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AP Chemistry Flashcards: Weak Acid And Base Equilibria

Study Weak Acid And Base Equilibria 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 Weak Acid And Base Equilibria, 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: Weak Acid And Base Equilibria

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QUESTION

What is the conjugate base of acetic acid, CH₃COOH?

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ANSWER

CH₃COO⁻. Loses a proton to form the conjugate base.

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All flashcards

Flashcard 1: What is the conjugate base of acetic acid, CH₃COOH?

Answer: CH₃COO⁻. Loses a proton to form the conjugate base.

Flashcard 2: How do you find [OH−][OH^-][OH−] from pOHpOHpOH?

Answer: [OH−]=10−pOH[OH^-] = 10^{-pOH}[OH−]=10−pOH. Inverse logarithm converts pOH back to concentration.

Flashcard 3: How is pKbpK_bpKb​ calculated from KbK_bKb​?

Answer: pKb=−log(Kb)pK_b = -\text{log}(K_b)pKb​=−log(Kb​). Negative logarithm converts base constant to pK scale.

Flashcard 4: Which is stronger: an acid with Ka=1.0×10−4K_a = 1.0 \times 10^{-4}Ka​=1.0×10−4 or Ka=1.0×10−6K_a = 1.0 \times 10^{-6}Ka​=1.0×10−6?

Answer: The acid with Ka=1.0×10−4K_a = 1.0 \times 10^{-4}Ka​=1.0×10−4 is stronger. Larger KaK_aKa​ value indicates greater acid dissociation.

Flashcard 5: Identify the effect of dilution on the pHpHpH of a weak acid solution.

Answer: Dilution increases pHpHpH slightly. Equilibrium shifts right, increasing ionization and pH.

Flashcard 6: What is the pHpHpH of a 0.050.050.05 M hydrochloric acid solution?

Answer: pH=1.3pH = 1.3pH=1.3. Strong acid completely dissociates: pH=−log⁡(0.05)=1.3pH = -\log(0.05) = 1.3pH=−log(0.05)=1.3.

Flashcard 7: Determine the pHpHpH of a solution if [H+]=1×10−3[H^+] = 1 \times 10^{-3}[H+]=1×10−3 M.

Answer: pH=3pH = 3pH=3. Use pH=−log⁡(1×10−3)=3pH = -\log(1 \times 10^{-3}) = 3pH=−log(1×10−3)=3.

Flashcard 8: What is the relationship between pHpHpH and pOHpOHpOH at 25o25^\text{o}25oC?

Answer: pH+pOH=14pH + pOH = 14pH+pOH=14. Sum of pH and pOH equals 14 at standard temperature.

Flashcard 9: Find KaK_aKa​ given pKa=4.76pK_a = 4.76pKa​=4.76.

Answer: Ka=1.74×10−5K_a = 1.74 \times 10^{-5}Ka​=1.74×10−5. Use Ka=10−pKa=10−4.76=1.74×10−5K_a = 10^{-pK_a} = 10^{-4.76} = 1.74 \times 10^{-5}Ka​=10−pKa​=10−4.76=1.74×10−5.

Flashcard 10: What is the general expression for the equilibrium constant KaK_aKa​ of a weak acid?

Answer: Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}Ka​=[HA][H+][A−]​. Shows equilibrium between weak acid and its dissociation products.

Flashcard 11: What is the expression for the percent ionization of a weak acid?

Answer: Percent ionization = \frac{[\text{H}^+]}{[\text{HA}]_0} \times 100\text{%}. Ratio of ionized acid to initial concentration times 100%.

Flashcard 12: What is the Bronsted-Lowry definition of a base?

Answer: A base is a proton acceptor. Accepts protons from acids in chemical reactions.

Flashcard 13: How is [H+][H^+][H+] calculated from pHpHpH?

Answer: [H+]=10−pH[H^+] = 10^{-pH}[H+]=10−pH. Inverse logarithm converts pH back to concentration.

Flashcard 14: What is the effect of temperature on the value of KwK_wKw​?

Answer: KwK_wKw​ increases with temperature. Higher temperature increases water dissociation.

Flashcard 15: What is the pHpHpH of a neutral solution at 25o25^\text{o}25oC?

