⚗️ Unit 4: Chemical Reactions

Master the language of chemistry — balancing equations, stoichiometry, titrations, and reaction types.

Exam Weight: 7–9% | Topics 4.1–4.9

📑 In This Unit

4.1 — Introduction for Reactions

4.2 — Net Ionic Equations

4.3 — Representations of Reactions

4.4 — Physical and Chemical Changes

4.5 — Stoichiometry

4.6 — Introduction to Titration

4.7 — Types of Chemical Reactions

4.8 — Introduction to Acid-Base Reactions

4.9 — Oxidation-Reduction (Redox) Reactions

📝 Practice Multiple-Choice Questions (10)

4.1 — Introduction for Reactions

A chemical equation is a symbolic representation of a chemical reaction showing reactants on the left and products on the right. Equations must be balanced to satisfy the Law of Conservation of Mass — atoms are neither created nor destroyed in a chemical reaction.

Balancing Chemical Equations

Write the unbalanced equation with correct formulas for all reactants and products.
Count atoms of each element on both sides.
Adjust coefficients (the numbers in front of formulas) to equalize atom counts. Never change subscripts.
Balance metals first, then nonmetals, then hydrogen, then oxygen last.
Verify: every element has the same count on both sides, and coefficients are in the simplest whole-number ratio.

Example: Balance: Fe₂O₃ + CO → Fe + CO₂
Fe: 2 on left, 1 on right → put 2 in front of Fe
O: 3 + 1 = 4 on left; need 4 on right → put 2 in front of CO₂, but check C
C: 1 on left, 2 on right → put 2 in front of CO: Fe₂O₃ + 2CO → 2Fe + 2CO₂
Check O: 3 + 2 = 5 on left; 4 on right — not balanced yet!
Try: Fe₂O₃ + 3CO → 2Fe + 3CO₂
Fe: 2 = 2; C: 3 = 3; O: 3 + 3 = 6 left; 6 right. Balanced!

State Symbols

(s) = solid | (l) = liquid | (g) = gas | (aq) = aqueous (dissolved in water)
State symbols are important for net ionic equations and for determining which species are electrolytes.

Conservation of Mass in calculations: The total mass of reactants always equals the total mass of products. If 10.0 g of reactants are used, 10.0 g of products form (even if one product is a gas that escapes).

4.2 — Net Ionic Equations

When ionic compounds are dissolved in water, they dissociate into ions. A net ionic equation shows only the species that actually participate in the reaction, removing the spectator ions (ions that appear on both sides unchanged).

Three Types of Equations

Molecular equation: Shows complete formulas of all reactants and products.
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
Complete ionic equation: Shows all strong electrolytes dissociated into ions.
Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
Net ionic equation: Remove spectator ions (Na⁺ and NO₃⁻ appear on both sides).
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Rules for writing net ionic equations:
• Dissociate strong electrolytes (soluble ionic compounds, strong acids, strong bases) into ions.
• Do NOT dissociate: weak acids, weak bases, insoluble compounds (precipitates), water, gases, or molecular compounds.
• Cancel spectator ions that appear identically on both sides.
• The net ionic equation must be balanced for both atoms and charge.

Example — Neutralization:
Molecular: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Complete ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)
Net ionic: H⁺(aq) + OH⁻(aq) → H₂O(l)
(This is the net ionic equation for ALL strong acid + strong base reactions)

4.3 — Representations of Reactions

Chemical reactions can be represented at three levels, and the AP exam expects you to move fluently between them:

Macroscopic: What you observe — color changes, gas bubbles, precipitate formation, temperature changes, odor.
Particulate: Diagrams showing individual atoms, molecules, and ions before and after the reaction. Must show correct stoichiometric ratios and conservation of atoms.
Symbolic: Balanced chemical equations with formulas, coefficients, and state symbols.

Interpreting particulate diagrams on the AP exam:
• Count atoms of each element before and after — they must be equal (conservation of mass).
• Identify the limiting reagent: the reactant that runs out first determines how much product forms.
• Leftover reactant molecules should appear in the “after” diagram.
• Products should show correct molecular formulas and ratios.
• Ions in solution should be drawn as separate particles surrounded by water molecules.

