Polyprotic Acids — Solve polyprotic acid problems (intro)...

SECTION 1

Historical Context & Motivation

Many of the acids you encounter every day — from phosphoric acid in soft drinks to carbonic acid in sparkling water — have more than one hydrogen that can be donated to water. These are called polyprotic acids, and understanding how they behave is essential to explaining everything from blood chemistry to ocean acidification. The study of acids with multiple ionizable protons unfolded over more than a century, driven by improvements in measurement techniques and theoretical models.

1834

Faraday Defines 'Acid'

Michael Faraday studied electrolysis and helped establish that acids conduct electricity, hinting that they produce charged species in water.

1884

Arrhenius Theory

Svante Arrhenius proposed that acids dissociate in water to release H⁺ ions, laying the groundwork for understanding how a single molecule can release more than one proton.

1909

pH Scale Introduced

Søren Sørensen developed the pH scale, giving chemists a practical way to measure and compare acidity — crucial for tracking stepwise proton donation.

1923

Brønsted-Lowry Theory

Johannes Brønsted and Thomas Lowry independently defined acids as proton donors, making the concept of multiple proton donations from one molecule much clearer.

1960s

Computer-Aided Equilibrium

With digital computing, scientists could solve complex equilibrium systems involving multiple dissociation steps, enabling precise prediction of dominant species at any pH.

The central question these advances raised is straightforward yet powerful: if a single acid molecule can lose more than one proton, how do we figure out how many protons are actually lost at a given pH, and which form of the acid is most abundant in solution? That is exactly what this lesson will teach you to do.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need a solid grasp of the key ideas that govern polyprotic acid behavior. Unlike a monoprotic acid (such as HCl), which donates only one proton, a polyprotic acid donates its protons in distinct, sequential steps. Each step has its own equilibrium constant, and each step is weaker than the one before it.

Stepwise Dissociation

Polyprotic acids lose protons one at a time, not all at once. Each removal is a separate equilibrium with its own K_a value.

Decreasing Kₐ Values

Each successive K_a is much smaller than the previous one. Removing a proton from a negatively charged ion is harder than from a neutral molecule.

Dominant Species

At any given pH, one form (fully protonated, partially deprotonated, or fully deprotonated) is present in the highest concentration. This is the dominant species.

pKₐ as a Boundary

At pH = pK_a, the protonated and deprotonated forms of that step are in equal concentration. The pK_a values act as boundary markers between dominant species regions.

✦ KEY TAKEAWAY

Think of a polyprotic acid as a person holding several tennis balls. They throw one ball at a time, not all at once. The first ball is easy to throw (large K_a1), the second is harder (smaller K_a2), and the third — if there is one — is the hardest (smallest K_a3). The pH of the solution tells you how many balls have already been thrown.

SECTION 3

Visualizing Stepwise Dissociation

The diagram below shows the stepwise dissociation of phosphoric acid (H₃PO₄), a triprotic acid with three dissociation steps. Notice how each arrow represents one proton being donated to water, and how the K_a decreases dramatically at each step.

Figure 1 — Phosphoric acid (H₃PO₄) loses its three protons in separate steps. Each K_a is roughly 100,000 times smaller than the last, illustrating why the first dissociation dominates the production of H⁺.

The dramatic decrease in K_a from one step to the next is not a coincidence — it reflects a fundamental physical reality. Once an acid has lost a proton, it carries a negative charge. Removing a positively charged proton from a negatively charged ion requires fighting an increasingly strong electrostatic attraction. This principle ensures that in most problems, only the first dissociation significantly affects the pH.

SECTION 4

Mathematical Framework

Every polyprotic acid problem relies on writing the correct equilibrium expressions for each dissociation step and recognizing which one matters most. Let's use a generic diprotic acid H₂A as our model. It has two dissociation steps.

FIRST DISSOCIATION

H₂A ⇌ H⁺ + HA⁻ Kₐ₁ = [H⁺][HA⁻] / [H₂A]

H₂A = fully protonated acid; HA⁻ = intermediate (conjugate base of step 1); K_a1 = acid dissociation constant for the first proton.