Answer: pH=7pH = 7pH=7. Equal concentrations of H+H^+H+ and OH−OH^-OH− at equilibrium.

Flashcard 16: What is the relationship between pKapK_apKa​ and the strength of an acid?

Answer: Lower pKapK_apKa​ means stronger acid. Lower pKapK_apKa​ corresponds to larger KaK_aKa​ and stronger acid.

Flashcard 17: Calculate the pOHpOHpOH of a solution with [OH−]=2×10−3[OH^-] = 2 \times 10^{-3}[OH−]=2×10−3 M.

Answer: pOH=2.7pOH = 2.7pOH=2.7. Use pOH=−log⁡(2×10−3)=2.7pOH = -\log(2 \times 10^{-3}) = 2.7pOH=−log(2×10−3)=2.7.

Flashcard 18: What is the pHpHpH of a 0.010.010.01 M sulfuric acid solution assuming complete dissociation?

Answer: pH=2pH = 2pH=2. Diprotic acid: [H+]=0.02[H^+] = 0.02[H+]=0.02 M, so pH=−log⁡(0.02)=2pH = -\log(0.02) = 2pH=−log(0.02)=2.

Flashcard 19: Calculate the pHpHpH for [OH−]=4×10−5[OH^-] = 4 \times 10^{-5}[OH−]=4×10−5 M.

Answer: pH=9.6pH = 9.6pH=9.6. Find pOH=4.4pOH = 4.4pOH=4.4, then pH=14−4.4=9.6pH = 14 - 4.4 = 9.6pH=14−4.4=9.6.

Flashcard 20: What is the expression for the equilibrium constant KbK_bKb​ of a weak base?

Answer: Kb=[BH+][OH−][B]K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}Kb​=[B][BH+][OH−]​. Shows equilibrium between weak base and its protonation products.

Flashcard 21: State the formula for calculating pHpHpH from [H+][H^+][H+] concentration.

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

Flashcard 22: How does pHpHpH change when a weak acid is diluted?

Answer: pHpHpH increases slightly. Dilution reduces concentration but increases degree of ionization.

Flashcard 23: What does a pHpHpH greater than 777 indicate?

Answer: The solution is basic. pH above 7 means hydroxide concentration exceeds hydronium.

Flashcard 24: What does a smaller KbK_bKb​ value indicate about the strength of a base?

Answer: The base is weaker. Lower KbK_bKb​ means less dissociation and weaker base.

Flashcard 25: Determine the pHpHpH of a 0.10.10.1 M NaOH solution.

Answer: pH=13pH = 13pH=13. Strong base: [OH−]=0.1[OH^-] = 0.1[OH−]=0.1 M, so pOH=1pOH = 1pOH=1, pH=13pH = 13pH=13.

Flashcard 26: What is the formula for the autoionization of water?

Answer: 2H2O⇌H3O++OH−2 \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-2H2​O⇌H3​O++OH−. Water molecules exchange protons to form ions.

Flashcard 27: What is the value of KwK_wKw​ at 25o25^\text{o}25oC?

Answer: Kw=1.0×10−14K_w = 1.0 \times 10^{-14}Kw​=1.0×10−14. Ion product of water at standard temperature.

Flashcard 28: What does a larger KaK_aKa​ value indicate about the strength of an acid?

Answer: The acid is stronger. Higher KaK_aKa​ means greater dissociation and stronger acid.

Flashcard 29: Determine the pKbpK_bpKb​ given Kb=4.5×10−4K_b = 4.5 \times 10^{-4}Kb​=4.5×10−4.

Answer: pKb=3.35pK_b = 3.35pKb​=3.35. Use pKb=−log⁡(4.5×10−4)=3.35pK_b = -\log(4.5 \times 10^{-4}) = 3.35pKb​=−log(4.5×10−4)=3.35.

Flashcard 30: Find the pKapK_apKa​ given Ka=1.8×10−5K_a = 1.8 \times 10^{-5}Ka​=1.8×10−5.

Answer: pKa=4.74pK_a = 4.74pKa​=4.74. Use pKa=−log⁡(Ka)=−log⁡(1.8×10−5)=4.74pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 4.74pKa​=−log(Ka​)=−log(1.8×10−5)=4.74.