AP Chemistry · Handsworth Secondary 2026–27

Unit 8 · Acids & Bases

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 8, 11–15% of the AP exam (one of the heaviest units), taught in 9 blocks. BC Chemistry 12 gave you pH, pOH, Kw, and strong-vs-weak; AP pushes into ICE-chart weak-acid pH, Ka·Kb=Kw, buffers & Henderson–Hasselbalch, and reading titration curves. Equilibrium (Unit 7) is the engine here, every acid–base problem is a K problem. The titration lab (CED Investigation 14) lives in this unit.

Blocks:  1 → A–B  ·  2 → C  ·  3 → D  ·  4 → E  ·  5 → F (lab)  ·  6 → G  ·  7 → H  ·  8 → I–J  ·  9 → K + review

Name:Block:Date:

A · Brønsted–Lowry & Conjugate Pairs CED 8.1 · LO 8.1.A

A Brønsted–Lowry acid donates a proton (H⁺); a base accepts one. Every acid–base reaction is a proton hand-off, so each reactant has a partner on the product side that differs by exactly one H⁺, a conjugate pair.

Core: acid → its conjugate base + H⁺. A conjugate pair differs by one proton (e.g. HF / F⁻, NH₄⁺ / NH₃). Water is amphiprotic: it acts as acid or base depending on its partner.
Autoionization: 2H₂O ⇌ H₃O⁺ + OH⁻ with K_w = [H₃O⁺][OH⁻] = 1.0×10⁻¹⁴ at 25 °C. Kw is temperature-dependent, neutral pH only equals 7.00 at 25 °C.
Worked example
In the reaction HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻, label each species and name the two conjugate pairs.
HCO₃⁻ accepts a proton → it is the base; H₂O donates → it is the acid.
HCO₃⁻ (base) gains H⁺ to become H₂CO₃ (its conjugate acid); H₂O (acid) loses H⁺ to become OH⁻ (its conjugate base).
Pairs: HCO₃⁻ / H₂CO₃  ·  H₂O / OH⁻

Practice A

  1. Write the conjugate base of each: HNO₂, H₂PO₄⁻, NH₄⁺.
  2. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, identify the acid, base, and both conjugate pairs.
  3. Explain why H₂O is called amphiprotic, giving one reaction where it is the acid and one where it is the base.

B · pH, pOH & Strong Acids/Bases CED 8.1–8.2 · LO 8.1.A, 8.2.A

pH and pOH are just log scales for the ion concentrations. Because a strong acid or base ionizes completely, you can read [H₃O⁺] or [OH⁻] straight off the formula, no equilibrium math.

Core: pH = −log[H₃O⁺] · pOH = −log[OH⁻] · pH + pOH = 14.00 (25 °C). Strong acid: [H₃O⁺] = initial acid conc. Strong base: [OH⁻] = initial conc (group I) or 2× initial conc (group II hydroxide).
Worked example
Find the pH of 0.010 M Ca(OH)₂.
Ca(OH)₂ is a group II strong base → [OH⁻] = 2 × 0.010 = 0.020 M
pOH = −log(0.020) = 1.70
pH = 14.00 − 1.70 = 12.30

Not tested: the strong-acid list to memorize is short, HCl, HBr, HI, HClO₄, H₂SO₄, HNO₃. Anything else you meet on the exam is weak unless told otherwise.

Practice B

  1. Find the pH of 0.0025 M HCl.
  2. Find the pH of 0.025 M KOH.
  3. A solution has [OH⁻] = 2.5×10⁻³ M. Find the pOH and pH, and state whether it is acidic or basic.

C · Weak Acids/Bases: Ka, Kb & ICE CED 8.3 · LO 8.3.A

A weak acid only partly ionizes, so [H₃O⁺] is much smaller than the acid concentration and you must solve an equilibrium. The acid-ionization constant Ka tells you how far it goes; an ICE table gets you the pH.

Core: for HA + H₂O ⇌ H₃O⁺ + A⁻,  K_a = [H₃O⁺][A⁻]/[HA] · pK_a = −log K_a.  Smaller Ka (larger pKa) = weaker acid. Percent ionization = ([H₃O⁺]/[HA]₀)×100%.
Worked example
Find the pH and percent ionization of 0.100 M acetic acid, Ka = 1.8×10⁻⁵.
ICE: [HA] 0.100→0.100−x · [H₃O⁺] 0→x · [A⁻] 0→x
Ka = x²/(0.100−x). Test the small-x approximation (Ka is tiny): x² ≈ (1.8×10⁻⁵)(0.100) = 1.8×10⁻⁶
x = [H₃O⁺] = 1.34×10⁻³ M → check: 1.34×10⁻³/0.100 = 1.3% < 5% ✓ approximation valid
pH = −log(1.34×10⁻³) = 2.87
pH = 2.87  ·  percent ionization = 1.3%

Not tested: you are not required to solve the full quadratic when the 5% rule passes; but if percent ionization comes out > 5% (dilute or stronger weak acids), drop the approximation and use the quadratic.

