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.

AP Chemistry

Introduction To Acid Base Reactions

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State the formula for calculating pOH.

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ANSWER

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

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Flashcard 1: State the formula for calculating pOH.

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

Flashcard 2: What is the conjugate base of H2SO4H_2SO_4?

Answer: The conjugate base of H2SO4H_2SO_4 is HSO4HSO_4^-. Sulfuric acid loses one proton to form bisulfate ion.

Flashcard 3: What is the KbK_b value indicative of a strong base?

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

Flashcard 4: What is the KbK_b expression for ammonia, NH3NH_3?

Answer: Kb=[NH4+][OH][NH3]K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}. Equilibrium expression for weak base ionization.

Flashcard 5: Identify the pHpH of a 0.1 M HNO3HNO_3 solution.

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

Flashcard 6: Find the pOH of a solution where pH=3pH = 3.

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

Flashcard 7: 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 8: Which equation represents the ion-product constant for water (KwK_w)?

Answer: Kw=[H+][OH]=1.0×1014K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25°C. This equilibrium constant applies to pure water at 25°C.

Flashcard 9: 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 10: What is the pH of a 0.01 M HCl solution?

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

Flashcard 11: What is the KaK_a expression for acetic acid, CH3COOHCH_3COOH?

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

Flashcard 12: What is the KaK_a value indicative of a strong acid?

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

Flashcard 13: What is the relationship between KaK_a and KbK_b for a conjugate acid-base pair?

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

Flashcard 14: What is the main characteristic of a strong base?

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

Flashcard 15: What is the term for the equilibrium constant for water?

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

Flashcard 16: 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^+ ions in aqueous solution. Arrhenius theory focuses on H+H^+ ion production in water.

Flashcard 17: Identify the Lewis definition of a base.

Answer: A Lewis base is an electron pair donor. Lewis theory focuses on electron pair interactions.

Flashcard 18: Identify the Lewis definition of a base.

Answer: A Lewis base is an electron pair donor. Lewis theory focuses on electron pair interactions.

Flashcard 19: 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^-] at standard temperature.

Flashcard 20: State the formula for calculating pH.

Answer: pH=log[H+]pH = -\text{log}[H^+]. Negative logarithm converts concentration to pH scale.

Flashcard 21: Find the pOH of a solution where pH=3pH = 3.

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

Flashcard 22: What is the KaK_a expression for acetic acid, CH3COOHCH_3COOH?

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

Flashcard 23: Calculate the pH given [H+]=1×108[H^+] = 1 \times 10^{-8} M.

Answer: pH=8pH = 8. Direct application: pH=log(1×108)=8pH = -\log(1 \times 10^{-8}) = 8.

Flashcard 24: Find the KaK_a value given pKa=3.5pK_a = 3.5.

Answer: Ka=3.16×104K_a = 3.16 \times 10^{-4}. Use formula: Ka=10pKa=103.5K_a = 10^{-pK_a} = 10^{-3.5}.

Flashcard 25: What is the KaK_a value indicative of a strong acid?

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

Flashcard 26: What is the relationship between pH and pOH at 25°C?

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

Flashcard 27: What is the KbK_b value indicative of a strong base?

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

Flashcard 28: Find the KaK_a value given pKa=3.5pK_a = 3.5.

Answer: Ka=3.16×104K_a = 3.16 \times 10^{-4}. Use formula: Ka=10pKa=103.5K_a = 10^{-pK_a} = 10^{-3.5}.

Flashcard 29: 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 30: What is the definition of a base according to the Arrhenius theory?

Answer: A base is a substance that increases the concentration of OHOH^- ions in aqueous solution. Arrhenius theory focuses on OHOH^- ion production in water.

Flashcard 31: Calculate the pH given [H+]=1×108[H^+] = 1 \times 10^{-8} M.

Answer: pH=8pH = 8. Direct application: pH=log(1×108)=8pH = -\log(1 \times 10^{-8}) = 8.

Flashcard 32: State the formula for calculating pH.

Answer: pH=log[H+]pH = -\text{log}[H^+]. Negative logarithm converts concentration to pH scale.

Flashcard 33: Calculate the pOH given [OH]=1×106[OH^-] = 1 \times 10^{-6} M.

Answer: pOH=6pOH = 6. Direct application: pOH=log(1×106)=6pOH = -\log(1 \times 10^{-6}) = 6.

