Master the math behind equilibria, pH, rate laws, and electrochemistry
This second-semester math review begins with exponent and logarithm skills, then develops equilibrium, acid–base, kinetics, thermodynamics, and electrochemical calculations. Work examples one step at a time without assuming a separate college math prerequisite.
Part 1 · Logarithms, equilibrium, and ICE tables
Logarithms and antilogarithms
A base-10 logarithm is the exponent on 10. Its inverse is raising 10 to a power.
log10(0.001) = −3. So, if log10(x) = −4.20, then x = 10−4.20 = 6.31×10−5.
Evaluate log10(1000).
Show worked answer
3, because 103 = 1000.
Common mistake: On a scientific calculator, distinguish log10 from natural log (ln).
Powers and small concentrations
Use exponent laws before entering products or quotients of very small equilibrium concentrations.
(2.0×10−4)²/(4.0×10−2) = 4.0×10−8/4.0×10−2 = 1.0×10−6.
Calculate (3.0×10−3)²/(9.0×10−2).
Show worked answer
9.0×10−6/9.0×10−2 = 1.0×10−4.
Common mistake: Use parentheses when squaring a number in scientific notation.
Write and evaluate Kc
Use balanced-equation coefficients as exponents; omit pure solids and liquids in the expression.
N2 + 3H2 ⇌ 2NH3. If [NH3] = 0.100 M, [N2] = 0.200 M, and [H2] = 0.500 M, Kc = (0.100)²/[(0.200)(0.500)³] = 0.400.
For A ⇌ 2B, [A] = 0.200 M and [B] = 0.300 M at equilibrium. Find Kc.
Show worked answer
Kc = [B]²/[A] = (0.300)²/(0.200) = 0.450.
Common mistake: Use equilibrium values—not initial amounts—in K expressions.
ICE tables and reaction changes
Initial–Change–Equilibrium tables convert one unknown reaction extent into equilibrium concentrations.
For N2O4 ⇌ 2NO2, start with 0.100 M N2O4 and 0 NO2. If x = 0.020 M dissociates: [N2O4]eq = 0.080 M and [NO2]eq = 0.040 M, so Kc = (0.040)²/0.080 = 0.020.
For A ⇌ 2B, starting at [A] = 0.200 M and [B] = 0, x = 0.030 M reacts. Find equilibrium [A] and [B].
Show worked answer
[A] = 0.200 − 0.030 = 0.170 M; [B] = 0 + 2(0.030) = 0.060 M.
Common mistake: A coefficient of 2 means the concentration change is ±2x, not ±x.
Reaction quotient Q versus K
Calculate Q with current—not necessarily equilibrium—concentrations, then compare it with K.
If Q = 0.030 and K = 0.100, then Q < K. The mixture shifts toward products to raise Q toward K.
If Q = 5.0 and K = 2.0, which direction is favored?
Show worked answer
Q > K: net change goes toward reactants until equilibrium is reached.
Common mistake: Q = K means the system is at equilibrium; it does not mean every concentration is equal.
Kp and Kc
For ideal gases, Kp = Kc(RT)Δngas, where Δngas is gas-product coefficient sum minus gas-reactant coefficient sum.
For 2NO2(g) ⇌ N2O4(g), Δn = 1−2 = −1. If Kc = 4.0 at 300 K, Kp = 4.0/(0.08206×300) ≈ 0.16.
For A(g) ⇌ B(g), what is the numerical Kp/Kc ratio under the ideal-gas approximation?
Show worked answer
Δn = 1−1 = 0; (RT)0 = 1, so Kp = Kc.
Common mistake: Count only gaseous species for Δngas; do not include solids or liquids.
Quadratic equations in equilibria
When x is not small compared with the initial concentration, solve the exact equation instead of dropping x.
For HA ⇌ H⁺ + A⁻, C = 0.100 M and Ka = 1.00×10−3: x²/(0.100−x) = 0.001 → x² + 0.001x − 0.000100 = 0. Positive root: x ≈ 9.51×10−3 M.