Practice C

  1. Find the pH of 0.20 M HClO, Ka = 4.0×10⁻⁷.
  2. Find the pH of 0.100 M NH₃, Kb = 1.8×10⁻⁵ (a weak base, solve for [OH⁻], then pOH → pH).
  3. A 0.250 M solution of HF (Ka = 6.8×10⁻⁴) is 5.1% ionized. Find [H₃O⁺] and the pH. (Check the 5% rule, does the approximation hold?)

D · Ka · Kb = Kw for a Conjugate Pair CED 8.3 · LO 8.3.A

A strong acid has a weak conjugate base, and vice versa. That trade-off is exact: for any conjugate acid–base pair, the two ionization constants multiply to Kw. This lets you get Kb of a base from the Ka of its conjugate acid (and back).

Core: K_a × K_b = K_w = 1.0×10⁻¹⁴ · equivalently pK_a + pK_b = 14.00 (25 °C). The Ka and Kb here belong to the same conjugate pair (e.g. HF and F⁻).
Worked example
Acetic acid has Ka = 1.8×10⁻⁵. Find Kb of acetate, CH₃COO⁻.
Kb = Kw/Ka = (1.0×10⁻¹⁴)/(1.8×10⁻⁵)
Kb(acetate) = 5.6×10⁻¹⁰  (very weak base, consistent with a moderately weak acid)

Practice D

  1. HF has Ka = 6.8×10⁻⁴. Find Kb of F⁻.
  2. NH₃ has Kb = 1.8×10⁻⁵. Find Ka of NH₄⁺ and its pKa.
  3. Acid HX has pKa = 9.30 and acid HY has pKa = 3.20. Which conjugate base is stronger? Explain using pKa + pKb = 14.

E · Acid–Base Reactions & How Buffers Work CED 8.4, 8.8 · LO 8.4.A, 8.8.A

Mix a strong acid and strong base and they react quantitatively to water; the pH comes from whatever is in excess. Mix a weak acid with a strong base and stop part-way and you build a buffer: a reservoir of both members of a conjugate pair that resists pH change.

Strong + strong: H⁺ + OH⁻ → H₂O. Find moles of each, subtract, divide leftover by total volume, take pH/pOH.
Buffer: contains large amounts of both HA and A⁻. Added base is mopped up by HA; added acid is mopped up by A⁻, so [H₃O⁺] barely moves. That is why blood (HCO₃⁻/H₂CO₃) holds pH ≈ 7.4.
Worked example
Mix 50.0 mL of 0.100 M HCl with 30.0 mL of 0.100 M NaOH. Find the pH.
mol H⁺ = 0.0500 L × 0.100 = 5.00×10⁻³ · mol OH⁻ = 0.0300 L × 0.100 = 3.00×10⁻³
H⁺ in excess: 5.00×10⁻³ − 3.00×10⁻³ = 2.00×10⁻³ mol, in 0.0800 L total
[H₃O⁺] = 2.00×10⁻³ / 0.0800 = 0.0250 M
pH = −log(0.0250) = 1.60

Practice E

  1. Mix 40.0 mL of 0.200 M NaOH with 40.0 mL of 0.100 M HNO₃. Find the pH.
  2. Which pair makes a buffer: (a) HCl + NaCl, (b) CH₃COOH + CH₃COONa, (c) NaOH + NaCl? Explain.
  3. Explain at the particulate level what happens when a few drops of strong acid are added to an acetic-acid/acetate buffer.

F · Henderson–Hasselbalch & Buffer pH CED 8.9 · LO 8.9.A

Once you accept that a buffer's ratio of conjugate base to acid barely changes, the equilibrium expression rearranges into one clean line: the buffer's pH is its pKa nudged by the log of that ratio.

Core: pH = pK_a + log([A⁻]/[HA]). When [A⁻] = [HA], the log is 0 and pH = pKa: the most stable buffer. Because it's a ratio, you can use moles instead of molarities (same volume cancels).
Worked example
A buffer is 0.20 M acetic acid and 0.30 M sodium acetate. Ka(acetic) = 1.8×10⁻⁵. Find the pH.
pKa = −log(1.8×10⁻⁵) = 4.74
pH = 4.74 + log(0.30/0.20) = 4.74 + log(1.5) = 4.74 + 0.18
pH = 4.92

Not tested: deriving the Henderson–Hasselbalch equation, and computing the pH change when acid or base is added to a buffer, are both excluded by the CED. Know how to use the equation and reason qualitatively about the shift.

