AP Chemistry Reference Sheet (2027), Explained

Using the sheet on exam day

You have the sheet for the whole exam, both sections, in the testing app and as a paper booklet you can do scratch work in. Learn its layout before exam day so you aren't hunting for an equation under time pressure.
The sheet never tells you when an equation applies. It lists $t_{1/2} = \frac{0.693}{k}$ without saying it's only for first-order reactions, and Henderson–Hasselbalch without saying it's only for buffers. Knowing the conditions is your job.
Watch letters that mean two things. On the sheet, $c$ is the speed of light, the concentration in $A = \varepsilon bc$ , the specific heat in $q = mc\Delta T$ , and a coefficient in $K_c$ . Bold $\boldsymbol{M}$ is molar mass and plain $M$ is molarity. $q$ is charge in Coulomb's law and $I = \frac{q}{t}$ , but heat in $q = mc\Delta T$ .
Use the metric prefix table to convert before you plug in: 500 nm is 500 × 10⁻⁹ m, and 25 kJ is 25,000 J. Pick the gas constant whose units match your problem: 0.08206 for liters and atm, 8.314 for joules.
Some things you need are not on the sheet at all, like $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ , Hess's law and $K_{sp}$ setups. See the list at the end of this page.

Constants and conversions

Planck's constant: $\displaystyle h = 6.626 \times 10^{-34}\ \text{J s}$; Use it in $E = h\nu$ to turn a photon's frequency into its energy in joules.
Speed of light: $\displaystyle c = 2.998 \times 10^{8}\ \text{m s}^{-1}$; Use it in $c = \lambda\nu$ to go between wavelength and frequency. Wavelength has to be in meters to match.
Avogadro's number: $\displaystyle 6.022 \times 10^{23}\ \text{mol}^{-1}$; Use it to go between moles and particles: atoms, molecules, ions or photons. Multiply moles by it to get a count, divide a count by it to get moles.
Gas constant, in joules: $\displaystyle R = 8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$; Use this value when energy is involved: $\Delta G^\circ = -RT\ln K$ and the Nernst equation. It gives an answer in joules, so convert if the rest of your work is in kJ.
Gas constant, in liters and atmospheres: $\displaystyle R = 0.08206\ \text{L atm mol}^{-1}\ \text{K}^{-1}$; Use this value in $PV = nRT$ when pressure is in atm and volume is in liters.
Standard temperature and pressure: $\displaystyle \text{STP} = 273.15\ \text{K and } 1.0\ \text{atm}$; STP means 0 °C and 1 atm. Don't mix it up with the 25 °C (298 K) used for standard thermodynamic values like $\Delta H^\circ$ .
Molar volume of an ideal gas at STP: $\displaystyle \text{Ideal gas at STP} = 22.4\ \text{L mol}^{-1}$; A shortcut for gas stoichiometry: at STP, 1 mol of any ideal gas takes up 22.4 L. Use it only at STP; anywhere else, use $PV = nRT$ .
Faraday's constant: $\displaystyle F = 96{,}485\ \text{coulombs} / 1\ \text{mol}\ e^-$; The charge on one mole of electrons. Use it in $\Delta G^\circ = -nFE^\circ$ , the Nernst equation, and electrolysis problems that turn coulombs into moles of electrons.
Hertz: $\displaystyle 1\ \text{hertz} = 1\ \text{s}^{-1}$; Frequency $\nu$ is in hertz, which is the same as per second. That's why $h\nu$ comes out in joules.
Pressure units: $\displaystyle 1\ \text{atm} = 760\ \text{mm Hg} = 760\ \text{torr}$; Convert pressures to atm before using $R = 0.08206$ . For example, 380 mm Hg is 0.500 atm.
Celsius to kelvin: $\displaystyle \text{K} = {}^\circ\text{C} + 273.15$; Every gas law and every thermodynamics equation with $T$ in it needs kelvin. A temperature change $\Delta T$ is the same size in °C and K.
Volt: $\displaystyle 1\ \text{volt} = \dfrac{1\ \text{joule}}{1\ \text{coulomb}}$; This is why $nFE^\circ$ comes out in joules: coulombs times volts is joules.
Ampere: $\displaystyle 1\ \text{ampere} = \dfrac{1\ \text{coulomb}}{1\ \text{second}}$; Current in amperes times time in seconds gives charge in coulombs, which you need in electrolysis problems.

Atomic structure

Light as photons, and the force between charged particles that explains periodic trends.

Energy of a photon

\displaystyle E = h\nu

What the symbols mean

$E$: energy of one photon; Unit: J
$h$: Planck's constant, 6.626 × 10⁻³⁴ J·s; Unit: J·s
$\nu$: frequency of the light (the Greek letter nu, not a v); Unit: s⁻¹ (Hz)

Use it when: Use it when you know the frequency of light and need the energy of one photon, or the other way around. It also links an electron's energy jump to the light it absorbs or emits.

Watch out: This gives the energy of one photon, not a mole of them. If the problem wants kJ/mol, multiply by Avogadro's number and divide by 1000.

Try it: What is the energy of one photon of light with a wavelength of 500. nm?