Flashcard 34: Which is a stronger base: NaOH or ammonia?

Answer: NaOH is a stronger base than ammonia. NaOH completely dissociates; ammonia only partially ionizes.

Flashcard 35: Name the conjugate base of hydrochloric acid (HCl).

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

Flashcard 36: State the formula for calculating pH.

Answer: pH=log[H+]pH = -\text{log}[H^+]. Negative logarithm converts concentration to pH scale.

Flashcard 37: What is the KbK_b value indicative of a strong base?

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

Flashcard 38: What is the term for the equilibrium constant for water?

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

Flashcard 39: Which ion is responsible for the acidity of a solution?

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

Flashcard 40: What is the pH of a 0.01 M NaOH solution?

Answer: pH=12pH = 12. Strong base: pOH=2pOH = 2, so pH=142=12pH = 14 - 2 = 12.

Flashcard 41: Which equation represents the ion-product constant for water (KwK_w)?

Answer: Kw=[H+][OH]=1.0×1014K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25°C. This equilibrium constant applies to pure water at 25°C.

Flashcard 42: 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^+ ions in aqueous solution. Arrhenius theory focuses on H+H^+ ion production in water.

Flashcard 43: Identify the Bronsted-Lowry definition of an acid.

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

Flashcard 44: What is the main characteristic of a strong base?

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

Flashcard 45: Find the KbK_b value given pKb=4.5pK_b = 4.5.

Answer: Kb=3.16×105K_b = 3.16 \times 10^{-5}. Use formula: Kb=10pKb=104.5K_b = 10^{-pK_b} = 10^{-4.5}.

Flashcard 46: Which ion is responsible for the acidity of a solution?

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

Flashcard 47: 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^+ ions in aqueous solution. Arrhenius theory focuses on H+H^+ ion production in water.

Flashcard 48: What is the term for the equilibrium constant for water?

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

Flashcard 49: What is the main characteristic of a strong acid?

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

Flashcard 50: What is the main characteristic of a strong acid?

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

Flashcard 51: What is the KbK_b expression for ammonia, NH3NH_3?

Answer: Kb=[NH4+][OH][NH3]K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}. Equilibrium expression for weak base ionization.

Flashcard 52: What is the KaK_a expression for acetic acid, CH3COOHCH_3COOH?

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

Flashcard 53: What is the main characteristic of a strong acid?

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

Flashcard 54: Identify the Lewis definition of an acid.

Answer: A Lewis acid is an electron pair acceptor. Lewis theory focuses on electron pair interactions.

Flashcard 55: Identify the pHpH of a 0.1 M KOHKOH solution.

Answer: pH=13pH = 13. Strong base: pOH=1pOH = 1, so pH=141=13pH = 14 - 1 = 13.

Flashcard 56: Name the conjugate base of hydrochloric acid (HCl).

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

Flashcard 57: Identify the pHpH of a 0.1 M HNO3HNO_3 solution.

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

Flashcard 58: 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^-] at standard temperature.

Flashcard 59: Which is a stronger base: NaOH or ammonia?

Answer: NaOH is a stronger base than ammonia. NaOH completely dissociates; ammonia only partially ionizes.

Flashcard 60: Calculate the pOH given [OH]=1×106[OH^-] = 1 \times 10^{-6} M.

Answer: pOH=6pOH = 6. Direct application: pOH=log(1×106)=6pOH = -\log(1 \times 10^{-6}) = 6.

Flashcard 61: Identify the Lewis definition of an acid.

Answer: A Lewis acid is an electron pair acceptor. Lewis theory focuses on electron pair interactions.

Flashcard 62: What is the main characteristic of a strong base?

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

Flashcard 63: What is the definition of a base according to the Arrhenius theory?

Answer: A base is a substance that increases the concentration of OHOH^- ions in aqueous solution. Arrhenius theory focuses on OHOH^- ion production in water.

Flashcard 64: What is the relationship between pH and pOH at 25°C?

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

Flashcard 65: What is the KaK_a value indicative of a strong acid?

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

Flashcard 66: Identify the Lewis definition of a base.

Answer: A Lewis base is an electron pair donor. Lewis theory focuses on electron pair interactions.

Flashcard 67: What is the KbK_b expression for ammonia, NH3NH_3?

Answer: Kb=[NH4+][OH][NH3]K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}. Equilibrium expression for weak base ionization.

Flashcard 68: 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 69: Identify the Lewis definition of an acid.

