Grade 12 Chemistry (SCH4U) Practice Quiz, With Worked Answers

The 28 questions at a glance

The questions follow the course's five strands. The table shows what each strand asks of a student, in plain terms, which questions practise it, and the curriculum expectations those questions touch.

Fourteen questions are multiple choice, and the answer explains why each wrong option fails. The other fourteen are short-answer or calculation questions, and the answer shows the formula, the numbers and the units.

Is SCH4U still the Grade 12 chemistry course in 2026-27?

Yes. The ministry's course page, updated in August 2026, still lists SCH4U as Chemistry, Grade 12, University, with the prerequisite Grade 11 Chemistry, University. The ministry says the 2008 science curriculum for Grade 10 and the 2008 science curriculum for Grades 11 and 12 both remain in effect, and every secondary science course from Grade 10 to Grade 12 is still based on those two documents. Grade 9 science is the one course that moved to a newer curriculum. We found no announcement of a new Grade 12 chemistry course as of 21 October 2026.

SCH4U is described in The Ontario Curriculum, Grades 11 and 12: Science, 2008 (Revised). The ministry describes SCH4U as a course in organic chemistry, the structure and properties of matter, energy changes and rates of reaction, equilibrium in chemical systems, and electrochemistry. The course has five content strands: organic chemistry, structure and properties of matter, energy changes and rates of reaction, chemical systems and equilibrium, and electrochemistry. A sixth strand, scientific investigation skills and career exploration, runs through all five; this quiz does not test it.

SCH3U is the only prerequisite for SCH4U. SCH4C is a separate Grade 12 college chemistry course with a different prerequisite, so this quiz does not cover it. Genie's SCH4U explainer already covers what the course is, who needs it and how it differs from SCH3U, so this page sticks to practice.

How the SCH4U exam counts from 2026-27

The rules changed this school year. The ministry's Growing Success page, updated on 14 October 2026, says: "Beginning in the 2026-2027 school year, most secondary courses will require written exams as a percentage of the final course mark."

From the 2026-27 school year, a Grade 11 or 12 student's final mark is 65% classroom work, 25% final evaluation and 10% attendance and participation.

For SCH4U, the 25% final evaluation is a written exam. The ministry's Grade 11 and 12 table puts English, mathematics and science on the same row, with a written exam and no culminating task. The same page exempts Grade 11 and 12 workplace preparation (E) and open (O) courses from mandatory written exams, so the rule covers SCH4U, a university preparation course, but not every Grade 12 science course. Written final exams from Grades 9 to 12 take place during the designated exam period.

The ministry policy lists no provincial exam for SCH4U. The one exception it makes to the school exam period is the EQAO Grade 9 mathematics assessment, which counts as the written exam for Grade 9 math. The page does not say who writes the paper for each course, so ask the teacher what the SCH4U exam will look like. Online and remote courses offered through school boards write the exam in person at the student's home school unless the board can provide secure online proctoring.

Our page on Ontario's new exam and attendance rules covers the rest of the change, and what changed in Ontario schools in September 2026 puts it next to the other changes this year.

Data used in this quiz

No ministry document we found says what data sheet or calculator an SCH4U exam allows, so this quiz carries its own data table. Every value comes from OpenStax Chemistry 2e, an openly licensed textbook, except the boiling points, which come from the NIST Chemistry WebBook. A school's own data sheet may round some values differently, and the answers say where that changes anything.

Standard conditions throughout: 25 °C, solutions at 1 mol/L, and 100 kPa for gases.

At 25 °C, Kw is 1.0 × 10⁻¹⁴, so pH + pOH = 14.00. Kw changes with temperature, which is why every pH question below says 25 °C.

Standard enthalpies of formation, ΔHf° (kJ/mol, 25 °C)

Standard reduction potentials, E° (V, 25 °C)

Many school tables print the copper value as +0.34 V. For the cells in this quiz, both values give the same answer to two decimal places.

How to use this quiz

Do all 28 on paper before reading a single answer. Cover the answer under each question with a sheet of paper and slide it down only once the question is done. Use a scientific calculator, keep the data table open, and skip the timer on the first pass.

Twenty-eight questions cannot predict an exam mark. What they can show is which strand is short, and which step inside it. Write the setup, the substitution and the units every time, even when the arithmetic feels obvious, because the working is what shows where a question went wrong.

Every question in this quiz is original and was written for this page. None is copied from a textbook, a board exam, a teacher's review package or the ministry's curriculum.

