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.

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.

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⁺.

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.

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⁺.

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.

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).

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.

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.

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.