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: Use Kohlrausch's law.
According to Kohlrausch's law, molar conductivity at infinite dilution is the sum of ionic conductivities of ions.
We need: \[ \lambda_m^\circ(CH_3COOH) = \lambda^\circ(CH_3COO^-)+\lambda^\circ(H^+) \]

Step 2: Write given molar conductivities in ionic form.
For barium acetate: \[ \lambda_m^\circ\left((CH_3COO)_2Ba\right) = 2\lambda^\circ(CH_3COO^-)+\lambda^\circ(Ba^{2+}) \] So, \[ x_1=2\lambda^\circ(CH_3COO^-)+\lambda^\circ(Ba^{2+}) \] For barium chloride: \[ \lambda_m^\circ(BaCl_2) = \lambda^\circ(Ba^{2+})+2\lambda^\circ(Cl^-) \] So, \[ x_2=\lambda^\circ(Ba^{2+})+2\lambda^\circ(Cl^-) \] For hydrochloric acid: \[ \lambda_m^\circ(HCl) = \lambda^\circ(H^+)+\lambda^\circ(Cl^-) \] So, \[ x_3=\lambda^\circ(H^+)+\lambda^\circ(Cl^-) \]

Step 3: Combine the expressions to get acetic acid.
Subtracting \(x_2\) from \(x_1\): \[ x_1-x_2 = 2\lambda^\circ(CH_3COO^-)-2\lambda^\circ(Cl^-) \] \[ \frac{x_1-x_2}{2} = \lambda^\circ(CH_3COO^-)-\lambda^\circ(Cl^-) \] Now adding \(x_3\): \[ \frac{x_1-x_2}{2}+x_3 = \lambda^\circ(CH_3COO^-)-\lambda^\circ(Cl^-) + \lambda^\circ(H^+)+\lambda^\circ(Cl^-) \] \[ \frac{x_1-x_2}{2}+x_3 = \lambda^\circ(CH_3COO^-)+\lambda^\circ(H^+) \]

Step 4: Final conclusion.
Therefore, \[ \lambda_m^\circ(CH_3COOH) = \boxed{\frac{x_1-x_2}{2}+x_3} \]