Search for an SCH4U practice test and most of the first page, checked on 21 October 2026, is old teacher PDFs, document-sharing sites, flashcard decks and Reddit threads. The top result is a practice exam from the 2007-08 school year. Where answers exist at all, the results we could open or read in a search snippet put them in a separate file, behind a sign-in or in a video, and most of the documents date from 2007 to 2022, before the 2026-27 exam rules. Old exams are still useful practice once you have finished this set.

Organic chemistry: questions 1 to 6

Question 1. Which is the correct IUPAC name for CH₃-CH(CH₃)-CH₂-CH₂-CH₃?

• (A) 4-methylpentane
• (B) 2-methylpentane
• (C) 2-methylbutane
• (D) hexane

Answer: (B). Find the longest continuous chain of carbon atoms. It has five carbons, so the parent name is pentane. There is one methyl branch. Number the chain from the end closest to the branch, which puts the methyl group on carbon 2. (A) numbers from the wrong end: the lowest possible locant wins, so 4 loses to 2. (C) counts only four carbons in the main chain and misses one. (D) is a different compound: hexane has the same formula, C₆H₁₄, but all six carbons sit in one straight chain with no branch.

Question 2. Name CH₃CH₂CH(OH)CH₂CH₃ and identify its functional group.

• (A) pentan-3-one, a ketone
• (B) pentan-2-ol, an alcohol
• (C) pentanal, an aldehyde
• (D) pentan-3-ol, an alcohol

Answer: (D). The main chain has five carbons, and the -OH group sits on carbon 3. The molecule is symmetrical, so carbon 3 is carbon 3 from either end. The -OH group makes it an alcohol, named with the ending -ol. The older form of the name, 3-pentanol, means the same compound. (A) is a ketone, which needs a C=O group on a carbon inside the chain; this molecule has an -OH group, so no C=O at all. (B) has the right family and the wrong locant. (C) is an aldehyde, which needs the C=O group at the end of the chain.

Question 3. What kind of compound is CH₃CH₂COOCH₃, and what is its name?

• (A) an ester, methyl propanoate
• (B) an ester, propyl methanoate
• (C) a carboxylic acid, butanoic acid
• (D) an ester, methyl ethanoate

Answer: (A). The -COO- group with a carbon on both sides marks an ester. Split the molecule at the single-bonded oxygen. The side with the C=O group, CH₃CH₂CO-, has three carbons counting the carbonyl carbon, so it comes from propanoic acid and is named propanoate. The other side, CH₃, comes from methanol and is named methyl. The alkyl part comes first in the name: methyl propanoate. (B) swaps the two parts; propyl methanoate is a different ester with the same formula. (C) also has the formula C₄H₈O₂, but in a carboxylic acid the second oxygen bonds to a hydrogen atom, and here it bonds to a carbon. (D) miscounts the acid side; ethanoate has two carbons.

Question 4. Butanoic acid reacts with methanol. Write the equation using condensed structural formulas, name the organic product, and name the other product.

Answer. CH₃CH₂CH₂COOH + CH₃OH → CH₃CH₂CH₂COOCH₃ + H₂O

The organic product is methyl butanoate, an ester, and the other product is water. The reaction is esterification: the -OH from the acid and the H from the alcohol's -OH leave as water, and the two remaining pieces join through an oxygen. A reaction where two molecules combine and water is produced is called a condensation reaction. Name the ester the same way as in Question 3. The alcohol gives the first word, methyl. The acid gives the second, butanoate. Check the atoms: five carbons, twelve hydrogens and three oxygens on each side.

Question 5. Ethene, CH₂=CH₂, reacts with bromine, Br₂. What type of reaction is this, and what is the product?

• (A) substitution; bromoethane and hydrogen bromide
• (B) addition; 1,1-dibromoethane
• (C) addition; 1,2-dibromoethane
• (D) elimination; ethyne

Answer: (C). The double bond opens, and one bromine atom adds to each of the two carbons it joined: CH₂=CH₂ + Br₂ → CH₂Br-CH₂Br. Adding atoms across a multiple bond is an addition reaction, and the product is 1,2-dibromoethane. (A) describes substitution, where a bromine atom replaces a hydrogen atom; with a double bond available, bromine adds instead. (B) gets the reaction type right but puts both bromines on one carbon, when each carbon of the double bond receives one. (D) runs the other way: elimination removes atoms to make a multiple bond, and here atoms are added.

