AP Chemistry — Cheatsheet
Formulas, exam-day tips, and key terms on one page.
Formulas & relationships
Avogadro's number
1 mole = 6.022 × 10²³ particles
Particles can be atoms, molecules, ions, or electrons — always state what you are counting.
Average atomic mass
avg mass = Σ (isotope mass × fractional abundance)
Fractional abundance is the percentage written as a decimal (e.g. 75% → 0.75).
Aufbau filling order
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s
Note the swap: 4s fills before 3d because it is slightly lower in energy.
Trend summary (top-right is the extreme)
→ across & ↑ up: radius ↓, ionization energy ↑, electronegativity ↑
Formal charge
formal charge = valence − nonbonding − ½ bonding
Count the atom's own valence electrons, subtract its lone-pair (nonbonding) electrons, then subtract half of its shared (bonding) electrons.
Bond type from electronegativity difference (ΔEN)
ΔEN ≳ 1.7 → ionic · 0.4–1.7 → polar covalent · < 0.4 → nonpolar covalent
These cutoffs are approximate guides, not hard walls — bonding is a continuum from pure sharing to full transfer.
Electron-domain geometry by domain count
2 → linear (180°) · 3 → trigonal planar (120°) · 4 → tetrahedral (109.5°)
This is the arrangement of ALL domains. The shape named by the atoms (molecular geometry) can differ when some domains are lone pairs.
Hybridization from electron domains
2 domains → sp (linear) · 3 → sp² (trigonal planar) · 4 → sp³ (tetrahedral)
Superscript = number of p orbitals mixed with the one s orbital; total hybrids = domains.
IMF strength ranking (per interaction)
ion–dipole > hydrogen bonding > dipole–dipole > London dispersion
A caution: in large molecules, many LDFs can add up to outweigh a single dipole–dipole attraction — compare total forces, not just the type.
Boiling condition
liquid boils when: vapor pressure = external pressure
Lower the external pressure and the boiling point drops; raise it (a pressure cooker) and the boiling point climbs.
Ideal gas law
PV = nRT
R = 0.08206 L·atm·mol⁻¹·K⁻¹. Always convert T to kelvin and match your pressure/volume units to R before plugging in.
Partial pressure from mole fraction
Pᵢ = Xᵢ × P_total, where Xᵢ = nᵢ ÷ n_total
Molarity
M = mol solute ÷ L solution
It is liters of *solution*, not liters of solvent — you dissolve the solute and then fill to the mark.
Building a net ionic equation
molecular → split strong electrolytes into ions → cancel spectators → net ionic
Only strong electrolytes split. Precipitates (s), gases (g), water, and weak acids stay written as whole formulas.
Grams ↔ moles
moles = mass ÷ molar mass
The universal on-ramp and off-ramp: grams → (÷ molar mass) → moles → (× mole ratio) → moles of target → (× molar mass) → grams.
Percent yield
percent yield = (actual yield ÷ theoretical yield) × 100%
A value above 100% signals an error (or impurity/wet product) — you cannot make more than theory allows.
Oxidation-number sum rule
Σ (oxidation numbers) = overall charge of the species
= 0 for a neutral compound; = the ion charge for a polyatomic ion. Solve for the unknown atom.
Molarity
M = mol ÷ L (so mol = M × V)
Volume must be in liters. 25.0 mL = 0.0250 L.
Titration relationships
mol titrant = M × V → mol unknown = mol titrant × (mole ratio) → M unknown = mol ÷ V
For a 1:1 reaction only, this simplifies to MₐVₐ = M_bV_b. When the ratio is not 1:1, you must apply it explicitly.
Average reaction rate
rate = −Δ[reactant] / Δt = +Δ[product] / Δt
For a A → b B, divide each term by its coefficient so all species give the same rate: −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt.
Factors that speed a reaction
↑ concentration, ↑ temperature, ↑ surface area, add catalyst → ↑ rate
Concentration, temperature, and surface area raise the collision rate; temperature and a catalyst also change the fraction of collisions that succeed.
General rate law
rate = k[A]ᵐ[B]ⁿ
m is the order in A, n the order in B, and (m + n) the overall order. Orders are usually 0, 1, or 2 and come only from data.
