You dissolve 0.10 M H₃PO₄ in water. A student sets up three simultaneous equilibrium expressions — one for each ionization — and solves them together to find the exact pH. What is the conceptual problem with this approach?
AThere is no problem; solving all three simultaneously gives the most accurate result
BThe student should use the Henderson-Hasselbalch equation instead of Ka expressions for polyprotic acids
CBecause Ka₁ >> Ka₂ >> Ka₃, the ionizations can be treated as sequential and independent; solving only Ka₁ with an ICE table gives pH accurate enough that subsequent ionizations contribute negligible H⁺
DOnly Ka₃ matters because it governs the final equilibrium state of the system
The enormous difference between successive Ka values (roughly 10⁵-fold for H₃PO₄) means each ionization essentially completes before the next begins. The H⁺ produced by the first step suppresses subsequent ionizations via Le Chatelier's principle, and Ka₂ and Ka₃ are already tiny. Treating all three as simultaneous creates a needlessly complex system; sequential treatment gives essentially the same answer. The simplification is justified by the physics, not just mathematical convenience.
Question 2 Multiple Choice
For a solution of H₂CO₃ (Ka₁ = 4.3 × 10⁻⁷, Ka₂ = 4.7 × 10⁻¹¹), which statement best describes the concentration of CO₃²⁻ at equilibrium?
A[CO₃²⁻] ≈ √(Ka₁ × C), where C is the initial H₂CO₃ concentration
B[CO₃²⁻] ≈ Ka₁, because the first ionization dominates the equilibrium
C[CO₃²⁻] ≈ Ka₂ ≈ 4.7 × 10⁻¹¹ M, nearly independent of the initial acid concentration
D[CO₃²⁻] cannot be estimated without solving all equilibria simultaneously
This counterintuitive result follows directly from the algebra of sequential equilibria. After solving Ka₁, [H⁺] ≈ [HCO₃⁻]. Substituting into the Ka₂ expression: Ka₂ = [H⁺][CO₃²⁻]/[HCO₃⁻] ≈ [H⁺][CO₃²⁻]/[H⁺] = [CO₃²⁻]. So [CO₃²⁻] ≈ Ka₂, regardless of the initial concentration. This elegant simplification is a direct consequence of Ka₁ >> Ka₂.
Question 3 True / False
For most diprotic and triprotic weak acids, the pH of the solution is determined almost entirely by the first ionization because Ka₁ >> Ka₂.
TTrue
FFalse
Answer: True
When Ka₁ >> Ka₂, the H⁺ already present from the first ionization suppresses the second ionization (Le Chatelier's principle), and Ka₂ itself is so small that even if it proceeded fully, its H⁺ contribution would be negligible. Solving the Ka₁ ICE table and verifying that Ka₂'s contribution is small (often ≈ Ka₂ itself) is the standard procedure, and the approximation holds well for most polyprotic acids.
Question 4 True / False
Sulfuric acid can be treated like other diprotic acids — solve Ka₁ first, then check if Ka₂ contributes negligibly to the final pH.
TTrue
FFalse
Answer: False
H₂SO₄ is a special case because its first ionization is strong (complete dissociation), not weak. Ka₁ is effectively infinite, so the first proton fully dissociates. The second proton has Ka₂ = 1.2 × 10⁻², which is not negligible — especially in dilute solutions where the bisulfate ion's additional dissociation significantly affects [H⁺]. You must explicitly account for Ka₂ in H₂SO₄ calculations, unlike with H₃PO₄ or H₂CO₃.
Question 5 Short Answer
Why is the concentration of the doubly-deprotonated species in a polyprotic acid solution approximately equal to Ka₂, and why is this result independent of the initial acid concentration?
Think about your answer, then reveal below.
Model answer: After solving the first ionization, [H⁺] ≈ [HA⁻]. Substituting these into the Ka₂ expression — Ka₂ = [H⁺][A²⁻]/[HA⁻] — gives Ka₂ ≈ [H⁺][A²⁻]/[H⁺] = [A²⁻]. Since the initial concentration cancels out of the ratio, [A²⁻] ≈ Ka₂ regardless of how concentrated the original acid was.
This result is one of the most useful shortcuts in polyprotic acid chemistry. It holds because the first ionization sets [H⁺] ≈ [HA⁻], creating a 1:1 ratio in the denominator of Ka₂ that exactly cancels [H⁺] from the numerator. The independence from initial concentration is counterintuitive but follows cleanly from the algebra — the concentration terms divide out.