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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.

AP Chemistry

Henderson Hasselbalch Equation

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QUESTION
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What does the Henderson-Hasselbalch equation assume about ionic strength?

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ANSWER

It is constant. Simplifies calculations by ignoring activity coefficients.

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Flashcard 1: What does the Henderson-Hasselbalch equation assume about ionic strength?

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

Flashcard 2: For a buffer with pKa=4.5pK_a = 4.5, what is pH when [A−]=0.2[A^-] = 0.2 M and [HA]=0.1[HA] = 0.1 M?

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

Flashcard 3: 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 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 relation between pH and pKapK_a when pH < pKapK_a?

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

Flashcard 6: If pKa=6.1pK_a = 6.1 and [A−]=0.25[A^-] = 0.25 M, [HA]=0.25[HA] = 0.25 M, what is the pH?

Answer: pH = 6.1. Equal concentrations make log term zero.

Flashcard 7: Calculate pH if pKa=9.3pK_a = 9.3, [A−]=0.05[A^-] = 0.05 M, [HA]=0.1[HA] = 0.1 M.

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

Flashcard 8: 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 9: If [A−]/[HA]=2[A^-]/[HA] = 2, find pH given pKa=7.2pK_a = 7.2.

Answer: pH = 7.5. log(2)=0.3\text{log}(2) = 0.3, so pH = 7.2 + 0.3.

Flashcard 10: Identify the meaning of [A−][A^-] in the Henderson-Hasselbalch equation.

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

Flashcard 11: 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 12: What is the effect of temperature on pKapK_a in the Henderson-Hasselbalch equation?

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

Flashcard 13: How is the Henderson-Hasselbalch equation derived?

Answer: From the acid dissociation constant expression. Taking negative log of the acid dissociation expression.

Flashcard 14: What is the relation between pH and pKapK_a when pH > pKapK_a?

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

Flashcard 15: What is the relation between pH and pKapK_a when pH < pKapK_a?

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

Flashcard 16: What is the impact on pH if [A−][A^-] decreases in the buffer solution?

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

Flashcard 17: How does the Henderson-Hasselbalch equation relate to buffers?

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

Flashcard 18: 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). Allows separation of the concentration ratio term.

Flashcard 19: What is the effect of temperature on pKapK_a in the Henderson-Hasselbalch equation?

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

Flashcard 20: Find the pH if pKa=4.75pK_a = 4.75, [A−]=0.1[A^-] = 0.1 M, [HA]=0.1[HA] = 0.1 M.

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

Flashcard 21: How is the Henderson-Hasselbalch equation derived?

Answer: From the acid dissociation constant expression. Taking negative log of the acid dissociation expression.

Flashcard 22: What does the Henderson-Hasselbalch equation assume about ionic strength?

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

Flashcard 23: If [A−]/[HA]=2[A^-]/[HA] = 2, find pH given pKa=7.2pK_a = 7.2.

Answer: pH = 7.5. log(2)=0.3\text{log}(2) = 0.3, so pH = 7.2 + 0.3.

Flashcard 24: Find pH for pKa=3.5pK_a = 3.5, [A−]=0.03[A^-] = 0.03 M, [HA]=0.01[HA] = 0.01 M.

Answer: pH = 4.0. log(3)≈0.5\text{log}(3) ≈ 0.5, so pH = 3.5 + 0.5.

Flashcard 25: What is the significance of the Henderson-Hasselbalch equation in titrations?

Answer: It helps estimate pH at various stages of a titration. Predicts pH changes during weak acid-base reactions.

Flashcard 26: How does the Henderson-Hasselbalch equation help in buffer preparation?

Answer: Determines component ratios for desired pH. Calculate required acid-to-base ratio for target pH.

Flashcard 27: How does the Henderson-Hasselbalch equation help in buffer preparation?

Answer: Determines component ratios for desired pH. Calculate required acid-to-base ratio for target pH.

Flashcard 28: How is the Henderson-Hasselbalch equation used in biological systems?

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

Flashcard 29: If [HA]=0.1[HA] = 0.1 M and [A−]=0.01[A^-] = 0.01 M, how does pH compare to pKapK_a?