Answer: First find the frequency: $\nu = \frac{c}{\lambda} = \frac{2.998 \times 10^8\ \text{m/s}}{500. \times 10^{-9}\ \text{m}} = 5.996 \times 10^{14}\ \text{s}^{-1}$ . Then $E = h\nu = (6.626 \times 10^{-34})(5.996 \times 10^{14}) = 3.97 \times 10^{-19}\ \text{J}$ .

Learn it: 3.12 Properties of Photons · 1.6 Photoelectron Spectroscopy · 3.11 Spectroscopy and the Electromagnetic Spectrum

Speed of light, wavelength and frequency

\displaystyle c = \lambda\nu

What the symbols mean

$c$: speed of light, 2.998 × 10⁸ m/s; Unit: m/s
$\lambda$: wavelength; Unit: m
$\nu$: frequency; Unit: s⁻¹ (Hz)

Use it when: Use it to switch between wavelength and frequency. Problems often give a wavelength in nm, so you need this before $E = h\nu$ .

Watch out: Forgetting to convert nanometers to meters. 1 nm = 10⁻⁹ m, so 500 nm is 5.00 × 10⁻⁷ m.

Learn it: 3.11 Spectroscopy and the Electromagnetic Spectrum · 3.12 Properties of Photons

Coulomb's law

\displaystyle F_{\text{coulombic}} \propto \dfrac{q_1 q_2}{r^2}

What the symbols mean

$F_{\text{coulombic}}$: the force of attraction or repulsion between two charges; Unit: N
$q_1, q_2$: the two charges, like a nucleus and an electron, or two ions; Unit: C
$r$: the distance between the charges; Unit: m
$\propto$: is proportional to: the sheet gives no constant, so use it to compare, not to calculate; Unit: none

Use it when: Use it to explain trends: why ionization energy goes up across a period (more protons, same shell), why atoms get bigger down a group, and why MgO has a stronger lattice than NaCl (bigger charges, smaller ions).

Watch out: Saying only one factor changes. In an explanation, name both: the charge (like the number of protons, or the ion charges) and the distance (like the shell number, or the ionic radii), and say which way each pushes the force.

Learn it: 1.5 Atomic Structure and Electron Configuration · 1.7 Periodic Trends · 2.3 Structure of Ionic Solids

Gases, liquids, and solutions

Gas laws, moles and mass, molarity, and measuring concentration with light.

Combined gas law

\displaystyle \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}

What the symbols mean

$P_1, P_2$: pressure before and after; Unit: atm, or any unit as long as both match
$V_1, V_2$: volume before and after; Unit: L, or any unit as long as both match
$T_1, T_2$: temperature before and after; Unit: K

Use it when: Use it when the same sample of gas (fixed moles) changes conditions and you need the new pressure, volume or temperature. If one variable stays constant, just leave it out of both sides.

Watch out: Using °C. A gas at 20 °C isn't half as hot as one at 40 °C, so the ratio only works in kelvin. Add 273.15 first.

Learn it: 3.4 Ideal Gas Law

Ideal gas law

\displaystyle PV = nRT

What the symbols mean

$P$: pressure; Unit: atm
$V$: volume; Unit: L
$n$: number of moles of gas; Unit: mol
$R$: gas constant, 0.08206 L·atm/(mol·K) with these units; Unit: L·atm/(mol·K)
$T$: temperature; Unit: K

Use it when: Use it when a problem gives three of pressure, volume, moles and temperature for one gas sample and asks for the fourth. Pair it with $n = \frac{m}{\boldsymbol{M}}$ to find a gas's molar mass.

Watch out: Mismatched units. With $R = 0.08206$ , pressure must be in atm, volume in liters and temperature in kelvin. Convert mm Hg, mL and °C before you plug in.

Try it: What is the pressure of 0.500 mol of gas in a 5.00 L container at 27 °C?

Answer: Convert to kelvin: 27 + 273.15 = 300.15 K. Then $P = \frac{nRT}{V} = \frac{(0.500)(0.08206)(300.15)}{5.00} = 2.46\ \text{atm}$ .

Learn it: 3.4 Ideal Gas Law · 3.6 Deviation from Ideal Gas Law

Partial pressure from mole fraction

\displaystyle P_A = P_{\text{total}} \times X_A, \text{ where } X_A = \dfrac{\text{moles A}}{\text{total moles}}

What the symbols mean

$P_A$: partial pressure of gas A, the pressure it would have alone in the container; Unit: atm
$P_{\text{total}}$: total pressure of the gas mixture; Unit: atm
$X_A$: mole fraction of A: its share of the total moles, between 0 and 1; Unit: none

Use it when: Use it in a gas mixture when you know the total pressure and the moles of each gas, and need one gas's pressure.

Watch out: Using mass fraction instead of mole fraction. Convert grams of each gas to moles first.

Learn it: 3.4 Ideal Gas Law

Total pressure of a gas mixture

\displaystyle P_{\text{total}} = P_A + P_B + P_C + \dots

What the symbols mean

$P_{\text{total}}$: total pressure of the mixture; Unit: atm
$P_A, P_B, P_C$: partial pressures of each gas in the mixture; Unit: atm

Use it when: Use it when gases share a container and you need the total pressure, or one gas's pressure from the total. A common case is a gas collected over water, where you subtract the water vapor pressure.

