Titration Curves of Polyprotic Acids (College Board AP® Chemistry): Study Guide
Titration curves of polyprotic acids
Polyprotic acids
Polyprotic acids have more than one ionizable hydrogen atom (acidic proton)
E.g. H2SO4, H3PO4
These acids ionize in steps, losing one proton at a time
Ionization steps for phosphoric acid, H3PO4
Step | Equation |
|---|---|
Step 1 | H3PO4 ⇌ H2PO4- + H+ |
Step 2 | H2PO4- ⇌ HPO42- + H+ |
Step 3 | HPO42- ⇌ PO43- + H+ |
Titrations of polyprotic acids have more than one equivalence point due to the series of ionization steps
Counting the equivalence points (steep rises) on the curve shows how many acidic protons have been titrated
When H3PO4 is titrated with a strong base such as NaOH, its protons react one at a time
At the first, second and third equivalence points, the major species are H2PO4-, HPO42- and PO43- respectively
Examiner Tips and Tricks
You do not need to calculate the concentration of each species along a polyprotic titration curve.
You do need to identify which species are present in large or small amounts at any point, and read each pKa from the curve
Between the equivalence points, there are mixtures of a conjugate acid and its conjugate base and are known as buffer regions
At the midpoint of each buffer region (each half-equivalence point), the two members of that conjugate pair have equal concentrations, so the pH equals the pKa for that proton
pKa1 at the first midpoint, pKa2 at the second and pKa3 at the third
E.g. at the second midpoint, [H2PO4-] = [HPO42-], so pH = pKa2
This is the half-equivalence point
The major species can be read from the curve at any point:
Before the first equivalence point: H3PO4 and H2PO4-
At the first equivalence point: H2PO4-
Between the first and second: H2PO4- and HPO42-
At the second equivalence point: HPO42-
Beyond it: HPO42- and PO43-
pH curve for addition of NaOH to H3PO4
At the midpoint of each buffer region, the pH equals the pKa for that proton (pKa1, pKa2, pKa3)
Worked Example
Use the titration curve of H3PO4 with NaOH above.
(a) How many equivalence points show protons of H3PO4 being removed?
(b) Which phosphorus species are present in large amounts after 1.5 H+ ions per molecule have been removed?
(c) Use the curve to estimate pKa2.
Answer (a):
The curve marks three equivalence points, at 1, 2 and 3 H+ ions removed per molecule, so H3PO4 has three acidic protons
The third rise is barely visible because HPO42− is a very weak acid
Answer (b):
After 1 H+ per molecule is removed, the main species is H2PO4−; after 2, it is HPO42−
1.5 H+ removed is halfway between, the second midpoint, so H2PO4− and HPO42− are present in equal amounts
Answer (c):
At the second midpoint [H2PO4−] = [HPO42−], so pH = pKa2
Reading the curve at 1.5 H+ removed gives pKa2 ≈ 7.4 (the data value is 7.21)
Examiner Tips and Tricks
pKa2 belongs to the second proton (H2PO4− ⇌ HPO42− + H+).
Only take it from Ka2. Do not take it from Ka1 or from Ka1 × Ka2.
Use pKa2 in the Henderson–Hasselbalch equation for an H2PO4−/HPO42− buffer.
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