AP Chemistry Quiz: Introduction To Acid Base Reactions

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Introduction To Acid Base ReactionsQuestion 1 of 20

In the reaction $\mathrm{C_2H_5NH_2(aq) + H_2O(l) \rightarrow C_2H_5NH_3^+(aq) + OH^-(aq)}$ , which species is the conjugate acid of $\mathrm{C_2H_5NH_2}$ ?

\mathrm{C_2H_5NH_2}

\mathrm{C_2H_5NH_3^+}

\mathrm{OH^-}

\mathrm{H_2O}

\mathrm{H_3O^+}

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AP Chemistry Quiz

AP Chemistry Quiz: Introduction To Acid Base Reactions

Practice Introduction To Acid Base Reactions in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Introduction To Acid Base Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In the reaction $\mathrm{C_2H_5NH_2(aq) + H_2O(l) \rightarrow C_2H_5NH_3^+(aq) + OH^-(aq)}$ , which species is the conjugate acid of $\mathrm{C_2H_5NH_2}$ ?

$\mathrm{C_2H_5NH_2}$
$\mathrm{C_2H_5NH_3^+}$ (correct answer)
$\mathrm{OH^-}$
$\mathrm{H_2O}$
$\mathrm{H_3O^+}$

Explanation: This question tests the ability to identify the conjugate acid of a base. In the reaction C₂H₅NH₂(aq) + H₂O(l) → C₂H₅NH₃⁺(aq) + OH⁻(aq), the conjugate acid forms by proton acceptance. C₂H₅NH₂ accepts H⁺, becoming C₂H₅NH₃⁺, its conjugate acid. Brønsted-Lowry conjugates involve proton gain. A tempting distractor is H₃O⁺, but it is not present, from assuming hydronium always forms. Add H⁺ to the base to identify its conjugate acid in products.

Question 2

For the reaction $\mathrm{HSO_4^-(aq) + H_2O(l) \rightarrow SO_4^{2-}(aq) + H_3O^+(aq)}$ , which species is the Brønsted–Lowry acid?

$\mathrm{SO_4^{2-}}$
$\mathrm{H_2O}$
$\mathrm{H_3O^+}$
$\mathrm{OH^-}$
$\mathrm{HSO_4^-}$ (correct answer)

Explanation: This question tests the ability to identify the Brønsted-Lowry acid in a reaction. In the reaction HSO₄⁻(aq) + H₂O(l) → SO₄²⁻(aq) + H₃O⁺(aq), the Brønsted-Lowry acid is the species that donates a proton. HSO₄⁻ donates a proton to H₂O, resulting in SO₄²⁻ and H₃O⁺, so HSO₄⁻ is the acid. The Brønsted-Lowry definition emphasizes proton donation, which HSO₄⁻ exhibits here. A tempting distractor is H₃O⁺, but it is incorrect because H₃O⁺ is a product and conjugate acid, not the initial donor, due to the misconception of identifying products as reactants. To spot the Brønsted-Lowry acid, trace the proton from reactant to product in the reaction equation.

Question 3

Consider the reaction:

$\mathrm{HNO_2(aq) + OH^-(aq) \rightarrow NO_2^-(aq) + H_2O(l)}$

Which choice correctly identifies the conjugate base of the acid in this reaction?

$\mathrm{H_2O}$
$\mathrm{NO_2^-}$ (correct answer)
$\mathrm{OH^-}$
$\mathrm{HNO_2}$
$\mathrm{H^+}$

Explanation: This question tests understanding of conjugate bases in Brønsted-Lowry acid-base reactions. A conjugate base is what remains after an acid donates a proton. In the reaction HNO₂(aq) + OH⁻(aq) → NO₂⁻(aq) + H₂O(l), HNO₂ acts as the acid by donating a proton to OH⁻. When HNO₂ loses this proton, it becomes NO₂⁻, which is therefore the conjugate base of HNO₂. Students who choose option D (HNO₂) confuse the acid itself with its conjugate base, not understanding that the conjugate base is the deprotonated form of the acid. To identify a conjugate base, find the acid in the reaction and determine what it becomes after losing one H⁺.

