AP Chemistry: Polyprotic Acid Equilibria

Understanding polyprotic acids is essential for mastering acid-base chemistry, as these molecules govern processes from blood pH regulation to industrial synthesis. Their unique behavior, involving the sequential release of multiple protons, requires a nuanced approach to equilibrium calculations that goes beyond simple monoprotic systems.

Defining Polyprotic Acids and Stepwise Dissociation

A polyprotic acid is an acid capable of donating more than one proton (H⁺ ion) per molecule. Common examples you will encounter include sulfuric acid (H₂SO₄), phosphoric acid (H₃PO₄), and carbonic acid (H₂CO₃). The dissociation does not happen all at once; instead, it proceeds through a series of stepwise equilibria. Each step has its own equilibrium constant, denoted as Ka₁, Ka₂, Ka₃, and so on.

For a generic diprotic acid, H₂A, the dissociations are: $${\text{H}}_2\text{A}(aq)\rightleftharpoons {\text{H}}^{+}(aq)+{\text{HA}}^{-}(aq)K_{a1}$$ $${\text{HA}}^{-}(aq)\rightleftharpoons {\text{H}}^{+}(aq)+{\text{A}}^{2-}(aq)K_{a2}$$ The key pattern is that $K_{a1}>K_{a2}>K_{a3}$. This makes intuitive sense: removing a proton from a neutral molecule (H₂A) is easier than removing one from a negatively charged ion (HA⁻), which experiences greater electrostatic attraction to the remaining proton.

The Hierarchy of Ka Values and Simplifying Assumptions

The large difference between successive Ka values (often by a factor of 10³ to 10⁶) is the most critical concept for problem-solving. Because $K_{a1}$ is so much larger, the first dissociation produces the vast majority of the H⁺ ions in solution. The second and third dissociations contribute a negligible additional amount of H⁺, but they are responsible for producing the fully deprotonated species (like A²⁻).

This leads to a powerful simplification: For most pH calculations of a polyprotic acid solution, you can treat the first dissociation as the sole significant source of H⁺. You approximate the system as a monoprotic acid (H₂A) with $K_a=K_{a1}$. However, you cannot ignore the later equilibria when you are asked to find the concentration of species like HA⁻ or A²⁻. Those concentrations are determined by the respective Ka expressions and the pH established by the first dissociation.

Calculating pH and Species Concentrations

Your approach depends on the acid type and the question asked. Let's work through the logic for different scenarios.

1. Strong First Dissociation: Sulfuric Acid (H₂SO₄)

Sulfuric acid is a special case because its first dissociation is complete (strong acid), while the second is weak ($K_{a2}=1.2\times 10^{-2}$). For a solution of H₂SO₄:

• The first step contributes [H⁺] equal to the initial acid concentration.
• The second step, HSO₄⁻ ⇌ H⁺ + SO₄²⁻, then occurs. You must solve this equilibrium using an ICE table where the initial [H⁺] is not zero—it's from the first step.

Example: Calculate the pH of 0.050 M H₂SO₄.

1. First dissociation: $[H^{+}]=0.050M$ from H₂SO₄ → H⁺ + HSO₄⁻.
2. Second dissociation: Set up ICE for HSO₄⁻ ⇌ H⁺ + SO₄²⁻ with initial [H⁺] = 0.050 M and initial [HSO₄⁻] = 0.050 M.
3. Solve the equilibrium expression: $K_{a2}=1.2\times 10^{-2}=\frac{(0.050+x)(x)}{(0.050-x)}$.
4. Solving for x gives a positive value, which is added to the initial [H⁺]. Final $[H^{+}]\approx 0.060M$, so pH ≈ 1.22.

2. Weak Stepwise Dissociation: Phosphoric Acid (H₃PO₄)

For acids like H₃PO₄ ($K_{a1}=7.5\times 10^{-3},K_{a2}=6.2\times 10^{-8},K_{a3}=4.8\times 10^{-13}$), all dissociations are weak. The pH is dominated by the first equilibrium.

