Most lost points come from “knew it, got it wrong”
Reviewing after an exam usually shows that many lost points weren’t about not knowing the material: the same few kinds of mistakes kept coming back. Listing them and checking each one during the exam is the easiest way to win those points back.
The common-mistakes checklist
Calculations and units
- mL not converted to L; °C not converted to K.
- The wrong R: with P in atm and V in L, R = 0.08206; for energy, 8.314 J/(mol·K).
- In ΔG° = −RT ln K, ΔG° must be in J; relate Ksp to solubility through the formula (MX₂: Ksp = 4s³).
- Significant figures that don’t match the data, or intermediate results rounded too early.
Signs and directions
- qrxn = −qsoln; if the temperature rises, ΔH < 0.
- Bond enthalpies: ΔH = Σ broken − Σ formed, not the other way round.
- Q < K goes forward and Q > K goes in reverse; E°cell = E°cathode − E°anode, and E° doesn’t change when you multiply by a coefficient.
Equilibrium and acids
- A catalyst changes neither K nor the equilibrium composition; only temperature changes K.
- Pure solids and pure liquids are left out of K expressions.
- Weak acids ionize only partly; when a weak acid is titrated with a strong base, the equivalence pH is above 7 and the half-equivalence pH equals pKa.
- A strong base like Ba(OH)₂ gives two OH⁻ per formula unit.
Structure and forces
- Boiling and melting overcome intermolecular forces, not covalent bonds.
- Polar bonds don’t make a polar molecule; check whether the shape is symmetrical.
- An ionic solid doesn’t conduct by itself; it conducts when molten or dissolved.
- When comparing properties, give all three links: structure → forces → property (4.3.3).
In 25.0 mL of 0.100 M HCl, n(HCl) = 25.0 × 0.100 = 2.50 mol.
Convert the volume to liters first: n = 0.0250 L × 0.100 M = 2.50 × 10⁻³ mol.
When using PV = nRT, 25 °C must be converted to 298 K.
Gas laws always use kelvins.
R = 8.314 J/(mol·K) can be used directly with P in atm and V in L.
With P in atm and V in L, use R = 0.08206 L·atm/(mol·K); 8.314 goes with Pa and m³, or with energy calculations.
When using ΔG° = −RT ln K, ΔG° must first be converted from kJ to J.
R is in J/(mol·K), so ΔG° must be in joules.
In calorimetry the solution’s temperature drops, so qsoln < 0 and ΔH for the reaction is also below 0.
The solution gives heat to the reaction, so the reaction is endothermic and ΔH > 0. qrxn = −qsoln.
From bond enthalpies, ΔH = Σ bonds formed − Σ bonds broken.
That’s backwards. Breaking bonds absorbs energy and forming them releases it: ΔH = Σ broken − Σ formed.
In a galvanic cell, electrons flow through the wire from the anode to the cathode.
Oxidation at the anode releases electrons, which flow to the cathode.
When Q > K, the reaction proceeds in the forward direction.
Q > K means there is “too much” product, so the reaction proceeds in reverse.
Adding a catalyst shifts the equilibrium forward and raises the yield.
A catalyst speeds up the forward and reverse reactions equally; it only shortens the time to reach equilibrium and changes neither the composition nor K.
Pure solids and pure liquids are left out of equilibrium expressions.
Their “concentrations” are constant and already built into K.
A 0.10 M weak acid has the same pH as a 0.10 M strong acid.
A weak acid ionizes only partly, so its [H⁺] is far below 0.10 M and its pH is higher.
At the half-equivalence point of a weak acid titrated with a strong base, pH = pKa.
There [HA] = [A⁻], so the Henderson-Hasselbalch equation gives pH = pKa.
Water’s boiling point is high because its O–H bonds break when it boils.
Boiling overcomes intermolecular forces; covalent bonds don’t break. Water boils high because of hydrogen bonding between molecules.
CO₂ has polar bonds, so CO₂ is a polar molecule.
CO₂ is linear, so its two C=O bond dipoles are equal and opposite and cancel; the molecule is nonpolar.
Down a group, atomic radius increases and first ionization energy decreases.
With more shells, the outer electrons are farther from the nucleus and held less tightly.
A solid NaCl crystal conducts electricity because it contains ions.
Ions in the solid can’t move freely; NaCl conducts only when molten or dissolved.
This site’s mistake notebook automatically collects the questions you get wrong in practice, grouped by unit, so you can redo them any time.
Example 1: a reversed sign from bond enthalpies
A student estimates ΔH for H₂(g) + Cl₂(g) → 2HCl(g) from bond enthalpies (H–H 436, Cl–Cl 243, H–Cl 432 kJ/mol) and writes ΔH = 2(432) − (436 + 243) = +185 kJ. Find and correct the mistake.
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Find the mistake. The student used “bonds formed − bonds broken,” which is backwards.
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Back to the principle. Breaking bonds absorbs energy (positive); forming bonds releases it (negative):
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Recalculate. One H–H and one Cl–Cl break, and two H–Cl form:
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Check. H₂ burning in Cl₂ is a vigorous exothermic reaction, so a negative ΔH makes sense.
Example 2: the wrong link between Ksp and solubility
A slightly soluble salt MX₂ has Ksp = 4.0 × 10⁻¹². A student writes s = √Ksp = 2.0 × 10⁻⁶ M. Find and correct the mistake.
