AP Chemistry · Handsworth Secondary 2026–27
Unit 8 · Study Guide
Dr. Ras Mulinta
Acids & Bases
Exam-focused review
A one-page map of what the Unit 8 test (and the AP exam) expects, a heavy unit at 11–15%. This is a checklist, not a re-teach: if a line doesn't click, go back to that section of the notes package. A periodic table and a Ka/Kb data sheet are provided on the exam.
Must be able to do
- 8.1 Identify Brønsted–Lowry acids/bases & conjugate pairs (differ by one H⁺); use K_w = [H₃O⁺][OH⁻] = 1.0×10⁻¹⁴ and pH + pOH = 14 at 25 °C.
- 8.2 pH/pOH of strong acids/bases by inspection, strong base group II gives [OH⁻] = 2 × conc.
- 8.3 Write Ka/Kb; find weak-acid (or weak-base) pH with an ICE table; compute percent ionization; check the 5% rule.
- 8.3 Relate a conjugate pair with K_a × K_b = K_w and pK_a + pK_b = 14.
- 8.4 / 8.8 Find pH after mixing strong + strong (excess reagent); recognize a buffer & explain how each member neutralizes added acid/base.
- 8.9 Use pH = pK_a + log([A⁻]/[HA]); at [A⁻] = [HA], pH = pKa.
- 8.5 Read titration curves: locate equivalence (mol titrant = mol analyte) & half-equivalence (pH = pKa); strong–strong equiv = pH 7, weak acid–strong base equiv > 7; count protons on polyprotic curves; pick an indicator with pKa ≈ equivalence pH.
- 8.4–8.5 Classify salts as acidic/basic/neutral by ion source and compute hydrolysis pH.
- 8.6–8.7 Link conjugate-base stability to acid strength; compare pH to pKa to decide whether HA or A⁻ predominates.
- 8.10–8.11 Reason about buffer capacity (concentration & ratio) and pH-dependent solubility (Le Châtelier, qualitative only).
The big idea that ties it together
Every acid–base problem is an equilibrium problem. One constant, K_w = K_a × K_b, links acids to their conjugate bases; the same Ka drives weak-acid pH (ICE), buffer pH (Henderson–Hasselbalch, where half-equivalence makes pH = pKa), salt hydrolysis, and the shape of every titration curve. Ask one question each time: what are the major species, and what equilibrium do they sit in?
Traps that cost marks
Six Unit-8 traps, the fix beside each.
- pH = −log(concentration) for a weak acid. That only works for a strong acid. For a weak acid you must run the ICE/Ka equilibrium first, [H₃O⁺] = x ≪ the stated concentration. Fix: if it has a Ka, do the equilibrium.
- Using pH + pOH = 14 everywhere. That holds only at 25 °C. At any other temperature Kw changes, so neutral pH ≠ 7 and the sum ≠ 14. Fix: if told Kw or a non-25 °C temperature, go back to Kw = [H₃O⁺][OH⁻].
- Calling the equivalence point pH 7. Only a strong–strong titration is pH 7. Weak acid + strong base → conjugate base left over → basic (pH > 7); weak base + strong acid → conjugate acid left over → acidic (pH < 7). Fix: name the species left at equivalence and hydrolyze it.
- Missing the pH = pKa gift. At the half-equivalence point [HA] = [A⁻], so pH = pKaa free Ka read straight off the curve. Fix: spot half-equivalence (or an equal-mole mixture) and write pH = pKa immediately.
- Writing a Ka for a strong acid or base. HCl, HNO₃, HBr, NaOH, KOH, Ba(OH)₂ dissociate completely, no equilibrium, no ICE table. Fix: get [H₃O⁺] or [OH⁻] by inspection (remember the ×2 for group-II hydroxides).
- Skipping the "x is small" check. The 0.100 M–x ≈ 0.100 M shortcut is valid only if ionization < 5%. For a fairly strong weak acid (Ka ~ 10⁻⁴) it can exceed 5% and the shortcut is wrong. Fix: compute x/C₀×100; if ≥ 5%, solve the full quadratic.
Don't waste time on (excluded by the CED)
Per-species concentrations along a polyprotic titration curve (know which species dominate & read each pKaskip the full math) · computing the pH change when acid/base is added to a buffer · deriving Henderson–Hasselbalch · computing solubility as a function of pH (reason qualitatively with Le Châtelier only). Know the patterns, skip the heavy edge-case math.
Quick self-check (answers in your head, then verify)
- pH of 0.050 M HNO₃?
- pH of 0.025 M KOH?
- pH of 0.15 M HCN, Ka = 4.9×10⁻¹⁰? (Is the 5% approximation safe?)
- Kb of F⁻ if Ka(HF) = 6.8×10⁻⁴?
- pH of a buffer that is 0.40 M NH₄⁺ / 0.20 M NH₃, given pKa(NH₄⁺) = 9.26?
- A weak acid–strong base titration's half-equivalence pH is 4.74. What is Ka? Is the equivalence-point pH above, below, or at 7?
- Classify 0.10 M NH₄Cl as acidic, basic, or neutral, and say why.
Check yourself: 1) pH = −log(0.050) = 1.30 2) pOH = −log(0.025) = 1.60 → pH = 12.40 3) x = √(4.9×10⁻¹⁰ × 0.15) = 8.6×10⁻⁶ → pH ≈ 5.07; ionization ≈ 0.006%, so the approximation is very safe 4) Kb = Kw/Ka = 1.0×10⁻¹⁴ / 6.8×10⁻⁴ = 1.5×10⁻¹¹ 5) pH = 9.26 + log(0.20/0.40) = 9.26 − 0.30 = 8.96 6) pH = pKa = 4.74 → Ka = 10⁻⁴·⁷⁴ = 1.8×10⁻⁵; equivalence pH is above 7 (weak acid leaves a basic conjugate base) 7) acidic: Cl⁻ is a spectator, NH₄⁺ is the conjugate acid of weak NH₃ and donates H⁺ to water.