Integrated rate laws & linear plots
Zero: [A] = [A]₀ − kt (plot [A] vs t) | First: ln[A] = ln[A]₀ − kt (plot ln[A] vs t) | Second: 1/[A] = 1/[A]₀ + kt (plot 1/[A] vs t)
Whichever plot is a straight line reveals the order; the slope gives k (−k for zero/first order, +k for second order).
First-order half-life
t½ = 0.693 / k
0.693 is ln 2. Because [A]₀ cancels out, first-order half-life is independent of starting concentration. (Zero order: t½ = [A]₀/2k; second order: t½ = 1/(k[A]₀) — these do depend on [A]₀.)
Spotting each species in a mechanism
Intermediate: appears first as a product, then a reactant. Catalyst: appears first as a reactant, then a product.
Both cancel out of the overall equation — the order in which they appear (made-then-used vs. used-then-remade) is what tells them apart.
Sign convention for enthalpy
ΔH = H(products) − H(reactants); exothermic ΔH < 0, endothermic ΔH > 0
Negative means energy leaves the system (released); positive means energy enters the system (absorbed).
Heat and temperature change
q = m·c·ΔT
q = heat (J), m = mass (g), c = specific heat capacity (J·g⁻¹·°C⁻¹), ΔT = T(final) − T(initial). For water, c = 4.18 J·g⁻¹·°C⁻¹.
Enthalpy from formation enthalpies
ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)
Each ΔH°f is multiplied by its coefficient. The ΔH°f of any element in its standard state is exactly 0.
Gibbs free energy
ΔG = ΔH − T·ΔS
T is absolute temperature in kelvin. Watch units: ΔH is usually kJ while ΔS is usually J·K⁻¹ — convert ΔS to kJ·K⁻¹ (divide by 1000) before subtracting.
Equilibrium-constant expression
For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
Products on top, reactants on the bottom, each raised to its coefficient. Omit any pure solid (s) or pure liquid (l).
Kp for gaseous equilibria
Kp = Kc (RT)^Δn, where Δn = (moles gas product) − (moles gas reactant)
Kp uses partial pressures instead of concentrations. If the number of gas moles is unchanged (Δn = 0), then Kp = Kc.
Reaction quotient Q vs. K
Q < K → shift right (forward); Q = K → at equilibrium; Q > K → shift left (reverse)
Q has the exact same form as K but uses the current (not necessarily equilibrium) concentrations. Comparing Q to K predicts the direction of change.
ICE relationship
Equilibrium concentration = Initial + Change (Change = ± coefficient × x)
Reactants are consumed (−), products are formed (+). The coefficient in front of x must match the balanced equation.
Temperature and K (heat-as-reagent rule)
Exothermic (ΔH < 0): heat is a product → raising T shifts left, K decreases. Endothermic (ΔH > 0): heat is a reactant → raising T shifts right, K increases.
Only a temperature change alters the numerical value of K. Concentration, volume, pressure, and catalyst changes never do.
Solubility product
For AₘBₙ(s) ⇌ m Aⁿ⁺(aq) + n Bᵐ⁻(aq): Ksp = [Aⁿ⁺]ᵐ [Bᵐ⁻]ⁿ
Only the dissolved ions appear, each raised to its coefficient. A larger Ksp (for the same ion-count stoichiometry) means a more soluble salt.
Ksp ↔ molar solubility
MX type: Ksp = s². MX₂ type: Ksp = 4s³. M₂X type: Ksp = 4s³.
Always rebuild the relationship from the balanced dissolution equation; do not assume Ksp = s² for every salt.
Definition of pH
pH = −log[H⁺]
The minus sign makes pH positive for typical solutions. Each whole pH unit is a factor of 10 in [H⁺]: pH 3 is ten times more acidic than pH 4.
The pH–pOH relationship
pH + pOH = 14 (at 25 °C, since Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴)
pOH = −log[OH⁻]. Once you know any one of pH, pOH, [H⁺], or [OH⁻], you can reach the other three.
Acid ionization constant
Ka = [H⁺][A⁻] / [HA] pKa = −log(Ka)
Larger Ka ↔ smaller pKa ↔ stronger acid. For a conjugate pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴.
Henderson–Hasselbalch equation
pH = pKa + log([A⁻] / [HA])
When [A⁻] = [HA] the log term is log(1) = 0, so pH = pKa. More conjugate base raises pH; more acid lowers it.