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Find the mistake. s = √Ksp only works for a 1 : 1 salt like AgCl. MX₂ dissolves to give one M²⁺ and two X⁻.
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Write the concentrations from the formula. MX₂(s) ⇌ M²⁺ + 2X⁻: [M²⁺] = s and [X⁻] = 2s.
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Substitute into Ksp.
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Solve for s.
- A solution with pH 3 has twice the [H⁺] of a solution with pH 5
A pH difference of 2 means a factor of 10² = 100 in [H⁺].
- Adding NaCl to a saturated AgCl solution lowers the Ksp of AgCl
At constant temperature Ksp is constant; what decreases is the solubility of AgCl.
- A reaction with ΔH < 0 and ΔS < 0 is thermodynamically favorable at low temperature
At low T the −TΔS term is small, so the negative ΔH decides ΔG.
- Raising the temperature speeds up both endothermic and exothermic reactions
Its effect on K depends on the reaction, but the rate always increases with temperature.
- Multiplying a cell reaction’s coefficients by 2 doubles E°cell
E° is an intensive property and doesn’t change with the coefficients (ΔG° doubles).
- One mole of any gas occupies 22.4 L at 25 °C and 1 atm
22.4 L is for 0 °C and 1 atm; at 25 °C it’s about 24.5 L.
- When two solutions are mixed in equal volumes, each solute’s concentration halves
The moles stay the same while the volume doubles (assuming the volumes add).
- Equal masses of CH₄ and O₂ contain the same number of molecules
Their molar masses differ (16.04 and 32.00 g/mol), so the numbers of molecules differ.
Common mistakes
Mistake 1: “Reading the answer to a missed question once is enough.” Understanding an answer isn’t the same as being able to produce it. Close the answer, redo the question yourself a day or two later, and count it as learned only when you get it right.
Mistake 2: “Careless mistakes don’t matter; I’ll just be more careful.” Carelessness usually follows a pattern (always forgetting a unit conversion, always flipping a sign). Write your patterns into your checklist and check them one by one in the exam.
Mistake 3: “Significant figures don’t matter as long as the number is right.” When a result’s digits clearly don’t match the data, the scoring guidelines may take off a point. Give final answers to the data’s significant figures.
Quick check
In the last two minutes before handing in, what is most worth checking?
These lose the most points and are the quickest to fix.
How many times greater is [H⁺] in a pH 2 solution than in a pH 4 solution?
Each pH unit is a factor of 10 in [H⁺].
What is the equilibrium expression for CaCO₃(s) ⇌ CaO(s) + CO₂(g)?
Pure solids are left out of the expression.
At 25 °C and 1 atm, one mole of an ideal gas occupies about:
V = RT/P = 0.08206 × 298.15 ÷ 1 ≈ 24.5 L.
When a half-reaction is multiplied by 2, its standard reduction potential E°:
E° is an intensive property.
Why does “human error” earn no credit as a source of error in a lab question?
Name the specific action and its effect on the measurement and the result.
AP-style practice
Below are a student’s answers to four parts, each with a mistake. Identify each mistake and give the correct answer.
(a) Find the pH of 0.020 M Ba(OH)₂. Student: [OH⁻] = 0.020 M, pOH = 1.70, pH = 12.30.
(b) A salt dissolves in 50.0 g of water and the temperature falls by 3.2 °C. Student: q = mcΔT = (50.0)(4.18)(−3.2) = −669 J, so dissolving is exothermic.
(c) Explain why H₂O has a higher boiling point than H₂S. Student: because O–H bonds are stronger than S–H bonds and harder to break when boiling.
(d) Doubling [A] doubles the initial rate (with everything else unchanged). Student: so the reaction is second order in A.
Model answer and scoring
(a) Mistake: each formula unit of Ba(OH)₂ gives 2 OH⁻. Correct: [OH⁻] = 0.040 M, pOH = 1.40, pH = 12.60.
(b) Mistake: −669 J is q for the solution, which gave heat to the dissolving process. Correct: qdissolving = +669 J, so dissolving is endothermic.
(c) Mistake: covalent bonds don’t break when a substance boils. Correct: H₂O molecules hydrogen bond to each other (O is small and highly electronegative), while H₂S molecules have mainly weaker dipole–dipole and dispersion forces; overcoming hydrogen bonds takes more energy, so H₂O boils at a higher temperature.
(d) Mistake: a rate proportional to [A] is first order. Correct: 2ⁿ = 2, so n = 1 and the reaction is first order in A.
Scoring: 1 point per part, for identifying the mistake and giving the correct answer with its reason.
Summary
- Common mistakes fall into four groups: calculations and units, signs and directions, equilibrium and acids, and structure and forces.
- To fix a mistake, find the step where it went wrong, then rework from the principle (the definition, equation or formula).
- Turn your own repeat mistakes into a checklist to review before the exam and check before handing in; redo missed questions from the notebook a few days later.
AP tip: After each practice set, write the reason for every lost point in one sentence (such as “forgot to double for Ba(OH)₂”). Reading those sentences before the exam prevents repeat mistakes better than ten new questions.
Sources: the bond enthalpies are commonly used average values; MX₂ is a hypothetical slightly soluble salt and its Ksp is illustrative.