4.4 — Physical and Chemical Changes

Understanding the difference between physical and chemical changes is fundamental to chemistry.

Feature	Physical Change	Chemical Change
Definition	Change in form or appearance; no new substance formed	New substance(s) produced with different properties
Bonds	Intermolecular forces may be overcome, but chemical bonds are NOT broken	Chemical bonds are broken and new bonds are formed
Reversibility	Usually easily reversible	Often difficult to reverse
Examples	Melting ice, boiling water, dissolving sugar, cutting paper	Combustion, rusting, cooking, acid-base reactions

Evidence of a chemical change:
• Color change (e.g., iron turning rust-colored)
• Gas production (bubbles forming)
• Precipitate formation (solid appearing in solution)
• Temperature change (exothermic or endothermic)
• Odor change
• Light or sound emission

Caution: Some physical changes also produce these signs (e.g., boiling produces bubbles, dissolving can release heat). Context matters!

AP exam distinction — dissolving:
• Dissolving an ionic compound (NaCl in water) involves breaking the crystal lattice and forming ion-dipole interactions. This is a physical change — no chemical bonds are broken. NaCl can be recovered by evaporation.
• Dissolving a metal in acid (Zn in HCl) is a chemical change — the Zn is oxidized and forms a new compound.

4.5 — Stoichiometry

Stoichiometry uses the mole ratios from a balanced chemical equation to calculate the amounts of reactants consumed and products formed. It is one of the most tested skills on the AP exam.

Formation of water from hydrogen and oxygen

The electrolysis of water demonstrates stoichiometric ratios: 2H₂ + O₂ → 2H₂O. (Wikimedia Commons)

The Mole Roadmap for Stoichiometry

Balance the chemical equation.
Convert given quantity to moles (if given in grams, use molar mass; if given as volume of solution, use molarity; if given as volume of gas at STP, divide by 22.4 L).
Use the mole ratio from the balanced equation to convert from moles of given substance to moles of desired substance.
Convert moles of desired substance to the requested unit (grams, liters, particles, etc.).

Example: How many grams of O₂ are needed to completely react with 10.0 g of H₂?
Equation: 2H₂ + O₂ → 2H₂O
Step 1: mol H₂ = 10.0 g / 2.016 g/mol = 4.96 mol
Step 2: mol O₂ = 4.96 mol H₂ × (1 mol O₂ / 2 mol H₂) = 2.48 mol
Step 3: mass O₂ = 2.48 mol × 32.00 g/mol = 79.4 g O₂

Limiting Reagent

When two or more reactants are present, the limiting reagent is the reactant that runs out first — it limits how much product can form. The other reactant(s) are in excess.

How to find the limiting reagent:
1. Convert each reactant to moles.
2. Divide each by its coefficient in the balanced equation.
3. The one with the smallest value is the limiting reagent.
4. Use the limiting reagent to calculate the theoretical yield.

Example: 5.0 mol H₂ and 2.0 mol O₂ react: 2H₂ + O₂ → 2H₂O
H₂: 5.0/2 = 2.5 | O₂: 2.0/1 = 2.0 ← smaller
O₂ is the limiting reagent. It produces 2.0 × 2 = 4.0 mol H₂O.

Percent Yield

% yield = (actual yield / theoretical yield) × 100%

• Theoretical yield: Maximum product calculated from stoichiometry using the limiting reagent.
• Actual yield: What you actually get in the lab (always ≤ theoretical yield).
• Percent yield is always ≤ 100% (if > 100%, there is an error or impurity).

4.6 — Introduction to Titration

A titration is a technique for determining the concentration of an unknown solution by reacting it with a solution of known concentration (the standard solution or titrant). It is most commonly used for acid-base reactions.

