Lessons · Equilibrium & acid–base

A buffer resists change: acetic acid + acetate

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

Dissolve $0.100\,\text{M}$ acetic acid and $0.100\,\text{M}$ sodium acetate in the same beaker. The acetate ion $\mathrm{C_2H_3O_2^-}$ from the salt is the acid's own conjugate base — so the solution now holds both halves of the equilibrium $\mathrm{HC_2H_3O_2} \rightleftharpoons \mathrm{H^+} + \mathrm{C_2H_3O_2^-}$ from the start. This is a buffer. It is still the ICE table — the same species ledger, the same solver — but the change column starts from $[\mathrm{C_2H_3O_2^-}]_0 = 0.100\,\text{M}$ , not zero. That pre-loaded product is a common ion: by Le Chatelier it pushes the ionization left, so hardly any acid ionizes and the pH lands near $\mathrm{p}K_a$ . What is it?

\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)	0.100	−x	0.1
$\mathrm{H}^{+}$ (aq)	0	+x	0.000018
$\mathrm{C_{2}H_{3}O_{2}}^{-}$ (aq)	0.100	+x	0.1

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

Mass actionx = 0.000018 M — the extent where Q = K_a.

$K_a = \dfrac{[\mathrm{H}^{+}][\mathrm{C_{2}H_{3}O_{2}}^{-}]}{[\mathrm{HC_{2}H_{3}O_{2}}]}$ =(0.000018)(0.1) / (0.1)=1.8×10⁻⁵= K_a ✓

Put the committed equilibrium concentrations back in and the quotient reproduces K_a to within 4.5×10⁻¹² — the solver found the extent, and an independent check re-solves it and agrees.

pH4.74≈ pK_a = 4.74

buffer ratio [A⁻]/[HA]1at equilibrium

acid ionized0.018%[H⁺] = 0.000018 M

Henderson–HasselbalchTake −log of K_a = [H⁺][A⁻]/[HA] and it rearranges topH = pK_a + log₁₀([A⁻]/[HA]) = 4.74 + log₁₀(1) = 4.74— the same value as −log₁₀[H⁺]. With equal amounts of acid and base the ratio is 1 andpH = pK_a. Henderson–Hasselbalch is nothing but mass action, logged.

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]
✓ pH = pK_a + log₁₀([A⁻]/[HA]) = −log₁₀[H⁺] [Henderson–Hasselbalch]

✗ Common misconception: “The added acetate is just a spectator salt, so the pH is the same as 0.100 M acetic acid on its own: pH = 2.88.”

The acetate is not a spectator here — it is the acid's conjugate base, a common ion. Le Chatelier: the pre-loaded C₂H₃O₂⁻ pushes $\mathrm{HC_{2}H_{3}O_{2}} \rightleftharpoons \mathrm{H}^{+} + \mathrm{C_{2}H_{3}O_{2}}^{-}$ to the left, so far less acid ionizes. The ledger shows the extent is only x = 0.000018 M (0.018% ionized) — about 74.1× smaller than the 0.00133 M the acid reaches alone. So [H⁺] is that much lower and pH = 4.74, not 2.88. That is what a buffer does: the reservoir of conjugate base holds the pH near pK_a.

Modeling assumptions — author-asserted, disclosed not discharged

model The sodium acetate dissociates completely, so $[\mathrm{C_2H_3O_2^-}]_0 = 0.100\,\text{M}$ before the acid equilibrium adjusts; the $\mathrm{Na^+}$ is a spectator.
model This single ionization is the only reaction that matters, and activities are approximated by molar concentrations (so $K_a$ and Henderson–Hasselbalch use concentrations).
model The $\mathrm{H^+}$ from water's own autoionization is negligible, so $[\mathrm{H^+}]_0 = 0$ 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:

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