Topic 14 · page 11 of 18

Diprotic acids and bases

A diprotic acid gives up two protons in two steps, so its titration has up to two equivalence points. The second is at exactly twice the volume of the first.

14 min read14.15

Read first: Four monoprotic curves

Two protons, two steps 14.15

A monoprotic acid such as HCl or CH₃COOH can donate one proton. A diprotic acid can donate two. It gives them up one at a time, so each step has its own equilibrium and its own acid dissociation constant.

Ethanedioic acid, H₂C₂O₄, is a weak diprotic acid:

The first proton is always easier to remove than the second, because the second must leave a species that already carries a negative charge. So $\text{p}K_{\text{a}1}$ is smaller than $\text{p}K_{\text{a}2}$. For ethanedioic acid they are 1.25 and 4.14.

A diprotic base is the mirror image. The carbonate ion, CO₃²⁻, can accept two protons, one at a time:

Where the equivalence points are 14.15

If a diprotic acid is titrated with a base, the base first removes one proton from every molecule, then the second. There are up to two equivalence points, and the second is at twice the volume of the first, because the second proton needs as much base again.

A strong diprotic acid 14.15

Sulfuric acid is treated as a strong diprotic acid, with both protons fully released. 25.0 cm³ of 0.100 mol dm⁻³ H₂SO₄ contains 5.00 × 10⁻³ mol of H⁺, so it needs 50.0 cm³ of 0.100 mol dm⁻³ NaOH. Its curve looks like the strong acid curve from page 9, with two differences.

• It starts lower, at pH 0.70, because [H⁺] is 0.200 mol dm⁻³.
• The single vertical section is at 50.0 cm³, centred on pH 7.00, and runs from about pH 3 to 11.

With both protons strong there is no separate first jump, so the curve shows one equivalence point, at twice the volume you would expect for a monoprotic acid.

A weak diprotic acid 14.15

Ethanedioic acid has equivalence points at 25.0 cm³ (all H₂C₂O₄ changed to HC₂O₄⁻) and 50.0 cm³ (all changed to C₂O₄²⁻).

• The second equivalence point is a clear jump, at pH 8.33.
• The first is hard to see. The two $\text{p}K_\text{a}$ values are only 2.9 units apart, so the two stages overlap and the pH rises smoothly through 25.0 cm³.
• Halfway through the second stage, at 37.5 cm³, [HC₂O₄⁻] = [C₂O₄²⁻] and the pH is 4.14, which is $\text{p}K_{\text{a}2}$. This region is a buffer, and the curve is flat there.

The further apart the two $\text{p}K_\text{a}$ values are, the sharper each jump.

A weak diprotic base 14.15

Carbonate has $\text{p}K_\text{a}$ values of 10.33 (for HCO₃⁻) and 6.35 (for H₂CO₃), and its curve has the same ideas upside down.

• The pH falls in two stages, with equivalence points at 25.0 and 50.0 cm³.
• At 12.5 cm³ half the CO₃²⁻ has been converted to HCO₃⁻, so the pH equals the $\text{p}K_\text{a}$ of HCO₃⁻, 10.33.
• At 37.5 cm³ half the HCO₃⁻ has been converted to H₂CO₃, so the pH equals the $\text{p}K_\text{a}$ of H₂CO₃, 6.35.
• The second jump (about pH 4.7 down to 3.2) is the clearer one, and methyl orange is the usual indicator for it. The first jump is short (about pH 8.7 down to 8.0), so the first end point is not sharp, and phenolphthalein only approximately marks it.

Check yourself

1. Define a diprotic acid, and write the two equations for the dissociation of ethanedioic acid, H₂C₂O₄.
2. 25.0 cm³ of 0.100 mol dm⁻³ sulfuric acid is titrated with 0.100 mol dm⁻³ NaOH. Treating the acid as fully dissociated, calculate the volume of NaOH at the equivalence point.
3. 20.0 cm³ of 0.0500 mol dm⁻³ sodium carbonate is titrated with 0.100 mol dm⁻³ hydrochloric acid. Calculate both equivalence volumes.
4. Why is the first equivalence point of ethanedioic acid with NaOH hard to see on the curve?
5. In the titration of ethanedioic acid with NaOH, the pH at 37.5 cm³ is 4.14. Explain why this equals a $\text{p}K_\text{a}$ and which one.
6. Which indicator suits the second equivalence point of the carbonate titration, and why?