How to solve this contradiction?

Question

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?

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.