Watch out: Forgetting a gas, especially water vapor when a gas is collected over water. Its partial pressure counts toward the total.

Learn it: 3.4 Ideal Gas Law

Moles from mass

\displaystyle n = \dfrac{m}{\boldsymbol{M}}

What the symbols mean

$n$: number of moles; Unit: mol
$m$: mass of the sample; Unit: g
$\boldsymbol{M}$: molar mass, from the periodic table (bold M on the sheet); Unit: g/mol

Use it when: Use it every time you go between grams and moles, which is the first step of almost every stoichiometry problem.

Watch out: Mixing up bold $\boldsymbol{M}$ (molar mass, g/mol) with plain $M$ (molarity, mol/L). The sheet uses both, so check which one the equation means.

Learn it: 1.1 Moles and Molar Mass · 4.5 Stoichiometry

Density

\displaystyle D = \dfrac{m}{V}

What the symbols mean

$D$: density; Unit: g/L for gases, g/mL for liquids and solids
$m$: mass; Unit: g
$V$: volume; Unit: L or mL

Use it when: Use it on its own, or with $PV = nRT$ and $n = \frac{m}{\boldsymbol{M}}$ to find a gas's density or molar mass at a given pressure and temperature.

Watch out: Mixing g/L and g/mL. Gas densities are usually given in g/L, and 1 g/mL = 1000 g/L, so keep track of which volume unit you used.

Learn it: 3.4 Ideal Gas Law

Kinetic energy

\displaystyle KE = \dfrac{1}{2}mv^2

What the symbols mean

$KE$: kinetic energy of a moving particle; Unit: J
$m$: mass of the particle; Unit: kg
$v$: velocity (speed) of the particle (a plain v, not the Greek nu); Unit: m/s

Use it when: Use it with kinetic molecular theory: at the same temperature, all gases have the same average kinetic energy, so lighter molecules must move faster.

Watch out: Thinking heavier molecules have more kinetic energy at the same temperature. Average KE depends only on temperature; a heavier gas just has a lower average speed.

Learn it: 3.5 Kinetic Molecular Theory

Molarity

\displaystyle M = \dfrac{n_{\text{solute}}}{\text{L}_{\text{solution}}}

What the symbols mean

$M$: molarity, the concentration of a solution; Unit: mol/L (M)
$n_{\text{solute}}$: moles of the dissolved substance; Unit: mol
$\text{L}_{\text{solution}}$: volume of the whole solution in liters; Unit: L

Use it when: Use it to find a concentration, or to find moles from a volume and concentration (moles = M × L). Titration and solution stoichiometry problems use it constantly.

Watch out: Using milliliters, or the volume of solvent instead of the total solution. Convert mL to L (divide by 1000) and use the final solution volume.

Learn it: 3.7 Solutions and Mixtures · 4.5 Stoichiometry · 4.6 Introduction to Titration

Beer–Lambert law

\displaystyle A = \varepsilon bc

What the symbols mean

$A$: absorbance, read from the spectrophotometer; Unit: none
$\varepsilon$: molar absorptivity: how strongly the substance absorbs at that wavelength; Unit: L/(mol·cm), also written M⁻¹ cm⁻¹
$b$: path length, the width of the cuvette the light passes through; Unit: cm
$c$: concentration (here c is not the speed of light); Unit: mol/L

Use it when: Use it when a lab measures absorbance to find a concentration. Absorbance is directly proportional to concentration, so a calibration graph of A against c is a straight line.

Watch out: Treating $c$ as the speed of light. In this equation it's concentration. Also remember $\varepsilon$ depends on the wavelength, so measurements must be at the same wavelength.

Learn it: 3.13 Beer-Lambert Law

Kinetics

The integrated rate laws for zero-, first- and second-order reactions, and half-life.

Zero-order integrated rate law

\displaystyle [\text{A}]_t - [\text{A}]_0 = -kt

What the symbols mean

$[\text{A}]_t$: concentration of reactant A at time t; Unit: mol/L
$[\text{A}]_0$: starting concentration of A; Unit: mol/L
$k$: rate constant; Unit: mol/(L·s), or M/s
$t$: time; Unit: s

Use it when: Use it for a zero-order reaction, where the rate doesn't depend on concentration. A graph of [A] against time is a straight line with slope $-k$ .

Watch out: Using it without checking the order. Only use it when [A] against time is the straight-line graph.

Learn it: 5.3 Concentration Changes Over Time

First-order integrated rate law

\displaystyle \ln[\text{A}]_t - \ln[\text{A}]_0 = -kt

What the symbols mean

$\ln[\text{A}]_t$: natural log of the concentration of A at time t; Unit: none
$\ln[\text{A}]_0$: natural log of the starting concentration; Unit: none
$k$: rate constant; Unit: s⁻¹
$t$: time; Unit: s

Use it when: Use it for a first-order reaction, including radioactive decay. A graph of ln[A] against time is a straight line with slope $-k$ .

