Assume the following:
Methanoic acid, ethanoic acid, and phenol are present in solution. It is assumed that without any chemical adulteration that sufficient heat will cause these organic acids to convert to formaldehyde, methanol, and phenol gas/vapor.
The lab objective is to prevent the formation of these orgnaic vapors.
When calcium carbonate is added to the solution of organic acids, the following compounds form: calcium methanoate salt from the methanoic acid; calcium ethanoate from the ethanoic acid; and calcium phenoxide salt from the phenol.
It is observed that very little formaldehyde observed in the gaseous side products.
If all 3 compounds were present in exactly the same molar concentration in solution, why is more methanoic acid (versus either ethanoic acid or phenol) converted to its respective "oate" salt?
I believe that it has to do with the relative acidity of the three acids. As you can see in the attached diagram the pKa's ...
The solution lies with the relative acidity of the three acids. As you can see in the attached diagram the pKa's of methanoic acid, ethanoic acid and phenol are 3.75, 4.7 and 10 respectively. This means that methanoic acid is 10 times more acidic than ethanoic acid. Ethanoic acid is 1000000 times more acidic than phenol. The reaction with calcium carbonate is a classic acid-base reaction as shown with the attached diagram. More acidic compounds thus react faster with calcium carbonate.