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Memory tricks for pH, buffers & acid-base theory

From the pH scale to titration curves to buffer calculations — acid-base chemistry is high-yield on every exam. These memory tricks lock in the definitions, strong acids, and calculation shortcuts you need under pressure.

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Memory Tricks

Proven mnemonics — fast to learn, hard to forget.

🧪 Acids & Bases — 9 Memory Tricks  ·  Click any card to expand · Save favorites · Switch to Flashcard or Quiz mode below
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pH Scale
pH 7 = Pure water = Perfect neutral
pH < 7 = acidic · pH = 7 = neutral · pH > 7 = basic (alkaline)
The pH scale runs 0–14. Pure water sits at exactly 7 — perfect neutral. Every unit below 7 is 10× more acidic; every unit above is 10× more basic. Battery acid sits near 0; drain cleaner near 14. Stomach acid is about pH 2; blood must stay between 7.35–7.45.
📖 Full Lesson → Difficulty: Beginner
pH formula
pH = -log[H⁺]. If [H⁺] = 10⁻³ M, then pH = 3. Each unit change = 10× change in [H⁺].
pOH
pOH = -log[OH⁻]. At 25°C: pH + pOH = 14. Find one, calculate the other instantly.
Common pH values
Lemon juice ≈ 2 · Vinegar ≈ 3 · Coffee ≈ 5 · Blood ≈ 7.4 · Baking soda ≈ 9 · Bleach ≈ 13.
Logarithmic scale
pH 3 is 10× more acidic than pH 4, and 100× more acidic than pH 5. This trips up students who treat pH as linear.
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Flashcard
🃏 pH Scale
pH scale — what is neutral?
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🃏 Answer
pH 7 = Pure water = Perfect neutral
pH formula — pH = -log[H⁺]. If [H⁺] = 10⁻³ M, then pH = 3. Each unit change = 10× change in [H⁺].
pOH — pOH = -log[OH⁻]. At 25°C: pH + pOH = 14. Find one, calculate the other instantly.
Common pH values — Lemon juice ≈ 2 · Vinegar ≈ 3 · Coffee ≈ 5 · Blood ≈ 7.4 · Baking soda ≈ 9 · Bleach ≈ 13.
Logarithmic scale — pH 3 is 10× more acidic than pH 4, and 100× more acidic than pH 5. This trips up students who treat pH as linear.
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Strong Acids
HCl HBr HI — "Have Big Ions"
Plus HNO₃ · H₂SO₄ · HClO₄ — the 6 strong acids that dissociate 100%
The 6 strong acids dissociate completely in water — no equilibrium, no Ka needed. Have Big Ions covers HCl, HBr, HI (the hydrohalic acids except HF). Add HNO₃ (nitric), H₂SO₄ (sulfuric), and HClO₄ (perchloric) to complete the set. Everything else is a weak acid.
📖 Full Lesson → Difficulty: Beginner
Why not HF?
HF is a weak acid — the H-F bond is extremely strong (F is small, holds tightly). Despite F being the most electronegative element, the bond strength prevents full dissociation.
H₂SO₄ is diprotic
First dissociation is complete (strong). Second dissociation is weak (HSO₄⁻ ⇌ H⁺ + SO₄²⁻). Only the first proton counts as "strong."
Strong acid calculation
For strong acids, [H⁺] = molarity of acid. 0.01 M HCl → [H⁺] = 0.01 M → pH = 2. No Ka needed.
Strong bases
Group 1 hydroxides (NaOH, KOH, LiOH) and Group 2 hydroxides (Ca(OH)₂, Ba(OH)₂) are the strong bases — dissociate completely.
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🃏 Strong Acids
Strong acids — which ones?
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🃏 Answer
HCl HBr HI — "Have Big Ions"
Why not HF? — HF is a weak acid — the H-F bond is extremely strong (F is small, holds tightly). Despite F being the most electronegative element, the bond strength prevents full dissociation.
H₂SO₄ is diprotic — First dissociation is complete (strong). Second dissociation is weak (HSO₄⁻ ⇌ H⁺ + SO₄²⁻). Only the first proton counts as "strong."
