Introduction to Acid-Base Reactions
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AP Chemistry › Introduction to Acid-Base Reactions
For the reaction $\mathrm{HNO_2(aq) + CN^-(aq) \rightarrow NO_2^-(aq) + HCN(aq)}$, which statement about proton transfer is correct?
$\mathrm{CN^-}$ donates $\mathrm{H^+}$ to $\mathrm{HNO_2}$
$\mathrm{HNO_2}$ donates $\mathrm{H^+}$ to $\mathrm{CN^-}$
$\mathrm{HCN}$ donates $\mathrm{H^+}$ to $\mathrm{NO_2^-}$
$\mathrm{NO_2^-}$ donates $\mathrm{H^+}$ to $\mathrm{HCN}$
$\mathrm{CN^-}$ is the acid because it is negative
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).
In the reaction $\mathrm{H_2S(aq) + OH^-(aq) \rightarrow HS^-(aq) + H_2O(l)}$, which pair are conjugates of each other?
$\mathrm{OH^-}$ and $\mathrm{H_2S}$
$\mathrm{H_2S}$ and $\mathrm{HS^-}$
$\mathrm{OH^-}$ and $\mathrm{HS^-}$
$\mathrm{H_2S}$ and $\mathrm{H_2O}$
$\mathrm{HS^-}$ 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.
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 $\mathrm{CO_3^{2-}}$?
$\mathrm{OH^-}$
$\mathrm{HCO_3^-}$
$\mathrm{H_3O^+}$
$\mathrm{CO_3^{2-}}$
$\mathrm{H_2CO_3}$
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.
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 $\mathrm{H_2O/OH^-}$
$\mathrm{NH_3/OH^-}$ and $\mathrm{H_2O/NH_4^+}$
$\mathrm{NH_4^+/OH^-}$ and $\mathrm{NH_3/H_2O}$
$\mathrm{NH_3/H_2O}$ and $\mathrm{NH_4^+/OH^-}$
$\mathrm{NH_4^+/NH_3}$ and $\mathrm{OH^-/H_2O}$
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.
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 $\mathrm{CH_3COOH}$?
$\mathrm{H_3O^+}$
$\mathrm{OH^-}$
$\mathrm{CH_3COO^-}$
$\mathrm{H_2O}$
$\mathrm{CH_3COOH_2^+}$
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.
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?
$\mathrm{HCl}$
$\mathrm{H_3O^+}$
$\mathrm{H^+}$
$\mathrm{H_2O}$
$\mathrm{Cl^-}$
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.
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 $\mathrm{C_2H_5NH_2}$?
$\mathrm{C_2H_5NH_2}$
$\mathrm{H_2O}$
$\mathrm{C_2H_5NH_3^+}$
$\mathrm{OH^-}$
$\mathrm{H_3O^+}$
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.
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?
$\mathrm{H_3O^+}$
$\mathrm{HSO_4^-}$
$\mathrm{SO_4^{2-}}$
$\mathrm{H_2O}$
$\mathrm{OH^-}$
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.
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?
Both $\mathrm{H_2O}$ molecules act only as bases
$\mathrm{H_3O^+}$ is the base and $\mathrm{OH^-}$ is the acid
One $\mathrm{H_2O}$ acts as an acid and the other as a base
$\mathrm{H_2O}$ is neither an acid nor a base because it is neutral
Both $\mathrm{H_2O}$ molecules act only as acids
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.
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?
$\mathrm{NO_2^-}$
$\mathrm{OH^-}$
$\mathrm{H^+}$
$\mathrm{HNO_2}$
$\mathrm{H_2O}$
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⁺.