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I was looking through the thermodynamic data chart at the back of my textbook and found that $\ce{S^{2-}_{(aq)}, F^{-}_{(aq)}, OH^{-}_{(aq)}}$ all have negative entropy values. But according to the Boltzmann equation, $S=k\ln W$, so if $S<0$, $W<1$. But since $W$ is the number of microstates, how can $W$ be less than 1?

I read in some other places than this is based on the entropy of $\ce{H^+_{(aq)}}$ being set to $0$, but why do we set this equal to $0$ and how does it follow Boltzmann's equation?

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    $\begingroup$ Boltzmann equation is applicable to compound as a whole. Entropy of particular ion is somewhat arbitrary anyway. $\endgroup$ Nov 4 '15 at 5:32
  • $\begingroup$ Googling "entropy of aqueous ions" will bring up some references. "Real" entropy would be measured relative to absolute zero. But measuring relative to say $\ce{H^+}$ in a 1 N solution at 20 degrees C would give you a stake in the ground against which you could measure other entropy values. My PChem is rusty at best. I'd also guess that choosing $\ce{H^+}$ as the stake in the ground has something to do with the standard electrode being the hydrogen electrode for electrochemical cells. The gist is that the tables are relative values - when the whole reaction is considered the bias drops out. $\endgroup$
    – MaxW
    Nov 4 '15 at 7:26
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    $\begingroup$ And for many solid-state tables (free energy of elements and alloys), the reference state is the stable state at 298K, and as for @MaxW's comments the values are relative to that. It is hard to perform measurements at absolute zero (tongue firmly in cheek). $\endgroup$
    – Jon Custer
    Nov 4 '15 at 15:56
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    $\begingroup$ It is especially hard to preform experiments in liquid water at absolute zero. $\endgroup$
    – MaxW
    Nov 4 '15 at 15:57
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    $\begingroup$ The reason that we can arbitrarily assign any entropy value to a chosen standard ion (in this case, $\ce{H3O+} \rm{(aq)}$, and the value assigned is $0$.) is that ions never appear on their own in chemistry equations due to charge conservation. If this were not the case we would not be able to use such a "standard". That's why we cannot apply the same to $\ce{H2}\rm{(g)}$ for example to "simplify" our calculation. $\endgroup$ Jan 27 '18 at 15:17

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