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AP Chemistry Flashcards: Introduction To Acid Base Reactions

Study Introduction To Acid Base Reactions in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Introduction To Acid Base Reactions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Chemistry Flashcards: Introduction To Acid Base Reactions

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QUESTION

Calculate the pH given [H+]=1×10−8[H^+] = 1 \times 10^{-8}[H+]=1×10−8 M.

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ANSWER

pH=8pH = 8pH=8. Direct application: pH=−log⁡(1×10−8)=8pH = -\log(1 \times 10^{-8}) = 8pH=−log(1×10−8)=8.

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Flashcard 1: Calculate the pH given [H+]=1×10−8[H^+] = 1 \times 10^{-8}[H+]=1×10−8 M.

Answer: pH=8pH = 8pH=8. Direct application: pH=−log⁡(1×10−8)=8pH = -\log(1 \times 10^{-8}) = 8pH=−log(1×10−8)=8.

Flashcard 2: What is the main characteristic of a strong acid?

Answer: A strong acid completely ionizes in solution. Nearly 100% ionization in aqueous solution.

Flashcard 3: What is the KaK_aKa​ value indicative of a strong acid?

Answer: A large KaK_aKa​ value indicates a strong acid. Large KaK_aKa​ means extensive ionization and stronger acidity.

Flashcard 4: What is the KaK_aKa​ expression for acetic acid, CH3COOHCH_3COOHCH3​COOH?

Answer: Ka=[CH3COO−][H+][CH3COOH]K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}Ka​=[CH3​COOH][CH3​COO−][H+]​. Equilibrium expression for weak acid ionization.

Flashcard 5: What is the relationship between KaK_aKa​ and KbK_bKb​ for a conjugate acid-base pair?

Answer: Ka×Kb=KwK_a \times K_b = K_wKa​×Kb​=Kw​. Fundamental relationship for conjugate acid-base pairs.

Flashcard 6: What is the pH of a 0.01 M HCl solution?

Answer: pH=2pH = 2pH=2. Strong acid: pH=−log⁡(0.01)=2pH = -\log(0.01) = 2pH=−log(0.01)=2.

Flashcard 7: What is the relationship between pH and pOH at 25°C?

Answer: pH+pOH=14pH + pOH = 14pH+pOH=14. Derived from the water equilibrium constant at 25°C.

Flashcard 8: What is the definition of an acid according to the Arrhenius theory?

Answer: An acid is a substance that increases the concentration of H+H^+H+ ions in aqueous solution. Arrhenius theory focuses on H+H^+H+ ion production in water.

Flashcard 9: Name the conjugate base of hydrochloric acid (HCl).

Answer: The conjugate base of HCl is Cl−Cl^-Cl−.. Formed by removing one proton from the acid.

Flashcard 10: What is the main characteristic of a strong base?

Answer: A strong base completely dissociates in solution. Nearly 100% dissociation in aqueous solution.

Flashcard 11: Which ion is responsible for the acidity of a solution?

Answer: The H+H^+H+ ion is responsible for the acidity of a solution. Higher [H+][H^+][H+] means lower pH and greater acidity.

Flashcard 12: Which ion is responsible for the basicity of a solution?

Answer: The OH−OH^-OH− ion is responsible for the basicity of a solution. Higher [OH−][OH^-][OH−] means higher pH and greater basicity.

Flashcard 13: Which is a stronger acid: HCl or acetic acid?

Answer: HCl is a stronger acid than acetic acid. HCl completely ionizes; acetic acid only partially ionizes.

Flashcard 14: What is the conjugate base of H2SO4H_2SO_4H2​SO4​?

Answer: The conjugate base of H2SO4H_2SO_4H2​SO4​ is HSO4−HSO_4^-HSO4−​. Sulfuric acid loses one proton to form bisulfate ion.

Flashcard 15: What is the KbK_bKb​ value indicative of a strong base?

Answer: A large KbK_bKb​ value indicates a strong base. Large KbK_bKb​ means extensive ionization and stronger basicity.