Question 6. How many structural isomers have the formula C₄H₁₀? Name each one, and explain which has the lower boiling point.

Answer. Two. Butane is the straight chain, CH₃CH₂CH₂CH₃. 2-methylpropane is the branched one, CH₃CH(CH₃)CH₃: a three-carbon chain with a methyl group on the middle carbon. Structural isomers share a molecular formula and differ in how the atoms are joined.

2-methylpropane has the lower boiling point. Both molecules are nonpolar, so the only forces between them are London dispersion forces. The branched molecule is more compact, so neighbouring molecules touch over a smaller area and the dispersion forces between them are weaker. Less energy is needed to separate them, so it boils at a lower temperature. OpenStax shows the same pattern for the three isomers of C₅H₁₂: the straight chain boils highest and the most compact one boils lowest.

Structure and properties of matter: questions 7 to 11

Question 7. How many unpaired electrons does a ground-state phosphorus atom have?

• (A) 3
• (B) 1
• (C) 5
• (D) 0

Answer: (A). Phosphorus has 15 electrons. Fill the orbitals in order by the aufbau principle: 1s² 2s² 2p⁶ 3s² 3p³. The three 3p orbitals have the same energy, and Hund's rule puts one electron in each before any pairs up. So the three 3p electrons are all unpaired. (B) pairs two of the 3p electrons, which Hund's rule rules out while an empty 3p orbital is available. (C) counts all five valence electrons, but the two 3s electrons share an orbital and are paired. (D) treats the atom as if every electron were paired.

Question 8. Predict the shape of a sulfur trioxide molecule, SO₃, and say whether it is polar.

• (A) trigonal pyramidal, polar
• (B) trigonal planar, polar
• (C) bent, polar
• (D) trigonal planar, nonpolar

Answer: (D). Count the regions of electron density around the sulfur atom. There are three, one for each S-O bond, and a double bond counts as one region. There is no lone pair on the sulfur, whichever Lewis structure your class uses. VSEPR theory says three regions spread out to 120° apart in a flat triangle, which is trigonal planar. Each S-O bond is polar, because oxygen is more electronegative than sulfur. In a symmetrical trigonal planar shape the three bond dipoles cancel, so the molecule is nonpolar. BF₃ is nonpolar for the same reason. (A) needs a lone pair on the central atom, as in NH₃. (B) has the right shape and ignores the cancelling. (C) needs two bonded atoms and at least one lone pair on the central atom, as in H₂O, and SO₃ has three bonded atoms and no lone pair.

Question 9. Which of these molecules is polar?

• (A) CO₂
• (B) NH₃
• (C) CH₄
• (D) BF₃

Answer: (B). In ammonia, the nitrogen atom has four electron pairs, and one of them is a lone pair. That makes the molecule a trigonal pyramid, with the three N-H bonds on one side. The bond dipoles do not cancel, so the molecule is polar. (A) CO₂ has two polar bonds, but the molecule is linear and the two bond dipoles point in opposite directions and cancel. (C) CH₄ is tetrahedral and (D) BF₃ is trigonal planar. Both are symmetrical, with identical bonds arranged so their dipoles cancel, so both are nonpolar. A molecule is polar only when it has at least one polar bond and a shape that leaves those bond dipoles uncancelled.

Question 10. Classify each solid as ionic, metallic, covalent network or molecular, and name what holds its particles together: sodium chloride, NaCl; copper, Cu; silicon dioxide, SiO₂; iodine, I₂.

Answer.

• NaCl is an ionic solid. Na⁺ and Cl⁻ ions are held together by ionic bonding.
• Cu is a metallic solid. Copper atoms are held together by metallic bonding.
• SiO₂ is a covalent network solid. Silicon and oxygen atoms are joined by covalent bonds through the whole crystal, with no separate molecules.
• I₂ is a molecular solid. Each I₂ molecule is held together by a covalent bond, and the molecules are held to each other only by weak Van der Waals forces.

The type of solid is what explains properties such as melting point, hardness and electrical conductivity, so the curriculum asks students to describe the properties that follow from each type. The key distinction is between the bonds inside a particle and the forces between particles. Melting iodine overcomes only the weak forces between I₂ molecules. Melting silicon dioxide means breaking covalent bonds.

Question 11. Rank propane (CH₃CH₂CH₃), ethanal (CH₃CHO) and ethanol (CH₃CH₂OH) from lowest to highest boiling point. Explain the order using the strongest intermolecular force in each.

