Lessons · Equilibrium & acid–base

The pH of a weak base: ammonia

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

Dissolve enough ammonia to make the solution $0.100\,\text{M}$ . Ammonia is a weak base: it has no $\mathrm{OH^-}$ of its own — instead it pulls a proton off a water molecule, $\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$ , and only a small fraction reacts, so the equilibrium sits far to the left. What is the pH? It is the ICE table once more — the same species ledger, initial → change → equilibrium — but with two new twists. First, water is the pure solvent: its activity is 1, so it never enters the equilibrium expression $K_b = [\mathrm{NH_4^+}][\mathrm{OH^-}]/[\mathrm{NH_3}]$ — exactly as a pure solid drops out of a $K_{sp}$ . Second, the extent gives $[\mathrm{OH^-}]$ , not $[\mathrm{H^+}]$ ; the bridge to pH is the ion-product of water, $[\mathrm{H^+}][\mathrm{OH^-}] = K_w = 1.0\times10^{-14}$ , which is why $\mathrm{pH} + \mathrm{pOH} = 14.00$ .

\mathrm{NH_{3}}\,\text{(aq)} + \mathrm{H_{2}O}\,\text{(l)} \rightleftharpoons \mathrm{NH_{4}}^{+}\,\text{(aq)} + \mathrm{OH}^{-}\,\text{(aq)}

K_b = \dfrac{[\mathrm{NH_{4}}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH_{3}}]}

=1.8×10⁻⁵at 25 °C

Species	Initial (M)	Change (M)	Equilibrium (M)
$\mathrm{NH_{3}}$ (aq)	0.100	−x	0.0987
$\mathrm{H_{2}O}$ (l)pure liquid · excluded from Q	—	—	—
$\mathrm{NH_{4}}^{+}$ (aq)	0	+x	0.00133
$\mathrm{OH}^{-}$ (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. Water is the pure solvent (activity 1), so it never enters the equilibrium expression.

Mass actionx = 0.00133 M = [OH⁻], the extent where Q = K_b.

$K_b = \dfrac{[\mathrm{NH_{4}}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH_{3}}]}$ =(0.00133)(0.00133) / (0.0987)=1.8×10⁻⁵= K_b ✓

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

[OH⁻]0.00133 M1.33% of the base ionized

pOH2.88= −log₁₀[OH⁻]

pH11.12= 14.00 − pOH

K_w bridgeThe extent gives [OH⁻] = 0.00133 M, not [H⁺]. Water ties them together:[H⁺][OH⁻] = K_w = 1×10⁻¹⁴, so[H⁺] = K_w/[OH⁻] = 0.0000000000075 M andpH + pOH = 11.12 + 2.88 = 14.00.

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_b [mass-action root]
✓ Water is excluded from Q; 0 < x < [NH₃]₀ — no concentration goes negative [extent physical]
✓ [H⁺] = K_w/[OH⁻]; pH = −log₁₀[H⁺]; pH + pOH = pK_w [K_w bridge]

✗ Common misconception: “The base is 0.100 M, so [OH-] = 0.100 M, pOH = -log(0.100) = 1.00, and pH = 14.00 - 1.00 = 13.00.”

That would be true for a strong base like NaOH, which delivers its OH⁻ completely. But ammonia has no OH⁻ of its own — it must take one from water, and the ledger shows the extent is only x = 0.00133 M, just 1.33% ionized. So [OH⁻] = 0.00133 M, pOH = 2.88, and through K_w the pH = 11.12 — not 13.00; the other 0.0987 M (98.67%) stays as intact NH₃. Notice the mirror: this base at 0.100 M lands as far above pH 7 as 0.100 M acetic acid (same K) lands below it.

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 $K_b$ and $K_w$ are written with concentrations.
model The $\mathrm{OH^-}$ from water's own autoionization ( $10^{-7}\,\text{M}$ ) is negligible beside the base's, so $[\mathrm{OH^-}]_0 = 0$ in the table; $K_w = 1.0\times10^{-14}$ is the 25 °C value.

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⁺] →