Question

The molar conductivity ($\Lambda_m^0$) at infinite dilution for HCl, NaCl and CH$_3$COONa at 25\(^{\circ}\)C are 426, 126 and 91 S cm\(^2\) respectively. The \(\Lambda_m^0\) for acetic acid at the same temperature will be

• A. 391 S cm\(^2\)
• B. 209 S cm\(^2\)
• C. 461 S cm\(^2\)
• D. 643 S cm\(^2\)

Hint:

For calculating the limiting molar conductivity of a weak electrolyte using Kohlrausch's Law, typically you add the \(\Lambda_m^0\) of two strong electrolytes whose ions, when combined and one common ion subtracted, yield the ions of the weak electrolyte. A common pattern is: $\Lambda_m^0(\text{Weak Acid}) = \Lambda_m^0(\text{Salt of Weak Acid}) + \Lambda_m^0(\text{Strong Acid}) - \Lambda_m^0(\text{Salt of Strong Acid})$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the limiting molar conductivity (\(\Lambda_m^0\)) for acetic acid (a weak electrolyte), given the limiting molar conductivities of three strong electrolytes (HCl, NaCl, CH$_3$COON(A). This requires the application of Kohlrausch's Law.

Step 2: Key Formula or Approach:
Kohlrausch's Law of Independent Migration of Ions states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the limiting molar conductivities of its individual ions. For a weak electrolyte like acetic acid ($CH_3COOH$), its \(\Lambda_m^0\) can be calculated from those of strong electrolytes that share its constituent ions.
We need to find a combination that gives:
\[ \Lambda_m^0(\text{CH}_3\text{COOH}) = \lambda^0_{\text{CH}_3\text{COO}^-} + \lambda^0_{\text{H}^+} \]
This can be achieved by:
\[ \Lambda_m^0(\text{CH}_3\text{COOH}) = \Lambda_m^0(\text{CH}_3\text{COONa}) + \Lambda_m^0(\text{HCl}) - \Lambda_m^0(\text{NaCl}) \]
Let's verify this:
$(\lambda^0_{\text{CH}_3\text{COO}^-} + \lambda^0_{\text{Na}^+}) + (\lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-}) - (\lambda^0_{\text{Na}^+} + \lambda^0_{\text{Cl}^-}) = \lambda^0_{\text{CH}_3\text{COO}^-} + \lambda^0_{\text{H}^+}$

Step 3: Detailed Explanation:
Given values:
- $\Lambda_m^0(\text{HCl}) = 426 \text{ S cm}^2 \text{ mol}^{-1}$
- $\Lambda_m^0(\text{NaCl}) = 126 \text{ S cm}^2 \text{ mol}^{-1}$
- $\Lambda_m^0(\text{CH}_3\text{COONa}) = 91 \text{ S cm}^2 \text{ mol}^{-1}$
Substitute these values into the derived equation:
\[ \Lambda_m^0(\text{CH}_3\text{COOH}) = 91 \text{ S cm}^2 + 426 \text{ S cm}^2 - 126 \text{ S cm}^2 \]
\[ \Lambda_m^0(\text{CH}_3\text{COOH}) = 517 \text{ S cm}^2 - 126 \text{ S cm}^2 \]
\[ \Lambda_m^0(\text{CH}_3\text{COOH}) = 391 \text{ S cm}^2 \text{ mol}^{-1} \]

Step 4: Final Answer:
The \(\Lambda_m^0\) for acetic acid is 391 S cm\(^2\).