Watch out: Using log (base 10) instead of ln (natural log). On your calculator, press ln, and undo it with $e^x$ , not $10^x$ .

Try it: A first-order reaction has $k = 0.0231\ \text{s}^{-1}$ . If $[\text{A}]_0 = 0.800\ \text{M}$ , what is [A] after 60.0 s?

Answer: $\ln[\text{A}]_t = \ln(0.800) - (0.0231)(60.0) = -0.223 - 1.386 = -1.609$ . Then $[\text{A}]_t = e^{-1.609} = 0.200\ \text{M}$ . That's two half-lives, since $t_{1/2} = \frac{0.693}{0.0231} = 30.0\ \text{s}$ .

Learn it: 5.3 Concentration Changes Over Time

Second-order integrated rate law

\displaystyle \dfrac{1}{[\text{A}]_t} - \dfrac{1}{[\text{A}]_0} = kt

What the symbols mean

$\frac{1}{[\text{A}]_t}$: one over the concentration of A at time t; Unit: L/mol
$\frac{1}{[\text{A}]_0}$: one over the starting concentration; Unit: L/mol
$k$: rate constant; Unit: L/(mol·s), or M⁻¹ s⁻¹
$t$: time; Unit: s

Use it when: Use it for a second-order reaction in one reactant. A graph of 1/[A] against time is a straight line with a positive slope equal to $k$ .

Watch out: Getting the sign wrong. This one is $+kt$ , not $-kt$ , because 1/[A] goes up as [A] goes down.

Learn it: 5.3 Concentration Changes Over Time

Half-life of a first-order reaction

\displaystyle t_{1/2} = \dfrac{0.693}{k}

What the symbols mean

$t_{1/2}$: half-life, the time for half the reactant to be used up; Unit: s
$k$: first-order rate constant; Unit: s⁻¹
$0.693$: ln 2, rounded; Unit: none

Use it when: Use it for first-order reactions to go between $k$ and the half-life. A first-order half-life stays the same no matter how much reactant is left.

Watch out: Using it for zero- or second-order reactions. The sheet doesn't say so, but this equation is only true for first order. For other orders the half-life changes as the concentration drops.

Learn it: 5.3 Concentration Changes Over Time

Equilibrium

Writing equilibrium constants from concentrations or gas pressures. The sheet's list of equilibrium constants names five: $K_c$ (molar concentrations), $K_p$ (gas pressures), $K_w$ (water), $K_a$ (acid) and $K_b$ (base).

Equilibrium constant in concentrations

\displaystyle K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}, \text{ where } a\text{A} + b\text{B} \rightleftarrows c\text{C} + d\text{D}

What the symbols mean

$K_c$: equilibrium constant written with molar concentrations; Unit: none
$[\text{A}], [\text{B}], [\text{C}], [\text{D}]$: concentrations at equilibrium; Unit: mol/L
$a, b, c, d$: coefficients from the balanced equation (here c is a coefficient, not concentration); Unit: none

Use it when: Use it to write the $K$ expression for any balanced reaction, to calculate $K$ from equilibrium concentrations, or to find a missing concentration. Put current, non-equilibrium values in the same form and you get $Q$ .

Watch out: Two classic slips: forgetting the coefficients as exponents, and putting pure solids and liquids in the expression. Leave out anything labeled (s) or (l), like solid CaCO₃ or liquid H₂O.

Try it: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium concentrations are [N₂] = 0.50 M, [H₂] = 0.20 M and [NH₃] = 0.10 M. Find $K_c$ .

Answer: $K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.10)^2}{(0.50)(0.20)^3} = \frac{0.010}{0.0040} = 2.5$ . Without the exponents you'd get 1.0, which is wrong.

Learn it: 7.3 Reaction Quotient and Equilibrium Constant · 7.4 Calculating the Equilibrium Constant

Equilibrium constant in partial pressures

\displaystyle K_p = \dfrac{(P_\text{C})^c(P_\text{D})^d}{(P_\text{A})^a(P_\text{B})^b}

What the symbols mean

$K_p$: equilibrium constant written with gas pressures; Unit: none
$P_\text{A}, P_\text{B}, P_\text{C}, P_\text{D}$: partial pressures of each gas at equilibrium; Unit: atm
$a, b, c, d$: coefficients from the balanced equation; Unit: none

Use it when: Use it for gas-phase equilibria when the problem gives partial pressures instead of concentrations.

Watch out: Including solids, liquids or dissolved species. $K_p$ uses only gases, each raised to its coefficient.

Learn it: 7.3 Reaction Quotient and Equilibrium Constant · 7.4 Calculating the Equilibrium Constant

Acids and bases

On the sheet these sit under Equilibrium: water's equilibrium, pH, acid and base constants, and the buffer equation.

Ionization of water

\displaystyle K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 25^\circ\text{C}

What the symbols mean

$K_w$: the equilibrium constant for water; Unit: none
$[\text{H}_3\text{O}^+]$: hydronium ion concentration; Unit: mol/L
$[\text{OH}^-]$: hydroxide ion concentration; Unit: mol/L

Use it when: Use it to find [OH⁻] from [H₃O⁺] or the other way around, in any water solution at 25 °C.