Practice F

  1. A buffer is 0.50 M HF and 0.50 M NaF (Ka = 6.8×10⁻⁴). Find the pH.
  2. A buffer is 0.40 M NH₄Cl and 0.20 M NH₃ (Ka of NH₄⁺ = 5.6×10⁻¹⁰). Find the pH.
  3. You want a buffer at pH = 4.74 using acetic acid (pKa 4.74). What ratio [A⁻]/[HA] do you need, and why?

G · Titration Curves: Equivalence & Half-Equivalence CED 8.5 · LO 8.5.A

A titration curve plots pH against volume of titrant. Its shape tells you the story. This is the unit's lab skill (CED Investigation 14) and a recurring FRQ, examiners love the equivalence vs. half-equivalence vs. endpoint distinction.

Equivalence point: moles titrant = moles analyte. For strong–strong, pH = 7.00 there. For weak acid–strong base, the conjugate base is left over, so the equivalence pH is > 7 (basic).
Half-equivalence point (halfway to equivalence): [HA] = [A⁻], so by Henderson–Hasselbalch pH = pK_a. Read pKa straight off the curve here. This region (around half-equiv) is the buffer region.
Worked example · ties to the titration lab
25.0 mL of 0.100 M acetic acid (Ka = 1.8×10⁻⁵) is titrated with 0.100 M NaOH. Find the pH at the equivalence point.
mol acid = 0.0250 L × 0.100 = 2.50×10⁻³; equal mol NaOH needs 25.0 mL → total volume 50.0 mL
All acid is converted to acetate: [CH₃COO⁻] = 2.50×10⁻³ / 0.0500 = 0.0500 M
Acetate is a weak base: Kb = Kw/Ka = 5.6×10⁻¹⁰; x = √(Kb·C) = √(5.6×10⁻¹⁰ × 0.0500) = 5.3×10⁻⁶ M = [OH⁻]
pOH = −log(5.3×10⁻⁶) = 5.28
pH = 14.00 − 5.28 = 8.72  (basic, as expected for a weak acid–strong base equivalence point)

Practice G

  1. On a weak acid–strong base curve, the pH at the half-equivalence point is 4.20. What is the acid's pKa and Ka?
  2. Why is the equivalence-point pH above 7 for a weak acid titrated with strong base, but exactly 7 for a strong acid–strong base titration?
  3. An indicator should change colour near the equivalence pH. For the acetic-acid titration above (equiv pH ≈ 8.7), would phenolphthalein (range 8.2–10) or methyl red (range 4.4–6.2) be the better choice? Explain.

H · Polyprotic Acids CED 8.5 · LO 8.5.A

A polyprotic acid donates more than one proton, each with its own Ka (Ka1 ≫ Ka2 ≫ …). Its titration curve shows one step per acidic proton, so the number of equivalence points counts the protons.

Core: H₂A → HA⁻ → A²⁻, with Ka1 > Ka2 (each successive proton is harder to remove from a more-negative ion). Two acidic protons → two equivalence points → two half-equivalence points (pH = pKa1 and pH = pKa2).
Worked example
A diprotic acid H₂A is titrated with NaOH. Its curve has two steps; the first half-equivalence point sits at pH 2.9 and the second at pH 7.2. Find pKa1 and pKa2.
At each half-equivalence point, pH = pKa for the proton being removed there.
pKa1 = 2.9  ·  pKa2 = 7.2  (Ka1 ≈ 1×10⁻³ ≫ Ka2 ≈ 6×10⁻⁸, as expected)

Not tested: computing the concentration of each species along a polyprotic titration curve is excluded by the CED. You are responsible for identifying which species dominate (large vs. small) at a point, and reading each pKanot full per-species math.

Practice H

  1. How many equivalence points appear on the titration curve of phosphoric acid, H₃PO₄? Why?
  2. For H₂CO₃ (Ka1 = 4.3×10⁻⁷, Ka2 = 4.7×10⁻¹¹), which proton comes off more easily, and what is pKa1?
  3. At a point between the first and second equivalence points of a diprotic acid, which species (H₂A, HA⁻, or A²⁻) is present in the largest amount?

I · Salt Hydrolysis: Acidic, Basic, or Neutral CED 8.4–8.5 · LO 8.4.A, 8.5.A

Dissolving a salt can change the pH. Decide by looking at each ion: is it the conjugate of a weak acid or base (it will react with water), or of a strong one (spectator, no effect)?

