Polyprotic Acids and Stepwise Ka Values

Introduction

Many important acids can give away more than one proton: sulfuric acid, carbonic acid, oxalic acid, phosphoric acid and citric acid are all examples. Such acids are called polyprotic . The key idea of this page is that the protons do not leave together. They leave one at a time, each step has its own equilibrium constant, and those constants are usually very different in size. Understanding this stepwise picture is the foundation for every speciation, buffer and titration calculation that follows in the unit.

Core explanation

Stepwise equilibria. A diprotic acid H₂A dissociates in two separate steps:

H₂A + H₂O ⇌ H₃O⁺ + HA⁻ Ka1 = [H₃O⁺][HA⁻] / [H₂A]

HA⁻ + H₂O ⇌ H₃O⁺ + A²⁻ Ka2 = [H₃O⁺][A²⁻] / [HA⁻]

A triprotic acid H₃A adds a third step with Ka3. Each constant describes one proton transfer and one conjugate pair. The anion HA⁻ is the conjugate base in the first step and the acid in the second.

Successive constants fall. For almost every polyprotic acid, Ka1 > Ka2 > Ka3. Typical values at 25 °C:

Acid Ka1 Ka2 Ka3 --- --- --- --- Oxalic, H₂C₂O₄ 5.9 × 10⁻² 6.4 × 10⁻⁵ — Carbonic (as dissolved CO₂) 4.3 × 10⁻⁷ 4.7 × 10⁻¹¹ — Ascorbic (vitamin C) 8.0 × 10⁻⁵ 1.6 × 10⁻¹² — Phosphoric, H₃PO₄ 7.5 × 10⁻³ 6.2 × 10⁻⁸ 4.8 × 10⁻¹³

For phosphoric acid the drop from one step to the next is roughly a factor of 10⁵. In pKa terms, pKa1 = 2.15, pKa2 = 7.20 and pKa3 = 12.32, so the steps are separated by about five pH units each.

Why the constants fall. Two effects combine. The main one is electrostatic: removing a positive proton from a neutral molecule is easier than removing it from an anion that already carries negative charge, because the negative charge attracts the departing proton. Each further step starts from a more negative species, so it is harder still. A smaller statistical effect also operates: H₂A has two protons that could leave but only one site that can accept a proton back, whereas HA⁻ has one proton to lose and two sites to regain it.

Overall constants. Multiplying stepwise expressions cancels the intermediate species:

Ka1 × Ka2 = [H₃O⁺]²[A²⁻] / [H₂A]

This overall constant is useful for relating the fully protonated and fully deprotonated forms directly, but it does not mean that two protons leave in one event, and it cannot be used to find the pH on its own.

Formulae

Ka(n) = [H₃O⁺][conjugate base of step n] / [acid of step n]; pKa = −log₁₀ Ka; overall β = Ka1 × Ka2 (× Ka3); pβ = pKa1 + pKa2 (+ pKa3).

Step-by-step reasoning

To set up any polyprotic acid problem:

1. Write the fully protonated form and remove one proton at a time, writing one equilibrium per step. 2. Attach the correct Ka to each step, in decreasing order of size. 3. Compare successive Ka values: a ratio of 10³ or more means the steps can usually be treated separately. 4. Decide which step controls the property you need (pH, a particular species, a buffer region).

Visual explanation

Imagine a staircase descending from H₃A through H₂A⁻ and HA²⁻ to A³⁻. Each tread sits at a pH equal to one pKa. On a pH line, draw ticks at 2.15, 7.20 and 12.32 for phosphoric acid: the space between ticks is where one species dominates, and each tick marks a hand-over between neighbours.

Real-world analogy

Think of peeling stickers off a charged balloon that pulls them back. The first sticker comes off easily. As the balloon gains charge with each removal, it clings harder to the stickers that remain, so each one needs more effort than the last — just as each proton is held more tightly than the one before.

Real-world example

Phosphoric acid is added to cola drinks as an acidulant. Only its first dissociation matters at the pH of the drink (about 2.5); the second and third protons stay attached. In contrast, phosphate in blood exists mostly as H₂PO₄⁻ and HPO₄²⁻, because physiological pH lies near pKa2.

Why?

Why do we need separate constants instead of one "strength" for the acid? Because each proton has a different tendency to leave. A single number would hide the fact that phosphoric acid behaves as a moderately strong acid for its first proton but as an extremely weak acid for its third.

Common misconception

"A diprotic acid releases twice as many H⁺ ions as a monoprotic acid of the same concentration." Usually false: the second proton is so weakly acidic that it contributes almost nothing to the pH of a solution of the acid alone.

Worked example

Question: Oxalic acid has Ka1 = 5.9 × 10⁻² and Ka2 = 6.4 × 10⁻⁵. Find pKa1, pKa2 and the overall constant for H₂C₂O₄ ⇌ 2H⁺ + C₂O₄²⁻.

Reasoning: pKa1 = −log(5.9 × 10⁻²) = 1.23. pKa2 = −log(6.4 × 10⁻⁵) = 4.19. Overall β = Ka1 × Ka2 = 5.9 × 10⁻² × 6.4 × 10⁻⁵ = 3.8 × 10⁻⁶, and pβ = 1.23 + 4.19 = 5.42.

Answer: pKa1 = 1.23, pKa2 = 4.19, overall constant 3.8 × 10⁻⁶ (pβ = 5.42). The Ka ratio is about 900, so the two steps overlap slightly more than for most polyprotic acids.

Quick check

1. Why is Ka2 of carbonic acid much smaller than Ka1? Answer: The second proton must leave a negatively charged ion, HCO₃⁻, which attracts it electrostatically, so it is released far less readily.

Exam focus

Always write polyprotic dissociations as separate steps with separate Ka expressions. Examiners reward correct identification of the acid and conjugate base in each step, and penalise "H₂A ⇌ 2H⁺ + A²⁻" presented as the only equilibrium. Remember that pKa values add when constants multiply.

Advanced insight

When the acidic groups are far apart in a large molecule, the electrostatic effect weakens and successive pKa values approach a minimum separation of log 4 ≈ 0.6, set by the statistical factor alone. Long-chain dicarboxylic acids such as adipic acid show pKa values of about 4.4 and 5.4, much closer together than those of oxalic acid, whose carboxyl groups are adjacent.

Summary

A polyprotic acid loses protons one at a time, each step with its own Ka. Successive constants fall, typically by several powers of ten, mainly because each proton leaves an increasingly negative species. Stepwise constants multiply to give overall constants, and pKa values add. The size of the gap between steps decides whether they can be treated independently.

Practice questions

1. Write the three stepwise dissociation equations for phosphoric acid. Answer: H₃PO₄ ⇌ H⁺ + H₂PO₄⁻; H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻; HPO₄²⁻ ⇌ H⁺ + PO₄³⁻. 2. Ascorbic acid has pKa1 = 4.10 and pKa2 = 11.80. Calculate Ka1 × Ka2. Answer: pβ = 4.10 + 11.80 = 15.90, so Ka1 × Ka2 = 10⁻¹⁵·⁹ ≈ 1.3 × 10⁻¹⁶. 3. Give two reasons why Ka2 is smaller than Ka1 for a diprotic acid. Answer: Electrostatic attraction between the leaving proton and the negative ion, and a statistical factor (fewer protons to lose and more sites to regain one). 4. Which acid in the table above has the steps closest together, and what does that imply? Answer: Oxalic acid (ratio about 900); its two dissociations overlap more, so they may need to be treated together in accurate work.