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 $\ce{CO2}$.

  • 3
    $\begingroup$ $\ce{NaHCO3}$ is less basic than $\ce{Na2CO3}$. Phenol is weakly acidic. $\endgroup$
    – DHMO
    Jan 30, 2017 at 11:29

1 Answer 1


The liberation of $\ce{CO2}$ from $\ce{Na2CO3}$ requires two protonation steps.

$$\begin{align} \ce{CO3^2- + HA &<=> HCO3- + A-}\\ \ce{HCO3- + HA &<=> CO2 ^ + H2O + A-} \end{align}$$

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:

$$\ce{PhOH + CO3^2- <=>> PhO- + HCO3-}$$

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 $\ce{CO2}$).


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