Flashcard 16: Identify the pHpHpH of a 0.1 M KOHKOHKOH solution.

Answer: pH=13pH = 13pH=13. Strong base: pOH=1pOH = 1pOH=1, so pH=14−1=13pH = 14 - 1 = 13pH=14−1=13.

Flashcard 17: Which ion forms when an acid donates a proton?

Answer: The conjugate base forms when an acid donates a proton. Result of proton donation in acid-base reactions.

Flashcard 18: Find the KaK_aKa​ value given pKa=3.5pK_a = 3.5pKa​=3.5.

Answer: Ka=3.16×10−4K_a = 3.16 \times 10^{-4}Ka​=3.16×10−4. Use formula: Ka=10−pKa=10−3.5K_a = 10^{-pK_a} = 10^{-3.5}Ka​=10−pKa​=10−3.5.

Flashcard 19: What is the term for the equilibrium constant for water?

Answer: The term is KwK_wKw​, the ion-product constant for water. Represents autoionization equilibrium of pure water.

Flashcard 20: Find the KbK_bKb​ value given pKb=4.5pK_b = 4.5pKb​=4.5.

Answer: Kb=3.16×10−5K_b = 3.16 \times 10^{-5}Kb​=3.16×10−5. Use formula: Kb=10−pKb=10−4.5K_b = 10^{-pK_b} = 10^{-4.5}Kb​=10−pKb​=10−4.5.

Flashcard 21: State the formula for calculating pOH.

Answer: pOH=−log[OH−]pOH = -\text{log}[OH^-]pOH=−log[OH−]. Negative logarithm converts hydroxide concentration to pOH scale.

Flashcard 22: Find the pOH of a solution where pH=3pH = 3pH=3.

Answer: pOH=11pOH = 11pOH=11. Use relationship: pH+pOH=14pH + pOH = 14pH+pOH=14.

Flashcard 23: Calculate the pOH given [OH−]=1×10−6[OH^-] = 1 \times 10^{-6}[OH−]=1×10−6 M.

Answer: pOH=6pOH = 6pOH=6. Direct application: pOH=−log⁡(1×10−6)=6pOH = -\log(1 \times 10^{-6}) = 6pOH=−log(1×10−6)=6.

Flashcard 24: Identify the pHpHpH of a 0.1 M HNO3HNO_3HNO3​ solution.

Answer: pH=1pH = 1pH=1. Strong acid: pH=−log⁡(0.1)=1pH = -\log(0.1) = 1pH=−log(0.1)=1.

Flashcard 25: What is the pH of a 0.01 M NaOH solution?

Answer: pH=12pH = 12pH=12. Strong base: pOH=2pOH = 2pOH=2, so pH=14−2=12pH = 14 - 2 = 12pH=14−2=12.

Flashcard 26: Identify the Bronsted-Lowry definition of a base.

Answer: A Bronsted-Lowry base is a proton (H+H^+H+) acceptor. Bronsted-Lowry theory emphasizes proton transfer reactions.

Flashcard 27: What are the products of a neutralization reaction?

Answer: A salt and water are the products of a neutralization reaction. General form: acid + base → salt + water.

Flashcard 28: What is the pH of a neutral solution at 25°C?

Answer: The pH of a neutral solution at 25°C is 7. At neutral pH, [H+]=[OH−][H^+] = [OH^-][H+]=[OH−] at standard temperature.

Flashcard 29: What is the KaK_aKa​ expression for acetic acid, CH3COOHCH_3COOHCH3​COOH?

Answer: Ka=[CH3COO−][H+][CH3COOH]K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}Ka​=[CH3​COOH][CH3​COO−][H+]​. Equilibrium expression for weak acid ionization.

Flashcard 30: What is the conjugate acid of OH−OH^-OH−?

Answer: The conjugate acid of OH−OH^-OH− is H2OH_2OH2​O. Hydroxide ion gains one proton to form water.