Half-equivalence point
at half-equivalence: [HA] = [A⁻] ⇒ pH = pKa
This is Henderson–Hasselbalch with log([A⁻]/[HA]) = log(1) = 0. It is the single most useful point for extracting pKa from a curve.
Gibbs free energy
ΔG = ΔH − TΔS
ΔG < 0 favored, ΔG > 0 not favored, ΔG = 0 at equilibrium. T is in kelvin, so the TΔS term always grows with temperature.
Free energy and the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ln K > 0 ⇒ ΔG° < 0 (products favored). K < 1 ⇒ ΔG° > 0 (reactants favored).
Standard cell potential
E°cell = E°cathode − E°anode
Both values are standard *reduction* potentials read straight from the table. Do NOT multiply a potential by the number of electrons or by a balancing coefficient — potential is an intensive property.
Free energy from cell potential
ΔG° = −nFE°
n = moles of electrons transferred, F = 96 485 C·mol⁻¹ (Faraday constant). E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous; E° < 0 ⇒ ΔG° > 0 ⇒ nonspontaneous.
Faraday's laws of electrolysis
moles e⁻ = Q ÷ F = (I·t) ÷ 96 485
Then scale by the half-reaction: e.g. Ag⁺ + e⁻ → Ag needs 1 mol e⁻ per mol Ag, while Al³⁺ + 3e⁻ → Al needs 3 mol e⁻ per mol Al.
The Nernst equation (at 298 K)
E = E° − (0.0592 / n) log Q
General form: E = E° − (RT/nF) ln Q. At 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it.
The PES energy balance
binding energy = photon energy − kinetic energy of ejected electron
Because every subshell holds its electrons with a characteristic tightness, each subshell produces its own peak at a fixed binding energy.
Coulomb's law (attraction)
F ∝ q₁q₂ / r²
q₁ is the effective nuclear charge (Zₑff), q₂ the electron’s charge, and r the distance between them. Doubling the distance quarters the force — the inverse-square term dominates.
Formal charge
formal charge = valence electrons − nonbonding electrons − ½(bonding electrons)
Count the atom's group valence, subtract all of its lone-pair electrons, then subtract half of the electrons it shares in bonds. The formal charges of a species must sum to its overall charge.
Average bond order (resonance hybrid)
bond order = (total bonds shared among the equivalent positions) ÷ (number of equivalent positions)
Equivalently, average the bond (single = 1, double = 2, triple = 3) for one linkage across all resonance structures. For NO₃⁻: 4 bonds over 3 equal N–O positions → 4/3 ≈ 1.33.
Sigma and pi bonds per bond type
single = 1σ · double = 1σ + 1π · triple = 1σ + 2π
To total a molecule: number of σ bonds = number of bonded atom pairs (every bonded connection has exactly one σ, including every X–H bond); number of π bonds = (double bonds) + 2 × (triple bonds).
Graham’s law of effusion
rate₁ / rate₂ = √(M₂ / M₁)
Lighter gas is faster. The heavier gas goes in the numerator under the root, so the lighter gas’s rate comes out larger.
Dalton’s law (partial pressures)
P_total = P₁ + P₂ + … , Pᵢ = Xᵢ × P_total , P_gas = P_total − P_water
The last form is for a gas collected over water: subtract the temperature-dependent vapor pressure of water to get the dry-gas pressure.
Beer–Lambert law
A = εbc
A is absorbance (unitless); ε is molar absorptivity (L·mol⁻¹·cm⁻¹); b is path length (cm); c is molar concentration (mol·L⁻¹). Rearranged, c = A ÷ (εb).
Dilution relationship
M₁V₁ = M₂V₂
Moles of solute are conserved on dilution. Subscript 1 is the concentrated stock, 2 is the diluted solution; any consistent volume unit works since it cancels.
Half-reaction balancing checklist
atoms (≠ O,H) → O with H₂O → H with H⁺ → charge with e⁻ → scale & add → (basic: + OH⁻ both sides, combine to H₂O, cancel)
Electrons always land on the more-positive side of a half-reaction: the reactant side for reduction, the product side for oxidation.
