I am getting confused about this the more problems I do on it. My understanding was that only strong acids and strong bases will react to produce water and a salt. Something like this:

Molecular Eq: $\ce{HCl(aq) + NaOH(aq) \rightarrow H2O(l) + NaCl}$
Net Ionic Eq: $\ce{H+(aq) +OH- (aq) \rightarrow H_2O(l)}$

Makes perfect sense, both the acid and base break apart and combine with each other. But then I ran across something like this:
Molecular Eq: $\ce{2 CH3CO2H (aq) + Ba(OH)2 (aq) → Ba(CH3CO2)2 (aq) + 2 H2O (l)}$
Net Ionic Eq: $\ce{2CH3CO2H (aq) + 2OH- (aq) → 2CH3CO2- (aq) + 2H2O (l)} $

The net ionic eq tells me that $\ce{CH3CO2H}$ will not break apart, but the molecular equation tells me that they will break apart and form a salt with $\ce{Ba}$.

I'm so confused by this. Why are they giving two opposing pieces of information? Am I missing something here?


1 Answer 1


The idea behind this is that a weak acid like $\ce{CH_3CO_2H}$ will break apart, but won't break apart completely. If you have some $\ce{CH_3CO_2H}$ in water, most of it will remain as $\ce{CH_3CO_2H}$, so in the net ionic equation it is written whole, but some will separate into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$ ions. This relationship is quantified by the $K_a$ of the acid. Essentially, the dissociation reaction $\ce{CH_3CO_2H -> CH_3CO_2^{-} + H^+}$goes both ways: $\ce{CH_3CO_2H}$ decomposes into $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, and $\ce{CH_3CO_2^-}$ reacts with $\ce{H^+}$ for form $\ce{CH_3CO_2H}$. Since the forward reaction happens at a rate proportional to the concentration of $\ce{CH_3CO_2H}$, and the backwards reaction happens at a rate proportional to the product of the concentrations of $\ce{CH_3CO_2^-}$ and $\ce{H^+}$, eventually the ratio $\frac{[\ce{CH_3CO_2^-}][\ce{H^+}]}{[\ce{CH_3CO_2H}]}$ becomes constant, and we call this constant the $K_a$ of the $\ce{CH_3CO_2H}$. For $\ce{CH_3CO_2H}, K_a=1.7\times 10^{-5}$.

Now its time to connect all this back to your question. When you add $\ce{Ba(OH)_2}$ to a solution of $\ce{CH_3CO_2H}$, the $\ce{OH^-}$ ions present from the dissociation of $\ce{Ba(OH)_2}$ react with the $\ce{H^+}$ ions from the (partial) dissociation of $\ce{CH_3CO_2H}$. This lowers the concentration of $\ce{H^+}$, and so more $\ce{CH_3CO_2H}$ dissociates to keep the ratio of concentrations constant at the $K_a$. Therefore, in the presence of enough $\ce{Ba(OH)_2}$, $\ce{CH_3CO_2H}$ dissociates almost completely, even though, on its own, $\ce{CH_3CO_2H}$ dissociates only to a very small extent.

  • 1
    $\begingroup$ I wouldn't say that the acetic acid dissociates almost completely, but that enough acetic acid dissociates to equal the positive charges of the barium (there may be an excess of acetic acid, which the excess will mainly be undissociated). $\endgroup$
    – LDC3
    Jul 12, 2015 at 19:54

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