Question 4

A reaction in water is shown below:

$\mathrm{H_2PO_4^-(aq) + H_2O(l) \rightarrow HPO_4^{2-}(aq) + H_3O^+(aq)}$

Which species is the Brønsted–Lowry acid in the reactants?

$\mathrm{HPO_4^{2-}}$
$\mathrm{H_3O^+}$
$\mathrm{H_2O}$
$\mathrm{H_2PO_4^-}$ (correct answer)
$\mathrm{PO_4^{3-}}$

Explanation: This question tests the identification of Brønsted-Lowry acids in aqueous reactions. A Brønsted-Lowry acid is a proton (H⁺) donor. In the reaction H₂PO₄⁻(aq) + H₂O(l) → HPO₄²⁻(aq) + H₃O⁺(aq), H₂PO₄⁻ loses a proton to become HPO₄²⁻, while H₂O gains a proton to become H₃O⁺. Therefore, H₂PO₄⁻ acts as the acid (proton donor) in the reactants. Students who choose option C (H₂O) incorrectly identify water as the acid, possibly because they're used to seeing water act as an acid in other reactions, not recognizing that water can act as either acid or base depending on what it reacts with. To identify the acid, look for which reactant loses H⁺ when comparing reactants to products.

Question 5

A student mixes aqueous solutions and observes the reaction:

$\mathrm{HCl(aq) + NH_3(aq) \rightarrow NH_4^+(aq) + Cl^-(aq)}$

Which choice lists the two conjugate acid–base pairs present?

$\mathrm{HCl/Cl^-}$ and $\mathrm{NH_3/NH_4^+}$ (correct answer)
$\mathrm{HCl/NH_4^+}$ and $\mathrm{NH_3/Cl^-}$
$\mathrm{HCl/NH_3}$ and $\mathrm{Cl^-/NH_4^+}$
$\mathrm{NH_4^+/NH_3}$ and $\mathrm{HCl/NH_3}$
$\mathrm{HCl/H^+}$ and $\mathrm{NH_3/OH^-}$

Explanation: This question tests the identification of conjugate acid-base pairs in a complete acid-base reaction. Conjugate acid-base pairs differ by exactly one proton (H⁺). In the reaction HCl(aq) + NH₃(aq) → NH₄⁺(aq) + Cl⁻(aq), HCl donates a proton to become Cl⁻, making HCl/Cl⁻ one conjugate pair. NH₃ accepts a proton to become NH₄⁺, making NH₃/NH₄⁺ the other conjugate pair. Students who choose option B incorrectly pair species from different sides that don't differ by one proton, such as HCl/NH₄⁺, showing confusion about what constitutes a conjugate pair. To identify all conjugate pairs, match each acid with what it becomes after losing H⁺, and each base with what it becomes after gaining H⁺.

Question 6

In aqueous solution, the following Brønsted–Lowry acid–base reaction occurs:

$\mathrm{HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)}$

Which choice correctly identifies the Brønsted–Lowry acid and base in the reactants?

Acid: $\mathrm{HCl}$ ; Base: $\mathrm{H_2O}$ (correct answer)
Acid: $\mathrm{H_3O^+}$ ; Base: $\mathrm{Cl^-}$
Acid: $\mathrm{Cl^-}$ ; Base: $\mathrm{H_3O^+}$
Acid: $\mathrm{HCl}$ ; Base: $\mathrm{Cl^-}$
Acid: $\mathrm{H_2O}$ ; Base: $\mathrm{HCl}$

Explanation: This question tests the ability to identify Brønsted-Lowry acids and bases in a reaction. According to Brønsted-Lowry theory, an acid is a proton (H⁺) donor and a base is a proton acceptor. In the reaction HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq), HCl donates a proton to H₂O, forming H₃O⁺ and Cl⁻. Therefore, HCl acts as the acid (proton donor) and H₂O acts as the base (proton acceptor). Students who choose option A incorrectly reverse the roles, possibly confusing which species gains versus loses the proton. To identify acids and bases correctly, look for which species loses H⁺ (acid) and which gains H⁺ (base) when comparing reactants to products.