Strong acid calculation — For strong acids, [H⁺] = molarity of acid. 0.01 M HCl → [H⁺] = 0.01 M → pH = 2. No Ka needed.
Strong bases — Group 1 hydroxides (NaOH, KOH, LiOH) and Group 2 hydroxides (Ca(OH)₂, Ba(OH)₂) are the strong bases — dissociate completely.
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Acid-Base Theory
BL (Brønsted-Lowry): Proton Donor/Acceptor · Lewis: Electron Pair
Brønsted-Lowry = H⁺ transfer · Lewis = electron pair donation
Two theories, two definitions. Brønsted-Lowry: acid = proton (H⁺) donor, base = proton acceptor. Lewis: acid = electron pair acceptor, base = electron pair donor. Lewis is broader — it includes reactions with no proton transfer at all (like BF₃ + NH₃). Every BL acid is also a Lewis acid, but not vice versa.
📖 Full Lesson → Difficulty: Intermediate
Conjugate pairs
In BL theory: HA ⇌ H⁺ + A⁻. HA and A⁻ are a conjugate acid-base pair. Strong acid → very weak conjugate base. Weak acid → stronger conjugate base.
Amphoteric species
Water is amphoteric — it can act as both acid and base (donates OR accepts H⁺). HSO₄⁻ is also amphoteric.
Lewis examples
BF₃ = Lewis acid (accepts electron pair into empty orbital). NH₃ = Lewis base (donates lone pair). No proton transfer occurs — BL theory can't describe this reaction.
Arrhenius (original)
Oldest definition: acid produces H⁺ in water, base produces OH⁻. Most limited — only works in aqueous solution. BL and Lewis are more general.
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🃏 Acid-Base Theory
Brønsted-Lowry vs Lewis acids and bases?
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🃏 Answer
BL (Brønsted-Lowry): Proton Donor/Acceptor · Lewis: Electron Pair
Conjugate pairs — In BL theory: HA ⇌ H⁺ + A⁻. HA and A⁻ are a conjugate acid-base pair. Strong acid → very weak conjugate base. Weak acid → stronger conjugate base.
Amphoteric species — Water is amphoteric — it can act as both acid and base (donates OR accepts H⁺). HSO₄⁻ is also amphoteric.
Lewis examples — BF₃ = Lewis acid (accepts electron pair into empty orbital). NH₃ = Lewis base (donates lone pair). No proton transfer occurs — BL theory can't describe this reaction.
Arrhenius (original) — Oldest definition: acid produces H⁺ in water, base produces OH⁻. Most limited — only works in aqueous solution. BL and Lewis are more general.
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Buffers
Buffer = Weak Acid + Its Salt (conjugate base)
Resists pH change by absorbing added H⁺ or OH⁻
A buffer contains a weak acid and its conjugate base (the salt). When acid is added — the base component neutralizes it. When base is added — the acid component neutralizes it. The buffer resists large pH changes. Blood (carbonic acid/bicarbonate) is the most important biological buffer system in your body.
📖 Full Lesson → Difficulty: Intermediate
Henderson-Hasselbalch
pH = pKa + log([A⁻]/[HA]). When [A⁻] = [HA], pH = pKa exactly. Best buffering capacity is within 1 pH unit of pKa.
How it works (add acid)
Add H⁺ → A⁻ + H⁺ → HA. The conjugate base absorbs the extra proton. pH barely changes because [H⁺] increase is minimal.
How it works (add base)
Add OH⁻ → HA + OH⁻ → A⁻ + H₂O. The weak acid donates a proton to neutralize the base. Again, pH barely changes.
Buffer capacity
Higher concentration of buffer components = greater capacity. A buffer is "exhausted" when one component is used up entirely — then pH changes rapidly.
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🃏 Buffers
Buffers — what are they made of?
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🃏 Answer
Buffer = Weak Acid + Its Salt (conjugate base)
Henderson-Hasselbalch — pH = pKa + log([A⁻]/[HA]). When [A⁻] = [HA], pH = pKa exactly. Best buffering capacity is within 1 pH unit of pKa.