SECOND DISSOCIATION

HA⁻ ⇌ H⁺ + A²⁻ Kₐ₂ = [H⁺][A²⁻] / [HA⁻]

HA⁻ = intermediate species; A²⁻ = fully deprotonated form; K_a2 = acid dissociation constant for the second proton. Typically K_a2 ≪ K_a1.

SIMPLIFICATION RULE

If Kₐ₁ / Kₐ₂ > 100, treat the first dissociation as a monoprotic problem to find [H⁺] and pH.

Because K_a2 is so much smaller, the second step contributes negligible additional H⁺. The concentration of the fully deprotonated form [A²⁻] ≈ K_a2 (see the shortcut explanation in Section 4 below).

DOMINANT SPECIES RULE

pH < pKₐ₁ → H₂A dominates | pKₐ₁ < pH < pKₐ₂ → HA⁻ dominates | pH > pKₐ₂ → A²⁻ dominates

Compare the solution pH to the pK_a values. The species that exists in the region surrounding the current pH is the dominant species. At pH = pK_a, two species are present in equal concentrations.

💡 Remember This Shortcut

For an introductory polyprotic acid problem where you dissolve the fully protonated form (like H₂SO₃ or H₂A), you can find the pH using only Kₐ₁ and an ICE table, just like a regular weak acid problem. Then separately note that [A²⁻] ≈ K_a2. This shortcut works because the ICE table gives [H⁺] ≈ [HA⁻] from the first dissociation. Substituting into the K_a2 expression: K_a2 = [H⁺][A²⁻] / [HA⁻] ≈ [H⁺][A²⁻] / [H⁺] = [A²⁻]. Therefore [A²⁻] ≈ K_a2. This simplification is valid only when [H⁺] ≈ [HA⁻], which holds when the first dissociation dominates (K_a1 / K_a2 > 100).

SECTION 5

Identifying the Dominant Species

The concept of dominant species is one of the most practical tools in acid-base chemistry. Given a polyprotic acid and a known pH, you can instantly determine which form of the acid is present in the greatest amount by comparing the pH to the pK_a values. The diagram below maps out the dominant species regions for carbonic acid (H₂CO₃), a diprotic acid that plays a critical role in blood buffering and ocean chemistry.

Figure 2 — Dominant species map for carbonic acid. The dashed lines at pK_a1 = 6.35 and pK_a2 = 10.33 divide the pH scale into three regions. Note that at blood pH (≈ 7.4), the bicarbonate ion HCO₃⁻ is the dominant species.

This map is incredibly useful. For example, if you know that blood has a pH of about 7.4, you can glance at the diagram and immediately see that 7.4 falls between pK_a1 and pK_a2, which means bicarbonate (HCO₃⁻) is the dominant species in blood. This is exactly why your body's primary blood buffer is the carbonic acid–bicarbonate system.

Common Polyprotic Acids and Their pKₐ Values

Acid	pKₐ₁	pKₐ₂	pKₐ₃
Sulfurous acid (H₂SO₃)	1.81	6.91	—
Carbonic acid (H₂CO₃)	6.35	10.33	—
Phosphoric acid (H₃PO₄)	2.15	7.20	12.35
Citric acid (H₃C₆H₅O₇)	3.13	4.76	6.40
Oxalic acid (H₂C₂O₄)	1.25	4.27	—

SECTION 6

Worked Example — Carbonic Acid

Let's solve a complete introductory polyprotic acid problem step by step. We'll find the pH and identify the dominant species when 0.10 M carbonic acid (H₂CO₃) is dissolved in water. Given: K_a1 = 4.3 × 10⁻⁷ and K_a2 = 4.7 × 10⁻¹¹. Note: The K_a1 value of 4.3 × 10⁻⁷ is the conventional apparent K_a1 for the carbonate system as it exists in water, which includes dissolved CO₂ in equilibrium with H₂CO₃. This is the value most commonly used in general chemistry and is the standard convention for problems involving the carbonate system.