Why is the approximation x ≪ 0.100 questionable here?
Show worked answer
x/C ≈ 0.00951/0.100 = 9.5%, greater than a common 5% approximation check.
Common mistake: Discard mathematically valid negative roots that imply impossible negative equilibrium concentrations.
Part 2 · pH, buffers, titrations, and solubility
Calculate pH from [H⁺]
pH = −log10aH⁺; introductory dilute-solution calculations commonly approximate activity by molar concentration.
With [H⁺] ≈ 3.00×10−4 M, pH ≈ −log10(3.00×10−4) = 3.523.
Find pH when [H⁺] ≈ 1.00×10−5 M.
Show worked answer
pH ≈ −log(1.00×10−5) = 5.000.
Common mistake: For log results, the concentration’s significant figures control the pH digits after the decimal.
Calculate [H⁺] from pH
Undo a negative base-10 log by raising 10 to −pH.
If pH = 4.20, [H⁺] ≈ 10−4.20 = 6.3×10−5 M.
A solution has pH 3.00. Estimate [H⁺].
Show worked answer
10−3.00 = 1.0×10−3 M.
Common mistake: Do not compute [H⁺] as 1/pH; the inverse is an antilogarithm.
pOH, Kw, and temperature
At 25 °C for dilute aqueous solutions, pH + pOH ≈ 14.00; also Kw ≈ 1.0×10−14. The sum changes with temperature.
If [OH⁻] ≈ 1.0×10−3 M at 25 °C, pOH = 3.00 and pH = 11.00.
At 25 °C, if pOH = 5.40, what is pH?
Show worked answer
pH = 14.00 − 5.40 = 8.60.
Common mistake: Do not automatically use 14.00 at arbitrary temperatures.
Weak-acid Ka calculation
Use an ICE table with Ka = [H⁺][A⁻]/[HA], then apply a valid approximation or quadratic.
For 0.100 M HA dissociating to x = 0.00100 M, Ka = (0.00100)²/(0.100−0.00100) ≈ 1.01×10−5.
If CHA = 0.200 M and x = 0.00200 M, calculate Ka.
Show worked answer
Ka = (0.00200)²/(0.200−0.00200) ≈ 2.02×10−5.
Common mistake: Do not assume every acid fully dissociates; a weak-acid equilibrium must be accounted for.
Buffers and Henderson–Hasselbalch
pH ≈ pKa + log10([A⁻]/[HA]) for a suitable dilute weak-acid/conjugate-base buffer.
With pKa = 4.76, [A⁻] = 0.200 M, [HA] = 0.100 M: pH ≈ 4.76 + log(2.00) = 5.06.
If [A⁻] = [HA], what is pH relative to pKa?
Show worked answer
The ratio is 1 and log(1) = 0, so pH ≈ pKa.
Common mistake: The ratio is base over acid, not acid over base. After a reaction, use the remaining amounts.
Titration mole relationships
At the equivalence point, balanced coefficients determine how much acid and base reacted.
25.00 mL of 0.1000 M monoprotic acid contains 0.002500 mol acid. Neutralizing with 0.1250 M NaOH requires V = 0.002500/0.1250 = 0.02000 L = 20.00 mL.
How much 0.100 M NaOH neutralizes 20.0 mL of 0.150 M monoprotic acid?
Show worked answer
n acid = 0.0200×0.150 = 0.00300 mol; V base = 0.00300/0.100 = 0.0300 L = 30.0 mL.
Common mistake: Equivalence does not always mean pH 7; weak-acid or weak-base systems have different equivalence-point pH.
Solubility product Ksp
Set up the ion-product expression from the dissolution stoichiometry and solve for molar solubility s.
AgCl(s) ⇌ Ag⁺ + Cl⁻. If Ksp = 1.8×10−10, Ksp = s² → s = 1.34×10−5 M in pure water.