Answer. Propane < ethanal < ethanol.

The three molecules have similar molar masses, 44, 44 and 46 g/mol, so their London dispersion forces are similar, and the difference comes from the other forces. Propane is nonpolar and has dispersion forces only. Ethanal has a polar C=O group, so it adds dipole-dipole attractions. Ethanol has an O-H group, so its molecules form hydrogen bonds with each other, the strongest of the three. The NIST Chemistry WebBook gives boiling points of about -42.0 °C for propane, 20.9 °C for ethanal and 78.4 °C for ethanol, which matches the ranking. Ranking by molar mass alone cannot separate propane from ethanal, which have almost the same molar mass. That method works within a family of similar molecules, and fails when the molecules have different polar groups.

Energy changes and rates of reaction: questions 12 to 17

Question 12. How much heat is needed to warm 50.0 g of water by 6.50 °C? Use c = 4.184 J/(g·°C).

• (A) 325 J
• (B) 209 J
• (C) 1.36 × 10³ J
• (D) 27.2 J

Answer: (C). Use Q = mcΔT. Q = 50.0 g × 4.184 J/(g·°C) × 6.50 °C = 1359.8 J. The mass and the temperature change each have three significant figures, so the answer is 1.36 × 10³ J, or 1.36 kJ. (A) leaves out the specific heat: 50.0 × 6.50 = 325. (B) leaves out the temperature change: 50.0 × 4.184 = 209. (D) leaves out the mass: 4.184 × 6.50 = 27.2. Writing the units on every number in the setup catches all three slips, because the answer only comes out in joules when all three are there.

Question 13. A student dissolves 4.00 g of sodium hydroxide, NaOH (molar mass 40.00 g/mol), in 100.0 mL of water in a coffee-cup calorimeter. The temperature rises from 22.0 °C to 32.4 °C. Take the mass of the solution as 104.0 g and its specific heat as 4.184 J/(g·°C), and assume no heat is lost to the cup or the air. Find the molar enthalpy of solution of NaOH.

Answer. ΔH = -45.3 kJ/mol.

• Temperature change: ΔT = 32.4 °C - 22.0 °C = 10.4 °C.
• Heat gained by the solution: Q = 104.0 g × 4.184 J/(g·°C) × 10.4 °C = 4525.4 J, or 4.53 kJ.
• Moles of NaOH: n = 4.00 g ÷ 40.00 g/mol = 0.100 mol.
• The solution gained the heat, so dissolving released it, and ΔH is negative: ΔH = -4.5254 kJ ÷ 0.100 mol = -45.3 kJ/mol.

The temperature change limits the answer to three significant figures. The mass of 104.0 g treats the water as 100.0 g, at a density of about 1.0 g/mL, plus the 4.00 g of solid. The negative sign is easy to drop. The temperature went up, so the process is exothermic, and an exothermic ΔH is negative.

Question 14. Use these two reactions to find ΔH for N₂(g) + 2O₂(g) → 2NO₂(g).

• N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ
• 2NO(g) + O₂(g) → 2NO₂(g), ΔH = -114.2 kJ

Choose one:

• (A) +294.8 kJ
• (B) -66.4 kJ
• (C) -294.8 kJ
• (D) +66.4 kJ

Answer: (D). Hess's law says the enthalpy change of a total process equals the sum of the enthalpy changes of its steps. Add the two equations as written. The 2NO(g) made in the first step is used up in the second, so it cancels, and what is left is N₂(g) + 2O₂(g) → 2NO₂(g). Add the two ΔH values: +180.6 kJ + (-114.2 kJ) = +66.4 kJ. The data table gives a check: ΔHf° of NO₂ is 33.2 kJ/mol, and two moles is 2 × 33.2 = 66.4 kJ. (A) flips the sign of the second step. (B) flips the sign of the total. (C) flips the sign of the first step. Only flip a step's sign when you reverse that equation, and neither equation needed reversing here.

Question 15. Use the data table to find ΔH for the complete combustion of one mole of propane, with water produced as a liquid.

Answer. ΔH = -2220.1 kJ per mole of propane.