Answer: A Lewis acid is an electron pair acceptor. Lewis theory focuses on electron pair interactions.

Flashcard 70: Find the KbK_b value given pKb=4.5pK_b = 4.5.

Answer: Kb=3.16×105K_b = 3.16 \times 10^{-5}. Use formula: Kb=10pKb=104.5K_b = 10^{-pK_b} = 10^{-4.5}.

Flashcard 71: Find the KaK_a value given pKa=3.5pK_a = 3.5.

Answer: Ka=3.16×104K_a = 3.16 \times 10^{-4}. Use formula: Ka=10pKa=103.5K_a = 10^{-pK_a} = 10^{-3.5}.

Flashcard 72: Which ion is responsible for the acidity of a solution?

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

Flashcard 73: What is the conjugate base of H2SO4H_2SO_4?

Answer: The conjugate base of H2SO4H_2SO_4 is HSO4HSO_4^-. Sulfuric acid loses one proton to form bisulfate ion.

Flashcard 74: Identify the Bronsted-Lowry definition of a base.

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

Flashcard 75: Calculate the pH given [H+]=1×108[H^+] = 1 \times 10^{-8} M.

Answer: pH=8pH = 8. Direct application: pH=log(1×108)=8pH = -\log(1 \times 10^{-8}) = 8.

Flashcard 76: Which is a stronger base: NaOH or ammonia?

Answer: NaOH is a stronger base than ammonia. NaOH completely dissociates; ammonia only partially ionizes.

Flashcard 77: Which ion is responsible for the basicity of a solution?

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

Flashcard 78: What is the pH of a 0.01 M NaOH solution?

Answer: pH=12pH = 12. Strong base: pOH=2pOH = 2, so pH=142=12pH = 14 - 2 = 12.

Flashcard 79: What is the pH of a 0.01 M HCl solution?

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

Flashcard 80: 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 81: Identify the pHpH of a 0.1 M HNO3HNO_3 solution.

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

Flashcard 82: Name the conjugate base of hydrochloric acid (HCl).

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

Flashcard 83: What is the pH of a 0.01 M NaOH solution?

Answer: pH=12pH = 12. Strong base: pOH=2pOH = 2, so pH=142=12pH = 14 - 2 = 12.

Flashcard 84: Find the pH of a solution where pOH=4pOH = 4.

Answer: pH=10pH = 10. Use relationship: pH+pOH=14pH + pOH = 14.

Flashcard 85: Which ion is responsible for the basicity of a solution?

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

Flashcard 86: Name the conjugate acid of ammonia (NH₃).

Answer: The conjugate acid of NH₃ is NH4+NH_4^+. Formed by adding one proton to the base.

Flashcard 87: What is the conjugate acid of OHOH^-?

Answer: The conjugate acid of OHOH^- is H2OH_2O. Hydroxide ion gains one proton to form water.

Flashcard 88: 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 89: 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^-] at standard temperature.

Flashcard 90: Identify the pHpH of a 0.1 M KOHKOH solution.

Answer: pH=13pH = 13. Strong base: pOH=1pOH = 1, so pH=141=13pH = 14 - 1 = 13.

Flashcard 91: State the formula for calculating pOH.

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

Flashcard 92: What is the relationship between KaK_a and KbK_b for a conjugate acid-base pair?

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

Flashcard 93: What is the relationship between KaK_a and KbK_b for a conjugate acid-base pair?

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

Flashcard 94: What is the conjugate acid of OHOH^-?

Answer: The conjugate acid of OHOH^- is H2OH_2O. Hydroxide ion gains one proton to form water.

Flashcard 95: What is the pH of a 0.01 M HCl solution?

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

Flashcard 96: Which equation represents the ion-product constant for water (KwK_w)?

Answer: Kw=[H+][OH]=1.0×1014K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25°C. This equilibrium constant applies to pure water at 25°C.

Flashcard 97: Identify the Bronsted-Lowry definition of an acid.

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

Flashcard 98: Identify the Bronsted-Lowry definition of an acid.

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

Flashcard 99: Find the KbK_b value given pKb=4.5pK_b = 4.5.

Answer: Kb=3.16×105K_b = 3.16 \times 10^{-5}. Use formula: Kb=10pKb=104.5K_b = 10^{-pK_b} = 10^{-4.5}.

Flashcard 100: What is the relationship between pH and pOH at 25°C?

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