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Step-by-Step Solution
Step 1: Identify the Required Molar Conductivity
We are asked to find the molar conductivity at infinite dilution of acetic acid, denoted as $ \Lambda_{HOAc}^0 $. From the question, the given data are:
$ \Lambda_{NaOAc}^0 = 91.0 \text{ S cm}^2/\text{mol} $
$ \Lambda_{HCl}^0 = 426.2 \text{ S cm}^2/\text{mol} $
The question asks which additional molar conductivity value is required to determine $ \Lambda_{HOAc}^0 $. The correct answer provided is $ \Lambda_{NaCl}^0 $.
Step 2: Apply Kohlrausch's Law of Independent Migration of Ions
Kohlrausch's law states that at infinite dilution, each ion contributes a certain fixed amount to the total molar conductivity, independently of other ions present. Therefore, for ionic solutions, we can write:
$$
\Lambda_{HOAc}^0 = \Lambda_{NaOAc}^0 + \Lambda_{HCl}^0 - \Lambda_{NaCl}^0
$$
The logic behind this relationship is as follows:
$ \Lambda_{NaOAc}^0 $ includes contributions from $ \text{Na}^+ $ and $ \text{CH}_3\text{COO}^- $ ions.
$ \Lambda_{HCl}^0 $ includes contributions from $ \text{H}^+ $ and $ \text{Cl}^- $ ions.
$ \Lambda_{NaCl}^0 $ includes contributions from $ \text{Na}^+ $ and $ \text{Cl}^- $ ions.
When we sum $ \Lambda_{NaOAc}^0 $ and $ \Lambda_{HCl}^0 $, the ionic contributions of $ \text{Na}^+ $, $ \text{CH}_3\text{COO}^- $, $ \text{H}^+ $, and $ \text{Cl}^- $ are all included. However, to correctly isolate $ \Lambda_{HOAc}^0 $ (which comes from $ \text{H}^+ $ and $ \text{CH}_3\text{COO}^- $), we must subtract the overlapping part due to $ \text{Na}^+ $ and $ \text{Cl}^- $, that is $ \Lambda_{NaCl}^0 $.
Step 3: Conclude the Additional Value Needed
From the above relationship, it is clear that $ \Lambda_{NaCl}^0 $ is the required additional molar conductivity to calculate $ \Lambda_{HOAc}^0 $. This supports the given correct answer:
$$
\Lambda_{HOAc}^0 = (\Lambda_{NaOAc}^0 + \Lambda_{HCl}^0) - \Lambda_{NaCl}^0
$$
Step 4: Final Answer
The additional value needed for calculating the molar conductivity of acetic acid at infinite dilution is $ \Lambda_{NaCl}^0 $.
