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This deck focuses on Introduction To Acids And Bases, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Study Introduction To Acids And Bases 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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Identify the pH range for acidic solutions.
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Acidic solutions have a pH less than 7. Higher [H+] than [OH−].
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This deck focuses on Introduction To Acids And Bases, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
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.
Answer: Acidic solutions have a pH less than 7. Higher [H+] than [OH−].
Answer: pKb=−log10Kb. Negative logarithm of base dissociation constant.
Answer: The conjugate base of H2O is OH−.. Loses a proton to form the conjugate base.
Answer: A Lewis acid is an electron pair acceptor. Receives an electron pair from a donor.
Answer: pH=−log10[H+]. Negative logarithm of hydrogen ion concentration.
Answer: Acidic solutions have a pH less than 7. Higher [H+] than [OH−].
Answer: pH+pOH=14 at 25°C. Based on the ion product of water.
Answer: The conjugate acid of OH− is H2O. Gains a proton to form the conjugate acid.
Answer: The common ion is OH−. Present in all aqueous base solutions.
Answer: pH=11. Strong base: pOH=3, so pH=14−3.
Answer: A strong acid completely ionizes in solution. 100% dissociation in aqueous solution.
Answer: pH=2. Strong acid completely ionizes: pH=−log(0.01).
Answer: The conjugate acid of NH3 is NH4+.. Gains a proton to form the conjugate acid.
Answer: An acid increases [H+] in aqueous solution. Releases hydrogen ions when dissolved.
Answer: Ka=[HA][H+][A−]. Equilibrium expression for acid dissociation.
Answer: NaOH is stronger than NH3. NaOH is strong, NH3 is weak.
Answer: A weak acid partially ionizes in solution. Incomplete dissociation in aqueous solution.
Answer: A Lewis base is an electron pair donor. Provides an electron pair to an acceptor.
Answer: NaOH is stronger than NH3. NaOH is strong, NH3 is weak.
Answer: The pH of a neutral solution is 7. Equal concentrations of H+ and OH−.
Answer: The conjugate base of H2SO4 is HSO4−. Loses one proton from diprotic acid.
Answer: A strong base completely dissociates in solution. 100% dissociation in aqueous solution.
Answer: A base increases [OH−] in aqueous solution. Releases hydroxide ions when dissolved.
Answer: Increasing [OH−] decreases pOH. pOH decreases as basicity increases.
Answer: The pH of pure water is 7. Neutral because [H+]=[OH−].
Answer: The pH of pure water is 7. Neutral because [H+]=[OH−].
Answer: The common ion is H+. Present in all aqueous acid solutions.
Answer: The conjugate acid of OH− is H2O. Gains a proton to form the conjugate acid.
Answer: Kw=1.0×10−14 at 25°C. Product of [H+] and [OH−] concentrations.
Answer: Kw increases with temperature. Higher temperature increases ion product.
Answer: A Lewis base is an electron pair donor. Provides an electron pair to an acceptor.
Answer: A Brønsted-Lowry base is a proton acceptor. Receives H+ from another species.
Answer: A Brønsted-Lowry base is a proton acceptor. Receives H+ from another species.
Answer: A weak base partially dissociates in solution. Incomplete dissociation in aqueous solution.
Answer: A strong base completely dissociates in solution. 100% dissociation in aqueous solution.
Answer: A base increases [OH−] in aqueous solution. Releases hydroxide ions when dissolved.
Answer: HCl is stronger than HF. HCl is a strong acid, HF is weak.
Answer: pOH=4. Calculated using pOH=−log[OH−].
Answer: HCl is stronger than HF. HCl is a strong acid, HF is weak.
Answer: Increasing [OH−] decreases pOH. pOH decreases as basicity increases.
Answer: pKa=−log10Ka. Negative logarithm of acid dissociation constant.
Answer: Kw=[H+][OH−]. Autoionization equilibrium expression for water.
Answer: Kw=[H+][OH−]. Autoionization equilibrium expression for water.
Answer: The common ion is H+. Present in all aqueous acid solutions.
Answer: The conjugate base of H2SO4 is HSO4−. Loses one proton from diprotic acid.
Answer: The conjugate acid of NH3 is NH4+.. Gains a proton to form the conjugate acid.
Answer: Basic solutions have a pH greater than 7. Higher [OH−] than [H+].
Answer: A Brønsted-Lowry acid is a proton donor. Gives up H+ to another species.
Answer: pH=2. Strong acid completely ionizes: pH=−log(0.01).
Answer: pKa=−log10Ka. Negative logarithm of acid dissociation constant.
Answer: The conjugate base of H2O is OH−.. Loses a proton to form the conjugate base.
Answer: An acid increases [H+] in aqueous solution. Releases hydrogen ions when dissolved.
Answer: Ka=[HA][H+][A−]. Equilibrium expression for acid dissociation.
Answer: pOH=4. Calculated using pOH=−log[OH−].
Answer: Increasing [H+] decreases pH. pH decreases as acidity increases.
Answer: Kb=[BOH][B+][OH−]. Equilibrium expression for base dissociation.
Answer: A Brønsted-Lowry acid is a proton donor. Gives up H+ to another species.
Answer: Basic solutions have a pH greater than 7. Higher [OH−] than [H+].
Answer: pKb=−log10Kb. Negative logarithm of base dissociation constant.
Answer: Increasing [H+] decreases pH. pH decreases as acidity increases.
Answer: The conjugate base of HCl is Cl−. Loses a proton to form the conjugate base.
Answer: The conjugate base of HCl is Cl−. Loses a proton to form the conjugate base.
Answer: The common ion is OH−. Present in all aqueous base solutions.
Answer: pH=−log10[H+]. Negative logarithm of hydrogen ion concentration.
Answer: A strong acid completely ionizes in solution. 100% dissociation in aqueous solution.
Answer: A Lewis acid is an electron pair acceptor. Receives an electron pair from a donor.
Answer: pH=11. Strong base: pOH=3, so pH=14−3.
Answer: pH=3. Calculated using pH=−log[H+].
Answer: A weak acid partially ionizes in solution. Incomplete dissociation in aqueous solution.
Answer: A weak base partially dissociates in solution. Incomplete dissociation in aqueous solution.
Answer: pH+pOH=14 at 25°C. Based on the ion product of water.
Answer: pH=3. Calculated using pH=−log[H+].
Answer: Kw=1.0×10−14 at 25°C. Product of [H+] and [OH−] concentrations.
Answer: Kw increases with temperature. Higher temperature increases ion product.
Answer: pOH=−log10[OH−]. Negative logarithm of hydroxide ion concentration.
Answer: Kb=[BOH][B+][OH−]. Equilibrium expression for base dissociation.