Question

What is molar conductivity of \(\text{CH}_3\text{COOH}\) at zero concentration if the molar conductivities of \(\text{H}_2\text{SO}_4\), \(\text{K}_2\text{SO}_4\) and \(\text{CH}_3\text{COOK}\) at zero concentrations are respectively x, y and z \(\Omega ^{-1}\text{cm}^2\text{mol}^{-1}\) ?

• A. \(\frac{(x-y)}{2}+z\)
• B. \((x-y+2z)\)
• C. \((x+y-z)\)
• D. \(\frac{(x-y)}{2}+2z\)

Hint:

Apply Kohlrausch law: write acetic acid as a combination of the three given electrolytes.

Answer 1

The Correct Option is A

Step 1: Understand the concept
Kohlrausch's law says the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions: \(\Lambda^\circ_m = \lambda^\circ_+ + \lambda^\circ_-\) (with stoichiometric numbers).

Step 2: Write the three given values
\(\Lambda^\circ(\text{H}_2\text{SO}_4) = 2\lambda_{\text{H}^+} + \lambda_{\text{SO}_4} = x\). \(\Lambda^\circ(\text{K}_2\text{SO}_4) = 2\lambda_{\text{K}^+} + \lambda_{\text{SO}_4} = y\). \(\Lambda^\circ(\text{CH}_3\text{COOK}) = \lambda_{\text{CH}_3\text{COO}^-} + \lambda_{\text{K}^+} = z\).

Step 3: Combine
Subtract: \(x - y = 2\lambda_{\text{H}^+} - 2\lambda_{\text{K}^+}\), so \(\lambda_{\text{H}^+} - \lambda_{\text{K}^+} = (x-y)/2\).

Step 4: Find acetic acid
\[ \Lambda^\circ(\text{CH}_3\text{COOH}) = \lambda_{\text{H}^+} + \lambda_{\text{CH}_3\text{COO}^-} = (\lambda_{\text{H}^+} - \lambda_{\text{K}^+}) + (\lambda_{\text{CH}_3\text{COO}^-} + \lambda_{\text{K}^+}) = \frac{x-y}{2} + z \]

Final Answer:
The limiting molar conductivity is (x-y)/2 + z. This is option (A). \[ \boxed{\text{(A) }\frac{x-y}{2}+z} \]