Question

The conductivity of \(0.01\ M\) solution of \(CH_3COOH\) at 298 K is \(1.65 \times 10^{-4}\ S cm^{-1}\). What is the \(pK_a\) value of the acid if \(\lambda^0(H^+)\) and \(\lambda^0(CH_3COO^-)\) are \(349.1\ Scm^2 mol^{-1}\) and \(40.9\ Scm^2 mol^{-1}\) respectively?

• A. 4.73
• B. 1.87
• C. 3.47
• D. 2.95

Hint:

For weak electrolytes, first calculate \(\Lambda_m\), then use \(\alpha=\frac{\Lambda_m}{\Lambda_m^0}\), and finally apply Ostwald dilution law.

Answer 1

The Correct Option is A

Step 1: Write the given data.
\[ \kappa = 1.65 \times 10^{-4}\ S cm^{-1} \] \[ C = 0.01\ M \] \[ \lambda^0(H^+) = 349.1\ Scm^2mol^{-1} \] \[ \lambda^0(CH_3COO^-) = 40.9\ Scm^2mol^{-1} \]

Step 2: Calculate limiting molar conductivity.
\[ \Lambda_m^0 = \lambda^0(H^+) + \lambda^0(CH_3COO^-) \] \[ \Lambda_m^0 = 349.1 + 40.9 = 390\ Scm^2mol^{-1} \]

Step 3: Calculate molar conductivity of solution.
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \] \[ \Lambda_m = \frac{1.65 \times 10^{-4} \times 1000}{0.01} \] \[ \Lambda_m = 16.5\ Scm^2mol^{-1} \]

Step 4: Find degree of dissociation.
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \] \[ \alpha = \frac{16.5}{390} \] \[ \alpha = 0.0423 \]

Step 5: Use Ostwald dilution law.
For weak acid:
\[ K_a = \frac{C\alpha^2}{1-\alpha} \]

Step 6: Substitute values.
\[ K_a = \frac{0.01(0.0423)^2}{1-0.0423} \] \[ K_a \approx 1.87 \times 10^{-5} \]

Step 7: Calculate \(pK_a\).
\[ pK_a = -\log K_a \] \[ pK_a = -\log(1.87 \times 10^{-5}) \] \[ pK_a \approx 4.73 \]
Final Answer:
\[ \boxed{4.73} \]