Watch out: Using 1.0 × 10⁻¹⁴ at other temperatures. $K_w$ grows as water warms, so pure water at a higher temperature has a pH below 7 and is still neutral.

Learn it: 8.1 Introduction to Acids and Bases · 8.2 pH and pOH of Strong Acids and Bases

pH plus pOH

\displaystyle \text{p}K_w = 14 = \text{pH} + \text{pOH} \text{ at } 25^\circ\text{C}

What the symbols mean

$\text{p}K_w$: negative log of $K_w$; Unit: none
$\text{pH}$: negative log of [H₃O⁺]; Unit: none
$\text{pOH}$: negative log of [OH⁻]; Unit: none

Use it when: Use it to switch between pH and pOH. For a base, it's often easiest to find pOH first, then subtract from 14.

Watch out: Stopping at pOH when the question asks for pH. For bases, check which one you reported.

Learn it: 8.1 Introduction to Acids and Bases · 8.2 pH and pOH of Strong Acids and Bases

pH and pOH

\displaystyle \text{pH} = -\log[\text{H}_3\text{O}^+], \quad \text{pOH} = -\log[\text{OH}^-]

What the symbols mean

$\text{pH}$: a log scale of acidity: lower means more acidic; Unit: none
$\text{pOH}$: the same scale for hydroxide; Unit: none
$\log$: base-10 log; Unit: none

Use it when: Use it to turn a concentration into pH or pOH. To go back, $[\text{H}_3\text{O}^+] = 10^{-\text{pH}}$ , which isn't printed on the sheet. For a strong acid, [H₃O⁺] equals the acid's concentration.

Watch out: Pressing ln instead of log, or dropping the minus sign. A pH should come out between about 0 and 14 for normal solutions; a negative answer for a dilute acid means a sign slip.

Try it: A solution has [OH⁻] = 2.0 × 10⁻⁵ M at 25 °C. What is its pH?

Answer: $\text{pOH} = -\log(2.0 \times 10^{-5}) = 4.70$ . Then $\text{pH} = 14 - 4.70 = 9.30$ .

Learn it: 8.1 Introduction to Acids and Bases · 8.2 pH and pOH of Strong Acids and Bases

Acid and base ionization constants

\displaystyle K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}, \quad K_b = \dfrac{[\text{OH}^-][\text{HB}^+]}{[\text{B}]}

What the symbols mean

$K_a$: acid ionization constant: bigger means a stronger acid; Unit: none
$K_b$: base ionization constant: bigger means a stronger base; Unit: none
$[\text{HA}], [\text{A}^-]$: concentrations of the weak acid and its conjugate base; Unit: mol/L
$[\text{B}], [\text{HB}^+]$: concentrations of the weak base and its conjugate acid; Unit: mol/L

Use it when: Use them to find the pH of a weak acid or weak base solution, usually with an ICE table. For a weak acid alone, $[\text{H}_3\text{O}^+] = [\text{A}^-] = x$ , so $K_a \approx \frac{x^2}{[\text{HA}]_0}$ .

Watch out: Using them for strong acids or bases, which ionize completely and don't need an equilibrium setup. Also, water is a liquid, so it never appears in these expressions.

Learn it: 8.3 Weak Acid and Base Equilibria

pKa and pKb

\displaystyle \text{p}K_a = -\log K_a, \quad \text{p}K_b = -\log K_b

What the symbols mean

$\text{p}K_a$: negative log of $K_a$; Unit: none
$\text{p}K_b$: negative log of $K_b$; Unit: none

Use it when: Use it to compare acid strengths quickly or to get $\text{p}K_a$ for the Henderson–Hasselbalch equation.

Watch out: Flipping the trend. A smaller $\text{p}K_a$ means a larger $K_a$ , so a stronger acid.

Learn it: 8.3 Weak Acid and Base Equilibria · 8.7 pH and pKa

Conjugate acid–base pairs

\displaystyle K_w = K_a \times K_b, \quad \text{p}K_w = \text{p}K_a + \text{p}K_b

What the symbols mean

$K_a$: $K_a$ of the acid in a conjugate pair; Unit: none
$K_b$: $K_b$ of its conjugate base; Unit: none
$K_w$: 1.0 × 10⁻¹⁴ at 25 °C; Unit: none

Use it when: Use it when you know $K_a$ for an acid like HF and need $K_b$ for its conjugate base F⁻, or the other way around.

Watch out: Using it for an acid and a base that aren't a conjugate pair. It only links HA with its own A⁻, or B with its own HB⁺.

Learn it: 8.3 Weak Acid and Base Equilibria

Henderson–Hasselbalch equation

\displaystyle \text{pH} = \text{p}K_a + \log\dfrac{[\text{A}^-]}{[\text{HA}]}

What the symbols mean

$\text{p}K_a$: $\text{p}K_a$ of the weak acid in the buffer; Unit: none
$[\text{A}^-]$: concentration (or moles) of the conjugate base; Unit: mol/L
$[\text{HA}]$: concentration (or moles) of the weak acid; Unit: mol/L

Use it when: Use it for a buffer, a solution with real amounts of both a weak acid and its conjugate base. When they're equal, the log term is zero and pH = pKa.