Yield and back-titration relationships
percent yield = (actual ÷ theoretical) × 100% | mol reacted with analyte = mol added − mol left over
Route every problem through moles: grams ÷ molar mass, or molarity × volume(L). Convert back to grams or molarity only at the very end.
Integrated rate laws as straight lines
Zero: [A] = −kt + [A]₀ (plot [A] vs t) | First: ln[A] = −kt + ln[A]₀ (plot ln[A] vs t) | Second: 1/[A] = kt + 1/[A]₀ (plot 1/[A] vs t)
In each case y = mx + b: the y-intercept is the starting value and the slope is ±k. Zero and first order give slope = −k; second order gives slope = +k.
Half-life by order
Zero: t½ = [A]₀ / (2k) First: t½ = 0.693 / k Second: t½ = 1 / (k[A]₀)
Only first-order t½ is independent of [A]₀ (0.693 = ln 2). If successive half-lives are equal → first order; if they get shorter → zero order; if they get longer → second order.
Arrhenius equation
k = A·e^(−Eₐ/RT)
R = 8.314 J·mol⁻¹·K⁻¹ (use Eₐ in J/mol). Two-temperature form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). A plot of ln k vs 1/T is linear with slope = −Eₐ/R.
Energy relationships on the diagram
Eₐ(forward) = E(transition state) − E(reactants) Eₐ(reverse) = E(transition state) − E(products) ΔH = Eₐ(forward) − Eₐ(reverse)
ΔH also equals E(products) − E(reactants). If Eₐ(forward) < Eₐ(reverse) the reaction is exothermic (ΔH < 0); if Eₐ(forward) > Eₐ(reverse) it is endothermic (ΔH > 0).
Heat and temperature change
q = m·c·ΔT with q(hot) + q(cold) = 0 (insulated)
q = heat (J), m = mass (g), c = specific heat (J·g⁻¹·°C⁻¹), ΔT = T(final) − T(initial). The hot object has ΔT < 0 (q < 0); the cold object has ΔT > 0 (q > 0).
Two cross-checks for ΔH°rxn
ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants); ΔH ≈ Σ(bonds broken) − Σ(bonds formed)
Each ΔH°f is weighted by its coefficient; ΔH°f of an element in its standard state is 0. The bond-enthalpy estimate is gas-phase only and approximate (tabulated bonds are averages).
Entropy change of a reaction
ΔS°rxn = Σ S°(products) − Σ S°(reactants)
S° values are in J·mol⁻¹·K⁻¹ and are always positive (even for elements). More moles of gas on the product side → ΔS > 0.
Gibbs free energy
ΔG = ΔH − T·ΔS
T is absolute temperature (kelvin). Convert ΔS from J·K⁻¹ to kJ·K⁻¹ before subtracting. ΔG < 0 spontaneous, ΔG > 0 nonspontaneous, ΔG = 0 at equilibrium.
Crossover (equilibrium) temperature
ΔG = 0 ⟹ T = ΔH / ΔS
Use ΔH in joules to match ΔS in J·K⁻¹ (or ΔH in kJ with ΔS in kJ·K⁻¹). For (+, +) the reaction turns spontaneous above T; for (−, −) it turns nonspontaneous above T.
Quadratic formula (for the exact ICE solve)
ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / 2a
Use when the small-x approximation fails the 5% test. Discard the root that gives a negative concentration or a value exceeding the initial amount.
Kp from Kc
Kp = Kc (RT)^Δn, Δn = (moles of gaseous product) − (moles of gaseous reactant)
R = 0.0821 L·atm·mol⁻¹·K⁻¹ and T is in kelvin. If Δn = 0 the factor is 1, so Kp = Kc. Only gas-phase species count toward Δn.
Solubility product and molar solubility
AₘBₙ(s) ⇌ m Aⁿ⁺ + n Bᵐ⁻: Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ. MX: Ksp = s². MX₂ or M₂X: Ksp = 4s³. MX₃: Ksp = 27s⁴.
The numerical prefix (4, 27, …) is (coefficient)^(coefficient) summed over the ions. To get s from Ksp for a 1:2 salt, take the cube root of Ksp/4.
Precipitation criterion
Q > Ksp → precipitate forms; Q = Ksp → just saturated; Q < Ksp → no precipitate
Q uses concentrations in the combined solution (after mixing dilutes them). Same Q-vs-K logic as reaction equilibria, applied to dissolving.