To find pH: Treat it as a monoprotic weak acid with $K_a=K_{a1}$. For a 0.10 M H₃PO₄ solution, you would set up: $K_{a1}=\frac{x^2}{0.10-x}$. Solve for x, which equals $[H^{+}]\approx [H_{2}P{O}_4^{-}]$. (Check the 5% rule; for this concentration, you likely need the quadratic formula).

To find [HPO₄²⁻] and [PO₄³⁻]: Use the pH from step one in the later equilibrium expressions.

• $[HP{O}_4^{2-}]$ is found from the $K_{a2}$ expression: $K_{a2}=\frac{[H^{+}][HP{O}_4^{2-}]}{[H_{2}P{O}_4^{-}]}$. Since $[H^{+}]\approx [H_{2}P{O}_4^{-}]$ from the first step, they approximately cancel, giving $[HP{O}_4^{2-}]\approx K_{a2}$.
• $[P{O}_4^{3-}]$ is found from $K_{a3}=\frac{[H^{+}][P{O}_4^{3-}]}{[HP{O}_4^{2-}]}$. Plug in known values for [H⁺] and [HPO₄²⁻] to solve.

3. Intermediate Species: Amphiprotic Ions (e.g., HCO₃⁻)

Solutions of an amphiprotic intermediate, like sodium bicarbonate (NaHCO₃), require a different formula. The H⁺ concentration for such a solution is approximately given by: $[H^{+}]\approx \sqrt{K_{a1}\cdot K_{a2}}$. For HCO₃⁻ (from H₂CO₃, $K_{a1}=4.3\times 10^{-7},K_{a2}=4.7\times 10^{-11}$), this gives $[H^{+}]=\sqrt{(4.3\times 10^{-7})(4.7\times 10^{-11})}=4.5\times 10^{-9}M$, or pH ≈ 8.35. This pH is independent of the concentration of the amphiprotic ion, provided the concentration is reasonably large compared to the Ka values.

Common Pitfalls

1. Adding Ka Values: A major mistake is thinking $K_{a(total)}=K_{a1}+K_{a2}$. This is incorrect. The overall constant for losing two protons is the product: $K_{a1}\times K_{a2}$.
2. Misapplying Simplifications: Using the "treat like monoprotic" shortcut for sulfuric acid will give a wildly incorrect answer because the first dissociation is strong. Always check if the first step is strong or weak.
3. Ignoring Later Equilibria for Species Concentration: You might correctly calculate the pH using only $K_{a1}$, but then state that $[A^{2-}]=0$. This is wrong. While the contribution to [H⁺] is negligible, $[A^{2-}]$ is defined by $K_{a2}$ and is not zero. Use the equilibrium expressions sequentially.
4. Forgetting the Charge Balance: In complex problems, writing the charge balance equation (sum of positive charges = sum of negative charges) is an invaluable tool for keeping track of all species, especially for solutions of amphiprotic ions or acid salts.

Summary

• Polyprotic acids dissociate in stepwise equilibria, with each step having a successively smaller acid dissociation constant ($K_a$).
• For pH calculation, the first dissociation ($K_{a1}$) is almost always dominant, allowing you to treat most polyprotic acids as monoprotic weak acids. The major exception is sulfuric acid, where the strong first dissociation must be accounted for before solving the second.
• To find the concentration of intermediate or fully deprotonated species (e.g., HA⁻, A²⁻), you must use the pH from the first step in the subsequent $K_{a2}$ and $K_{a3}$ expressions.
• The pH of a solution of an amphiprotic intermediate (like HCO₃⁻) is approximately the average of $p{K}_{a1}$ and $p{K}_{a2}$, given by $[H^{+}]\approx \sqrt{K_{a1}\cdot K_{a2}}$.
• Avoid the critical errors of adding Ka values or completely ignoring the existence of species produced in later dissociation steps.