Which solution is more conductive

Which of the following solutions is more conductive:


1) 0.1 M sodium hydroxide and 0.1 M hydrochloric acid mixture (effectively NaCl?)


2) 0.1 M sodium hydroxide and 0.1 M acetic acid mixture (effectively NaCH3COO?)


I know the answer is the first one but I can't explain why. I have been taught (high school chemistry) that conductivity of ions is simply relative to [ions]. If that were true, both solutions would be equally conductive. I know that strictly speaking, based on the initial acids and bases, 1) would be more conductive, but I can't understand this because there should be none of the original reactants present anyways. This came up on a test, and I can't justify 1) being correct based on what I have learned.




Conductivity of different ions carrying the same amount of charge can vary as per the comment of Ivan. In this question, acetic acid is a weak acid so the final acetic anion will combine with the hydrogen ion from the autodissociation of water molecule to form acetic acid molecules. This is known as the hydrolysis of $\ce{NaAc}$. Hydrochloric acid, on the other hand, is a strong acid so the hydrolysis won't happen. Hydrolysis is the main reason that $\ce{NaAc}$ solution is basic while $\ce{NaCl}$ is neutral. 



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> There should be none of the original reactants present anyways.
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> 


This is one fallacy in your reasoning. There are some $\ce{CH3COOH} \rm(aq)$ formed due to hydrolysis in the final solution since it is weak acid.


To summarize, in the first case the conducting particles are equal amount of $\ce{Na+}$ and $\ce{Cl-}$, while in the latter they are $\ce{Na+}$, $\ce{Ac-}$ and $\ce{OH-}$, with the concentration of the first equal the sum of the other two. This is what I believe you need to know. Beyond that without looking up the conductivity table or analyzing the structures of the ions we **cannot deduce** whose conductance is higher.


In paricular, given a strong acid $\ce{HA}$ and a weak acid $\ce{HB}$, saying that for the solution of the same concentration, $\ce{NaA}$ will be more conductive than $\ce{NaB}$ is **simply wrong** and there **are counterexamples**.


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