Typical titration setup — titrant is added from a buret to the analyte in a flask until the endpoint is reached. (Wikimedia Commons)

Key Terminology

Analyte: The solution of unknown concentration (in the flask).
Titrant: The solution of known concentration (in the buret).
Equivalence point: The point at which the moles of titrant exactly react with the moles of analyte (stoichiometric ratio is satisfied).
Endpoint: The point at which the indicator changes color. Ideally, the endpoint is very close to the equivalence point.
Indicator: A substance that changes color near the equivalence point (e.g., phenolphthalein turns pink in base).

Titration calculation:
At the equivalence point: moles of acid = moles of base (for 1:1 stoichiometry)
M_acid × V_acid = M_base × V_base

Example: 25.0 mL of unknown HCl requires 18.5 mL of 0.200 M NaOH to reach the equivalence point. Find [HCl].
mol NaOH = 0.0185 L × 0.200 M = 0.00370 mol
HCl + NaOH → NaCl + H₂O (1:1 ratio), so mol HCl = 0.00370 mol
[HCl] = 0.00370 / 0.0250 = 0.148 M

For non-1:1 ratios:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
At equivalence: mol H₂SO₄ = (1/2) × mol NaOH
Always use the balanced equation to determine the correct stoichiometric ratio.

Back Titration

A back titration is used when the analyte cannot be easily titrated directly. You add a known excess of one reagent, let it react, then titrate the excess with another standard solution. The difference tells you how much analyte was present.

4.7 — Types of Chemical Reactions

Reactions can be classified into categories that help predict products. Knowing these patterns is essential for the AP exam free-response questions.

Reaction Type	General Form	Example	How to Recognize
Synthesis	A + B → AB	2Na + Cl₂ → 2NaCl	Two or more substances combine to form one product
Decomposition	AB → A + B	2H₂O₂ → 2H₂O + O₂	One compound breaks into two or more substances
Single Replacement	A + BC → AC + B	Zn + CuSO₄ → ZnSO₄ + Cu	An element replaces another in a compound (use activity series)
Double Replacement	AB + CD → AD + CB	AgNO₃ + NaCl → AgCl + NaNO₃	Ions switch partners; often forms a precipitate, gas, or water
Combustion	C_xH_y + O₂ → CO₂ + H₂O	CH₄ + 2O₂ → CO₂ + 2H₂O	Organic compound reacts with O₂; produces CO₂ and H₂O

Precipitation Reactions & Solubility Rules

A precipitation reaction occurs when two aqueous ionic solutions mix and produce an insoluble solid (precipitate). Memorize these solubility rules:

Soluble (dissolve in water):
• All Group 1 (Li⁺, Na⁺, K⁺) and NH₄⁺ salts
• All nitrates (NO₃⁻) and acetates (CH₃COO⁻)
• Most chlorides, bromides, iodides — except Ag⁺, Pb²⁺, Hg₂²⁺
• Most sulfates — except BaSO₄, PbSO₄, CaSO₄, SrSO₄

Insoluble (form precipitates):
• Most hydroxides (OH⁻) — except Group 1 and Ba(OH)₂, Ca(OH)₂ (slightly)
• Most carbonates (CO₃²⁻), phosphates (PO₄³⁻), and sulfides (S²⁻) — except Group 1 and NH₄⁺

Activity Series for Single Replacement

A more active metal will displace a less active metal from a solution. If the free element is above the element in the compound on the activity series, the reaction proceeds. Otherwise, no reaction occurs.

4.8 — Introduction to Acid-Base Reactions

Acid-base reactions are among the most important in chemistry. The AP exam focuses on the Brønsted-Lowry definitions.

Brønsted-Lowry Definitions

Acid: A proton (H⁺) donor.
Base: A proton (H⁺) acceptor.
In every acid-base reaction, there is a conjugate acid-base pair: the acid donates a proton to become its conjugate base; the base accepts a proton to become its conjugate acid.