Conjugate relationship
Kb = Kw / Ka (and Ka · Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C)
To find how basic a conjugate base A⁻ is, divide Kw by the Ka of its parent acid HA. A weaker acid (small Ka) has a stronger conjugate base (large Kb).
Henderson–Hasselbalch (solved for the ratio)
pH = pKa + log([A⁻] / [HA]) ⇒ [A⁻] / [HA] = 10^(pH − pKa)
The exponent pH − pKa drives the design: it is 0 (ratio 1) at pH = pKa, +1 (ratio 10) one unit above, and −1 (ratio 0.1) one unit below.
Free energy from cell potential
ΔG° = −nFE°
n = moles of electrons transferred in the balanced reaction, F = 96 485 C·mol⁻¹. E° > 0 ⇒ ΔG° < 0 ⇒ spontaneous. E° is intensive — never scale it by n or by balancing coefficients; the n out front is what accounts for reaction size.
Free energy from the equilibrium constant
ΔG° = −RT ln K
R = 8.314 J·mol⁻¹·K⁻¹, T in kelvin. K > 1 ⇒ ΔG° < 0 (products favored); K < 1 ⇒ ΔG° > 0 (reactants favored).
Voltage and equilibrium (298 K shortcut)
log K = nE° / 0.0592
Derived by equating ΔG° = −nFE° with ΔG° = −RT ln K at 25 °C. A modest E° of a few tenths of a volt produces an enormous K — voltage and equilibrium are exponentially related.
Faraday's laws of electrolysis
mol e⁻ = Q ÷ F = (I·t) ÷ 96 485
Then divide by the electrons-per-ion from the half-reaction: Cu²⁺ + 2e⁻ → Cu needs 2 mol e⁻ per mol Cu; Ag⁺ + e⁻ → Ag needs 1.
The Nernst equation (298 K)
E = E° − (0.0592 / n) log Q
General form E = E° − (RT/nF) ln Q; at 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it; Q = 1 recovers E = E°.
Concentration cell (identical electrodes, 298 K)
E = −(0.0592 / n) log Q, with E° = 0
Q = [ion]dilute ÷ [ion]concentrated < 1, so log Q < 0 and E > 0. The concentrated half-cell is always the cathode; electrons flow to shrink the gradient.
On the exam
- On the AP exam, moles are the hub of almost every quantitative problem. Train the reflex: grams → (÷ molar mass) → moles → (× ratio) → moles of target → (× molar mass or Avogadro) → answer.
- AP loves connecting configuration to the periodic table: the s-block is groups 1–2, the d-block is the transition metals, and the p-block is groups 13–18. An element’s valence configuration is readable straight off its position.
- Whenever a free-response asks you to *justify* a trend, name the mechanism — “higher Zₑff” or “additional shell / more shielding.” Stating the trend alone rarely earns the point; the reasoning does.
- On free-response Lewis questions, always tally your electrons against your step-1 total before moving on — an off-by-two count is the single most common way to lose the point. Then verify formal charges sum to the overall charge of the species.
- The AP exam constantly links a property to a structural cause. Never just name the bond type — state the mechanism: "conducts when molten because the ions become mobile" or "malleable because the nondirectional electron sea allows cores to slide."
- Keep two questions separate on the exam: "How many total domains?" fixes the electron-domain geometry and ideal angle; "How many of those are lone pairs?" fixes the named molecular shape and how far the real angle is squeezed below ideal.
- The exam's favorite two-step: first use domain count for both geometry AND hybridization (they come from the same number), then judge polarity by asking whether the shape lets the bond dipoles cancel. Symmetry with identical outer atoms → nonpolar; asymmetry or lone pairs → polar.
- On free-response, always name the IMF *and* the reason. "Water has hydrogen bonding, which is stronger than the dipole–dipole forces in H₂S, so water has the higher boiling point" earns the point; just saying "water boils higher" does not.
- Reading a phase diagram: moving right (heating) at constant pressure crosses from solid to liquid to gas; moving up (compressing) at constant temperature can push a gas into a liquid or solid. Locate the triple point and critical point first — they anchor the whole graph.
