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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.

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.

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 $Ba(OH)_2$ 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.

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.

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

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

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 $Ba(OH)_2$ 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.