Why is phenol soluble in sodium carbonate but not in sodium bicarbonate?
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 $\ce{CO2}$).