Watch out: Using it when there's no buffer, like a weak acid alone or a strong acid. It also only works with the ratio base over acid; flipping it gives the wrong sign on the log.

Try it: A buffer has 0.20 M acetate (A⁻) and 0.10 M acetic acid (HA). The $\text{p}K_a$ of acetic acid is 4.74. What is the pH?

Answer: $\text{pH} = 4.74 + \log\frac{0.20}{0.10} = 4.74 + 0.30 = 5.04$ . More base than acid, so the pH is above the $\text{p}K_a$ .

Learn it: 8.9 Henderson-Hasselbalch Equation · 8.8 Properties of Buffers

Thermodynamics

Heat, enthalpy, entropy and free energy. The sheet chains $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -RT\ln K = -nFE^\circ$ into one block; here each relation gets its own entry, and the last one is under Electrochemistry.

Heat and temperature change

\displaystyle q = mc\Delta T

What the symbols mean

$q$: heat absorbed or released (here q is heat, not charge); Unit: J
$m$: mass of the substance being heated, often the water or solution; Unit: g
$c$: specific heat capacity (water is 4.18 J/(g·°C)); Unit: J/(g·°C)
$\Delta T$: temperature change, final minus initial; Unit: °C or K (same size)

Use it when: Use it in calorimetry to find the heat gained or lost by water or a solution from its temperature change.

Watch out: Getting the sign of the reaction's heat wrong. If the water warms up, $q$ for the water is positive, so the reaction released heat and its $\Delta H$ is negative.

Try it: 50.0 g of water (c = 4.18 J/(g·°C)) warms by 6.00 °C. How much heat did it absorb?

Answer: $q = mc\Delta T = (50.0)(4.18)(6.00) = 1.25 \times 10^3\ \text{J}$ , or 1.25 kJ.

Learn it: 6.4 Heat Capacity and Calorimetry · 6.3 Heat Transfer and Thermal Equilibrium

Enthalpy change from enthalpies of formation

\displaystyle \Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_{f\ \text{products}} - \sum \Delta H^\circ_{f\ \text{reactants}}

What the symbols mean

$\Delta H^\circ_{\text{reaction}}$: standard enthalpy change of the reaction: negative means exothermic; Unit: kJ/mol
$\Delta H^\circ_f$: standard enthalpy of formation of each substance, from a table; Unit: kJ/mol
$\sum$: add up, after multiplying each by its coefficient; Unit: none

Use it when: Use it when a problem gives a table of $\Delta H^\circ_f$ values and asks for the reaction's $\Delta H^\circ$ .

Watch out: Forgetting to multiply by coefficients, or flipping the order. It's products minus reactants. Elements in their standard state, like O₂(g), have $\Delta H^\circ_f = 0$ .

Learn it: 6.8 Enthalpy of Formation

Entropy change from standard entropies

\displaystyle \Delta S^\circ_{\text{reaction}} = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}

What the symbols mean

$\Delta S^\circ_{\text{reaction}}$: standard entropy change of the reaction; Unit: J/(mol·K)
$S^\circ$: standard (absolute) entropy of each substance, from a table; Unit: J/(mol·K)

Use it when: Use it when a problem gives $S^\circ$ values and asks for $\Delta S^\circ$ of a reaction.

Watch out: Setting $S^\circ = 0$ for elements. Unlike $\Delta H^\circ_f$ , an element's $S^\circ$ is not zero, so include every substance. Also note it's in J, not kJ.

Learn it: 9.2 Absolute Entropy and Entropy Change

Free energy change from free energies of formation

\displaystyle \Delta G^\circ_{\text{reaction}} = \sum \Delta G^\circ_{f\ \text{products}} - \sum \Delta G^\circ_{f\ \text{reactants}}

What the symbols mean

$\Delta G^\circ_{\text{reaction}}$: standard Gibbs free energy change: negative means thermodynamically favorable; Unit: kJ/mol
$\Delta G^\circ_f$: standard free energy of formation of each substance, from a table; Unit: kJ/mol

Use it when: Use it when a table gives $\Delta G^\circ_f$ values. If it gives $\Delta H^\circ$ and $\Delta S^\circ$ instead, use $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ .

Watch out: Forgetting the coefficients, just like with enthalpy. Elements in their standard state have $\Delta G^\circ_f = 0$ .

Learn it: 9.3 Gibbs Free Energy and Thermodynamic Favorability

Free energy from enthalpy and entropy

\displaystyle \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

What the symbols mean

$\Delta G^\circ$: standard free energy change: negative means thermodynamically favorable; Unit: kJ/mol
$\Delta H^\circ$: standard enthalpy change; Unit: kJ/mol
$T$: temperature; Unit: K
$\Delta S^\circ$: standard entropy change; Unit: J/(mol·K)

Use it when: Use it to decide if a reaction is favorable at a given temperature, or to find the temperature where it switches (set $\Delta G^\circ = 0$ , so $T = \frac{\Delta H^\circ}{\Delta S^\circ}$ ).