- When a problem gives grams but asks about a reaction or concentration, convert to moles first — moles are the currency of both molarity (mol ÷ L) and stoichiometry. Watch that volume is in liters and that "solution" volume, not solvent volume, goes in the denominator.
- AP free-response almost always asks for the *net ionic* equation, and it must be balanced for mass and charge — spectators earn no credit. Memorize the core solubility rules so you can decide instantly which species split and which stay solid.
- On limiting-reactant free-response, show the moles-of-product each reactant would make and explicitly state which is smaller. The reactant giving the *smaller* product amount is limiting, and that smaller number is your theoretical yield.
- To name agents, first find what is oxidized and what is reduced, then flip the labels: the reduced species is the oxidizing agent, the oxidized species is the reducing agent. Mixing these up is the single most common redox error on the AP exam.
- Every solution-stoichiometry problem runs the same loop: convert volume + molarity to moles, cross the balanced equation with the mole ratio, then convert back using the target’s volume or molar mass. Keep volumes in liters throughout and the arithmetic stays clean.
- When asked to *justify* a rate change, name the mechanism: more frequent collisions, a greater fraction of collisions exceeding Eₐ, or a lower Eₐ (catalyst only). Naming the factor without the mechanism rarely earns the point.
- The orders in a rate law come from **experimental data only** — never from the stoichiometric coefficients of the overall balanced equation. Reading exponents off the coefficients is the single most common kinetics mistake on the exam.
- A **constant** half-life is the signature of first order; a linear ln[A]-vs-t plot confirms it. If half-life instead grows as the reaction proceeds, suspect second order; if a plain [A]-vs-t plot is linear, it is zero order.
- To recover the RDS from data: if the experimental rate law is rate = k[A]²[B], the slow step must consume 2 A and 1 B (or an equivalent set once an intermediate is expressed via a fast equilibrium). Mechanism, RDS, and rate law must all agree.
- A quick reflex for the exam: "exo = exit," heat exits the system, ΔH < 0, surroundings warm. "Endo = into," heat goes into the system, ΔH > 0, surroundings cool. The temperature change you feel in the beaker is the *opposite* sign of the system's ΔH.
- Free-response calorimetry problems almost always want the heat *of the reaction*, not just of the water. Compute q(water) with m·c·ΔT, then flip the sign to get q(reaction), and finally divide by moles to report ΔH in kJ·mol⁻¹.
- Three routes, one answer: Hess's law with given steps, the ΔH°f formula with a table, and bond enthalpies for gas-phase estimates. If a problem hands you ΔH°f values, reach for Σproducts − Σreactants first — it is the fastest and the most exam-common.
- Memorize the four-case table: (−, +) spontaneous at all T; (+, −) never; (−, −) spontaneous at low T; (+, +) spontaneous at high T. When ΔH and ΔS pull the same direction, temperature is irrelevant; when they conflict, T > ΔH/ΔS marks the switch point.
- On the exam, state the reason with the comparison. "Q < K, so the reaction proceeds forward" earns the point; "it goes forward" alone often does not. Also remember: reversing a reaction inverts K (K becomes 1/K), and multiplying an equation through by n raises K to the nth power.
- To decide whether a precipitate forms, compute the ion product Q for the trial concentrations and compare to Ksp: Q > Ksp means the solution is supersaturated and a precipitate forms; Q < Ksp means it stays dissolved; Q = Ksp is exactly saturated. It is the same Q-versus-K logic from Lesson 1, applied to dissolving.
- Identify conjugate pairs by counting protons: the acid always has exactly one more H⁺ than its conjugate base. HCO₃⁻ can be the conjugate base of H₂CO₃ or the conjugate acid of CO₃²⁻ — amphoteric species show up often on the exam.
- Do not confuse concentration with strength. A concentrated weak acid can still have a modest pH, and a dilute strong acid can be nearly neutral. Ka/pKa describes *how completely* an acid ionizes; molarity describes *how much* acid is present.
- A buffer works best when pKa is within about 1 unit of the target pH, i.e. the [A⁻]/[HA] ratio stays between 1:10 and 10:1. To build a buffer for a given pH, choose a weak acid whose pKa is close to that pH.
- Polyprotic acids (H₂CO₃, H₃PO₄) donate protons one at a time, each with its own Ka, so their curves show *multiple* equivalence points — one steep jump per ionizable proton, with Ka1 > Ka2 > Ka3 since each successive H⁺ is harder to pull off a more negative ion.