Rules: cation from a strong base (Na⁺, K⁺, Ca²⁺) = spectator. Anion from a strong acid (Cl⁻, NO₃⁻) = spectator. Anion of a weak acid (e.g. CH₃COO⁻, F⁻) is a weak base → basic salt. Cation that is a conjugate acid of a weak base (e.g. NH₄⁺) → acidic salt.
Worked example
Find the pH of 0.10 M NH₄Cl. Kb(NH₃) = 1.8×10⁻⁵.
Cl⁻ is a spectator (conjugate of strong HCl). NH₄⁺ is the conjugate acid of weak NH₃ → acidic salt.
Ka(NH₄⁺) = Kw/Kb = (1.0×10⁻¹⁴)/(1.8×10⁻⁵) = 5.6×10⁻¹⁰
x = [H₃O⁺] = √(Ka·C) = √(5.6×10⁻¹⁰ × 0.10) = 7.5×10⁻⁶ M
pH = −log(7.5×10⁻⁶) = 5.13  (acidic, as predicted)

Practice I

  1. Classify each as acidic, basic, or neutral: NaCl, KF, NH₄NO₃, NaCH₃COO.
  2. Find the pH of 0.0500 M sodium acetate (Ka of acetic acid = 1.8×10⁻⁵).
  3. Explain why NaCl dissolves to give a neutral solution but NaF gives a basic one.

J · Structure ↔ Strength & pH vs pKa CED 8.6–8.7 · LO 8.6.A, 8.7.A

Acid strength is a structure story: the more stable the conjugate base, the stronger the acid. And comparing solution pH to an acid's pKa tells you which form (protonated or deprotonated) predominates, the key idea behind choosing a titration indicator.

Strength rules: more electronegative / more electron-withdrawing groups stabilize the conjugate base → stronger acid. Resonance delocalization (carboxylic acids) also stabilizes it. Carboxylic acids = common weak acids; amines/ammonia & carboxylate ions = common weak bases.
pH vs pKa: when pH < pK_a the acid form (HA) dominates; when pH > pK_a the base form (A⁻) dominates; at pH = pKa they're equal. Pick an indicator whose pKa ≈ the equivalence pH.
Worked example
Acetic acid has pKa = 4.74. In a solution buffered at pH 6.0, which form predominates, CH₃COOH or CH₃COO⁻?
pH 6.0 > pKa 4.74, so the base form is favoured.
CH₃COO⁻ (the deprotonated, base form) predominates.

Practice J

  1. Trichloroacetic acid (CCl₃COOH) is much stronger than acetic acid (CH₃COOH). Explain using conjugate-base stability.
  2. A weak acid has pKa = 3.5. At pH 2.0, which form (HA or A⁻) predominates?
  3. An indicator has pKa = 5.0. Is it a good choice for an equivalence point near pH 9? Why or why not?

K · Buffer Capacity & pH-Dependent Solubility CED 8.10–8.11 · LO 8.10.A, 8.11.A

Two closing ideas. Buffer capacity is how much acid or base a buffer can absorb before it breaks. And the solubility of certain salts depends on pH, a Le Châtelier consequence when one ion is a weak acid/base or hydroxide.

Capacity: raising the concentration of both buffer components (same ratio) keeps the pH but increases capacity. A buffer with more HA than A⁻ resists added base better; more A⁻ than HA resists added acid better.
pH & solubility: if a salt's anion is a weak base (F⁻, CO₃²⁻, OH⁻), adding H₃O⁺ removes it, shifting dissolution forward → more soluble in acid. Salts of strong-acid anions (Cl⁻, NO₃⁻) are pH-insensitive.
Worked example
Is CaF₂ more soluble in pure water or in acidic solution? Explain with Le Châtelier.
CaF₂ ⇌ Ca²⁺ + 2F⁻. F⁻ is the conjugate base of weak HF, so added H₃O⁺ converts F⁻ → HF.
Removing F⁻ pulls the dissolution equilibrium to the right.
CaF₂ is more soluble in acidic solution.

Not tested: actually computing solubility as a function of pH is excluded by the CED, reason qualitatively with Le Châtelier only.

Practice K

  1. Buffer X is 0.10 M HA / 0.10 M A⁻; buffer Y is 1.0 M HA / 1.0 M A⁻. Same pH? Same capacity? Explain.
  2. Will Mg(OH)₂ dissolve more readily in acidic or basic solution? Explain.
  3. A buffer has [A⁻] much greater than [HA]. Does it resist added acid or added base more effectively?
AP Chemistry · Unit 8, Acids & Bases · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 8.1–8.11, LOs 8.1.A–8.11.A) and BC Chemistry 12. Titration lab = CED Investigation 14. Exclusion statements drawn from the CED. Ka/Kb values are standard 25 °C reference values; verify against the data table provided on your exam.