• Balanced equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).
• Use ΔH = Σ nΔHf°(products) - Σ nΔHf°(reactants).
• Products: 3(-393.51) + 4(-285.83) = -1180.53 - 1143.32 = -2323.85 kJ.
• Reactants: (-103.8) + 5(0) = -103.8 kJ. Oxygen is an element in its standard state, so its ΔHf° is 0.
• ΔH = -2323.85 - (-103.8) = -2220.05 kJ.
• Round for addition and subtraction: the answer keeps as many decimal places as the least precise value, and -103.8 has one. So ΔH = -2220.1 kJ, or about -2.22 × 10³ kJ.

Two slips to watch for. One is using the value for water vapour, -241.82 kJ/mol, which gives about -2044 kJ; the question says liquid water, and the state matters. The other is subtracting in the wrong order, products from reactants, which flips the sign.

Question 16. A potential energy diagram shows the reactants at 50 kJ, the top of the curve at 135 kJ and the products at 10 kJ. Find the activation energy of the forward reaction, ΔH for the forward reaction, and the activation energy of the reverse reaction. Is the forward reaction endothermic or exothermic?

Answer.

• Forward activation energy: 135 kJ - 50 kJ = 85 kJ, measured from the reactants up to the peak.
• ΔH: 10 kJ - 50 kJ = -40 kJ, measured from the reactants to the products.
• Reverse activation energy: 135 kJ - 10 kJ = 125 kJ, measured from the products up to the peak.

The products sit lower than the reactants, so the forward reaction is exothermic. A useful check: the reverse activation energy equals the forward activation energy minus ΔH, and 85 - (-40) = 125. A catalyst would lower the peak by providing an alternative mechanism with a lower activation energy. It would lower both activation energies and leave ΔH unchanged.

Question 17. Which change would NOT increase the rate of a reaction between a solid and a solution?

• (A) lowering the concentration of the solution
• (B) raising the temperature
• (C) crushing the solid into a powder
• (D) adding a suitable catalyst

Answer: (A). Collision theory says a reaction happens when particles collide with enough energy and the right orientation. Lowering the concentration means fewer reactant particles in each millilitre, so fewer collisions each second, and the rate drops. The other three all speed it up. (B) Raising the temperature makes particles move faster, so they collide more often and a larger share of collisions have enough energy to react. (C) Crushing the solid exposes more surface, so more particles are available to collide. (D) A catalyst increases the reaction rate without being consumed in the reaction, by offering a pathway with a lower activation energy.

Chemical systems and equilibrium: questions 18 to 23

Question 18. In a titration, 25.00 mL of 0.0800 mol/L sulfuric acid, H₂SO₄, is exactly neutralised by 32.00 mL of sodium hydroxide solution. Find the concentration of the NaOH.

Answer. [NaOH] = 0.125 mol/L.

• Balanced equation: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l).
• Moles of acid: 0.02500 L × 0.0800 mol/L = 2.000 × 10⁻³ mol.
• Each mole of H₂SO₄ needs two moles of NaOH: 2 × 2.000 × 10⁻³ = 4.000 × 10⁻³ mol of NaOH.
• Concentration: 4.000 × 10⁻³ mol ÷ 0.03200 L = 0.125 mol/L.

The acid concentration has three significant figures, so the answer does too. The slip this question is built to catch is the 1:1 ratio. Sulfuric acid gives two hydrogen ions per formula unit, and treating it like HCl gives 0.0625 mol/L, exactly half the right answer. Always write the balanced equation first.

Question 19.

• (a) Write the equilibrium constant expression for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), and for CaCO₃(s) ⇌ CaO(s) + CO₂(g).
• (b) For H₂(g) + I₂(g) ⇌ 2HI(g), Keq = 49 at a certain temperature. A flask starts with 1.00 mol/L of H₂ and 1.00 mol/L of I₂ and no HI. Find all three equilibrium concentrations.

Answer.

(a) Keq = [SO₃]² / ([SO₂]²[O₂]). Each concentration is raised to the power of its coefficient, products over reactants. For the second reaction, Keq = [CO₂]. Pure solids are left out of the expression, so neither CaCO₃ nor CaO appears.

(b) Set up the changes. H₂ and I₂ each fall by x, and HI rises by 2x. At equilibrium, [H₂] = [I₂] = 1.00 - x and [HI] = 2x.

• Keq = (2x)² / ((1.00 - x)(1.00 - x)) = 49.
• Both sides are perfect squares, so take the square root: 2x / (1.00 - x) = 7.0.
• Solve: 2x = 7.0 - 7.0x, so 9.0x = 7.0 and x = 0.778.
• [HI] = 2x = 1.56 mol/L, and [H₂] = [I₂] = 1.00 - 0.778 = 0.222 mol/L.

