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AP Chemistry Flashcards: Ph And Pk

Study Ph And Pk 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 Ph And Pk, 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: Ph And Pk

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QUESTION

What does a high pKa_aa​ indicate about the strength of an acid?

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ANSWER

Weak acid. High pKa_aa​ means low ionization tendency.

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Flashcard 1: What does a high pKa_aa​ indicate about the strength of an acid?

Answer: Weak acid. High pKa_aa​ means low ionization tendency.

Flashcard 2: What is the pH of a neutral solution at 25°C?

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

Flashcard 3: If KaK_aKa​ is small, is the acid strong or weak?

Answer: Weak acid. Small KaK_aKa​ means limited ionization.

Flashcard 4: Find the pOH of a solution with pH = 2.

Answer: pOH = 12. Using the relationship pH + pOH = 14.

Flashcard 5: Calculate the pH of a 0.1 M HCl solution.

Answer: pH = 1. HCl completely ionizes; [H+^++] = 0.1 M.

Flashcard 6: What does a high pKa_aa​ indicate about the strength of an acid?

Answer: Weak acid. High pKa_aa​ means low ionization tendency.

Flashcard 7: What is the pKb_bb​ of a strong base?

Answer: Low pKb_bb​. Strong bases have very negative pKb_bb​ values.

Flashcard 8: What is the pH of a 0.1 M acetic acid solution with Ka=1.8×10−5K_a = 1.8 \times 10^{-5}Ka​=1.8×10−5?

Answer: pH ≈ 2.88. Using the weak acid approximation formula.

Flashcard 9: Identify the term for the negative logarithm of the base dissociation constant.

Answer: pKb_bb​. Standard notation for base dissociation constant.

Flashcard 10: Calculate the pKb_bb​ for a base with Kb=4.5×10−8K_b = 4.5 \times 10^{-8}Kb​=4.5×10−8.

Answer: pKb_bb​ ≈ 7.35. Using pKb=−log⁡(4.5×10−8)\text{pK}_b = -\log(4.5 \times 10^{-8})pKb​=−log(4.5×10−8).

Flashcard 11: What is the formula for calculating pOH?

Answer: pOH=−log10[OH−]\text{pOH} = -\text{log}_{10}[\text{OH}^-]pOH=−log10​[OH−]. Negative log base 10 of hydroxide ion concentration.

Flashcard 12: What is the formula for the ion product of water (KwK_wKw​) at 25°C?

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

Flashcard 13: What does a low pKb_bb​ indicate about the strength of a base?

Answer: Strong base. Low pKb_bb​ means high ionization tendency.

Flashcard 14: Calculate the [OH−^-−] for a solution with pOH = 8.

Answer: [OH−^-−] = 1.0×10−81.0 \times 10^{-8}1.0×10−8 M. Using [OH−^-−] = 10−pOH10^{-\text{pOH}}10−pOH relationship.

Flashcard 15: What is the relationship between KaK_aKa​ and pKapK_apKa​?

Answer: pKa=−log10(Ka)\text{pK}_a = -\text{log}_{10}(K_a)pKa​=−log10​(Ka​). Takes the negative logarithm of the acid constant.

Flashcard 16: Define the term 'acid dissociation constant' (KaK_aKa​).

Answer: KaK_aKa​ is the equilibrium constant for the dissociation of an acid. Measures the extent of acid ionization in solution.

Flashcard 17: What is the formula for the base dissociation constant (KbK_bKb​)?

Answer: KbK_bKb​ is the equilibrium constant for the dissociation of a base. Measures the extent of base ionization in solution.

Flashcard 18: Calculate the [H+^++] for a solution with pH = 6.

Answer: [H+^++] = 1.0×10−61.0 \times 10^{-6}1.0×10−6 M. Using [H+^++] = 10−pH10^{-\text{pH}}10−pH relationship.

Flashcard 19: Identify the pH range of a basic solution.

Answer: pH > 7. Higher OH−^-− concentration than neutral solution.

Flashcard 20: Determine the pOH of a solution with [OH−^-−] = 1.0×10−41.0 \times 10^{-4}1.0×10−4 M.

Answer: pOH = 4. Using pOH=−log⁡(1.0×10−4)=4\text{pOH} = -\log(1.0 \times 10^{-4}) = 4pOH=−log(1.0×10−4)=4.

Flashcard 21: What is the pH of a 0.01 M NH3_33​ solution (Kb=1.8×10−5K_b = 1.8 \times 10^{-5}Kb​=1.8×10−5)?

Answer: pH ≈ 11.13. Using weak base approximation with given KbK_bKb​.

Flashcard 22: Calculate the pKa_aa​ for an acid with Ka=3.2×10−4K_a = 3.2 \times 10^{-4}Ka​=3.2×10−4.

Answer: pKa_aa​ ≈ 3.5. Using pKa=−log⁡(3.2×10−4)\text{pK}_a = -\log(3.2 \times 10^{-4})pKa​=−log(3.2×10−4).

Flashcard 23: If KaK_aKa​ is large, is the acid strong or weak?

Answer: Strong acid. Large KaK_aKa​ means extensive ionization.

Flashcard 24: Determine the pH of a 0.01 M NaOH solution.

Answer: pH = 12. NaOH gives [OH−^-−] = 0.01 M, so pOH = 2.

Flashcard 25: Which has a higher pH: 0.1 M HCl or 0.1 M CH3_33​COOH?

Answer: 0.1 M CH3_33​COOH. Acetic acid is weak, HCl is strong acid.

Flashcard 26: Calculate KbK_bKb​ if Ka=1.0×10−5K_a = 1.0 \times 10^{-5}Ka​=1.0×10−5 for a conjugate acid-base pair.

Answer: Kb=1.0×10−9K_b = 1.0 \times 10^{-9}Kb​=1.0×10−9. Using Ka×Kb=1.0×10−14K_a \times K_b = 1.0 \times 10^{-14}Ka​×Kb​=1.0×10−14.

Flashcard 27: What is the pH of pure water at 25°C?

Answer: pH = 7. Autoionization gives equal H+^++ and OH−^-−.

Flashcard 28: Calculate the pH of a solution with [H+^++] = 1.0×10−31.0 \times 10^{-3}1.0×10−3 M.

Answer: pH = 3. Using pH=−log⁡(1.0×10−3)=3\text{pH} = -\log(1.0 \times 10^{-3}) = 3pH=−log(1.0×10−3)=3.

Flashcard 29: State the relationship between pH and pOH in water at 25°C.

Answer: pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14. Based on the ion product of water at 25°C.

Flashcard 30: If pKa_aa​ is 4, what is the strength of the acid?

Answer: Weak acid. pKa_aa​ = 4 indicates moderate acid strength.