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.

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.

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.

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.

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.

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.

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.

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⁺.

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.

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⁺.