Check by substituting back: 1.56² ÷ 0.222² ≈ 49. Keq = 49 has only two significant figures, so the strictly rounded answers are 1.6 mol/L and 0.22 mol/L. Keep three figures in the working and round at the end, or follow your teacher's rule.

Question 20. The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic. From the data table, ΔH = 2(-395.72) - 2(-296.83) = -197.78 kJ, about -197.8 kJ. Predict the shift in the equilibrium and the effect on Keq for (i) raising the temperature and (ii) compressing the gases into a smaller volume at constant temperature.

• (A) (i) shifts right, Keq increases; (ii) shifts right, Keq increases
• (B) (i) shifts left, Keq unchanged; (ii) shifts left, Keq unchanged
• (C) (i) shifts left, Keq decreases; (ii) shifts right, Keq unchanged
• (D) (i) shifts left, Keq decreases; (ii) no shift, Keq unchanged

Answer: (C). (i) For an exothermic reaction, treat heat as a product. Raising the temperature adds heat, so the equilibrium shifts left to use it up. A temperature change is the one change that alters the value of Keq. Here there is less SO₃ at the new equilibrium, so Keq decreases. (ii) Compressing the gases raises every concentration, and the system shifts toward the side with fewer moles of gas. The left side has three moles of gas and the right side has two, so it shifts right. Keq stays the same, because only temperature changes it. (A) treats heat as a reactant and lets compression change Keq. (B) is wrong on both parts: it forgets that temperature changes Keq, and on compression it shifts left, toward the side with more moles of gas. (D) misses that the two sides have different numbers of gas moles.

Question 21. What is the pH of 0.0025 mol/L sodium hydroxide, NaOH, at 25 °C? Use Kw = 1.0 × 10⁻¹⁴.

• (A) 2.60
• (B) 11.40
• (C) 12.40
• (D) 10.40

Answer: (B). NaOH is a strong base, so it ionises completely and [OH⁻] = 0.0025 mol/L. Then pOH = -log(0.0025) = 2.60, and pH = 14.00 - 2.60 = 11.40. The same answer comes the long way: [H₃O⁺] = Kw ÷ [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 0.0025 = 4.0 × 10⁻¹² mol/L, and -log(4.0 × 10⁻¹²) = 11.40. The concentration has two significant figures, so the pH has two decimal places. (A) is the pOH, reported as if it were the pH. A basic solution at 25 °C has a pH above 7, which rules it out at a glance. (C) uses 0.025 mol/L, ten times too concentrated, which raises the pH by 1. (D) uses 0.00025 mol/L, ten times too dilute, which lowers the pH by 1.

Question 22. Find the pH of 0.100 mol/L acetic acid at 25 °C. Its Ka is 1.8 × 10⁻⁵.

Answer. pH = 2.87.

• Acetic acid is a weak acid, so only a small share of it ionises: CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq).
• Ka = [H₃O⁺][CH₃COO⁻] / [CH₃COOH]. Let x = [H₃O⁺] = [CH₃COO⁻], so Ka = x² / (0.100 - x).
• Because Ka is small, assume x is tiny next to 0.100: x² = 1.8 × 10⁻⁵ × 0.100 = 1.8 × 10⁻⁶, and x = 1.34 × 10⁻³ mol/L.
• pH = -log(1.34 × 10⁻³) = 2.87.

Check the assumption: 1.34 × 10⁻³ ÷ 0.100 = 1.3% ionised, so x really is small next to 0.100. Solving the full quadratic instead gives a pH of 2.88. The two differ only in the last decimal place, so state which method you used. The slip to avoid is treating acetic acid as a strong acid: -log(0.100) gives a pH of 1.00, far too low.

Question 23. Calcium fluoride, CaF₂, has Ksp = 4.0 × 10⁻¹¹ at 25 °C. Find its molar solubility in pure water.

Answer. About 2.2 × 10⁻⁴ mol/L.

• CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq).
• If s mol/L dissolves, [Ca²⁺] = s and [F⁻] = 2s.
• Ksp = [Ca²⁺][F⁻]² = s(2s)² = 4s³.
• 4s³ = 4.0 × 10⁻¹¹, so s³ = 1.0 × 10⁻¹¹ and s = 2.15 × 10⁻⁴ mol/L.