How it works (add acid) — Add H⁺ → A⁻ + H⁺ → HA. The conjugate base absorbs the extra proton. pH barely changes because [H⁺] increase is minimal.
How it works (add base) — Add OH⁻ → HA + OH⁻ → A⁻ + H₂O. The weak acid donates a proton to neutralize the base. Again, pH barely changes.
Buffer capacity — Higher concentration of buffer components = greater capacity. A buffer is "exhausted" when one component is used up entirely — then pH changes rapidly.
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Ka & Kb
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
Conjugate pair Ka and Kb always multiply to Kw at 25°C
For any conjugate acid-base pair, Ka × Kb = Kw (1.0 × 10⁻¹⁴ at 25°C). This means: strong acid (large Ka) → very weak conjugate base (tiny Kb). Know Ka for an acid and you can instantly calculate Kb for its conjugate base — no table lookup needed.
📖 Full Lesson → Difficulty: Intermediate
pKa + pKb = 14
Taking -log of both sides: pKa + pKb = pKw = 14 at 25°C. If pKa = 4.74 (acetic acid), then pKb of acetate = 14 - 4.74 = 9.26.
Strong acids
HCl Ka ≈ 10⁷ (essentially infinite). Conjugate base Cl⁻ has Kb ≈ 10⁻²¹ — essentially zero. Cl⁻ is such a weak base it doesn't affect pH at all.
Weak acid calculation
Acetic acid Ka = 1.8×10⁻⁵. pH of 0.1M solution: use ICE table or approximate [H⁺] = √(Ka × C) = √(1.8×10⁻⁶) ≈ 0.00134 M → pH ≈ 2.87.
Kw itself
Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C. In pure water: [H⁺] = [OH⁻] = 1.0×10⁻⁷ M → pH = 7. Kw increases with temperature — pure water at 37°C has pH slightly below 7.
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🃏 Ka & Kb
How are Ka, Kb and Kw related?
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🃏 Answer
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
pKa + pKb = 14 — Taking -log of both sides: pKa + pKb = pKw = 14 at 25°C. If pKa = 4.74 (acetic acid), then pKb of acetate = 14 - 4.74 = 9.26.
Strong acids — HCl Ka ≈ 10⁷ (essentially infinite). Conjugate base Cl⁻ has Kb ≈ 10⁻²¹ — essentially zero. Cl⁻ is such a weak base it doesn't affect pH at all.
Weak acid calculation — Acetic acid Ka = 1.8×10⁻⁵. pH of 0.1M solution: use ICE table or approximate [H⁺] = √(Ka × C) = √(1.8×10⁻⁶) ≈ 0.00134 M → pH ≈ 2.87.
Kw itself — Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C. In pure water: [H⁺] = [OH⁻] = 1.0×10⁻⁷ M → pH = 7. Kw increases with temperature — pure water at 37°C has pH slightly below 7.
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Titration
Equivalence Point ≠ Neutral — depends on the pair
Strong/strong = pH 7 · Weak acid/strong base = pH > 7 · Strong acid/weak base = pH < 7
The equivalence point is where moles of acid = moles of base — NOT necessarily pH 7. Strong acid + strong base = pH 7 (neutral salt). Weak acid + strong base = pH > 7 (basic, because conjugate base hydrolyzes). Strong acid + weak base = pH < 7 (acidic). This is one of the most tested concepts in acid-base chemistry.
📖 Full Lesson → Difficulty: Intermediate
Half-equivalence point
Halfway to equivalence: [HA] = [A⁻] → pH = pKa. This is the best point for buffer activity and is how you determine pKa experimentally.
Titration curve shape
Strong/strong: steep S-curve, sharp equivalence point. Weak acid/strong base: flatter S-curve, buffer region visible, equivalence above pH 7.
Indicator choice
Indicator must change color at or near the equivalence point. Phenolphthalein (8–10) = weak acid titrations. Methyl orange (3–4) = weak base titrations.