For MX2(s) ⇌ M²⁺ + 2X⁻, express Ksp in terms of s in pure water.
Show worked answer
[M²⁺] = s, [X⁻] = 2s, so Ksp = s(2s)² = 4s³.
Common mistake: A 1:2 salt does not generally give Ksp = s².
Common-ion effect
If the solution already contains a substantial concentration of one dissolution ion, the other ion is usually much less soluble.
For AgCl with Ksp = 1.8×10−10 in approximately 0.100 M Cl⁻: [Ag⁺] ≈ Ksp/[Cl⁻] = 1.8×10−9 M.
What happens to the molar solubility of AgCl when [Cl⁻] increases, all else equal?
Show worked answer
It decreases because the ion product approaches Ksp at a lower [Ag⁺].
Common mistake: Check when a negligible-x approximation is reasonable.
Precipitation prediction using Qsp
Use current ion concentrations to compute Qsp and compare it with Ksp.
With [Ag⁺] = 1.0×10−5 M and [Cl⁻] = 1.0×10−4 M, Qsp = 1.0×10−9. Since Qsp > Ksp(AgCl) = 1.8×10−10, precipitation is favored.
If Qsp < Ksp, what is the mixture’s saturation status?
Show worked answer
It is unsaturated relative to the solid; precipitation is not favored at the stated concentrations.
Common mistake: For mixtures, use ion concentrations after mixing and dilution.
Part 3 · Kinetics and rate laws
Method of initial rates
Compare two experiments that differ in only one reactant concentration to find an empirical order.
If doubling [A] causes rate to become four times larger, 4 = 2m, so m = 2 (second order in A).
If tripling [B] triples the rate while everything else is fixed, what is the order in B?
Show worked answer
3 = 3n → n = 1.
Common mistake: Reaction coefficients usually do not equal rate-law exponents for an overall multi-step reaction.
First-order integrated rate laws
For a first-order process, ln[A]t = ln[A]0 − kt and t1/2 = ln2/k.
For k = 0.0231 min−1, t1/2 = 0.693/0.0231 ≈ 30.0 min.
If k = 0.0462 min−1, calculate the first-order half-life.
Show worked answer
t1/2 = 0.693/0.0462 = 15.0 min.
Common mistake: Use ln (natural log), not log10, in the integrated first-order equation.
Arrhenius equation and temperature
Convert activation energy to joules per mole when using R = 8.314 J·mol−1·K−1, and use Kelvin.
ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). With Ea = 50,000 J/mol, T1 = 300 K, T2 = 310 K, ln ratio ≈ 0.647, giving k2/k1 ≈ 1.91.
For positive activation energy, does k increase or decrease as T increases?
Show worked answer
It generally increases according to the Arrhenius relationship.
Common mistake: Do not put kJ/mol into a formula when R is in J/(mol·K).
Part 4 · Thermodynamics and electrochemistry
Gibbs free energy ΔG = ΔH − TΔS
Subtract a temperature-times-entropy term from enthalpy, after putting their energy units on the same scale.
With ΔH = −50.0 kJ/mol, ΔS = −100 J/(mol·K) = −0.100 kJ/(mol·K), T = 298 K: ΔG = −50.0 − 298(−0.100) = −20.2 kJ/mol.
If ΔH = +20.0 kJ/mol and ΔS = +0.0800 kJ/(mol·K) at 300 K, find ΔG.
Show worked answer
ΔG = 20.0 − 300(0.0800) = −4.0 kJ/mol.
Common mistake: A negative ΔG indicates thermodynamic favorability at the stated conditions, not how fast a reaction happens.
Link ΔG° to K
At a fixed temperature, standard Gibbs free energy and the equilibrium constant satisfy ΔG° = −RT ln K.
If K = 10.0 at 298 K, ΔG° = −(8.314 J/mol·K)(298 K)ln(10.0) ≈ −5.71 kJ/mol.
What is ΔG° if K = 1 at any temperature?
