Question

Following limiting molar conductivities are given as $\lambda^{o}_{m}\left(H_{2}SO_{4}\right) \,= \,x S\, cm^{2} mol^{-1}$ $\lambda^{o}_{m}\left(K_{2}SO_{4}\right) = y S\,cm^{2} mol^{-1}$ $\lambda^{o}_{m}\left(CH_{3}COOK\right) =z S\, cm^{2} mol^{-1} $ $\lambda^{o}_{m}\left(in \,S \,cm^{2} mol^{-1}\right)$ for $CH_3COOH$ will be-

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

Answer 1

The Correct Option is D

To determine the limiting molar conductivity of acetic acid $ \left(CH_3COOH \right) \right)$, we can use the principles of Kohlrausch's Law of Independent Migration of Ions. According to this law, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its ions.

Given: $ \lambda^{o}_{m}(H_{2}SO_{4}) = x \, S \, cm^{2} \, mol^{-1} $ $ \lambda^{o}_{m}(K_{2}SO_{4}) = y \, S \, cm^{2} \, mol^{-1} $ $ \lambda^{o}_{m}(CH_{3}COOK) = z \, S \, cm^{2} \, mol^{-1} $

We can write the ionic contributions for these substances as follows:

• $ \lambda^{o}_{m}(H_{2}SO_{4}) = 2\lambda^{o}(H^+) + \lambda^{o}(SO_4^{2-}) $
• $ \lambda^{o}_{m}(K_{2}SO_{4}) = 2\lambda^{o}(K^+) + \lambda^{o}(SO_4^{2-}) $
• $ \lambda^{o}_{m}(CH_{3}COOK) = \lambda^{o}(CH_3COO^-) + \lambda^{o}(K^+) $

We need to find $ \lambda^{o}_{m}(CH_3COOH) $, which can be expressed as: $ \lambda^{o}_{m}(CH_3COOH) = \lambda^{o}(CH_3COO^-) + \lambda^{o}(H^+) $

From the expressions above, combine equations to eliminate ions and find the desired expression for acetic acid:

1. Subtract the second equation from the first: $ \lambda^{o}_{m}(H_{2}SO_{4}) - \lambda^{o}_{m}(K_{2}SO_{4}) = [2\lambda^{o}(H^+) + \lambda^{o}(SO_4^{2-})] - [2\lambda^{o}(K^+) + \lambda^{o}(SO_4^{2-})] $
2. This simplifies to: $ 2\lambda^{o}(H^+) - 2\lambda^{o}(K^+) = x - y $
3. Divide this entire equation by 2: $ \lambda^{o}(H^+) - \lambda^{o}(K^+) = \frac{x - y}{2} $
4. Now, we utilize $ \lambda^{o}_{m}(CH_{3}COOK) = \lambda^{o}(CH_3COO^-) + \lambda^{o}(K^+) $: Rearrange to solve for $ \lambda^{o}(CH_3COO^-) $: $ \lambda^{o}(CH_3COO^-) = \lambda^{o}_{m}(CH_{3}COOK) - \lambda^{o}(K^+) $
5. Add the expressions for $ \lambda^{o}(H^+) - \lambda^{o}(K^+) $ and $ \lambda^{o}(CH_3COO^-) $: $$ \lambda^{o}(CH_3COO^-) + \lambda^{o}(H^+) = \left(\lambda^{o}_{m}(CH_{3}COOK) - \lambda^{o}(K^+)\right) + \left(\lambda^{o}(H^+) - \lambda^{o}(K^+)\right) $$ Simplify to: $$ \lambda^{o}_{m}(CH_3COOH) = \frac{x - y}{2} + z $$

Therefore, the correct answer is: $$ \frac{x - y}{2} + z $$