Finding pH of 0.10 M H₂CO₃

Step 1 — Check the Simplification RuleCalculate K_a1 / K_a2 = (4.3 × 10⁻⁷) / (4.7 × 10⁻¹¹) ≈ 9,100. Since this is much greater than 100, we can treat the first dissociation as a simple monoprotic acid problem and ignore the second dissociation's contribution to [H⁺].

Kₐ₁ / Kₐ₂ ≈ 9,100 ≫ 100 → Use Kₐ₁ only

Step 2 — Set Up ICE Table for First DissociationWrite the equilibrium: H₂CO₃ ⇌ H⁺ + HCO₃⁻. Initial concentrations: [H₂CO₃] = 0.10 M, [H⁺] = 0, [HCO₃⁻] = 0. Change: −x, +x, +x. Equilibrium: (0.10 − x), x, x.

Step 3 — Write and Solve the Kₐ₁ ExpressionK_a1 = x² / (0.10 − x) = 4.3 × 10⁻⁷. Since K_a1 is very small compared to 0.10, we can approximate 0.10 − x ≈ 0.10. Then x² = (4.3 × 10⁻⁷)(0.10) = 4.3 × 10⁻⁸. Taking the square root: x = √(4.3 × 10⁻⁸) ≈ 2.07 × 10⁻⁴ M.

[H⁺] = x ≈ 2.07 × 10⁻⁴ M

Step 4 — Calculate pHpH = −log[H⁺] = −log(2.07 × 10⁻⁴) ≈ 3.68.

pH ≈ 3.68

Step 5 — Identify the Dominant SpeciesCompare pH = 3.68 to the pK_a values. Using the given K_a1 = 4.3 × 10⁻⁷: pK_a1 = −log(4.3 × 10⁻⁷) ≈ 6.37. The reference table lists pK_a1 = 6.35, a slight difference due to rounding of the K_a1 value. Either way, pK_a2 = −log(4.7 × 10⁻¹¹) ≈ 10.33. Since pH (3.68) < pK_a1 (6.35–6.37), the fully protonated form is dominant.

Dominant species: H₂CO₃

Step 6 — Find [CO₃²⁻] (Bonus Insight)A useful shortcut: when you dissolve a diprotic acid, the concentration of the fully deprotonated species equals K_a2. Here is why: the ICE table for the first dissociation gives [H⁺] ≈ [HCO₃⁻] = x ≈ 2.07 × 10⁻⁴ M. Substituting into the K_a2 expression: K_a2 = [H⁺][CO₃²⁻] / [HCO₃⁻]. Because [H⁺] ≈ [HCO₃⁻] from step 1, these terms cancel, giving [CO₃²⁻] ≈ K_a2 = 4.7 × 10⁻¹¹ M. This shortcut is valid because [H⁺] ≈ [HCO₃⁻] holds when the first dissociation dominates (K_a1 / K_a2 ≫ 100). This confirms the second dissociation barely occurs.

[CO₃²⁻] ≈ 4.7 × 10⁻¹¹ M

SECTION 7

Monoprotic vs. Polyprotic: Key Differences

Students often wonder: "Why can't I just treat a polyprotic acid exactly like a monoprotic acid?" In many ways you can — and that's the beauty of the simplification rule. But there are important differences to keep in mind.

Monoprotic vs. Polyprotic Acid Comparison

Feature	Monoprotic Acid	Polyprotic Acid
Number of ionizable H⁺	1	2 or more (diprotic, triprotic, etc.)
Equilibrium expressions	1 Kₐ expression	Separate Kₐ for each step (Kₐ₁, Kₐ₂, ...)
pH calculation	One ICE table	Usually one ICE table (Kₐ₁ only), with [A²⁻] ≈ Kₐ₂
Dominant species question	Either HA or A⁻	Multiple possible forms (H₂A, HA⁻, A²⁻)
Titration curve	1 equivalence point	Multiple equivalence points (1 per proton)

✦ KEY TAKEAWAY

Solving an introductory polyprotic acid problem is like eating a multi-course meal: each course (dissociation) arrives separately, and the first course is by far the most filling. You can figure out how full you are (the pH) by focusing mostly on the appetizer (K_a1). The later courses (K_a2, K_a3) add flavor but barely change how full you feel.