Calculation shortcut
M₁V₁ = M₂V₂ at equivalence point. If 25 mL of 0.1M NaOH neutralizes HCl: 25×0.1 = V₂×M₂. Solve for unknown concentration or volume.
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🃏 Titration
Titration — is the equivalence point always neutral?
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🃏 Answer
Equivalence Point ≠ Neutral — depends on the pair
Half-equivalence point — Halfway to equivalence: [HA] = [A⁻] → pH = pKa. This is the best point for buffer activity and is how you determine pKa experimentally.
Titration curve shape — Strong/strong: steep S-curve, sharp equivalence point. Weak acid/strong base: flatter S-curve, buffer region visible, equivalence above pH 7.
Indicator choice — Indicator must change color at or near the equivalence point. Phenolphthalein (8–10) = weak acid titrations. Methyl orange (3–4) = weak base titrations.
Calculation shortcut — M₁V₁ = M₂V₂ at equivalence point. If 25 mL of 0.1M NaOH neutralizes HCl: 25×0.1 = V₂×M₂. Solve for unknown concentration or volume.
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Polyprotic
H₂SO₄ loses H⁺ in steps — first strong, second weak
H₂CO₃ · H₂SO₄ · H₃PO₄ — each proton has its own Ka, getting smaller
Polyprotic acids lose protons one at a time. Each successive Ka is much smaller than the last — it gets harder to remove each proton because the ion becomes more negative. For most calculations, only Ka₁ matters because Ka₂ is so much smaller. H₂CO₃ (carbonic acid) is the most tested polyprotic acid.
📖 Full Lesson → Difficulty: Advanced
Carbonic acid system
H₂CO₃ ⇌ H⁺ + HCO₃⁻ (Ka₁ = 4.3×10⁻⁷) · HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (Ka₂ = 4.8×10⁻¹¹). Critical for blood pH buffering — CO₂ drives Ka₁ equilibrium.
Phosphoric acid
H₃PO₄ has 3 protons. Ka₁ = 7.5×10⁻³ · Ka₂ = 6.2×10⁻⁸ · Ka₃ = 2.2×10⁻¹³. Each is about 10,000× smaller — each proton much harder to remove.
Intermediate species
HCO₃⁻ and H₂PO₄⁻ are amphoteric — they can donate OR accept a proton. pH of intermediate = (pKa₁ + pKa₂)/2 — a useful shortcut.
Simplification rule
If Ka₁/Ka₂ > 1000, treat as monoprotic using only Ka₁. The second dissociation contributes negligibly to [H⁺].
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🃏 Polyprotic
Polyprotic acids — how do they lose protons?
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🃏 Answer
H₂SO₄ loses H⁺ in steps — first strong, second weak
Carbonic acid system — H₂CO₃ ⇌ H⁺ + HCO₃⁻ (Ka₁ = 4.3×10⁻⁷) · HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (Ka₂ = 4.8×10⁻¹¹). Critical for blood pH buffering — CO₂ drives Ka₁ equilibrium.
Phosphoric acid — H₃PO₄ has 3 protons. Ka₁ = 7.5×10⁻³ · Ka₂ = 6.2×10⁻⁸ · Ka₃ = 2.2×10⁻¹³. Each is about 10,000× smaller — each proton much harder to remove.
Intermediate species — HCO₃⁻ and H₂PO₄⁻ are amphoteric — they can donate OR accept a proton. pH of intermediate = (pKa₁ + pKa₂)/2 — a useful shortcut.
Simplification rule — If Ka₁/Ka₂ > 1000, treat as monoprotic using only Ka₁. The second dissociation contributes negligibly to [H⁺].
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ICE Tables
ICE (I=Initial concentrations, C=Change using stoichiometry, E=Equilibrium concentrations) — Initial, Change, Equilibrium
Set up every weak acid/base equilibrium problem the same way
ICE tables are the universal tool for weak acid/base problems. Initial — write starting concentrations. Change — use +x and -x based on stoichiometry. Equilibrium — add Initial + Change. Then plug into Ka or Kb expression and solve for x. The 5% rule: if x/initial < 5%, the approximation x ≈ √(Ka × C) is valid.
