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8.3.3Acids and Bases

Titration Curves of Polyprotic Acids — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 5 parts Lesson 20 of 21 Unit quiz

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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5Stepwise Ionisation

Because protons are removed one after another, the titration curve is built from several linked weak-acid curves.

Polyprotic acid
  • Polyprotic acid An acid with more than one ionisable hydrogen, for example H₂CO₃ (diprotic) and H₃PO₄ (triprotic).
One proton at a time
  • One proton at a time Each ionisation is a separate equilibrium with its own Ka, and the protons come off in a fixed order.
Decreasing strength
  • Decreasing strength Ka1 > Ka2 > Ka3 because removing a proton from an increasingly negative ion is progressively harder.
Phosphoric acid
  • Phosphoric acid H₃PO₄ ⇌ H⁺ + H₂PO₄⁻, then H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻, then HPO₄²⁻ ⇌ H⁺ + PO₄³⁻.

Part 2 of 5Features of the Curve

1
Early additions of NaOH convert H₃PO₄ into H₂PO₄⁻, creating the first buffer region around pKa1.
2
The first equivalence point is reached when all H₃PO₄ has become H₂PO₄⁻, using one mole of OH⁻ per mole of acid.
3
Further base converts H₂PO₄⁻ into HPO₄²⁻, forming a second buffer region centred on pKa2.
4
The second equivalence point is reached after two moles of OH⁻ per mole of original acid have been added.
5
A third step to PO₄³⁻ is possible in principle, but Ka3 is so small that the last proton is barely seen in water.

Part 3 of 5Reading Values from the Curve

Reading Values from the Curve
Equivalence volumes
  • Equivalence volumes Successive equivalence volumes are equally spaced for a fully distinguishable polyprotic acid,
    • because each step uses the same moles of base.
First half-equivalence
  • First half-equivalence Halfway to the first equivalence point [H₃PO₄] = [H₂PO₄⁻], so pH = pKa1.
Second half-equivalence
  • Second half-equivalence Halfway between the first and second equivalence points [H₂PO₄⁻] = [HPO₄²⁻], so pH = pKa2.
Rule of thumb
  • Rule of thumb The nth half-equivalence point lies at a volume of (n - ½) times the first equivalence volume, and its pH equals pKan.
RegionDominant speciespH relationship
StartH₃PO₄Set by Ka1
First half-equivalenceH₃PO₄ = H₂PO₄⁻pH = pKa1
First equivalenceH₂PO₄⁻Between pKa1 and pKa2
Second half-equivalenceH₂PO₄⁻ = HPO₄²⁻pH = pKa2

Part 4 of 5Worked Example

Question

  • 20.0 mL of 0.100 mol/L H₃PO₄ is titrated with 0.100 mol/L NaOH.
  • Ka1 = 7.5 × 10⁻³ and Ka2 = 6.2 × 10⁻⁸.
  • Find the volume of NaOH at the first and second equivalence points and the pH at the first two half-equivalence points.

  1. Moles of H₃PO₄ = 0.100 × 0.0200 = 2.00 × 10⁻³ mol.
  2. First equivalence needs 2.00 × 10⁻³ mol of OH⁻, which is 20.0 mL;
    • second equivalence needs 4.00 × 10⁻³ mol, which is 40.0 mL.
  3. The first half-equivalence point is at 10.0 mL and the second is at 30.0 mL.
  4. pKa1 = -log(7.5 × 10⁻³) = 2.12, so the pH at 10.0 mL is 2.12.
  5. pKa2 = -log(6.2 × 10⁻⁸) = 7.21, so the pH at 30.0 mL is 7.21.

Answer

  • Equivalence at 20.0 mL and 40.0 mL;
    • pH = 2.12 at 10.0 mL and 7.21 at 30.0 mL.

Part 5 of 5Exam Checklist & Tips

8.3.3 – Key Points for the Exam

  • Polyprotic acid An acid that can donate more than one proton per molecule.
  • Amphiprotic species A species such as H₂PO₄⁻ that can act as either an acid or a base.
  • Successive Ka values Equilibrium constants for each proton removal, always decreasing in size.
  • Buffer region A flat section of the curve where a conjugate acid-base pair is present.

Common Mistakes

  • Treating all the protons as released in a single step with one Ka.
  • Forgetting that the second equivalence point needs twice as much base as the first, not the same amount again from zero.
  • Reading pKa2 at the first equivalence point instead of at the midpoint between the first and second equivalence points.
  • Expecting a clear third equivalence point for H₃PO₄ in water when Ka3 is extremely small.

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