Ksp has two significant figures, so the answer is 2.2 × 10⁻⁴ mol/L. The solid does not appear in the Ksp expression. The slips to watch for are forgetting to double the fluoride concentration, or forgetting to square it. Writing Ksp = s² gives 6.3 × 10⁻⁶ mol/L, which is wrong by a factor of more than 30.

Electrochemistry: questions 24 to 28

Question 24. In the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which statement is correct?

• (A) Zn is oxidised and is the reducing agent; Cu²⁺ is reduced and is the oxidising agent.
• (B) Zn is reduced and is the oxidising agent; Cu²⁺ is oxidised and is the reducing agent.
• (C) Zn is oxidised and is the oxidising agent; Cu²⁺ is reduced and is the reducing agent.
• (D) Cu²⁺ is oxidised to Cu, and Zn is the oxidising agent.

Answer: (A). Track the oxidation numbers. Zinc goes from 0 to +2, so it loses two electrons and is oxidised. Copper goes from +2 to 0, so it gains two electrons and is reduced. The substance that is oxidised gives its electrons away and so reduces the other one, which makes zinc the reducing agent. Cu²⁺ takes those electrons, which makes it the oxidising agent. (B) has both processes backwards. (C) has the processes right and the agent names swapped. (D) calls a gain of electrons oxidation.

Question 25. Balance this reaction in acidic solution using the half-reaction method: MnO₄⁻(aq) + Fe²⁺(aq) → Mn²⁺(aq) + Fe³⁺(aq).

Answer. MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

• Oxidation half-reaction: Fe²⁺ → Fe³⁺ + e⁻.
• Reduction half-reaction, step by step. Start with MnO₄⁻ → Mn²⁺. Balance oxygen with water: MnO₄⁻ → Mn²⁺ + 4H₂O. Balance hydrogen with H⁺: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O. Balance charge with electrons. The left side is +7 and the right side is +2, so add five electrons to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
• The reduction uses five electrons and the oxidation gives one, so multiply the oxidation by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻.
• Add the two half-reactions. The five electrons cancel, leaving the equation above.

Check the charge: +17 on each side (-1 + 10 + 8 on the left, +2 + 15 on the right). Check the atoms: one Mn, five Fe, four O and eight H on each side. Manganese goes from an oxidation number of +7 to +2, which matches the five electrons it gains.

Question 26. A standard galvanic cell has a zinc electrode in 1 mol/L Zn²⁺ and a copper electrode in 1 mol/L Cu²⁺, joined by a salt bridge. Find E°cell, name the anode and the cathode, and give the direction of electron flow in the wire.

Answer. E°cell = +1.10 V. Zinc is the anode and copper is the cathode. Electrons flow through the wire from zinc to copper.

• Copper has the higher reduction potential, +0.337 V against -0.7618 V for zinc, so Cu²⁺ is reduced at the copper electrode, the cathode. Zinc is oxidised at the zinc electrode, the anode.
• E°cell = E°cathode - E°anode = 0.337 V - (-0.7618 V) = 1.0988 V, which is 1.10 V. Using the +0.34 V many school tables print gives 0.34 + 0.76 = 1.10 V as well.
• Cell notation, anode on the left: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s).

In a galvanic cell, oxidation occurs at the anode and reduction at the cathode. The anode is the negative terminal and the cathode is the positive one, and electrons leave the zinc, travel through the wire and arrive at the copper. Inside the cell, ions in the salt bridge move to keep each half-cell's charge balanced. The positive E°cell tells you the reaction runs on its own.

Question 27. Will copper metal react with 1 mol/L silver nitrate solution under standard conditions?

• (A) No. E°cell = -0.463 V.
• (B) Yes. E°cell = +1.137 V, and Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).
• (C) Yes. E°cell = +0.463 V, and Cu(s) + Ag⁺(aq) → Cu²⁺(aq) + Ag(s).
• (D) Yes. E°cell = +0.463 V, and Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).

Answer: (D). Silver has the higher reduction potential, so Ag⁺ is reduced and copper is oxidised. E°cell = E°cathode - E°anode = 0.7996 V - 0.337 V = +0.463 V. A positive E°cell means the reaction is spontaneous. Balance the electrons: each copper atom gives two and each silver ion takes one, so two silver ions are needed. The silver half-reaction is doubled, but its E° stays at 0.7996 V. A reduction potential is never multiplied by the coefficient. (A) subtracts in the wrong order. (B) adds the two potentials instead of subtracting. (C) has the right voltage and an equation that is not balanced for charge, +1 on the left and +2 on the right. A copper wire left in silver nitrate solution slowly grows a grey coat of silver while the solution turns blue, which is this reaction happening.