📖 Full Lesson → Difficulty: Intermediate
Example setup
0.1M acetic acid: HA ⇌ H⁺ + A⁻. I: 0.1, 0, 0. C: -x, +x, +x. E: 0.1-x, x, x. Ka = x²/(0.1-x) ≈ x²/0.1 = 1.8×10⁻⁵. x = [H⁺] = 0.00134M. pH = 2.87.
5% approximation rule
If x/initial < 5%, the approximation (0.1-x ≈ 0.1) is valid and saves quadratic math. Check: 0.00134/0.1 = 1.34% — approximation valid here.
When to use quadratic
If x/initial > 5%, use the quadratic formula: x² + Ka·x - Ka·C = 0. Or iterate — plug approximation back in until stable.
Weak bases
Same ICE setup but use Kb. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Solve for [OH⁻], then pOH, then pH = 14 - pOH.
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🃏 ICE Tables
ICE
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🃏 Answer
ICE (I=Initial concentrations, C=Change using stoichiometry, E=Equilibrium concentrations) — Initial, Change, Equilibrium
Example setup — 0.1M acetic acid: HA ⇌ H⁺ + A⁻. I: 0.1, 0, 0. C: -x, +x, +x. E: 0.1-x, x, x. Ka = x²/(0.1-x) ≈ x²/0.1 = 1.8×10⁻⁵. x = [H⁺] = 0.00134M. pH = 2.87.
5% approximation rule — If x/initial < 5%, the approximation (0.1-x ≈ 0.1) is valid and saves quadratic math. Check: 0.00134/0.1 = 1.34% — approximation valid here.
When to use quadratic — If x/initial > 5%, use the quadratic formula: x² + Ka·x - Ka·C = 0. Or iterate — plug approximation back in until stable.
Weak bases — Same ICE setup but use Kb. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Solve for [OH⁻], then pOH, then pH = 14 - pOH.
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Salt Hydrolysis
Salt from strong + strong = neutral · weak + strong = NOT neutral
Salts of weak acids or bases hydrolyze — their solutions are NOT pH 7
When a salt dissolves, its ions may react with water (hydrolyze). NaCl — Na⁺ from strong base, Cl⁻ from strong acid — neither reacts with water → pH 7. CH₃COONa — acetate from weak acid — acetate hydrolyzes (A⁻ + H₂O ⇌ HA + OH⁻) → basic solution. NH₄Cl — ammonium from weak base — hydrolyzes → acidic solution.
📖 Full Lesson → Difficulty: Intermediate
Quick prediction rule
Strong acid + strong base salt → neutral (pH 7). Weak acid + strong base salt → basic. Strong acid + weak base salt → acidic. Weak acid + weak base → compare Ka vs Kb to determine.
Acetate example
Sodium acetate (CH₃COONa) in water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Produces OH⁻ → basic solution. pH of 0.1M solution ≈ 8.87.
Ammonium example
NH₄Cl in water: NH₄⁺ ⇌ NH₃ + H⁺. Produces H⁺ → acidic solution. pH of 0.1M solution ≈ 5.12.
Exam tip
Identify the parent acid and base of the salt. If either parent is weak, the solution won't be neutral. This determines which way pH shifts.
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🃏 Salt Hydrolysis
Salt hydrolysis — when is a salt solution not neutral?
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🃏 Answer
Salt from strong + strong = neutral · weak + strong = NOT neutral
Quick prediction rule — Strong acid + strong base salt → neutral (pH 7). Weak acid + strong base salt → basic. Strong acid + weak base salt → acidic. Weak acid + weak base → compare Ka vs Kb to determine.
Acetate example — Sodium acetate (CH₃COONa) in water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Produces OH⁻ → basic solution. pH of 0.1M solution ≈ 8.87.
Ammonium example — NH₄Cl in water: NH₄⁺ ⇌ NH₃ + H⁺. Produces H⁺ → acidic solution. pH of 0.1M solution ≈ 5.12.
Exam tip — Identify the parent acid and base of the salt. If either parent is weak, the solution won't be neutral. This determines which way pH shifts.
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