Polyprotic Acids

Core Concept

A polyprotic acid can donate more than one proton, but it does so in separate, stepwise equilibria, each with its own Ka. Successive protons are always harder to remove, so Ka1 > Ka2 > Ka3, often by factors of 10⁵ or more. The practical upshot: for most calculations, the first ionization alone sets the pH, because later ionizations contribute negligibly to [H₃O⁺]. Sulfuric acid is the important exception — its first proton is strong and its second is moderately weak.

Key Ideas

• Stepwise ionization: H₃PO₄ → H₂PO₄⁻ → HPO₄²⁻ → PO₄³⁻, each step with its own Ka.
• Ka ordering: Ka1 ≫ Ka2 ≫ Ka3; each successive proton is harder to pull off a more negatively charged species.
• pH is set by Ka1: because Ka2 is so much smaller, the second step's added H₃O⁺ is negligible (unless Ka values are close, as in H₂SO₄).
• Amphiprotic intermediates (H₂PO₄⁻, HCO₃⁻, HSO₄⁻) can both donate and accept.
• A²⁻ from Ka2 equals Ka2 (approximately): for a diprotic acid H₂A, [A²⁻] ≈ Ka2 when the first ionization dominates.

Equations and Variables

• Diprotic acid, step 1: H₂A + H₂O ⇌ H₃O⁺ + HA⁻, Ka1 = [H₃O⁺][HA⁻]/[H₂A]
• Step 2: HA⁻ + H₂O ⇌ H₃O⁺ + A²⁻, Ka2 = [H₃O⁺][A²⁻]/[HA⁻]
• Triprotic (phosphoric): Ka1 = 7.5 × 10⁻³, Ka2 = 6.2 × 10⁻⁸, Ka3 = 4.2 × 10⁻¹³
• Sulfuric: first step complete; Ka2 = 1.2 × 10⁻² (HSO₄⁻ ⇌ H⁺ + SO₄²⁻)
• Total ionization is NOT the sum of Kas; it is the product of the individual steps.

How It Works

1. Write each ionization as its own equilibrium with its own arrow and Ka.
2. Recognize the charge effect: after step 1 the remaining ion is negatively charged, so it holds its remaining proton more tightly; after step 2 it is more negative still. This is why Ka falls sharply.
3. To find pH, solve step 1 alone (treat as a monoprotic weak acid with Ka = Ka1) using an ICE table.
4. The H₃O⁺ produced in step 1 suppresses step 2 (common-ion effect), making its contribution negligible.
5. If you need the concentration of the fully deprotonated ion A²⁻, it is approximately equal to Ka2 (for a diprotic acid).
6. For H₂SO₄, add the complete first ionization (giving [H₃O⁺] = C plus the first H⁺) to the partial second ionization via an ICE table on HSO₄⁻.

Worked Example

Find the pH of 0.10 M H₂CO₃ (Ka1 = 4.3 × 10⁻⁷, Ka2 = 4.7 × 10⁻¹¹).

Because Ka1 ≫ Ka2, use only step 1:

H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻, Ka1 = x²/(0.10 − x)

x ≈ √(4.3 × 10⁻⁷ × 0.10) = √(4.3 × 10⁻⁸) = 2.07 × 10⁻⁴ M

Check: (2.07 × 10⁻⁴ / 0.10) × 100% = 0.21% < 5% ✓

pH = −log(2.07 × 10⁻⁴) = 3.68

The second ionization contributes only about Ka2 = 4.7 × 10⁻¹¹ M of extra H₃O⁺ — utterly negligible against 2.07 × 10⁻⁴ M.

How it works

1. Write each ionization as its own equilibrium with its own arrow and Ka.
2. Recognize the charge effect: after step 1 the remaining ion is negatively charged, so it holds its remaining proton more tightly; after step 2 it is more negative still. This is why Ka falls sharply.
3. To find pH, solve step 1 alone (treat as a monoprotic weak acid with Ka = Ka1) using an ICE table.
4. The H₃O⁺ produced in step 1 suppresses step 2 (common-ion effect), making its contribution negligible.
5. If you need the concentration of the fully deprotonated ion A²⁻, it is approximately equal to Ka2 (for a diprotic acid).
6. For H₂SO₄, add the complete first ionization (giving [H₃O⁺] = C plus the first H⁺) to the partial second ionization via an ICE table on HSO₄⁻.

Common confusions

• "Ka2 adds equally to pH." — The second ionization is typically ~10⁵ times smaller, so it is negligible for pH; it matters only if Ka1 and Ka2 are close (H₂SO₄).
• "H₂SO₄ is like other diprotic acids." — Its first proton is strong (complete) and its second is weak; you must handle both.
• "[A²⁻] = [H₃O⁺]." — For a diprotic acid, [A²⁻] ≈ Ka2, which is far smaller than [H₃O⁺] from step 1.
• "The overall K equals Ka1 + Ka2." — The overall ionization constant for losing two protons is the product Ka1 × Ka2, not the sum.
• "HPO₄²⁻ is neutral in water." — It is amphiprotic and basic in pure water because its Kb exceeds its Ka.

Quick review

• Stepwise ionization with separate Kas; Ka1 > Ka2 > Ka3.
• pH set by Ka1; later steps negligible (except H₂SO₄).
• [A²⁻] ≈ Ka2 for a diprotic acid.
• H₂SO₄: strong first proton + weak second (Ka2 = 1.2 × 10⁻²).
• Overall two-proton K = Ka1 × Ka2.