Watch out: Mixing kJ and J. $\Delta H^\circ$ is usually in kJ and $\Delta S^\circ$ in J/K, so divide $\Delta S^\circ$ by 1000 first. And use kelvin, not °C.

Try it: For N₂ + 3H₂ → 2NH₃, $\Delta H^\circ = -92.2\ \text{kJ/mol}$ and $\Delta S^\circ = -198.7\ \text{J/(mol·K)}$ . Find $\Delta G^\circ$ at 298 K.

Answer: Convert: $\Delta S^\circ = -0.1987\ \text{kJ/(mol·K)}$ . Then $\Delta G^\circ = -92.2 - (298)(-0.1987) = -92.2 + 59.2 = -33.0\ \text{kJ/mol}$ , so it's favorable at 298 K.

Learn it: 9.3 Gibbs Free Energy and Thermodynamic Favorability

Free energy and the equilibrium constant

\displaystyle \Delta G^\circ = -RT\ln K

What the symbols mean

$\Delta G^\circ$: standard free energy change; Unit: J/mol
$R$: gas constant, 8.314 J/(mol·K); Unit: J/(mol·K)
$T$: temperature; Unit: K
$K$: equilibrium constant; Unit: none

Use it when: Use it to go between $\Delta G^\circ$ and $K$ . Negative $\Delta G^\circ$ means $K > 1$ (products favored); positive means $K < 1$ .

Watch out: Two at once: using $R = 0.08206$ instead of 8.314, and leaving $\Delta G^\circ$ in kJ. With 8.314, $\Delta G^\circ$ must be in J. It's also ln, not log.

Learn it: 9.5 Free Energy and Equilibrium

Electrochemistry

Cell potential, free energy and electrolysis. Faraday's constant is in the constants list above.

Free energy and cell potential

\displaystyle \Delta G^\circ = -nFE^\circ

What the symbols mean

$\Delta G^\circ$: standard free energy change; Unit: J/mol
$n$: number of moles of electrons transferred in the balanced redox reaction; Unit: mol e⁻
$F$: Faraday's constant, 96,485 C/mol e⁻; Unit: C/mol e⁻
$E^\circ$: standard cell potential; Unit: V

Use it when: Use it to go between a cell's standard potential and $\Delta G^\circ$ . A positive $E^\circ$ gives a negative $\Delta G^\circ$ : a favorable, galvanic cell.

Watch out: Using moles of reactant for $n$ . It's the moles of electrons transferred, found by balancing the half-reactions. Don't multiply $E^\circ$ by coefficients either; potentials don't scale.

Try it: A zinc–copper cell transfers 2 electrons and has $E^\circ = 1.10\ \text{V}$ . Find $\Delta G^\circ$ .

Answer: $\Delta G^\circ = -(2)(96{,}485)(1.10) = -2.12 \times 10^5\ \text{J/mol}$ , or −212 kJ/mol.

Learn it: 9.9 Cell Potential and Free Energy

Current

\displaystyle I = \dfrac{q}{t}

What the symbols mean

$I$: current; Unit: A
$q$: charge (here q is charge, not heat); Unit: C
$t$: time; Unit: s

Use it when: Use it in electrolysis: current × time gives coulombs, Faraday's constant turns coulombs into moles of electrons, and the half-reaction turns those into moles of metal.

Watch out: Leaving time in minutes or hours. Convert to seconds first, since 1 A = 1 C/s.

Try it: A current of 2.00 A runs for 965 s through a Cu²⁺ solution. What mass of copper plates out? (Cu²⁺ + 2e⁻ → Cu, molar mass 63.55 g/mol)

Answer: $q = It = (2.00)(965) = 1930\ \text{C}$ , which is $\frac{1930}{96{,}485} = 0.0200\ \text{mol}\ e^-$ . Two electrons per Cu gives 0.0100 mol Cu, and $0.0100 \times 63.55 = 0.636\ \text{g}$ .

Learn it: 9.11 Electrolysis and Faraday's Law

Nernst equation

\displaystyle E_{\text{cell}} = E^\circ_{\text{cell}} - \dfrac{RT}{nF}\ln Q

What the symbols mean

$E_{\text{cell}}$: cell potential under the actual conditions; Unit: V
$E^\circ_{\text{cell}}$: standard cell potential (1 M solutions, 1 atm gases); Unit: V
$R$: gas constant, 8.314 J/(mol·K); Unit: J/(mol·K)
$T$: temperature; Unit: K
$n$: moles of electrons transferred; Unit: mol e⁻
$F$: Faraday's constant, 96,485 C/mol e⁻; Unit: C/mol e⁻
$Q$: reaction quotient for the cell reaction; Unit: none

Use it when: Use it, or just its logic, when concentrations aren't standard. If $Q < 1$ , $\ln Q$ is negative and $E_{\text{cell}}$ is bigger than $E^\circ_{\text{cell}}$ ; if $Q > 1$ , it's smaller.

Watch out: Getting the direction backward. Many questions only ask whether the voltage goes up or down, so reason from $Q$ : more product makes $Q$ bigger and the voltage lower.

Learn it: 9.10 Cell Potential Under Nonstandard Conditions

Not on the sheet: know these

The exam expects you to know these without being given them.