Question 28. An iron nail is coated with zinc (galvanised), and the coating gets scratched down to the iron. Explain why the iron still does not corrode at the scratch. What would happen if the nail were coated with copper instead and scratched?

Answer. Zinc's reduction potential, -0.7618 V, is lower than iron's, -0.447 V. That makes zinc the stronger reducing agent, the metal more easily oxidised. Where zinc and iron touch in moist air, the zinc is oxidised in place of the iron and gives up electrons that keep the iron from losing its own. The zinc is used up first, which is why this is called sacrificial or cathodic protection. It keeps working at a scratch, because the zinc around the scratch still touches the iron.

Copper is the reverse case. Its reduction potential, +0.337 V, is higher than iron's. Where a copper coating is scratched, iron is the more easily oxidised metal of the pair, so the iron at the scratch corrodes and the copper gives it no protection. It does worse than nothing: where the two metals touch in moist air they form a galvanic cell with iron as the anode, and that speeds up the corrosion of the exposed iron. A copper coating protects iron only while it is unbroken.

Answer key at a glance

Use this table to mark the quiz by strand. The skill column is what a wrong answer points to.

What to do with your score

Count the wrong answers by strand, not in total. Twenty-eight questions spread over five strands is five or six per strand, which is enough to show a pattern and too few to grade anyone.

A few wrong in each strand. The chemistry is probably fine. Go back to each miss and find the step that broke: a sign on ΔH, a mole ratio, a coefficient left off an exponent, a calculator in the wrong mode. Those are the cheapest marks to win back.

One strand clearly weaker. That is the useful result, and it is what the quiz is for. Take that strand back to the class notes and the textbook chapter, then redo the questions in it a day later without looking at the working.

Organic weak, especially questions 1 to 3. Naming comes before reactions. A reaction question that gives its reactants as names cannot be started by a student who cannot turn those names into structures.

Equilibrium weak, especially questions 19, 22 and 23. Check the setup before the algebra. All three start with writing the expression and the change in each concentration, and a wrong setup cannot be rescued by careful arithmetic.

Electrochemistry weak, especially questions 24 and 26. Learn the link once: oxidation happens at the anode, and the substance oxidised is the reducing agent. Questions 24 and 26 both test that link.

Nearly everything wrong. Stop doing practice tests and find out what is missing underneath. Grade 12 chemistry uses the toolkit Grade 11 builds, and a quick check of the basics is Genie's periodic table quiz, which has 25 questions with answers. If the trouble is how to prepare rather than the chemistry, our guide on how to study for a test covers the method. The same format is used for Grade 10 math in Genie's MPM2D practice quiz.

Where the course goes next

Ontario's diploma requires 2 credits in science, and chemistry is not compulsory. Whether a student needs SCH4U depends on the university programs they are aiming for, so check each program's own admission page for its exact list. Applications to Ontario universities go through OUAC, and our OUAC application guide explains the process.

Most students weighing SCH4U are weighing the other Grade 12 sciences too. Our SBI4U explainer covers Grade 12 biology, and the SCH4U explainer gives three real examples of what programs ask for.

When a teacher helps more than another practice test

A practice quiz shows which strand is short. It cannot show why, and the why decides what to do next.

A student who misses Question 13 might not understand calorimetry. Or they might understand it perfectly and drop the negative sign at the last line. A student who misses Question 22 might not know what Ka means, or might know it and treat acetic acid as a strong acid out of habit. Those look the same on a mark sheet and need different teaching.

A teacher who reads the working can see which it was. Genie's online chemistry tutor page says a teacher can work through "lab calculations, written lab questions, past unit tests, and exam review using the student's own course material". Genie works only with Canadian Certified Teachers, licensed by a provincial governing body such as the Ontario College of Teachers. Before a teacher sees a student profile, their certificate number and issuing body are checked against the provincial registry, with a live government-ID match. Genie calls this the Trust Engine. Students are matched by course, pace and personality, and parents get a per-lesson record of what was covered and what their child understood.

A certified chemistry teacher can also chart which courses a student needs to reach a target program, then focus the teaching on the subject areas that get them there. Our page on what a provincially certified teacher is explains what that certification means.