Question 7

For the reaction $\mathrm{HNO_2(aq) + CN^-(aq) \rightarrow NO_2^-(aq) + HCN(aq)}$ , which statement about proton transfer is correct?

$\mathrm{HNO_2}$ donates $\mathrm{H^+}$ to $\mathrm{CN^-}$ (correct answer)
$\mathrm{CN^-}$ donates $\mathrm{H^+}$ to $\mathrm{HNO_2}$
$\mathrm{NO_2^-}$ donates $\mathrm{H^+}$ to $\mathrm{HCN}$
$\mathrm{CN^-}$ is the acid because it is negative
$\mathrm{HCN}$ donates $\mathrm{H^+}$ to $\mathrm{NO_2^-}$

Explanation: This question tests understanding of proton transfer in acid-base reactions. In the reaction HNO₂(aq) + CN⁻(aq) → NO₂⁻(aq) + HCN(aq), we must identify the correct proton transfer. HNO₂ loses a proton to become NO₂⁻, while CN⁻ gains that proton to become HCN, so HNO₂ donates H⁺ to CN⁻. Students often choose option E thinking negative ions must be acids, but charge doesn't determine acid-base behavior—proton transfer does, and negative species like CN⁻ often act as bases by accepting protons. To determine proton transfer direction, compare reactants to products to see which species lost H⁺ (acid) and which gained H⁺ (base).

Question 8

In the reaction $\mathrm{H_2S(aq) + OH^-(aq) \rightarrow HS^-(aq) + H_2O(l)}$ , which pair are conjugates of each other?

$\mathrm{H_2S}$ and $\mathrm{H_2O}$
$\mathrm{OH^-}$ and $\mathrm{HS^-}$
$\mathrm{H_2S}$ and $\mathrm{HS^-}$ (correct answer)
$\mathrm{OH^-}$ and $\mathrm{H_2S}$
$\mathrm{HS^-}$ and $\mathrm{H_2O}$

Explanation: This question tests understanding of conjugate acid-base pairs. In the reaction H₂S(aq) + OH⁻(aq) → HS⁻(aq) + H₂O(l), conjugate pairs are species that differ by one proton with one as reactant and one as product. H₂S donates a proton to become HS⁻, making H₂S/HS⁻ a conjugate acid/base pair. OH⁻ accepts that proton to become H₂O, making OH⁻/H₂O another conjugate pair. Students often incorrectly pair two reactants like OH⁻ and H₂S (option D), but conjugates must be on opposite sides of the equation. To identify conjugate pairs, look for species differing by one H⁺ with one in reactants and one in products.

Question 9

Consider the reaction $\mathrm{H_3O^+(aq) + CO_3^{2-}(aq) \rightarrow HCO_3^-(aq) + H_2O(l)}$ . Which option correctly identifies the conjugate acid of $\mathrm{CO_3^{2-}}$ ?

$\mathrm{CO_3^{2-}}$
$\mathrm{OH^-}$
$\mathrm{H_3O^+}$
$\mathrm{HCO_3^-}$ (correct answer)
$\mathrm{H_2CO_3}$

Explanation: This question tests identification of conjugate acid-base relationships. In the reaction H₃O⁺(aq) + CO₃²⁻(aq) → HCO₃⁻(aq) + H₂O(l), the conjugate acid of a base is formed when the base accepts a proton. CO₃²⁻ (carbonate) accepts a proton from H₃O⁺ to become HCO₃⁻ (bicarbonate), making HCO₃⁻ the conjugate acid of CO₃²⁻. Students might incorrectly choose H₂CO₃ (option A) thinking they should add two protons since carbonate has a 2- charge, but conjugate pairs differ by only one proton. To find a conjugate acid, add exactly one H⁺ to the base.

