Why is phenol soluble in sodium carbonate but not in sodium bicarbonate? The original question was basically to predict which of sodium carbonate or sodium bicarbonate can be used to distinguish between phenol and benzene.

I was sure that phenol neither reacts nor dissolves in $\ce{NaHCO3}$ but I was not sure about the $\ce{Na2CO3}$.  By searching on Google I found that phenol is soluble in sodium carbonate. I am curious to know why this is, even though phenol doesn't liberate $\ce{CO2}$.
The liberation of $\ce{CO2}$ from $\ce{Na2CO3}$ requires two protonation steps.

$$\begin{align}
\ce{CO3^2- + HA &amp;&lt;=&gt; HCO3- + A-}\\
\ce{HCO3- + HA &amp;&lt;=&gt; CO2 ^ + H2O + A-}
\end{align}$$

Neglecting the concentrations, the position of these equilibria will depend on the acid strength of $\ce{HA}$. If $\ce{HA}$ is a stronger acid than $\ce{HCO3-}$, for example, the first equilibrium will favour the products.

This is the case for phenol, which has a $\mathrm{p}K_\mathrm{a}$ of $9.95$, whereas $\ce{HCO3-}$ has a $\mathrm{p}K_\mathrm{a}$ of $10.32$. Therefore, as long as you don't have too much phenol, it will exist in the soluble phenolate form:

$$\ce{PhOH + CO3^2- &lt;=&gt;&gt; PhO- + HCO3-}$$

The solubility of phenol does not necessitate that it releases carbon dioxide.



Now, the second protonation. If you want to quantitatively determine whether $\ce{CO2}$ will actually be liberated, you need more data than the $\mathrm{p}K_\mathrm{a}$ values, since quite a significant amount of $\ce{CO2}$ can exist in the aqueous form. However, one thing is certain: if you want to produce $\ce{CO2 (g)}$ you first have to protonate $\ce{HCO3-}$ to form $\ce{H2CO3}$ (which dissociates to give $\ce{CO2 (aq)}$).

This is not possible with phenol, since $\ce{H2CO3}$ has a $\mathrm{p}K_\mathrm{a}$ of $6.3$ (even after correcting for dissolved $\ce{CO2}$).