Show worked answer
ln(1) = 0, so ΔG° = 0.
Common mistake: Use ln rather than log10 and convert J to kJ before reporting kJ/mol.
Cell potential and ΔG°
For a galvanic cell, E°cell = E°cathode − E°anode; ΔG° = −nFE°cell.
If n = 2 mol electrons per mol reaction and E° = +1.10 V, ΔG° = −2(96485 C/mol)(1.10 J/C) ≈ −212 kJ/mol reaction.
For n = 1 and E° = +0.800 V, find ΔG°.
Show worked answer
−(1)(96485)(0.800) = −77,188 J/mol ≈ −77.2 kJ/mol reaction.
Common mistake: n is the number of moles of electrons per mole of balanced cell reaction; do not confuse it with a concentration.
Nernst equation
At 25 °C, E = E° − (0.05916 V/n)log10Q for an idealized cell with appropriately defined activities.
With E° = 1.10 V, n = 2, and Q = 10: E = 1.10 − (0.05916/2)log(10) ≈ 1.07 V.
What is E for E° = 1.10 V, n = 2, Q = 100 at 25 °C?
Show worked answer
E = 1.10 − (0.05916/2)×2 = 1.04 V.
Common mistake: The 0.05916 factor is specific to approximately 25 °C; the general equation uses RT/(nF).
Part 5 · Solution properties and decay
Freezing-point depression
Ideal dilute-solution freezing-point change is ΔTf = iKfm, where m is molality and i is an ideal particle factor.
For ideal NaCl dissociation, i = 2, water Kf = 1.86 °C·kg/mol, m = 0.500 mol/kg: ΔTf = 2(1.86)(0.500) = 1.86 °C; freezing point ≈ −1.86 °C.
For a nonelectrolyte at m = 0.500 mol/kg in water, find ΔTf.
Show worked answer
i = 1, ΔTf = 1(1.86)(0.500) = 0.930 °C.
Common mistake: Molality (mol/kg solvent) is not the same as molarity (mol/L solution), and real electrolytes may have nonideal i.
Osmotic pressure
For a sufficiently dilute ideal solution, osmotic pressure π ≈ iMRT.
For 0.100 M sucrose (i = 1) at 298 K, R = 0.08206 L·atm·mol−1·K−1: π = (1)(0.100)(0.08206)(298) = 2.45 atm.
Under the same assumptions, what is π for 0.200 M sucrose at 298 K?
Show worked answer
π ≈ 1(0.200)(0.08206)(298) = 4.89 atm.
Common mistake: Use the appropriate R units to match the pressure unit you want.
Half-life and exponential decay (where included)
Radioactive decay is exponential: N = N0(1/2)t/t1/2. Some courses cover this in General Chemistry I instead.
After two half-lives, a 100. g radioactive sample has 100.×(1/2)² = 25.0 g of the original nuclide remaining.
What remains from 80.0 g of a radionuclide after three half-lives?
Show worked answer
80.0×(1/2)³ = 10.0 g.
Common mistake: Half-life refers to the amount of the specific undecayed radionuclide; it does not mean total sample mass necessarily vanishes.
How to know which equation to use
Ask what changes: Equilibrium concentration? Use K/Q and an ICE table. Hydronium or hydroxide? Use pH/pOH. Weak acid mixed with its conjugate base? Consider the buffer approximation after any stoichiometric reaction. Reaction speed? Identify a rate law before applying kinetic formulas. Electricity and spontaneity? Link E, Q, and ΔG. Freezing or osmotic effects? Count dissolved particles under appropriate conditions.
If you get an implausible answer, go back through the foundation skills and check logarithm keys, Kelvin conversions, units, and signs. Return to General Chemistry I Math if mole or concentration conversions are the actual obstacle.
Many universities organize these topics differently; your syllabus determines course placement. Introductory pH, K, and Nernst forms here use standard dilute/ideal approximations; more advanced chemistry uses activities and appropriate standard states.