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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.

AP Chemistry

Ph And Pk

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QUESTION
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Calculate KbK_b if Ka=1.0×10−5K_a = 1.0 \times 10^{-5} for a conjugate acid-base pair.

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ANSWER

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

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Flashcard 1: Calculate KbK_b if Ka=1.0×10−5K_a = 1.0 \times 10^{-5} for a conjugate acid-base pair.

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

Flashcard 2: Calculate the pH of a 0.05 M H2_2SO4_4 solution.

Answer: pH ≈ 1. H2_2SO4_4 is diprotic; [H+^+] = 0.10 M.

Flashcard 3: Identify the pH range of a basic solution.

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

Flashcard 4: What is the pKb_b of a strong base?

Answer: Low pKb_b. Strong bases have very negative pKb_b values.

Flashcard 5: Identify the pH range of an acidic solution.

Answer: pH < 7. Higher H+^+ concentration than neutral solution.

Flashcard 6: What is the formula for the base dissociation constant (KbK_b)?

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

Flashcard 7: Calculate the pH of a 0.1 M HCl solution.

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

Flashcard 8: Find the pOH of a solution with pH = 2.

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

Flashcard 9: What does a high pKa_a indicate about the strength of an acid?

Answer: Weak acid. High pKa_a means low ionization tendency.

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

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

Flashcard 11: Calculate the [H+^+] for a solution with pH = 6.

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

Flashcard 12: What is the relationship between KaK_a and pKapK_a?

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

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 pairs.

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

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

Flashcard 15: What is the pH of a 0.1 M acetic acid solution with Ka=1.8×10−5K_a = 1.8 \times 10^{-5}?

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

Flashcard 16: State the relationship between pH and pOH in water at 25°C.

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

Flashcard 17: Identify the term for the negative logarithm of the base dissociation constant.

Answer: pKb_b. Standard notation for base dissociation constant.

Flashcard 18: What is the pH of pure water at 25°C?

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

Flashcard 19: Define the term 'acid dissociation constant' (KaK_a).

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

Flashcard 20: What is the formula for calculating pH?

Answer: pH=−log10[H+]\text{pH} = -\text{log}_{10}[\text{H}^+]. Negative log base 10 of hydrogen ion concentration.

Flashcard 21: Which has a higher pH: 0.1 M HCl or 0.1 M CH3_3COOH?

Answer: 0.1 M CH3_3COOH. Acetic acid is weak, HCl is strong acid.

Flashcard 22: Calculate the pH of a 0.001 M HNO3_3 solution.

Answer: pH = 3. HNO3_3 completely ionizes; [H+^+] = 0.001 M.

Flashcard 23: Determine the pH of a 0.01 M NaOH solution.

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

Flashcard 24: Calculate the pKa_a for an acid with Ka=3.2×10−4K_a = 3.2 \times 10^{-4}.

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

Flashcard 25: Determine the pH of a 0.01 M NaOH solution.

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

Flashcard 26: What is the pH of a 1.0 M solution of a strong base?

Answer: pH = 14. Strong base gives [OH−^-] = 1.0 M, pOH = 0.

Flashcard 27: Find the pH of a 0.1 M solution of a strong acid.

Answer: pH = 1. Strong acids completely ionize in solution.

Flashcard 28: If KaK_a is small, is the acid strong or weak?

Answer: Weak acid. Small KaK_a means limited ionization.

Flashcard 29: If KaK_a is large, is the acid strong or weak?

Answer: Strong acid. Large KaK_a means extensive ionization.

Flashcard 30: What is the formula for calculating pOH?

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

Flashcard 31: What does a high pKa_a indicate about the strength of an acid?

Answer: Weak acid. High pKa_a means low ionization tendency.

Flashcard 32: Calculate the pKb_b for a base with Kb=4.5×10−8K_b = 4.5 \times 10^{-8}.

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

Flashcard 33: What is the pH of a 0.01 M NH3_3 solution (Kb=1.8×10−5K_b = 1.8 \times 10^{-5})?

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

Flashcard 34: State the relationship between KbK_b and pKbpK_b.

Answer: pKb=−log⁡10(Kb)pK_b = -\log_{10}(K_b). Takes the negative logarithm of the base constant.

Flashcard 35: What is the pH of a neutral solution at 25°C?

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

Flashcard 36: What is the formula for the ion product of water (KwK_w) at 25°C?

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

Flashcard 37: Calculate the pH of a solution with [H+^+] = 1.0×10−31.0 \times 10^{-3} M.

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

Flashcard 38: What does a low pKb_b indicate about the strength of a base?

Answer: Strong base. Low pKb_b means high ionization tendency.

Flashcard 39: Determine the pOH of a solution with [OH−^-] = 1.0×10−41.0 \times 10^{-4} M.

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

Flashcard 40: If KaK_a is large, is the acid strong or weak?

Answer: Strong acid. Large KaK_a means extensive ionization.

Flashcard 41: Find the pH of a solution with pOH = 5.

Answer: pH = 9. Using the relationship pH + pOH = 14.

Flashcard 42: If pKa_a is 4, what is the strength of the acid?

Answer: Weak acid. pKa_a = 4 indicates moderate acid strength.