- Do not confuse ΔG° with ΔG. ΔG° = −RT ln K compares the *standard state* to equilibrium; a large positive ΔG° just means a small K. The actual ΔG (which is 0 at equilibrium) depends on the real concentrations through ΔG = ΔG° + RT ln Q.
- The AP exam frequently tests the "do not multiply potentials" trap. When you scale a half-reaction to balance electrons, you multiply the atoms and the electrons — but never the E° value. Potential is energy *per charge*, an intensive property that is independent of how much reaction you run.
- The classic Faraday problem gives current and time and asks for mass. Chain the units without skipping steps: A × s = C, then C ÷ 96 485 = mol e⁻, then ÷ electrons-per-ion = mol product, then × molar mass = grams. Carrying units through each arrow prevents the most common errors.
- A concentration cell is the purest test of the Nernst idea: identical electrodes, same E° of 0, so E depends entirely on the concentration difference through −(0.0592/n) log Q. Electrons flow to equalize the two sides, and the voltage dies the instant the concentrations match.
- AP frequently shows two spectra and asks which belongs to the larger atom or which subshell was affected by ionization. Anchor every answer in the two rules: peak position = binding energy (tighter hold, higher Zₑff, shifts left), peak area = electron count. Removing an electron shrinks the corresponding peak’s height.
- When a free-response asks you to *justify* a ranking, cite both Coulomb levers explicitly: state whether Zₑff or r changed and in which direction, then conclude about the force. “Higher Zₑff, same shell → stronger attraction → higher IE / smaller radius” is the sentence that earns the point.
- On free response, never justify equal bond lengths by saying the molecule "switches" between structures — that earns no credit. State that the bonding is delocalized into one resonance hybrid with a single fractional bond order. And always confirm your chosen Lewis structure by showing formal charges that sum to the species' charge.
- Two exam reflexes: (1) σ bonds equal the number of bonded pairs, so σ + π together must equal the total bond count you drew — use that to self-check. (2) When asked to explain a solid's property, name the model explicitly — "delocalized electron sea/band" for metals, "continuous covalent network" for diamond, "mobile ions only when molten" for ionic — and connect it directly to the observed property.
- For real-gas questions, tie the deviation to a *cause*: high pressure → molecular volume no longer negligible (real V larger); low temperature → intermolecular attractions reduce wall impacts (real P smaller). Naming the mechanism, not just "it deviates," is what earns the point.
- For a Beer’s-law free-response, expect a calibration graph of absorbance vs concentration: its slope is εb, so a steeper line means a larger molar absorptivity. Read an unknown by finding its absorbance on the line and dropping to the concentration axis — and always confirm the path length is 1.00 cm before using ε directly.
- On the AP exam a redox equation earns full credit only when it is balanced for mass *and* charge. Always end by summing the charges on each side — if they are not equal, you mis-scaled the electrons or lost an H⁺/H₂O when combining the halves.
- Multi-step AP problems reward a clear moles ledger: write moles of every species, label each with its source, and apply the balanced mole ratio at each junction. Do all conversions to grams or molarity only after the moles are settled — that keeps limiting-reactant, gravimetric, and back-titration problems on the same reliable track.
- On free-response, justify the order by naming **which plot is linear**, then read k off the slope with the right sign (−k for zero/first order, +k for second order) and give **units by overall order** (M·s⁻¹, s⁻¹, M⁻¹·s⁻¹). A bare "second order" without pointing to the linear 1/[A] plot usually loses the justification point.
- To annotate a diagram fast: mark the **peak as the transition state**, arrow **up to it from reactants (Eₐ forward)** and **up to it from products (Eₐ reverse)**, and read **ΔH from reactant level to product level**. Check with ΔH = Eₐ(forward) − Eₐ(reverse). Count humps: two humps means a two-step mechanism with an intermediate in the valley.
- On the free-response section, calorimetry usually feeds Hess or ΔH°f: measure q(reaction) with the calorimeter, divide by moles to get a molar ΔH, then use it as one step or cross-check it against Σ ΔH°f(products) − Σ ΔH°f(reactants). Always report the final sign — exothermic is negative.