Strong vs. Weak Acids and Bases

	Strong (100% dissociation)	Weak (partial dissociation)
Acids	HCl, HBr, HI, HNO₃, H₂SO₄, HClO₃, HClO₄	CH₃COOH (acetic), HF, H₂CO₃, H₃PO₄, HCN, HNO₂
Bases	NaOH, KOH, LiOH, Ca(OH)₂, Ba(OH)₂, Sr(OH)₂	NH₃, CH₃NH₂, pyridine, CO₃²⁻, HCO₃⁻

Memorize the 7 strong acids! Everything else is weak. A strong acid/base dissociates 100% in water. A weak acid/base establishes an equilibrium with only partial ionization.

Neutralization reactions:
• Strong acid + Strong base: H⁺(aq) + OH⁻(aq) → H₂O(l) (net ionic for all such reactions)
• Weak acid + Strong base: HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l)
• Strong acid + Weak base: H⁺(aq) + B(aq) → BH⁺(aq)

Gas-evolving reactions (memorize these patterns):
• Acid + carbonate → salt + H₂O + CO₂(g)
• Acid + sulfite → salt + H₂O + SO₂(g)
• Acid + sulfide → salt + H₂S(g)

4.9 — Oxidation-Reduction (Redox) Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between species. One substance loses electrons (oxidized) while another gains electrons (reduced). These reactions are fundamental to electrochemistry, metabolism, and industrial processes.

Redox reaction showing electron transfer: oxidation (loss of e⁻) and reduction (gain of e⁻). (Wikimedia Commons)

OIL RIG:
• Oxidation Is Loss (of electrons) — oxidation number increases
• Reduction Is Gain (of electrons) — oxidation number decreases

Oxidizing agent: Gets reduced (gains electrons); causes oxidation in the other species.
Reducing agent: Gets oxidized (loses electrons); causes reduction in the other species.

Rules for Assigning Oxidation Numbers

Free elements have oxidation number = 0 (e.g., Na, O₂, P₄)
Monatomic ions: oxidation number = charge (e.g., Na⁺ = +1, Cl⁻ = −1)
Hydrogen is usually +1 (except in metal hydrides like NaH, where it is −1)
Oxygen is usually −2 (except in peroxides like H₂O₂, where it is −1, and in OF₂ where it is +2)
Fluorine is always −1
The sum of oxidation numbers in a neutral compound = 0; in a polyatomic ion = the ion’s charge

Example — Identifying oxidation and reduction:
2Fe₂O₃ + 3C → 4Fe + 3CO₂

Fe in Fe₂O₃: let Fe = x, O = −2 → 2x + 3(−2) = 0 → x = +3
Fe in product: 0 (free element)
Fe goes from +3 to 0: reduction (gained electrons). Fe₂O₃ is the oxidizing agent.

C in reactant: 0 (free element)
C in CO₂: let C = x, O = −2 → x + 2(−2) = 0 → x = +4
C goes from 0 to +4: oxidation (lost electrons). C is the reducing agent.

Activity Series

The activity series ranks metals by their tendency to be oxidized (lose electrons). A metal higher on the series can displace a metal lower on the series from a solution of its ions.

Top (most easily oxidized): Li, K, Ca, Na, Mg, Al, Zn, Fe, Ni, Sn, Pb, H₂, Cu, Ag, Pt, Au (least easily oxidized)

A metal above H₂ can react with acids to produce H₂ gas.
A metal below H₂ does not react with most acids (Cu, Ag, Au).

📝 Practice Multiple-Choice Questions

Test your knowledge of Unit 4. Click “Show Answer” to reveal the correct choice and explanation.

1. When the equation __Fe + __O₂ → __Fe₂O₃ is balanced with the smallest whole-number coefficients, the coefficient for Fe is:

(A) 1
(B) 2
(C) 3
(D) 4

Answer: (D)
4Fe + 3O₂ → 2Fe₂O₃. Check: Fe: 4 = 2(2) = 4; O: 3(2) = 6 = 2(3) = 6. The coefficient for Fe is 4.