Standard cell potential from half-cells
$\displaystyle E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
Problems give a table of standard reduction potentials and expect you to combine them. Use both values as reduction potentials, and never multiply them by coefficients. A positive answer means a galvanic cell.
Learn it: 9.8 Galvanic (Voltaic) and Electrolytic Cells · 9.9 Cell Potential and Free Energy
Enthalpy from bond enthalpies
$\displaystyle \Delta H^\circ = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$
Breaking bonds takes energy and forming them releases it. Note this is reactants minus products, the opposite order from the formation-enthalpy equation on the sheet.
Learn it: 6.7 Bond Enthalpies
Hess's law
To combine reactions: reverse a reaction and you flip the sign of its $\Delta H$ ; multiply a reaction by a number and you multiply its $\Delta H$ by the same number. Then add the $\Delta H$ values.
Learn it: 6.9 Hess's Law
Heat lost equals heat gained
$\displaystyle q_{\text{system}} = -q_{\text{surroundings}}$
In calorimetry, the heat the water gains is the heat the reaction loses. So $\Delta H$ for the reaction is $-q_{\text{water}}$ divided by the moles that reacted.
Learn it: 6.3 Heat Transfer and Thermal Equilibrium · 6.4 Heat Capacity and Calorimetry
Q compared with K
If $Q < K$ , the reaction shifts forward toward products. If $Q > K$ , it shifts in reverse. If $Q = K$ , it's at equilibrium.
Learn it: 7.3 Reaction Quotient and Equilibrium Constant · 7.10 Reaction Quotient and Le Châtelier's Principle
K for reversed or multiplied reactions
$\displaystyle K_{\text{reverse}} = \dfrac{1}{K}, \quad K_{\text{times } n} = K^n$
Reverse a reaction and its new $K$ is the reciprocal. Multiply all coefficients by $n$ and raise $K$ to the $n$ . Add two reactions and multiply their $K$ values.
Learn it: 7.6 Properties of the Equilibrium Constant
Solubility product and molar solubility
$\displaystyle \text{CaF}_2\text{: } K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = (s)(2s)^2 = 4s^3$
$K_{sp}$ follows the same rules as any $K$ : the solid is left out, and coefficients become exponents. Write each ion's concentration in terms of the molar solubility $s$ to go between them.
Learn it: 7.11 Introduction to Solubility Equilibria · 7.12 Common-Ion Effect
Dilution
$\displaystyle M_1V_1 = M_2V_2$
Adding water doesn't change the moles of solute, so molarity times volume stays the same. Use it to find a new concentration after diluting.
Learn it: 3.7 Solutions and Mixtures · 4.6 Introduction to Titration
Percent composition and empirical formulas
$\displaystyle \text{mass percent} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100\%$
To find an empirical formula, turn each mass percent into grams (assume 100 g), then into moles, then divide by the smallest number of moles to get the whole-number ratio.
Learn it: 1.3 Elemental Composition of Pure Substances
pH equals pKa at half-equivalence
$\displaystyle \text{pH} = \text{p}K_a$
Halfway to the equivalence point in a weak acid titration, half the acid has become its conjugate base, so [HA] = [A⁻]. Read the pH there off the curve and you have the $\text{p}K_a$ .
Learn it: 8.5 Acid-Base Titrations · 8.9 Henderson-Hasselbalch Equation

Using the sheet on exam day

Constants and conversions

Atomic structure

Energy of a photon

What the symbols mean

Speed of light, wavelength and frequency

What the symbols mean

Coulomb's law

What the symbols mean

Gases, liquids, and solutions

Combined gas law

What the symbols mean

Ideal gas law

What the symbols mean

Partial pressure from mole fraction

What the symbols mean

Total pressure of a gas mixture

What the symbols mean

Moles from mass

What the symbols mean

Density

What the symbols mean

Kinetic energy

What the symbols mean

Molarity

What the symbols mean

Beer–Lambert law

What the symbols mean

Kinetics

Zero-order integrated rate law

What the symbols mean

First-order integrated rate law

What the symbols mean

Second-order integrated rate law

What the symbols mean

Half-life of a first-order reaction

What the symbols mean

Equilibrium

Equilibrium constant in concentrations

What the symbols mean

Equilibrium constant in partial pressures

What the symbols mean

Acids and bases

Ionization of water

What the symbols mean

pH plus pOH

What the symbols mean

pH and pOH

What the symbols mean

Acid and base ionization constants

What the symbols mean

pKa and pKb

What the symbols mean

Conjugate acid–base pairs

What the symbols mean

Henderson–Hasselbalch equation

What the symbols mean

Thermodynamics

Heat and temperature change

What the symbols mean

Enthalpy change from enthalpies of formation

What the symbols mean

Entropy change from standard entropies

What the symbols mean

Free energy change from free energies of formation

What the symbols mean

Free energy from enthalpy and entropy

What the symbols mean

Free energy and the equilibrium constant

What the symbols mean

Electrochemistry

Free energy and cell potential

What the symbols mean

Current

What the symbols mean

Nernst equation

What the symbols mean

Not on the sheet: know these

Standard cell potential from half-cells

Enthalpy from bond enthalpies