In the previous few posts, we have been discussing acid-base titrations involving pH calculations and analyzing the titrations curves.

A typical acid-base titration curve can be obtained, for example, based on the reaction of HCl and NaOH:

 

 

Check the original article for more details and explanations, and once the concepts are clear, let’s discuss the titration of a polyprotic acid.

For example, H₂SO₃ is a diprotic acid, and its dissociation can be divides in two parts each associated with a different dissociation constant:

 

H₂SO₃(aq) + H₂O(l) ⇆ HSO₃^–(aq) + H₃O⁺(aq) K_a1 = 1.6 x 10^-2

HSO₃^–(aq) + H₂O(l) ⇆ SO₃^2-(aq) + H₃O⁺(aq) K_a2 = 6.4 x 10^-8

 

When titrated with a base, one of the protons reacts with the hydroxide first to form the conjugate base HSO₃^–, which then reacts with the base forming the SO₃^2- conjugate base.

Let’s say we are titrating 25.0 mL of 0.100 M H₂SO₃ with 0.100 M NaOH. The first part of the titration represents the reaction between H₂SO₃ and NaOH, and the second part is for the reaction between the conjugate base HSO₃^– and NaOH.

 

H₂SO₃(aq) + OH^–(aq) ⇆ HSO₃^–(aq) + H₂O(l)

HSO₃^–(aq) + OH^–(aq) ⇆ SO₃^2–(aq) + H₂O(l)

 

The reason for the subsequent reaction with NaOH is that H₂SO₃ is a much stronger acid than its conjugate base HSO₃^– as seen from the K_a values.

So, the shape of the titration curve is going to be like merging two separate ones for monoprotic acids :

 

 

The pH before starting the titration is low because there is only acid present in the solution. H₂SO₃ is a weak acid and to calculate the pH, we follow the steps discussed earlier in the acid-base chapter. As the base is added, the pH increases, and up until the first equivalence point, we have a buffer solution composed of the H₂SO₃/HSO₃^– system.

 

 

The pH at the first half equivalence point is equal to the pK_a of sulfurous acid. Remember, this is because the pH of buffer solutions, when [HA] = [A^–] becomes equal to the pK_a since the logarithm term becomes zero.

 

\[{\rm{pH}}\;{\rm{ = }}\;{\rm{p}}{K_{{\rm{a1}}}}\;{\rm{ + }}\;\cancel{{{\rm{log}}\frac{{{\rm{[HS}}{{\rm{O}}_{\rm{3}}}^{\rm{–}}{\rm{]}}}}{{{\rm{[}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}{\rm{]}}}}}}\; = \,{\rm{p}}{K_{{\rm{a1}}}}\]

 

At the first equivalence point, pH is below 7 because of the acidic nature of HSO₃^–.

The pH at the second half equivalence point corresponds to the pK_a of the conjugate base HSO₃^–. Notice that for both equivalence points, the same volume of the NaOH (25.0 mL) is needed because the HSO₃^– is produced from H₂SO₃ in a 1:1 mole-ratio.

The information from titration curves, and especially the correlation between the half equivalence point and the pH can be used to estimate the pKa values for the polyprotic acid and identify an unknown acid.

These should be the key points about the titration of polyprotic acids. Check the previous posts on titrations for more information and calculations:

 

• Strong Acid–Strong Base Titrations
• Titration of a Weak Acid by a Strong Base
• Titration of a Weak Base by a Strong Acid

 

Check Also

• Buffer Solutions
• The Henderson–Hasselbalch Equation
• The pH of a Buffer Solution
• Preparing a Buffer with a Specific pH
• The Common Ion Effect
• The pH and pK_a Relationship
• Strong Acid–Strong Base Titrations
• Titration of a Weak Acid by a Strong Base
• Titration of a Weak Base by a Strong Acid
• Buffer Solutions Practice Problems
• K_sp and Molar Solubility
• The Effect of a Common Ion on Solubility
• The Effect of pH on Solubility
• Will a Precipitate Form? K_sp and Q
• K_sp and Molar Solubility Practice Problems