Question 10

In the reaction $\mathrm{NH_3(aq) + H_2O(l) \rightarrow NH_4^+(aq) + OH^-(aq)}$ , which option correctly identifies the conjugate acid–base pairs?

$\mathrm{NH_3/NH_4^+}$ and $\mathrm{H_2O/OH^-}$ (correct answer)
$\mathrm{NH_4^+/NH_3}$ and $\mathrm{OH^-/H_2O}$
$\mathrm{NH_3/OH^-}$ and $\mathrm{H_2O/NH_4^+}$
$\mathrm{NH_3/H_2O}$ and $\mathrm{NH_4^+/OH^-}$
$\mathrm{NH_4^+/OH^-}$ and $\mathrm{NH_3/H_2O}$

Explanation: This question tests your understanding of conjugate acid-base pairs in Brønsted-Lowry theory. In the reaction NH₃(aq) + H₂O(l) → NH₄⁺(aq) + OH⁻(aq), conjugate pairs differ by exactly one proton. NH₃ accepts a proton to become NH₄⁺, making NH₃/NH₄⁺ a conjugate base/acid pair. H₂O donates a proton to become OH⁻, making H₂O/OH⁻ a conjugate acid/base pair. A common error is pairing reactants with reactants or products with products (option D pairs NH₃/H₂O and NH₄⁺/OH⁻), but conjugate pairs must be on opposite sides of the equation. To identify conjugate pairs, find species that differ by one H⁺ with one on the reactant side and one on the product side.

Question 11

Consider the reaction $\mathrm{CH_3COOH(aq) + H_2O(l) \rightarrow H_3O^+(aq) + CH_3COO^-(aq)}$ . Which option correctly identifies the conjugate base of $\mathrm{CH_3COOH}$ ?

$\mathrm{H_3O^+}$
$\mathrm{CH_3COO^-}$ (correct answer)
$\mathrm{H_2O}$
$\mathrm{OH^-}$
$\mathrm{CH_3COOH_2^+}$

Explanation: This question tests your ability to identify conjugate bases in acid-base reactions. In the reaction CH₃COOH(aq) + H₂O(l) → H₃O⁺(aq) + CH₃COO⁻(aq), the conjugate base of an acid is what remains after the acid donates a proton. CH₃COOH (acetic acid) donates H⁺ to become CH₃COO⁻ (acetate ion), making CH₃COO⁻ the conjugate base of CH₃COOH. A common mistake is choosing H₂O (option C) because water can act as a base, but here water is acting as the base that accepts the proton from acetic acid, not the conjugate base of acetic acid. To find a conjugate base, remove one H⁺ from the acid.

Question 12

Consider the Brønsted–Lowry reaction: $\mathrm{HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)}$ . Which species acts as the Brønsted–Lowry base?

$\mathrm{Cl^-}$
$\mathrm{H_2O}$ (correct answer)
$\mathrm{HCl}$
$\mathrm{H_3O^+}$
$\mathrm{H^+}$

Explanation: This question tests the ability to identify the Brønsted-Lowry base in an acid-base reaction. In the reaction HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq), the Brønsted-Lowry base is the species that accepts a proton (H⁺). HCl donates a proton to H₂O, forming H₃O⁺ and Cl⁻, so H₂O acts as the base by accepting the proton. According to the Brønsted-Lowry definition, acids donate protons and bases accept them, confirming H₂O's role here. A tempting distractor is Cl⁻, but it is incorrect because Cl⁻ is the conjugate base and does not accept a proton in this reaction, reflecting the misconception of confusing products with reactants in proton transfer. To identify the Brønsted-Lowry base, always look for the species that gains a proton in the forward reaction.

Question 13

In the reaction $\mathrm{H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}$ , which statement correctly identifies the Brønsted–Lowry roles of water?

One $\mathrm{H_2O}$ acts as an acid and the other as a base (correct answer)
Both $\mathrm{H_2O}$ molecules act only as acids
Both $\mathrm{H_2O}$ molecules act only as bases
$\mathrm{H_3O^+}$ is the base and $\mathrm{OH^-}$ is the acid
$\mathrm{H_2O}$ is neither an acid nor a base because it is neutral

Explanation: This question tests the understanding of amphoteric behavior in water's autoionization. In the reaction H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq), water acts both as acid and base. One H₂O donates H⁺ to another, forming H₃O⁺ and OH⁻, showing amphoterism. Brønsted-Lowry allows substances to be both depending on context. A tempting distractor is both H₂O act only as acids, but it is incorrect because one accepts the proton, from the misconception that water only donates in ionization. Recognize amphoteric species by checking if they can donate or accept protons in different reactions.

Question 14

Consider the reaction:

$\mathrm{NH_3(aq) + H_2O(l) \rightarrow NH_4^+(aq) + OH^-(aq)}$

Which species is the conjugate acid of $\mathrm{NH_3}$ in this reaction?

$\mathrm{OH^-}$
$\mathrm{NH_4^+}$ (correct answer)
$\mathrm{H_2O}$
$\mathrm{NH_3}$
$\mathrm{H_3O^+}$

Explanation: This question tests understanding of conjugate acid-base pairs in Brønsted-Lowry theory. A conjugate acid is formed when a base accepts a proton (H⁺). In the reaction NH₃(aq) + H₂O(l) → NH₄⁺(aq) + OH⁻(aq), NH₃ acts as a base by accepting a proton from H₂O. When NH₃ gains this proton, it becomes NH₄⁺, which is therefore the conjugate acid of NH₃. Students who choose option D might confuse the original base (NH₃) with its conjugate acid, not recognizing that conjugate acids and bases differ by exactly one proton. To identify conjugate pairs, remember that a base plus one H⁺ equals its conjugate acid.

Question 15

The reaction below occurs in aqueous solution:

$\mathrm{H_3O^+(aq) + CO_3^{2-}(aq) \rightarrow HCO_3^-(aq) + H_2O(l)}$

Which species acts as the Brønsted–Lowry base in the reactants?

$\mathrm{H_3O^+}$
$\mathrm{HCO_3^-}$
$\mathrm{CO_3^{2-}}$ (correct answer)
$\mathrm{H_2O}$
$\mathrm{H^+}$

Explanation: This question tests the identification of Brønsted-Lowry bases in acid-base reactions. A Brønsted-Lowry base is a proton (H⁺) acceptor. In the reaction H₃O⁺(aq) + CO₃²⁻(aq) → HCO₃⁻(aq) + H₂O(l), H₃O⁺ donates a proton to CO₃²⁻, which accepts it to form HCO₃⁻. Therefore, CO₃²⁻ acts as the base (proton acceptor) in the reactants. Students who choose option A (H₃O⁺) incorrectly identify the acid as the base, possibly confusing the roles or not recognizing that H₃O⁺ is always an acid (proton donor) in aqueous solutions. To identify the base among reactants, look for which species gains a proton when comparing reactants to products.

Question 16

Consider the reaction $\mathrm{H_2PO_4^-(aq) + OH^-(aq) \rightarrow HPO_4^{2-}(aq) + H_2O(l)}$ . Which species is the conjugate base of $\mathrm{H_2PO_4^-}$ ?

$\mathrm{HPO_4^{2-}}$ (correct answer)
$\mathrm{OH^-}$
$\mathrm{H_2O}$
$\mathrm{H_3PO_4}$
$\mathrm{H_3O^+}$

Explanation: This question tests the ability to identify the conjugate base of an acid. In the reaction H₂PO₄⁻(aq) + OH⁻(aq) → HPO₄²⁻(aq) + H₂O(l), the conjugate base forms after donation. H₂PO₄⁻ donates H⁺, leaving HPO₄²⁻ as conjugate base. Conjugates differ by H⁺. A tempting distractor is OH⁻, but it is the base, not conjugate of H₂PO₄⁻, from role mixing. Subtract H⁺ from the acid to find its conjugate base in the reaction.

Question 17

Consider the reaction $\mathrm{HClO_4(aq) + H_2O(l) \rightarrow H_3O^+(aq) + ClO_4^-(aq)}$ . Which species is the conjugate base of the acid?

$\mathrm{H_2O}$
$\mathrm{H_3O^+}$
$\mathrm{ClO_4^-}$ (correct answer)
$\mathrm{HClO_4}$
$\mathrm{Cl^-}$

Explanation: This question tests the ability to identify the conjugate base of the acid. In the reaction HClO₄(aq) + H₂O(l) → H₃O⁺(aq) + ClO₄⁻(aq), the conjugate base remains after acid proton donation. HClO₄ donates H⁺, leaving ClO₄⁻ as its conjugate base. Brønsted-Lowry conjugates differ by H⁺. A tempting distractor is H₂O, but it is the base, not the conjugate of HClO₄, due to role confusion. To find the conjugate base, remove H⁺ from the acid and locate it in the products.

Question 18

For the reaction $\mathrm{HBr(aq) + CO_3^{2-}(aq) \rightarrow HCO_3^-(aq) + Br^-(aq)}$ , which species is the Brønsted–Lowry base?

$\mathrm{HCO_3^-}$
$\mathrm{H^+}$
$\mathrm{Br^-}$
$\mathrm{CO_3^{2-}}$ (correct answer)
$\mathrm{HBr}$

Explanation: This question tests the ability to identify the Brønsted-Lowry base. In the reaction HBr(aq) + CO₃²⁻(aq) → HCO₃⁻(aq) + Br⁻(aq), the base accepts a proton. CO₃²⁻ accepts H⁺ from HBr, becoming HCO₃⁻, so it is the base. Brønsted-Lowry bases gain protons. A tempting distractor is HCO₃⁻, but it is the conjugate acid, not the base, due to product confusion. Look for the reactant that gains H⁺ to form a product with one more proton.

Question 19

Consider the reaction $\mathrm{HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)}$ . Which choice correctly identifies the Brønsted–Lowry base in the forward reaction?

$\mathrm{H_2O}$
$\mathrm{H_2CO_3}$
$\mathrm{OH^-}$
$\mathrm{HCO_3^-}$ (correct answer)
$\mathrm{H_2O}$ (because it produces $\mathrm{OH^-}$ )

Explanation: This question tests the identification of Brønsted-Lowry bases in reactions producing hydroxide ions. In the reaction HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq), water donates a proton to HCO₃⁻, making water the acid in this reaction. HCO₃⁻ accepts the proton to become H₂CO₃, making HCO₃⁻ the Brønsted-Lowry base. The production of OH⁻ indicates that water acted as an acid, not a base. A common misconception is thinking water is the base (choice A or E) because the reaction produces OH⁻, but the base is determined by which species accepts a proton, not by what products form. To identify the base in any reaction, look for the species that gains H⁺ in the forward direction.

Question 20

Consider the reaction $\mathrm{HSO_4^-(aq) + H_2O(l) \rightarrow SO_4^{2-}(aq) + H_3O^+(aq)}$ . In the forward direction, which species acts as the Brønsted–Lowry base?

$\mathrm{HSO_4^-}$
$\mathrm{H_3O^+}$
$\mathrm{SO_4^{2-}}$
$\mathrm{H_2O}$ (correct answer)
$\mathrm{SO_4^{2-}}$ (as an Arrhenius base only)

Explanation: This question tests the ability to identify the Brønsted-Lowry base in an acid-base reaction. In the reaction HSO₄⁻(aq) + H₂O(l) → SO₄²⁻(aq) + H₃O⁺(aq), HSO₄⁻ donates a proton to water, making HSO₄⁻ the acid. Water accepts this proton to form H₃O⁺, making H₂O the Brønsted-Lowry base in the forward reaction. The base is always the proton acceptor in Brønsted-Lowry theory. A common misconception is thinking SO₄²⁻ is the base (choice C) because it has a negative charge, but SO₄²⁻ is a product, not a reactant, and bases are identified by their role in accepting protons, not by their charge. To identify the base, look for the species that gains a proton in the forward reaction.