Why do polyprotic acids become less acidic as they lose protons?

1 Answer
Jan 30, 2018

The more stable the conjugate base is, the more acidic the conjugate acid is. The conjugate base will be more stable when the charge density is lower.

The more protons you take off, the higher the charge density, ergo the less acidic the conjugate acid becomes.


Great question. There are multiple factors that affect acidity. I would say that it is primarily to do with the stability of the conjugate base. The more stable the conjugate base, the more acidic the conjugate acid will be.

What makes the base more stable you ask? Well, charge density is important. This is the electrical charge per length, area or volume. Something that stabilises charge is how far it is spread out. So the larger the volume, the more stable the charge will be. E.g. #I^(-)# will be more stable than #Cl^(-)# because iodide is larger.

In general, the lower the charge density, the more stable the species will be. So, all other things being equal, an anion with a 1- charge will be more stable than one with a 2- charge. Importantly, you can't just compare two completely different anions, because a 2- anion can potentially be more stable than a 1- anion if there are other factors affording it stability.

Let's look at sulfuric acid. It is a diprotic acid that has two dissociation steps

#H_2SO_4rightleftharpoonsH^(+)+HSO_4^(-)# #K_a>1#

#HSO_4^(-)rightleftharpoonsH^(+)+SO_4^(2-)# #K_a=1.2*10^-2#

In this case, we can compare the two conjugate bases because they are similar and say that #HSO_4^(-)# will be more stable than #SO_4^(2-)# because it has a lower charge density. Therefore, #H_2SO_4# will be more acidic than #HSO_4^(-)#, shown by the larger #K_a# value.

Here is a great website that explains this and some factors that affect charge stability


#K_a# source: