AP Chemistry Quiz: Introduction To Acids And Bases

What this quiz covers

This quiz focuses on Introduction To Acids And Bases, 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.

Question 1

1. Brønsted–Lowry acid, because it donates $\mathrm{H^+}$ to water.
2. Brønsted–Lowry base, because it accepts $\mathrm{H^+}$ from water. (correct answer)
3. conjugate acid of $\mathrm{CH_3COOH}$.
4. Lewis acid, because it donates an electron pair to water.
5. spectator ion, because it does not change into a different species.

Explanation: This question tests introduction to acids and bases using the Brønsted-Lowry definition to identify bases as proton acceptors. In the reaction CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq), the acetate ion CH₃COO⁻ gains a proton from water to become CH₃COOH, while water loses that proton to become OH⁻. Since CH₃COO⁻ accepts a proton, it acts as a Brønsted-Lowry base. Choice C (conjugate acid of CH₃COOH) is incorrect because CH₃COO⁻ is actually the conjugate base of CH₃COOH, not its conjugate acid. When analyzing acid-base reactions, track the proton transfer: the species gaining H⁺ is the base, the species losing H⁺ is the acid.

Question 2

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

Explanation: This question tests introduction to acids and bases focusing on identifying conjugate base relationships. In the reaction H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq), H₂PO₄⁻ donates a proton to water, becoming HPO₄²⁻. The conjugate base of any acid is what remains after the acid loses one proton. Since H₂PO₄⁻ loses H⁺ to form HPO₄²⁻, HPO₄²⁻ is the conjugate base of H₂PO₄⁻. Choice E (PO₄³⁻) is incorrect because it would require losing two protons from H₂PO₄⁻, not just one. Remember that conjugate acid-base pairs always differ by exactly one proton, and the conjugate base has one fewer H⁺ than its conjugate acid.

Question 3

1. $\mathrm{HSO_4^-}$ acts as a Brønsted–Lowry base because it accepts $\mathrm{H^+}$.
2. $\mathrm{HSO_4^-}$ acts as a Brønsted–Lowry acid because it donates $\mathrm{H^+}$. (correct answer)
3. $\mathrm{HSO_4^-}$ is a spectator ion and does not participate in proton transfer.
4. $\mathrm{HSO_4^-}$ is an Arrhenius base because it produces $\mathrm{H^+}$ in water.
5. $\mathrm{HSO_4^-}$ is neither an acid nor a base because it is negatively charged.

Explanation: This question assesses the skill of introduction to acids and bases, using the Brønsted–Lowry definition of acids as proton donors. In the reaction HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq), proton transfer is from HSO₄⁻ to H₂O. HSO₄⁻ donates H⁺ to become SO₄²⁻, confirming its role as the acid. This matches choice B, describing HSO₄⁻ as a Brønsted–Lowry acid. A tempting distractor is A, suggesting it acts as a base by accepting H⁺, but the reaction shows donation, not acceptance. Always track H⁺ transfer; the donor is the acid, and it forms its conjugate base.

Question 4

1. $\mathrm{H_2O}$ and $\mathrm{SO_3^{2-}}$
2. $\mathrm{H_3O^+}$ and $\mathrm{SO_3^{2-}}$
3. $\mathrm{HSO_3^-}$ and $\mathrm{SO_3^{2-}}$ (correct answer)
4. $\mathrm{H_2O}$ and $\mathrm{HSO_3^-}$
5. $\mathrm{HSO_3^-}$ and $\mathrm{H_3O^+}$

Explanation: This question tests introduction to acids and bases focusing on conjugate acid-base pairs in the Brønsted-Lowry framework. In the reaction HSO₃⁻(aq) + H₂O(l) ⇌ SO₃²⁻(aq) + H₃O⁺(aq), a conjugate pair differs by exactly one H⁺. HSO₃⁻ loses one H⁺ to become SO₃²⁻, making HSO₃⁻/SO₃²⁻ a conjugate acid-base pair (HSO₃⁻ is the acid, SO₃²⁻ is its conjugate base). Similarly, H₂O gains one H⁺ to become H₃O⁺, making H₂O/H₃O⁺ the other conjugate pair. Students might incorrectly choose option B (H₃O⁺ and SO₃²⁻) because these are both products, but they don't form a conjugate pair since they don't differ by exactly one proton. The strategy is to find species that differ by exactly one H⁺, where one is a reactant and the other is a product.

Question 5

1. a Brønsted–Lowry acid because it donates $\mathrm{H^+}$
2. a Brønsted–Lowry base because it accepts $\mathrm{H^+}$ (correct answer)
3. a conjugate acid because it forms $\mathrm{HF}$
4. an Arrhenius acid because it produces $\mathrm{H^+}$
5. a spectator ion because it is a halide

Explanation: This question assesses the skill of introduction to acids and bases, applying the Brønsted–Lowry definition where a base accepts a proton. In the reaction H⁺(aq) + F⁻(aq) → HF(aq), proton transfer is to F⁻ from H⁺. F⁻ accepts the H⁺ to form HF, making it the base. This determines choice B as correct. A tempting distractor is A, suggesting acid behavior, but no proton is donated by F⁻ here. Always track H⁺ transfer; the acceptor in the reaction is the base, forming its conjugate acid.

Question 6

1. A Brønsted–Lowry acid (correct answer)
2. A Brønsted–Lowry base
3. A conjugate base
4. A spectator species
5. A Lewis base because it accepts an electron pair from $\mathrm{HS^-}$

Explanation: This question tests introduction to acids and bases by classifying water's role using Brønsted-Lowry definitions. In the reaction HS⁻(aq) + H₂O(l) ⇌ H₂S(aq) + OH⁻(aq), we track proton movement in the forward direction. HS⁻ gains an H⁺ to become H₂S, while H₂O loses an H⁺ to become OH⁻. Since water donates a proton to HS⁻, water acts as a Brønsted-Lowry acid in this reaction. Students might choose option B thinking water always acts as a base, but water is amphiprotic and its role depends on what it's reacting with. The strategy is to check whether water gains H⁺ (base) or loses H⁺ (acid) by comparing H₂O to its product form.

Question 7

1. $\mathrm{CH_3COOH}$
2. $\mathrm{CH_3COO^-}$
3. $\mathrm{Na^+}$
4. $\mathrm{H_2O}$ (correct answer)
5. $\mathrm{OH^-}$

Explanation: This question tests introduction to acids and bases by identifying the Brønsted-Lowry acid in a hydrolysis reaction. In CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq), we need to identify which species donates a proton in the forward reaction. CH₃COO⁻ gains an H⁺ to become CH₃COOH (acting as base), while H₂O loses an H⁺ to become OH⁻, making water the proton donor and thus the Brønsted-Lowry acid. Students might incorrectly choose CH₃COOH (option A) because acetic acid is typically an acid, but in this reaction CH₃COOH is the product formed when the base accepts a proton. Remember to focus on the forward reaction and identify which reactant loses H⁺—that's your acid.

Question 8

1. $\mathrm{H_3O^+}$
2. $\mathrm{H_2O}$
3. $\mathrm{Cl^-}$ (correct answer)
4. $\mathrm{HCl}$
5. $\mathrm{H^+}$

Explanation: This question tests introduction to acids and bases focusing on conjugate base identification in the Brønsted-Lowry framework. In the reaction HCl + H₂O → H₃O⁺ + Cl⁻, we need to find what remains after the acid donates its proton. HCl donates an H⁺ to water, leaving behind Cl⁻, while water accepts that H⁺ to become H₃O⁺. The conjugate base of an acid is what remains after proton donation, so Cl⁻ is the conjugate base of HCl. A common mistake is choosing H₃O⁺ (choice A) because students associate it with acids, but H₃O⁺ is actually the conjugate acid of water in this reaction. To find a conjugate base, identify the acid first, then determine what species remains after it loses one H⁺.

Question 9

1. $\mathrm{H^+}$
2. $\mathrm{NO_2^-}$
3. $\mathrm{H_2O}$
4. $\mathrm{HNO_2}$ (correct answer)
5. $\mathrm{OH^-}$

Explanation: This question assesses the skill of introduction to acids and bases, focusing on the Brønsted–Lowry acid as the proton donor. In the reaction HNO₂(aq) + OH⁻(aq) → NO₂⁻(aq) + H₂O(l), proton transfer occurs from HNO₂ to OH⁻. HNO₂ loses H⁺ to become NO₂⁻, confirming it as the acid. This determines choice D as correct. A tempting distractor is B, OH⁻, but it accepts the proton, acting as the base, not the acid. Always track H⁺ transfer; the acid is the species that donates the proton in the reaction.

Question 10

1. $\mathrm{HF}$ and $\mathrm{H_3O^+}$
2. $\mathrm{H_2O}$ and $\mathrm{F^-}$
3. $\mathrm{HF}$ and $\mathrm{F^-}$ (correct answer)
4. $\mathrm{H_2O}$ and $\mathrm{H_3O^+}$
5. $\mathrm{F^-}$ and $\mathrm{H_3O^+}$

Explanation: This question tests introduction to acids and bases focusing on identifying conjugate acid-base pairs. In the reaction HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F⁻(aq), HF donates a proton to become F⁻, making HF/F⁻ a conjugate acid-base pair. Similarly, H₂O accepts a proton to become H₃O⁺, making H₂O/H₃O⁺ another conjugate pair. The question asks for one conjugate pair, and HF/F⁻ is correct because they differ by exactly one proton. Choice D (H₂O/H₃O⁺) is also a conjugate pair but represents the other half of the reaction. Remember that conjugate pairs always appear on opposite sides of the equation and differ by exactly one H⁺.

Question 11

1. $\mathrm{HCl}$
2. $\mathrm{H_2O}$ (correct answer)
3. $\mathrm{H_3O^+}$
4. $\mathrm{Cl^-}$
5. $\mathrm{H^+}$

Explanation: This question tests introduction to acids and bases using the Brønsted-Lowry definition to identify bases as proton acceptors. In the reaction HCl(g) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq), HCl donates a proton to water, with water accepting that proton to become H₃O⁺. Since water accepts the proton, it acts as the Brønsted-Lowry base, while HCl is the acid. Choice D (Cl⁻) is incorrect because it's the conjugate base of HCl, not the base in this reaction - it's a product, not a reactant accepting protons. When identifying acids and bases in reactions, focus on which reactant species gains or loses protons during the reaction process.

Question 12

1. $\mathrm{NH_4Cl}$ because it contains both ions
2. $\mathrm{NH_4^+}$
3. $\mathrm{NH_3}$ (correct answer)
4. $\mathrm{HCl}$
5. $\mathrm{Cl^-}$

Explanation: This question assesses the skill of introduction to acids and bases, using the Brønsted–Lowry definition where the base accepts a proton. In the reaction HCl(g) + NH₃(g) → NH₄⁺Cl⁻(s), proton transfer occurs from HCl to NH₃. NH₃ accepts H⁺ to form NH₄⁺, confirming it as the base. This determines choice B as correct. A tempting distractor is C, Cl⁻, but it is the conjugate base produced, not the initial base. Always track H⁺ transfer; the base is the proton acceptor in the reaction.

Question 13

1. $\mathrm{NH_4^+}$
2. $\mathrm{CO_3^{2-}}$
3. $\mathrm{H_2PO_4^-}$ (correct answer)
4. $\mathrm{Na^+}$
5. $\mathrm{Cl^-}$

Explanation: This question assesses the skill of introduction to acids and bases, focusing on the Brønsted–Lowry concept of amphiprotic species that can both donate and accept protons. To identify amphiprotic behavior, look for species with a proton to donate and the ability to accept one, like H₂PO₄⁻. H₂PO₄⁻ can donate H⁺ to become HPO₄²⁻ or accept H⁺ to form H₃PO₄. This determines C as the correct amphiprotic species. A tempting distractor is B, CO₃²⁻, but it can only accept H⁺ effectively, lacking a proton to donate. Always track H⁺ transfer; amphiprotic species must be capable of both donating and accepting protons in reactions.

Question 14

1. $\mathrm{NH_3}$ and $\mathrm{OH^-}$
2. $\mathrm{H_2O}$ and $\mathrm{OH^-}$
3. $\mathrm{NH_3}$ and $\mathrm{NH_4^+}$ (correct answer)
4. $\mathrm{H_2O}$ and $\mathrm{NH_4^+}$
5. $\mathrm{NH_4^+}$ and $\mathrm{OH^-}$

Explanation: This question assesses the skill of introduction to acids and bases, emphasizing conjugate acid-base pairs that differ by one proton. In the reaction NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), proton transfer happens from H₂O to NH₃, forming NH₄⁺ and OH⁻. The pair NH₃ and NH₄⁺ differ by H⁺, with NH₄⁺ as the acid and NH₃ as its conjugate base. This identifies D as the correct pair of conjugate partners. A tempting distractor is A, NH₃ and OH⁻, but they are not conjugates since they do not differ by a single H⁺ and belong to different pairs. Always track H⁺ transfer; conjugate pairs are linked by the gain or loss of one proton.

Question 15

1. $\mathrm{HCl}$
2. $\mathrm{NaCl}$
3. $\mathrm{NH_3}$ (correct answer)
4. $\mathrm{Cl^-}$ (from $\mathrm{NaCl}$) because it always accepts $\mathrm{H^+}$ strongly
5. $\mathrm{Na^+}$ (from $\mathrm{NaCl}$) because it produces $\mathrm{OH^-}$ directly

Explanation: This question assesses the skill of introduction to acids and bases, using the Brønsted–Lowry definition where a base accepts a proton. Among the substances, identify proton transfer potential: NH₃ has a lone pair on nitrogen, allowing it to accept H⁺ from water. This makes NH₃ the Brønsted–Lowry base when dissolved in water. Thus, choice C is correct based on its ability to participate in proton acceptance. A tempting distractor is D, Cl⁻, but it is a very weak base and does not strongly accept H⁺ compared to NH₃. Always track H⁺ transfer; bases are species with lone pairs that can accept protons in aqueous solutions.

