Affinity

Lessons · Equilibrium & acid–base

One acid, three protons: the pH of phosphoric acid

ICE ledger machine-checked — every dissolved row is initial + ν·x, one extent xKa1 data-sourced (openstax-chemistry-2e)3 modeling assumptions (disclosed)

Phosphoric acid, H3PO4\mathrm{H_3PO_4}, has three acidic protons — it is triprotic. But it does not give them up all at once: it ionizes in stages, each with its own equilibrium constant, and each KaK_a is about 10510^5 smaller than the last (Ka1=7.5×10−3K_{a1} = 7.5\times10^{-3}, Ka2=6.2×10−8K_{a2} = 6.2\times10^{-8}, Ka3=4.2×10−13K_{a3} = 4.2\times10^{-13}). So the first ionization is the only one that releases an appreciable amount of H+\mathrm{H^+} — and it is just the weak-acid ICE table again. Each later stage is the same solve, run on the concentrations the stage before it left behind.

H3PO4⇌H++H2PO4−\mathrm{H_{3}PO_{4}} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H_{2}PO_{4}}^{-}
Ka1=[H+][H2PO4−][H3PO4]K_{a1} = \dfrac{[\mathrm{H}^{+}][\mathrm{H_{2}PO_{4}}^{-}]}{[\mathrm{H_{3}PO_{4}}]}=7.5×10⁻³at 25 °C
SpeciesInitial (M)Change (M)Equilibrium (M)
H3PO4\mathrm{H_{3}PO_{4}} (aq)0.1−x0.0761
H+\mathrm{H}^{+} (aq)0+x0.0239
H2PO4−\mathrm{H_{2}PO_{4}}^{-} (aq)0+x0.0239

machine-checkedThe Change row is ν·x — the very same extent ledger ci = ci,0 + νi·xas any reaction. This is the first of 3 ionizations — the one that releases nearly all of the H⁺.

Mass actionx = 0.0239 M — the extent where Q = Ka1.

Ka1=[H+][H2PO4−][H3PO4]K_{a1} = \dfrac{[\mathrm{H}^{+}][\mathrm{H_{2}PO_{4}}^{-}]}{[\mathrm{H_{3}PO_{4}}]}=(0.0239)(0.0239) / (0.0761)=7.5×10⁻³= Ka1 ✓

Put the committed equilibrium concentrations back in and the quotient reproduces Ka1 to within 2.9×10⁻¹² — the solver found the extent, and an independent check re-solves it and agrees.

pH1.62set almost entirely by the first ionization
[H+]0.0239 M23.9% of the 1st proton ionizes
ionizes in3 stagesKa1 ≫ Ka2 ≫ Ka3
The later ionizationsEach is the same ICE solve — run on what the stage before it left behind.
StageEquilibriumKaextent xproduct formed
2H₂PO₄⁻ ⇌ H+ + HPO₄²⁻6.2×10⁻⁸6.2×10⁻⁸[HPO₄²⁻] = 6.2×10⁻⁸
3HPO₄²⁻ ⇌ H+ + PO₄³⁻4.2×10⁻¹³1.09×10⁻¹⁸[PO₄³⁻] = 1.09×10⁻¹⁸

The payoffThe first ionization gives [H+] = 0.0239 M (23.9%); each later stage adds a vanishing amount. And because [H+] ≈ [H₂PO₄⁻] after stage 1, the second stage's Ka collapses to just its product: [HPO₄²⁻] = 6.2×10⁻⁸ M ≈ Ka2 = 6.2×10⁻⁸. The fully-stripped PO₄³⁻ is essentially absent (1.09×10⁻¹⁸ M).

Every species at equilibrium:[H₃PO₄] = 7.61×10⁻² M · [H₂PO₄⁻] = 2.39×10⁻² M · [HPO₄²⁻] = 6.2×10⁻⁸ M · [PO₄³⁻] = 1.09×10⁻¹⁸ M
VerificationProven at build time — not asserted.
  • ✓ Every equilibrium concentration = initial + ν·x [ICE identity]
  • ✓ The extent x re-solved independently — numerically, to high precision — reproduces Q = Ka1 [mass-action root]
  • ✓ 0 < x < [H₃PO₄]₀ — no concentration goes negative [extent physical]
  • ✓ pH = −log₁₀[H⁺] [log consistent]
✗ Common misconception: “Phosphoric acid has three ionizable protons, so all three ionize substantially and [H+] is about three times what a monoprotic acid would give.”

Only the first proton comes off appreciably. Because each Ka is about 105smaller than the last, the second ionization barely proceeds (x2 = 6.2×10⁻⁸ M) and the third is vanishing (PO₄³⁻ ≈1.09×10⁻¹⁸ M). So [H+] = 0.0239 Mcomes almost entirely from stage 1 — the extra protons add nothing measurable, and pH = 1.62, not one-third lower. A triprotic acid is not “three times” a monoprotic one; it is one moderately weak acid followed by two that essentially do not ionize.

Modeling assumptions — author-asserted, disclosed not discharged
  • model The three ionizations are solved in succession — each stage on the equilibrium concentrations the previous stage left — rather than simultaneously. This is exact in the limit of well-separated KaK_a (here each is ∼105\sim10^5 apart), the standard treatment.
  • model Water's own autoionization is neglected: [H+]0=0[\mathrm{H^+}]_0 = 0 before the acid ionizes (valid because the acid's [H+][\mathrm{H^+}] is far above 10−710^{-7} M).
  • model Activities are approximated by molar concentrations — the ideal-dilute-solution model, which is why each KaK_a is written with concentrations.

Concepts in this lesson

Linked into the Chemical Atlas where an entry exists; the rest fill in as the Atlas grows.

Practice this

The lesson goes deep on one scenario; the gym builds fluency by repetition. Drill these: