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AP Chemistry Flashcards: Henderson Hasselbalch Equation

Study Henderson Hasselbalch Equation 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 Henderson Hasselbalch Equation, 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: Henderson Hasselbalch Equation

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QUESTION

How does dilution affect the pH of a buffer according to the Henderson-Hasselbalch equation?

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ANSWER

pH remains relatively constant. Ratio stays constant when both components dilute equally.

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Flashcard 1: How does dilution affect the pH of a buffer according to the Henderson-Hasselbalch equation?

Answer: pH remains relatively constant. Ratio stays constant when both components dilute equally.

Flashcard 2: What happens to pH if [A−]=[HA][A^-] = [HA][A−]=[HA] in the Henderson-Hasselbalch equation?

Answer: pH equals pKapK_apKa​. When the ratio equals 1, the log term becomes zero.

Flashcard 3: What assumption is made about [A−][A^-][A−] and [HA][HA][HA] in the Henderson-Hasselbalch equation?

Answer: They are equilibrium concentrations. Assumes no significant change from initial values.

Flashcard 4: What does the term 'buffer capacity' refer to?

Answer: The amount of acid or base the buffer can neutralize. Measures resistance to pH change upon acid/base addition.

Flashcard 5: What is the primary use of the Henderson-Hasselbalch equation?

Answer: To calculate the pH of buffer solutions. Combines weak acid-base pairs to resist pH changes.

Flashcard 6: Identify the meaning of [A−][A^-][A−] in the Henderson-Hasselbalch equation.

Answer: The concentration of the conjugate base. The deprotonated form of the weak acid.

Flashcard 7: What is the effect of temperature on pKapK_apKa​ in the Henderson-Hasselbalch equation?

Answer: pKapK_apKa​ can change with temperature. Temperature affects equilibrium constant values.

Flashcard 8: If [HA]=0.1[HA] = 0.1[HA]=0.1 M and [A−]=0.01[A^-] = 0.01[A−]=0.01 M, how does pH compare to pKapK_apKa​?

Answer: pH < pKapK_apKa​. More acid than base means pH below pKapK_apKa​.

Flashcard 9: How does the Henderson-Hasselbalch equation relate to buffers?

Answer: It calculates the pH of a buffer given pKapK_apKa​, [A−][A^-][A−], and [HA][HA][HA]. Uses component concentrations to predict buffer pH.

Flashcard 10: For a buffer with pKa=6.3pK_a = 6.3pKa​=6.3, what is pH when [A−]=0.5[A^-] = 0.5[A−]=0.5 M, [HA]=0.25[HA] = 0.25[HA]=0.25 M?

Answer: pH = 6.6. log(2)=0.3\text{log}(2) = 0.3log(2)=0.3, so pH = 6.3 + 0.3.

Flashcard 11: Identify the meaning of [HA][HA][HA] in the Henderson-Hasselbalch equation.

Answer: The concentration of the weak acid. The protonated form that can donate hydrogen ions.

Flashcard 12: In the Henderson-Hasselbalch equation, what does a higher pKapK_apKa​ indicate?

Answer: A weaker acid. Larger pKapK_apKa​ corresponds to smaller KaK_aKa​ value.

Flashcard 13: Calculate pH if pKa=9.3pK_a = 9.3pKa​=9.3, [A−]=0.05[A^-] = 0.05[A−]=0.05 M, [HA]=0.1[HA] = 0.1[HA]=0.1 M.

Answer: pH = 9.0. log(0.5)=−0.3\text{log}(0.5) = -0.3log(0.5)=−0.3, so pH = 9.3 - 0.3.

Flashcard 14: What is the logarithmic form of the Henderson-Hasselbalch equation?

Answer: pH=pKa+log([A−][HA])pH = pK_a + \text{log}(\frac{[A^-]}{[HA]})pH=pKa​+log([HA][A−]​). Standard form using base-10 logarithm.

Flashcard 15: What is the relation between pH and pKapK_apKa​ when pH > pKapK_apKa​?

Answer: The solution has more conjugate base than acid. Higher pH indicates more base than acid present.

Flashcard 16: What does the Henderson-Hasselbalch equation assume about ionic strength?

Answer: It is constant. Simplifies calculations by ignoring activity coefficients.

Flashcard 17: Determine the pH if pKa=7.8pK_a = 7.8pKa​=7.8, [A−]=0.3[A^-] = 0.3[A−]=0.3 M, [HA]=0.1[HA] = 0.1[HA]=0.1 M.

Answer: pH = 8.3. log(3)≈0.5\text{log}(3) ≈ 0.5log(3)≈0.5, so pH = 7.8 + 0.5.

Flashcard 18: Which condition is assumed in the Henderson-Hasselbalch equation?

Answer: The solution is a buffer solution. Contains both weak acid and conjugate base in equilibrium.

Flashcard 19: What is the impact on pH if [A−][A^-][A−] decreases in the buffer solution?

Answer: pH decreases. Less conjugate base shifts equilibrium toward lower pH.

Flashcard 20: What is the effect on pH if [A−][A^-][A−] increases relative to [HA][HA][HA]?

Answer: pH increases. Larger numerator in ratio increases log term value.

Flashcard 21: Find the pH if pKa=4.75pK_a = 4.75pKa​=4.75, [A−]=0.1[A^-] = 0.1[A−]=0.1 M, [HA]=0.1[HA] = 0.1[HA]=0.1 M.

Answer: pH = 4.75. Equal concentrations make the log term zero.

Flashcard 22: For a buffer with pKa=4.5pK_a = 4.5pKa​=4.5, what is pH when [A−]=0.2[A^-] = 0.2[A−]=0.2 M and [HA]=0.1[HA] = 0.1[HA]=0.1 M?

Answer: pH = 4.8. log(2)=0.3\text{log}(2) = 0.3log(2)=0.3, so pH = 4.5 + 0.3.

Flashcard 23: Calculate the pH if pKa=5.4pK_a = 5.4pKa​=5.4, [A−]=0.2[A^-] = 0.2[A−]=0.2 M, [HA]=0.2[HA] = 0.2[HA]=0.2 M.

Answer: pH = 5.4. Equal concentrations make the log term zero.

Flashcard 24: State the formula for the Henderson-Hasselbalch equation.

Answer: pH=pKa+log([A−][HA])pH = pK_a + \text{log}(\frac{[A^-]}{[HA]})pH=pKa​+log([HA][A−]​). Relates buffer pH to acid strength and component ratio.

Flashcard 25: What does the ratio [A−]/[HA][A^-]/[HA][A−]/[HA] signify in the Henderson-Hasselbalch equation?

Answer: Relative amounts of conjugate base and weak acid. Determines whether solution is acidic or basic.

Flashcard 26: State the logarithmic property used in the Henderson-Hasselbalch equation.

Answer: log(a/b)=log(a)−log(b)\text{log}(a/b) = \text{log}(a) - \text{log}(b)log(a/b)=log(a)−log(b). Allows separation of the concentration ratio term.

Flashcard 27: What does pKapK_apKa​ represent in the Henderson-Hasselbalch equation?

Answer: The negative logarithm of the acid dissociation constant. Higher pKapK_apKa​ means weaker acid dissociation.

Flashcard 28: Find the pH if pKa=8.0pK_a = 8.0pKa​=8.0, [A−]=0.1[A^-] = 0.1[A−]=0.1 M, [HA]=0.05[HA] = 0.05[HA]=0.05 M.

Answer: pH = 8.3. log(2)=0.3\text{log}(2) = 0.3log(2)=0.3, so pH = 8.0 + 0.3.

Flashcard 29: What is the relation between pH and pKapK_apKa​ when pH < pKapK_apKa​?

Answer: The solution has more weak acid than conjugate base. Lower pH indicates more acid than base present.

Flashcard 30: How is the Henderson-Hasselbalch equation used in biological systems?

Answer: To maintain pH in physiological systems. Blood pH regulation uses bicarbonate buffer system.