- A reliable free-response routine: compute ΔH°rxn and ΔS°rxn from tables, decide the sign case, and if ΔH and ΔS conflict solve T = ΔH/ΔS for the crossover (keep units consistent — ΔH in J with ΔS in J·K⁻¹). Then answer the temperature question by naming which side of the crossover is spontaneous.
- On the free-response, the point is often awarded for the justification, not just the number. After a small-x solve, explicitly write the ratio (e.g. "x/[initial] = 1% < 5%, so the approximation is valid"). If it fails, say so and switch to the quadratic — graders reward recognizing the boundary.
- Two exam reflexes for solubility problems: (1) after mixing solutions, recompute every concentration in the combined volume before finding Q — dilution is the most common oversight; (2) always carry the ion coefficient as an exponent, so a common ion or trial ion with coefficient 2 is squared. Both slips move the answer by orders of magnitude.
- On the exam, first sort each salt ion into "spectator" or "hydrolyzes." Group 1/Group 2 cations and the anions of strong acids (Cl⁻, Br⁻, I⁻, NO₃⁻, ClO₄⁻) are spectators. Any leftover ion traced to a weak acid or weak base is the one that sets the pH.
- Free-response titration questions reward naming the region before calculating. Say "this is the buffer region, so I use Henderson–Hasselbalch" or "this is past equivalence, so I use excess [OH⁻]." Choosing the right tool for the region earns the method points even before the arithmetic.
- Memorize the triangle: E° ↔ ΔG° ↔ K, joined by ΔG° = −nFE° and ΔG° = −RT ln K. Given any one vertex you can reach the other two. The single most common error is scaling E° by n or by a coefficient — E° is intensive and never changes; the n lives in the −nFE° expression, not inside the voltage.
- On the AP exam a concentration cell is instantly recognizable: same metal on both sides, same ion, E° = 0. Do not panic that the voltage looks like it should be zero — plug into E = −(0.0592/n) log Q and let the concentration ratio do the work. The concentrated half-cell is always the cathode.
How to get a 5
- Always show the full setup (formula → substituted numbers → answer) with correct units and sig figs — AP awards the substitution point even if the final arithmetic slips.
- For every "explain/justify," tie your reasoning to particle-level behavior (collisions, IMFs, Coulombic attraction, Zeff) — vague macroscopic statements don't earn the point.
- Master acid-base/equilibrium: buffers, ICE tables, titration curves, and Ksp appear on nearly every exam and are point-dense.
- On multiple choice, use dimensional analysis and proportional reasoning to eliminate answers fast rather than grinding exact calculations.
Key terms
Ideal gas law equation and R value — PV = nRT; R = 0.0821 L·atm·mol⁻¹·K⁻¹ (or 8.314 J·mol⁻¹·K⁻¹).
Definition of a buffer — A solution of a weak acid and its conjugate base (or weak base + conjugate acid) that resists pH change when small amounts of acid or base are added.
Henderson–Hasselbalch equation — pH = pKa + log([A⁻]/[HA]). pH = pKa when [A⁻] = [HA].
Relationship: Kc and Kp — Kp = Kc(RT)^Δn, where Δn = moles gaseous products − moles gaseous reactants.
Sign of ΔG for spontaneity — ΔG < 0 spontaneous; ΔG > 0 nonspontaneous; ΔG = 0 at equilibrium. ΔG = ΔH − TΔS.
ΔG° relationship to K — ΔG° = −RT ln K. K > 1 → ΔG° < 0; K < 1 → ΔG° > 0.
What does a catalyst do? — Lowers activation energy via an alternate pathway; speeds forward and reverse rates equally; does NOT change ΔH, ΔG, or K.
Photoelectron spectroscopy (PES): peak position and height — Peak position (binding energy) indicates the subshell/energy level; peak height (area) indicates the number of electrons in that subshell.
Periodic trend: atomic radius — Decreases across a period (increasing Zeff), increases down a group. Cations smaller than parent atom; anions larger.
Strong acids to memorize — HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄ (and HClO₃) — they ionize completely in water.
Oxidation vs reduction (LEO/GER) — Oxidation = Loss of Electrons (oxidation number increases, at the anode); Reduction = Gain of Electrons (number decreases).
Hess's Law / ΔH°rxn from formation — ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants). ΔH°f of an element in its standard state = 0.