Polyprotic Acids: pH, Amphiprotic Ions and Species Fractions
Carbonic acid, phosphoric acid and sulphurous acid each release more than one proton. Questions on them look long, but they usually reduce to three small ideas: use only the first dissociation for the pH of the acid, use the average of two pKa values for an amphiprotic salt such as NaHCO₃, and use fractions to find which species dominates at a given pH. This page works through all three for IIT-JAM, GATE, CSIR-NET and CUET-PG level problems.
The key relations
pH of H₂A (Ka1 ≫ Ka2): [H⁺] = (−Ka1 + √(Ka1² + 4·Ka1·C)) / 2 ≈ √(Ka1·C)
pH of the amphiprotic ion HA⁻: pH ≈ ½ (pKa1 + pKa2)
Fraction of species at a given [H⁺]: α(H₂A) = [H⁺]²/D, α(HA⁻) = Ka1[H⁺]/D, α(A²⁻) = Ka1Ka2/D, where D = [H⁺]² + Ka1[H⁺] + Ka1Ka2
What the terms mean
- Ka1, Ka2 – stepwise dissociation constants. Ka1 is always larger than Ka2, because it is harder to remove a proton from a negatively charged ion.
- C – the analytical (formal) concentration of the acid, in mol L⁻¹.
- Amphiprotic species – a species that can both give and accept a proton, such as HCO₃⁻ or H₂PO₄⁻.
- α (alpha) – the fraction of the total acid present as one particular species. The fractions add up to 1.
The shortcut for the amphiprotic ion holds when the concentration is not very small and Ka1 and Ka2 are well separated. It is an approximation; if the question gives full data and asks for an exact value, follow the question.
Data used below (typical textbook values at 25 °C)
| Acid | Ka1 | Ka2 | Ka3 |
|---|---|---|---|
| H₂CO₃ (carbonic) | 4.3 × 10⁻⁷ (pKa1 = 6.37) | 4.7 × 10⁻¹¹ (pKa2 = 10.33) | – |
| H₃PO₄ (phosphoric) | 7.5 × 10⁻³ (pKa1 = 2.12) | 6.2 × 10⁻⁸ (pKa2 = 7.21) | 4.8 × 10⁻¹³ (pKa3 = 12.32) |
Different books give slightly different values. If your question or textbook gives its own constants, use those.
Worked examples
Example 1: pH of 0.10 M H₂CO₃. Treat only the first step, because Ka2 is smaller than Ka1 by a factor of about 10⁴.
[H⁺] ≈ √(Ka1 · C) = √(4.3 × 10⁻⁷ × 0.10) = √(4.3 × 10⁻⁸) = 2.07 × 10⁻⁴ M.
pH = −log(2.07 × 10⁻⁴) = 3.68.
Exact quadratic gives [H⁺] = 2.07 × 10⁻⁴ M as well (pH 3.68), because x is tiny compared with 0.10, so the approximation is fine.
[CO₃²⁻] ≈ Ka2 = 4.7 × 10⁻¹¹ M. This is a useful fact: in a solution of a diprotic acid with no other source of H⁺ or A²⁻, [A²⁻] is about equal to Ka2.
Example 2: pH of NaHCO₃ solution. HCO₃⁻ is amphiprotic.
pKa1 = −log(4.3 × 10⁻⁷) = 6.37; pKa2 = −log(4.7 × 10⁻¹¹) = 10.33.
pH ≈ ½ (6.37 + 10.33) = ½ × 16.70 = 8.35.
The pH does not depend on the concentration of NaHCO₃ in this approximation, which is why it is a favourite short question. A slightly basic result is as expected: hydrolysis of HCO₃⁻ (to give OH⁻) is stronger than its acid dissociation.
Example 3: pH of 0.10 M H₃PO₄. Phosphoric acid has a fairly large Ka1, so the simple square-root formula is not very accurate here. Use the quadratic.
x² / (0.10 − x) = 7.5 × 10⁻³, so x² + 7.5 × 10⁻³ x − 7.5 × 10⁻⁴ = 0.
x = (−7.5 × 10⁻³ + √(5.625 × 10⁻⁵ + 3.0 × 10⁻³)) / 2 = (−0.0075 + 0.05528) / 2 = 0.0239 M.
pH = −log(0.0239) = 1.62.
Compare the shortcut: √(7.5 × 10⁻³ × 0.10) = 0.0274 M, pH = 1.56. The difference (0.06 pH units) comes from ignoring the drop in [H₃PO₄]. Because [H₂PO₄⁻] ≈ 0.0239 M and [H⁺] ≈ 0.0239 M, the second step gives only a very small extra H⁺ (about Ka2 ≈ 6.2 × 10⁻⁸ M), so ignoring it is correct.
Example 4: which phosphate species dominates at pH 7.00?
[H⁺] = 1.0 × 10⁻⁷. D = [H⁺]³ + Ka1[H⁺]² + Ka1Ka2[H⁺] + Ka1Ka2Ka3.
Fractions (calculated with the constants in the table):
H₃PO₄ = 8.2 × 10⁻⁶ ≈ 0
H₂PO₄⁻ = 0.617
HPO₄²⁻ = 0.383
PO₄³⁻ = 1.8 × 10⁻⁶ ≈ 0
Sum = 0.617 + 0.383 = 1.000. At pH 7.00, which is just below pKa2 = 7.21, H₂PO₄⁻ is a little ahead. At pH = 7.21 the two would be equal. This is why the H₂PO₄⁻/HPO₄²⁻ pair is the buffer near neutral pH.
Example 5: amphiprotic phosphate ions. Using the same rule: for NaH₂PO₄, pH ≈ ½ (2.12 + 7.21) = 4.67; for Na₂HPO₄, pH ≈ ½ (7.21 + 12.32) = 9.76. Always use the two pKa values on either side of the ion in question.
Common mistakes
- Adding the H⁺ from both steps. For a typical diprotic acid, the second step adds a negligible amount when Ka1 ≫ Ka2.
- Using ½(pKa1 + pKa2) for the acid itself. That formula is for the intermediate ion (HA⁻), not for H₂A.
- Choosing the wrong pair of constants. H₂PO₄⁻ lies between pKa1 and pKa2; HPO₄²⁻ lies between pKa2 and pKa3.
- Using the square-root shortcut when Ka1 is large (phosphoric acid at 0.10 M) and x is not small. Check x against C; if x is more than about 5% of C, use the quadratic.
- Treating the pH of an amphiprotic salt as 7. It is usually not 7.
Where this appears
| Exam | Typical use |
|---|---|
| IIT-JAM, CUET-PG | pH of polyprotic acids, pH of amphiprotic salts, buffer pairs. Check the latest syllabus. |
| GATE, CSIR-NET | Speciation, fractions at a given pH, and links to titration curves. Check the current official notification. |
Use the pH tool to check [H⁺] and pH once you have worked out the concentration of the species that sets the pH.
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