SECTION 8

Connection to Advanced Theory

The introductory approach you've learned here — treating only the first dissociation and using pK_a boundaries to identify dominant species — is a powerful starting point. As you advance, you'll encounter situations where the simplification breaks down and more sophisticated techniques are needed.

Introductory vs. Advanced Polyprotic Acid Analysis

Topic	Introductory Level (This Lesson)	Advanced Level (Future Topics)
pH calculation	Use Kₐ₁ only with an ICE table	Solve simultaneous equilibria; alpha (α) fraction equations
Dominant species	Compare pH to pKₐ values	Calculate exact fractions using alpha expressions: α₀, α₁, α₂
Titration	Recognize multiple equivalence points	Calculate pH at each equivalence point and half-equivalence point
Buffers	Understand that pKₐ = pH at equal concentrations	Design polyprotic buffers; use Henderson-Hasselbalch for each step
Special cases	Assume Kₐ₁ ≫ Kₐ₂	Handle cases where Kₐ values are close (e.g., sulfuric acid's second dissociation)

One particularly important advanced topic is the pH of an amphoteric species — the intermediate form like HA⁻ that can act as either an acid or a base. In future courses, you'll learn that the pH of a solution of the intermediate is approximately (pK_a1 + pK_a2) / 2. For now, just know that this connection exists and that your understanding of stepwise dissociation lays the groundwork for it.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

Phosphoric acid (H₃PO₄) has three dissociation constants: K_a1 = 7.5 × 10⁻³, K_a2 = 6.2 × 10⁻⁸, K_a3 = 4.8 × 10⁻¹³. Explain why K_a1 > K_a2 > K_a3. Why does each successive dissociation become harder?

PROBLEM 2 — BASIC CALCULATION

Calculate the pH of a 0.050 M solution of sulfurous acid (H₂SO₃). K_a1 = 1.5 × 10⁻² and K_a2 = 6.3 × 10⁻⁸. Use only the first dissociation.

PROBLEM 3 — INTERMEDIATE

A solution of oxalic acid (H₂C₂O₄) has a measured pH of 3.50. Given pK_a1 = 1.25 and pK_a2 = 4.27, identify the dominant species at this pH. Then determine what species would dominate at pH = 0.80 and pH = 5.00.

PROBLEM 4 — APPLIED

A lake sample has a pH of 8.20 and contains dissolved carbonic acid species. Given pK_a1 = 6.35 and pK_a2 = 10.33 for carbonic acid, identify the dominant carbonate species. A factory accidentally dumps a strong acid into the lake, lowering the pH to 5.50. What is the new dominant species, and how does this affect dissolved CO₂?

PROBLEM 5 — CRITICAL THINKING

Citric acid (H₃C₆H₅O₇) has pK_a1 = 3.13, pK_a2 = 4.76, and pK_a3 = 6.40. Notice that these pK_a values are much closer together than those of phosphoric acid (2.15, 7.20, 12.35). Would you expect the simplification of treating only K_a1 to be more or less accurate for citric acid than for phosphoric acid? Justify your reasoning.

SUMMARY

Lesson Summary

Polyprotic acids donate protons in stepwise dissociations, each governed by its own equilibrium constant (Kₐ₁, Kₐ₂, Kₐ₃). Each successive K_a is much smaller than the last because removing a proton from an increasingly negative ion is harder. When Kₐ₁ / Kₐ₂ > 100, you can simplify the problem by treating only the first dissociation with an ICE table to find the pH — exactly like a monoprotic weak acid problem.

To identify the dominant species at any given pH, compare the pH to the pKₐ values: if pH < pK_a1, the fully protonated form dominates; if pK_a1 < pH < pK_a2, the intermediate form dominates; if pH > pK_a2, the fully deprotonated form dominates. A bonus shortcut: when dissolving the fully protonated diprotic acid, [A²⁻] ≈ Kₐ₂ (because [H⁺] ≈ [HA⁻] from the first dissociation, so these terms cancel in the Kₐ₂ expression). These tools prepare you for more advanced topics like polyprotic titrations and buffer design.