Question 16

1. $\mathrm{H_2O}$
2. $\mathrm{H_3O^+}$
3. $\mathrm{HCO_3^-}$ (correct answer)
4. $\mathrm{CO_3^{2-}}$
5. Both $\mathrm{HCO_3^-}$ and $\mathrm{H_2O}$

Explanation: This question assesses the skill of introduction to acids and bases, focusing on the Brønsted–Lowry definition where an acid donates a proton and a base accepts one. In the reaction HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq), proton transfer occurs from HCO₃⁻ to H₂O. HCO₃⁻ loses an H⁺ to become CO₃²⁻, indicating it is the proton donor. Therefore, HCO₃⁻ is the Brønsted–Lowry acid, corresponding to choice C. A tempting distractor is E, both HCO₃⁻ and H₂O, but H₂O is accepting the proton here, acting as the base, not an acid. Always track H⁺ transfer; the acid becomes its conjugate base after donating the proton.

Question 17

1. $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COO^-}$ (correct answer)
2. $\mathrm{H_2O}$ and $\mathrm{CH_3COO^-}$
3. $\mathrm{CH_3COOH}$ and $\mathrm{H_3O^+}$
4. $\mathrm{H_2O}$ and $\mathrm{CH_3COOH}$
5. $\mathrm{H_3O^+}$ and $\mathrm{CH_3COO^-}$

Explanation: This question tests the skill of introduction to acids and bases, emphasizing conjugate pairs where the acid and its conjugate base differ by one proton. In the reaction, identify pairs by comparing formulas before and after proton transfer. CH3COOH donates H+ to become CH3COO-, making them a conjugate acid-base pair. H3O+ and H2O form another pair, but the question asks for one specific pair. A tempting distractor is CH3COOH and H3O+ (choice C), but they are both acids, not conjugates. To find conjugate pairs, compare species differing by H+; acid becomes its conjugate base after donation.

Question 18

1. A Brønsted–Lowry acid (proton donor)
2. A Brønsted–Lowry base (proton acceptor) (correct answer)
3. An Arrhenius acid because it contains hydrogen
4. A spectator ion because its charge does not change
5. A Lewis acid because it donates an electron pair

Explanation: This question tests introduction to acids and bases using the Brønsted-Lowry definition to classify species behavior. In the reaction HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq), we track proton movement to identify acids and bases. HCO₃⁻ gains an H⁺ to become H₂CO₃, which means HCO₃⁻ accepts a proton, making it a Brønsted-Lowry base. The H⁺ comes from H₂O, which loses it to become OH⁻, confirming water acts as the acid here. Students might choose option A thinking HCO₃⁻ donates a proton because it contains hydrogen, but in this specific reaction it's accepting, not donating. The key is to compare the reactant form (HCO₃⁻) with its product form (H₂CO₃) to see whether H⁺ was gained (base) or lost (acid).

Question 19

1. $\mathrm{H_2O}$
2. $\mathrm{H_3O^+}$
3. $\mathrm{CH_3COOH}$
4. $\mathrm{OH^-}$
5. $\mathrm{CH_3COO^-}$ (correct answer)

Explanation: This question tests introduction to acids and bases by identifying conjugate base relationships in Brønsted-Lowry theory. In the reaction CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq), the conjugate base of an acid is what remains after the acid donates its proton. CH₃COOH loses one H⁺ to become CH₃COO⁻, making CH₃COO⁻ the conjugate base of CH₃COOH. The proton is transferred to H₂O, which becomes H₃O⁺. Students might incorrectly choose H₂O (option B) thinking it's related to CH₃COOH, but H₂O is actually the base in this reaction that forms its own conjugate acid (H₃O⁺). Remember: to find a conjugate base, remove one H⁺ from the acid; to find a conjugate acid, add one H⁺ to the base.

Question 20

1. Only $\mathrm{H_3O^+}$
2. Only $\mathrm{NO_3^-}$
3. Only $\mathrm{H_2O}$
4. $\mathrm{H_3O^+}$ and $\mathrm{H_2O}$
5. $\mathrm{H_2O}$ and $\mathrm{NO_3^-}$ (correct answer)

Explanation: This question assesses the skill of introduction to acids and bases, using the Brønsted–Lowry definition of bases as proton acceptors. Among the species, H₂O and NO₃⁻ can accept H⁺: H₂O to H₃O⁺ via lone pairs, NO₃⁻ to HNO₃ due to its charge. H₃O⁺, however, donates H⁺ rather than accepts. This selects choice E as correct. A tempting distractor is D, including H₃O⁺ and H₂O, but H₃O⁺ is an acid, not a base. Always track H⁺ transfer; bases have features like lone pairs or negative charges that enable proton acceptance.