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AP Chemistry › Unit 4

AP Chemistry Unit 4: Chemical Reactions

7 to 9% of the exam · 180 practice questions · 5 free-response questions

This unit shifts from physical changes to chemical ones. You will write balanced molecular, complete ionic, and net ionic equations, classify reactions as acid-base, redox, or precipitation, and do stoichiometry, including titration and limiting reactant problems. Many exam questions ask you to justify why a reaction belongs to a type, using the equation as evidence. The most common calculation error is comparing grams instead of moles to find the limiting reactant. Students also leave spectator ions in net ionic equations or forget to balance charge.

Practice this unit

White belt · 60 questionsRegular · 60 questionsBlack belt · 60 questionsFree response

White belt covers core terms and foundational ideas, Regular is set at exam level, and Black belt has harder multi-step questions.

Topics in this unit

Free-response questions

Sample questions

These are taken from the Regular set, with the answer and explanation shown.

Topic 4.2 Net Ionic Equations

1. Na₂CO₃(aq) is mixed with excess HCl(aq), and bubbles of CO₂ form. Which is the net ionic equation?

  1. ACO₃²⁻(aq) + H⁺(aq) → H₂O(l) + CO₂(g)
  2. B2 Na⁺(aq) + 2 Cl⁻(aq) → 2 NaCl(s)
  3. CNa₂CO₃(aq) + 2 H⁺(aq) → 2 Na⁺(aq) + H₂O(l) + CO₂(g)
  4. DCO₃²⁻(aq) + 2 H⁺(aq) → H₂O(l) + CO₂(g)

Answer: D. Na₂CO₃ is a soluble sodium salt, so it is written as ions, and Na⁺ is a spectator. Cl⁻ is also a spectator. Carbonate takes two H⁺ to make H₂O and CO₂, and the charges balance (−2 + 2 = 0).

Common trap: Using one H⁺. Then H and charge do not balance: the left has 1 H and charge −1, the right has 2 H and charge 0.

Topic 4.6 Introduction to Titration

Titration of a solid monoprotic acid with NaOH

A student dissolves samples of an unknown solid monoprotic acid, HA, in about 50 mL of water and titrates each one with 0.1000 M NaOH(aq) to the equivalence point: HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l).

TrialMass of HA (g)Initial buret reading (mL)Final buret reading (mL)
10.40840.2020.20
20.510520.2045.20
30.30631.0516.05

Hypothetical student data for practice.

2. Based on the data, what volume of NaOH(aq) was added to reach the equivalence point in Trial 2?

  1. A20.20 mL
  2. B25.00 mL
  3. C45.20 mL
  4. D65.40 mL

Answer: B. Volume delivered = final reading − initial reading = 45.20 mL − 20.20 mL = 25.00 mL. Trial 2 started where Trial 1 ended.

Common trap: Choosing 45.20 mL from the final reading alone. That includes the 20.20 mL used in Trial 1.

Topic 4.8 Introduction to Acid-Base Reactions

3. In the reaction HCN(aq) + OH⁻(aq) ⇌ CN⁻(aq) + H₂O(l), which species is the conjugate base of the acid?

  1. ACN⁻
  2. BH₂O
  3. COH⁻
  4. DHCN

Answer: A. HCN is the acid because it donates H⁺ to OH⁻. Removing that H⁺ from HCN leaves CN⁻, its conjugate base. H₂O is the conjugate acid of OH⁻.

Common trap: Choosing OH⁻ because it is a base. OH⁻ is the base reactant, not the conjugate base of HCN.

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