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Here is the overall formula:

$$\ce{CH3COONa + HCl <=> NaCl + CH3COOH}$$ (sodium acetate + hydrochloric acid $\ce{<=>}$sodium chloride + acetic acid)

Which way does the above reaction naturally proceed? I've tried Googling this, but different sources seem to suggest contradictory answers, and I can't really tell what's correct.

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Mostly the reaction will proceed from left to right, unless you have ridiculous amount of acetic acid. Hydrochloric acid is much stronger, so it suppresses dissociation of acetic acid.

However, this is only part of the truth. In fact, in water solutions many compounds (partially) dissociates:

$HCl = H^+ + Cl^-$

The process is strongly endothermic in vacuum, but subsequent interaction of ions with water makes it mostly energy neutral or even highly exothermic for some compounds. This means, that in you case the solution would actually contain $H^+$, $OH^-$ (very little), $CH_3COO^-$, $CH_3COOH$, $HCl$ (almost none, may be ignored in most cases), $Na^+$. The actual reaction would be

$CH_3COO^- + H^+ = CH_3COOH$

The actual direction of the reaction would depend on initial concentraion and guarded by equilibrium constant.

The keywords for googling/textbook reading are 'electrolitic dissociation', 'ionic reaction', 'chemical equilibrium', 'equilibrium constant'. Generally, the procedure of determining which concentration of which ions the system will have requires solution of system of equations, including equilibrium constants, electroneutrality equation and material balance equations. The system often may have no analytic solutions, in which case it is usually approximated by simpler system, dropping some elements of sums in the system.

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  • $\begingroup$ So let's say my initial conditions are reversed and I am starting with the right hand side CH3COOH and NaCl. In this scenario I should expect a small shift to the left to produce a minuscule amount of HCL and CH3COONa, but the majority of the reactants would remain as CH3COOH and NaCl? $\endgroup$ – JST May 28 '14 at 7:33
  • $\begingroup$ Or more accurately, the NaCl would fully dissociate into Na+ and Cl- ions, and a small amount of CH3COOH would dissociate into H+ and CH3COO- ions, and some of the CH3COO- ions would combine with Na+ to give CH3COONa. Thus my solution would contain Na+, Cl-, CH3COOH, H+, CH3COO-, CH3COONa, and (almost none) HCl? $\endgroup$ – JST May 28 '14 at 7:37
  • $\begingroup$ @JST $CH3COONa$ is generally accepted to be fully dissociated while in solution like most (but not all, exceptions do exist) salts. Ignoring this part, you are correct. $\endgroup$ – permeakra May 28 '14 at 8:03
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As denoted by the equilibrium arrows, the reaction proceeds in both directions.

Initially, the net direction depends on the concentrations of the chemical species (Le Chatelier's principle). For example, if there's a lot of acetic acid, the reaction will go left. If there's lots of acetate, the reaction will go right ("lots" is a relative term, see the equilibrium constant for acetic acid: $K_a = 1.8\times10^{-5}$). The net reaction will proceed in this direction until the equilibrium concentrations are achieved. However, keep in mind that even at this stage the reaction still occurs in both directions, but there will be a net direction (one direction is favoured over the other).

Once equilibrium occurs, there is no net direction but, again, the reaction still occurs in both directions (at the same rate now).

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  • $\begingroup$ I only put in equilibrium arrows since I don't know which way it goes or if it is indeed an equilibrium. Various online sources I read showed the arrow going either one direction or another. $\endgroup$ – JST May 28 '14 at 5:52
  • $\begingroup$ Well you got lucky, because it is an equilibrium! The net reaction direction really depends on the initial conditions. $\endgroup$ – canadianer May 28 '14 at 5:57
  • $\begingroup$ +1, but in the interest of clarity, I would specify that the equilibrium here overwhelmingly favors the formation of products. $\ce{HCl}$ is many orders of magnitude more acidic than $\ce{CH3CO2H}$, so the reverse reaction should be negligible. $\endgroup$ – Greg E. May 28 '14 at 6:26
  • $\begingroup$ Perhaps it should be in the answer, but the asker didn't specify what species they're starting with and their amounts. Or should we assume that the compounds on the left are present initially? Is it also safe to assume that the amount of HCl present would have an effect on the reaction such that it is effectively irreversible? $\endgroup$ – canadianer May 28 '14 at 6:38
  • $\begingroup$ @canadianer, I'm not suggesting you make any assumptions about initial concentrations, I just think it would be valuable to mention that the equilibrium ratio of $\ce{CH3CO2H}$ to $\ce{HCl}$ here should be on the order of roughly $10^{12}:1$ (assuming stoichiometric amounts). The second paragraph of your answer reads to me as though the reaction from right to left is a lot more significant than it normally would be in practice. Maybe that's just me. In any case, only a friendly critique, no offense intended. $\endgroup$ – Greg E. May 28 '14 at 7:51

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