PolyProtic Titration Graphs

 

Although the mathematics of polyprotic titrations are beyond the scope of this text, we can glean some information from their graphs. 

Let’s start with what it means for an acid to be polyprotic.

By the Bronsted/Lowry definition, an acid is a compound that can donate an \(H^{+1}\) ion to water or to a base. Any compound that can donate more than one \(H^{+1}\) ion is polyprotic. That means that \(H_2SO_4, ~H_2CO_3, ~and ~H_3PO_4\) are all polyprotic acids. This can be seen in the formula where more than on H is found at the beginning.

This is, in fact, why inorganic chemists write the formula of acetic acid as \(HC_2H_3O_2\) with the “H’s” split. Acetic acid is monoprotic, that is, it can only donate ONE H ion, so we write the formula with only ONE H at the beginning.

Just because an acid CAN donate 2 (or more) \(H^{+1}\) ions, doesn’t mean that those ions will be donated at the same time. Polyprotic acids can be thought of as two (or more) consecutive acids. So, sulfurous acid (\(H_2SO_3\)) reacts with water in a two step process:

Step 1: \(H_2SO_3 + H_2O \rightleftharpoons H_3O^+ + HSO_3^{-1}\)

Step 2: \(HSO_3^{-1} + H_2O \rightleftharpoons H_3O^+ + SO_3^{-2}\)

The same two step idea applies during a titration. When a stong base,like NaOH, is added to a solution of sulfurous acid, the hydroxide (\(OH^{-1}\)) first reacts with (and removes) ONE \(H^+\) ion from each acid:

\(H_2SO_3 + OH^{-1} \rightarrow H_2O + HSO_3^{-1}\)

Then, any additional NaOH added will remove the second \(H^+\) from the acid:

\(HSO_3^{-1} + OH^{-1} \rightarrow H_2O + SO_3^{-2}\)

This “two-step” process is visible on a titration graph. The graph will have two equivalence points: the first when the moles of base = the moles of acid. That portion of the process will involve removing ONE \(H^+\) ion from each acid molecule. After that point, the base will begin removing the second \(H^+\) ion from the acid, leading to a second equivalence point.

The graph below was created to show the general shape of a diprotic acid titration graph.

In addition to the two equivalence points, there are two maximum buffer points, since there are (essentially) two different acids (\(H_2SO_3 ~and~ HSO_3^{-1}\))

Notice that the volume required to reach the second equivalence point is exactly twice the volume needed to reach the first equivalence point. That should make sense, since the number of \(H^+\) ions removed in each case is the same.

The graph rises from left to right. Although that seems obvious, it points out several things:

• The first acid (in this case \(H_2SO_3\)) is stronger, and therefore more acidic, than the second acid (\(HSO_3^{-1}\)).
• \(K_{a1}\) (the \(K_a\) for \(H_2SO_3\)) > \(K_{a2}\) (the \(K_a\) for \(HSO_3^{-1}\))