2. What is the net ionic equation for the reaction between Pb(NO₃)₂(aq) and KI(aq)?

(A) Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2KNO₃(aq)
(B) Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
(C) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)
(D) K⁺(aq) + NO₃⁻(aq) → KNO₃(aq)

Answer: (B)
PbI₂ is insoluble (Pb²⁺ with I⁻). The spectator ions K⁺ and NO₃⁻ are removed. Net ionic: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).

3. Which of the following is evidence of a chemical change?

(A) Ice melting at 0°C
(B) Sugar dissolving in water
(C) A nail rusting when exposed to air and water
(D) Alcohol evaporating from an open container

Answer: (C)
Rusting is a chemical change: iron reacts with oxygen and water to form iron oxide (Fe₂O₃). A new substance with different properties is formed. The other options are physical changes.

4. If 5.0 mol H₂ and 2.0 mol O₂ react according to 2H₂ + O₂ → 2H₂O, what is the theoretical yield of H₂O?

(A) 2.0 mol
(B) 4.0 mol
(C) 5.0 mol
(D) 7.0 mol

Answer: (B)
Find limiting reagent: H₂: 5.0/2 = 2.5; O₂: 2.0/1 = 2.0. O₂ is limiting (smaller value). Mol H₂O = 2.0 mol O₂ × (2 mol H₂O / 1 mol O₂) = 4.0 mol H₂O.

5. A student obtains 15.0 g of product in a reaction where the theoretical yield is 20.0 g. What is the percent yield?

(A) 25.0%
(B) 50.0%
(C) 75.0%
(D) 133%

Answer: (C)
% yield = (actual / theoretical) × 100 = (15.0 / 20.0) × 100 = 75.0%.

6. 30.0 mL of 0.100 M NaOH is required to neutralize 20.0 mL of an HCl solution. What is the molarity of the HCl?

(A) 0.0667 M
(B) 0.100 M
(C) 0.150 M
(D) 0.300 M

Answer: (C)
mol NaOH = 0.0300 L × 0.100 M = 0.00300 mol. HCl + NaOH is 1:1, so mol HCl = 0.00300. [HCl] = 0.00300 / 0.0200 = 0.150 M.

7. When solutions of Na₂CO₃ and CaCl₂ are mixed, a precipitate forms. Which compound precipitates?

(A) NaCl
(B) CaCO₃
(C) Na₂CO₃
(D) No precipitate forms

Answer: (B)
According to solubility rules, most carbonates are insoluble (except alkali metals and NH₄⁺). CaCO₃ is insoluble and forms a white precipitate. NaCl is soluble.

8. In the reaction: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), which species is the reducing agent?

(A) Zn
(B) HCl
(C) ZnCl₂
(D) H₂

Answer: (A)
Zn goes from 0 to +2 (oxidized — loses electrons). The reducing agent is the species that gets oxidized. Zn donates electrons to H⁺, which is reduced from +1 to 0 (H₂ gas).

9. What is the oxidation number of Cr in K₂Cr₂O₇?

(A) +3
(B) +5
(C) +6
(D) +7

Answer: (C)
K = +1 (Group 1), O = −2. 2(+1) + 2(Cr) + 7(−2) = 0 → 2 + 2Cr − 14 = 0 → 2Cr = 12 → Cr = +6.

10. According to the activity series, which of the following metals can displace Cu²⁺ from a CuSO₄ solution?

(A) Ag
(B) Au
(C) Pt
(D) Zn

Answer: (D)
Zn is above Cu in the activity series, so it is more easily oxidized and can displace Cu²⁺: Zn + CuSO₄ → ZnSO₄ + Cu. Ag, Au, and Pt are below Cu and cannot displace it.

💡 Study Tip for Unit 4

Unit 4 brings together many core skills. Make sure you can balance equations quickly, write net ionic equations (memorize the strong acids and solubility rules), and solve stoichiometry problems with limiting reagents. Practice titration calculations until they are second nature. For redox, focus on assigning oxidation numbers and identifying the oxidizing/reducing agents. These skills are essential for Units 7–9 (equilibrium, acids/bases, and electrochemistry).

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