Lessons · Equilibrium & acid–base

Watching the pH climb: titrating acetic acid with NaOH

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

Add sodium hydroxide, a strong base, drop by drop to 25.0 mL of 0.100 M acetic acid and watch the pH. Each drop neutralizes a little acid, $\mathrm{HC_2H_3O_2 + OH^- \rightarrow C_2H_3O_2^- + H_2O}$ — but the pH at every moment is still set by the acetic-acid equilibrium $\mathrm{HC_2H_3O_2 \rightleftharpoons H^+ + C_2H_3O_2^-}$ . So the whole curve is just the ICE table marched: a weak acid at the start, a buffer through the middle (flattest right where half the acid is neutralized — there $\mathrm{pH} = \mathrm{p}K_a$ ), a jump through the equivalence point (where all the acid has become acetate, a weak base — so the equivalence pH is above 7), and excess strong base after.

\mathrm{HC_{2}H_{3}O_{2}} \rightleftharpoons \mathrm{H}^{+} + \mathrm{C_{2}H_{3}O_{2}}^{-}

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

Species	Initial (M)	Change (M)	Equilibrium (M)
$\mathrm{HC_{2}H_{3}O_{2}}$ (aq)	0.1	−x	0.0987
$\mathrm{H}^{+}$ (aq)	0	+x	0.00133
$\mathrm{C_{2}H_{3}O_{2}}^{-}$ (aq)	0	+x	0.00133

machine-checkedThe Change row is ν·x — the very same extent ledger c_i = c_i,0 + ν_i·xas any reaction. This is the starting solution, before any base is added — the leftmost point of the curve below.

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

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

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

The titration curveEvery point is the ICE re-solved — the ledger marched as NaOH is added.

The base drives the neutralization $\mathrm{HC_{2}H_{3}O_{2}} + \mathrm{OH^-} \rightarrow \mathrm{C_{2}H_{3}O_{2}}^{-} + \mathrm{H_2O}$ , but the pH at every drop is still set by the acid equilibrium above.

startpH 2.88at 0 mL

½-eq · pH = pKₐpH 4.75at 12.5 mL

equivalencepH 8.72at 25 mL

Read the curveThe gentle stretch is the buffer region — flattest at 12.5 mL(half the acid neutralized), where pH ≈ pK_a = 4.74 — the buffer identity, exact under Henderson–Hasselbalch; the exact solve gives 4.75, a hair above pK_abecause a little more acid ionizes (the small gap is just rounding either side of their shared value). The steep jump is the equivalence point at 25 mL: all the acid has become C₂H₃O₂⁻ — a weak base — so the pH there is 8.72, well above 7.

start (0 mL)pH 2.88the pure weak acid

half-equivalencepH 4.75≈ pK_a = 4.74

equivalencepH 8.72basic — the acid is now a weak base

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 = K_a [mass-action root]
✓ 0 < x < [HC₂H₃O₂]₀ — no concentration goes negative [extent physical]
✓ every curve point's pH re-derived by its region; pH ≈ pK_a at half-equivalence; equivalence is basic [curve re-derived]

✗ Common misconception: “At the equivalence point the acid is exactly neutralized, so the pH is 7.”

That holds only for a strong acid + strong base. Here the acid is weak: at equivalence every HC₂H₃O₂ has been converted to its conjugate base C₂H₃O₂⁻, which is itself a weak base — it takes a proton back from water, releasing OH⁻. So the solution is basic: the ledger's weak-base solve gives pH 8.72 at the equivalence point, not 7. A pH of 7 would need the product to be a spectator; C₂H₃O₂⁻ is not one.

Modeling assumptions — author-asserted, disclosed not discharged

model Each region is solved by the dominant equilibrium — the weak acid / buffer before equivalence, the conjugate base's hydrolysis at equivalence, excess $\mathrm{OH^-}$ after — the standard treatment; the strong base is taken as fully dissociated, so the added $\mathrm{OH^-}$ moles are exact.
model Volumes are additive (the total is the acid volume plus the added titrant volume), and activities are approximated by molar concentrations — the ideal-dilute-solution model.

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:

Weak-acid pH: solve the ICE table for [H⁺] →