm^2 \, mol^{-1}} = 0.05 \]

### Step 3: Calculate the concentration of dissociated acetic acid
The concentration of acetic acid (C) is given as 0.007 M. The concentration of dissociated acetic acid (Cα) is:
\[ C\alpha = 0.007 \times 0.05 = 0.00035 \, M \]

### Step 4: Calculate the dissociation constant (Ka)
The dissociation constant (Ka) for acetic acid can be calculated using the formula:
\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \]
Since the concentration of \(H^+\) and \(CH_3COO^-\) is equal to \(C\alpha\), and the concentration of undissociated acetic acid is:
\[ [CH_3COOH] = C(1 - \alpha) = 0.007(1 - 0.05) = 0.007 \times 0.95 = 0.00665 \, M \]
Now substituting the values into the Ka formula:
\[ K_a = \frac{(0.00035)(0.00035)}{0.00665} = \frac{1.225 \times 10^{-7}}{0.00665} \approx 1.84 \times 10^{-5} \, M \]

### Final Answer
The dissociation constant (Ka) of acetic acid is approximately \(1.84 \times 10^{-5} \, M\).

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