Affinity

Lessons · Equilibrium & acid–base

The pH of a weak acid: acetic acid

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

Dissolve enough acetic acid — the acid in vinegar — to make the solution 0.100 M0.100\,\text{M}. Acetic acid is a weak acid: put it in water and only a small fraction of its molecules hand a proton to a water molecule, so the ionization HC2H3O2⇌H++C2H3O2−\mathrm{HC_2H_3O_2} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2H_3O_2^-} settles at an equilibrium short of completion. What is the pH? The move is the ICE table — the very same species ledger as any reaction, initial → change → equilibrium — but instead of running to a limiting reagent, the extent stops where the forward and reverse rates balance: where the reaction quotient equals KaK_a.

HC2H3O2⇌H++C2H3O2−\mathrm{HC_{2}H_{3}O_{2}} \rightleftharpoons \mathrm{H}^{+} + \mathrm{C_{2}H_{3}O_{2}}^{-}
Ka=[H+][C2H3O2−][HC2H3O2]K_a = \dfrac{[\mathrm{H}^{+}][\mathrm{C_{2}H_{3}O_{2}}^{-}]}{[\mathrm{HC_{2}H_{3}O_{2}}]}=1.8×10⁻⁵at 25 °C
SpeciesInitial (M)Change (M)Equilibrium (M)
HC2H3O2\mathrm{HC_{2}H_{3}O_{2}} (aq)0.100−x0.0987
H+\mathrm{H}^{+} (aq)0+x0.00133
C2H3O2−\mathrm{C_{2}H_{3}O_{2}}^{-} (aq)0+x0.00133

machine-checkedThe Change row is ν·x — the very same extent ledger ci = ci,0 + νi·xas any reaction. What differs is only where the extent stops.

Mass actionx = 0.00133 M — the extent where Q = Ka.

Ka=[H+][C2H3O2−][HC2H3O2]K_a = \dfrac{[\mathrm{H}^{+}][\mathrm{C_{2}H_{3}O_{2}}^{-}]}{[\mathrm{HC_{2}H_{3}O_{2}}]}=(0.00133)(0.00133) / (0.0987)=1.8×10⁻⁵= Ka ✓

Put the committed equilibrium concentrations back in and the quotient reproduces Ka to within 3.3×10⁻¹² — the solver found the extent, and an independent check re-solves it and agrees.

[H+]0.00133 M1.33% of the acid ionized
pH2.88= −log10[H+]
un-ionized0.0987 Mstays as intact HC₂H₃O₂
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 = Ka [mass-action root]
  • ✓ 0 < x < [HC₂H₃O₂]₀ — no concentration goes negative [extent physical]
  • ✓ pH = −log₁₀[H⁺] [log consistent]
✗ Common misconception: “The acid is 0.100 M, so [H+] = 0.100 M and pH = -log(0.100) = 1.00.”

That would be true for a strong acid, which ionizes completely. But the ledger shows the extent is only x = 0.00133 M — just 1.33%of the acid ionizes. So [H+] = 0.00133 M and pH = 2.88, not 1.00; the other 0.0987 M (98.67%) stays as intact HC₂H₃O₂ molecules. “Weak” means the equilibrium barely moves off the left.

Modeling assumptions — author-asserted, disclosed not discharged
  • model The solution has reached equilibrium, and this single ionization is the only reaction that matters (one dominant equilibrium).
  • model Activities are approximated by molar concentrations — the ideal-dilute-solution model, which is why KaK_a is written with concentrations.
  • model The H+\mathrm{H^+} from water's own autoionization (10−7 M10^{-7}\,\text{M}) is negligible beside the acid's, so [H+]0=0[\mathrm{H^+}]_0 = 0 in the table.

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: