Question

What is the molar conductivity of \(CH_3CO_2H\) at infinite dilution?
Given that, \[ \lambda_m^\circ\left((CH_3CO_2)_2Ba\right)=x_1\ S\ cm^2\ mol^{-1} \] \[ \lambda_m^\circ(BaCl_2)=x_2\ S\ cm^2\ mol^{-1} \] \[ \lambda_m^\circ(HCl)=x_3\ S\ cm^2\ mol^{-1} \]

• A. \(\dfrac{x_1-x_2}{2}+x_3\)
• B. \(\dfrac{x_1-x_3}{2}+x_2\)
• C. \(\dfrac{x_2-x_3}{2}+x_1\)
• D. \(x_1+x_3-x_2\)

Hint:

For weak electrolytes like \(CH_3COOH\), molar conductivity at infinite dilution is calculated using Kohlrausch's law by combining strong electrolyte conductivities.

Answer 1

The Correct Option is A

Step 1: State what we need.
We want \( \lambda_m^\circ(CH_3COOH) = \lambda^\circ(CH_3COO^-) + \lambda^\circ(H^+) \) by Kohlrausch law.
Step 2: Extract ionic conductivity from barium acetate.
\( \lambda_m^\circ((CH_3COO)_2Ba) = x_1 \). Barium acetate dissociates as \( Ba^{2+} + 2CH_3COO^- \): \[ x_1 = \lambda^\circ(Ba^{2+}) + 2\lambda^\circ(CH_3COO^-) \Rightarrow \lambda^\circ(CH_3COO^-) = \frac{x_1 - \lambda^\circ(Ba^{2+})}{2} \]
Step 3: Get barium ion conductivity from BaCl2.
\( x_2 = \lambda^\circ(Ba^{2+}) + 2\lambda^\circ(Cl^-) \).
Step 4: Get H+ conductivity from HCl.
\( x_3 = \lambda^\circ(H^+) + \lambda^\circ(Cl^-) \).
Step 5: Combine the expressions.
Substituting and cancelling \( Cl^- \) terms: \[ \lambda_m^\circ(CH_3COOH) = \frac{x_1 - x_2}{2} + x_3 \]
Step 6: Match with options.
Option 1 gives \( \frac{x_1 - x_2}{2} + x_3 \). \[ \boxed{\dfrac{x_1 - x_2}{2} + x_3} \]