Answer: pH < pKapK_a. More acid than base means pH below pKapK_a.

Flashcard 30: What is the effect on pH if [A−][A^-] increases relative to [HA][HA]?

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

Flashcard 31: What is the effect on pH if [A−][A^-] increases relative to [HA][HA]?

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

Flashcard 32: For a buffer with pKa=6.3pK_a = 6.3, what is pH when [A−]=0.5[A^-] = 0.5 M, [HA]=0.25[HA] = 0.25 M?

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

Flashcard 33: Calculate pH if pKa=9.3pK_a = 9.3, [A−]=0.05[A^-] = 0.05 M, [HA]=0.1[HA] = 0.1 M.

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

Flashcard 34: 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 35: For a buffer with pKa=4.5pK_a = 4.5, what is pH when [A−]=0.2[A^-] = 0.2 M and [HA]=0.1[HA] = 0.1 M?

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

Flashcard 36: State the formula for the Henderson-Hasselbalch equation.

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

Flashcard 37: What does pKapK_a represent in the Henderson-Hasselbalch equation?

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

Flashcard 38: 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). Allows separation of the concentration ratio term.

Flashcard 39: For a buffer with pKa=6.3pK_a = 6.3, what is pH when [A−]=0.5[A^-] = 0.5 M, [HA]=0.25[HA] = 0.25 M?

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

Flashcard 40: If [HA]=0.1[HA] = 0.1 M and [A−]=0.01[A^-] = 0.01 M, how does pH compare to pKapK_a?

Answer: pH < pKapK_a. More acid than base means pH below pKapK_a.

Flashcard 41: What happens to pH if [A−]=[HA][A^-] = [HA] in the Henderson-Hasselbalch equation?

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

Flashcard 42: What assumption is made about [A−][A^-] and [HA][HA] in the Henderson-Hasselbalch equation?

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

Flashcard 43: In the Henderson-Hasselbalch equation, what does a higher pKapK_a indicate?

Answer: A weaker acid. Larger pKapK_a corresponds to smaller KaK_a value.

Flashcard 44: If pKa=6.1pK_a = 6.1 and [A−]=0.25[A^-] = 0.25 M, [HA]=0.25[HA] = 0.25 M, what is the pH?

Answer: pH = 6.1. Equal concentrations make log term zero.

Flashcard 45: What does the ratio [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 46: What is the impact on pH if [A−][A^-] decreases in the buffer solution?

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

Flashcard 47: What does the ratio [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 48: What is the logarithmic form of the Henderson-Hasselbalch equation?

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

Flashcard 49: How does the Henderson-Hasselbalch equation relate to buffers?

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

Flashcard 50: 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 51: Identify the meaning of [HA][HA] in the Henderson-Hasselbalch equation.

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

Flashcard 52: Identify the meaning of [A−][A^-] in the Henderson-Hasselbalch equation.

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

Flashcard 53: What is the logarithmic form of the Henderson-Hasselbalch equation?

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

Flashcard 54: Find the pH if pKa=8.0pK_a = 8.0, [A−]=0.1[A^-] = 0.1 M, [HA]=0.05[HA] = 0.05 M.

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

Flashcard 55: In the Henderson-Hasselbalch equation, what does a higher pKapK_a indicate?

Answer: A weaker acid. Larger pKapK_a corresponds to smaller KaK_a value.

Flashcard 56: What assumption is made about [A−][A^-] and [HA][HA] in the Henderson-Hasselbalch equation?

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

Flashcard 57: Identify the meaning of [HA][HA] in the Henderson-Hasselbalch equation.

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

Flashcard 58: 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 59: Find the pH if pKa=8.0pK_a = 8.0, [A−]=0.1[A^-] = 0.1 M, [HA]=0.05[HA] = 0.05 M.

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

Flashcard 60: Calculate the pH if pKa=5.4pK_a = 5.4, [A−]=0.2[A^-] = 0.2 M, [HA]=0.2[HA] = 0.2 M.

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

Flashcard 61: How is the Henderson-Hasselbalch equation used in biological systems?

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

Flashcard 62: What does pKapK_a represent in the Henderson-Hasselbalch equation?

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

Flashcard 63: What happens to pH if [A−]=[HA][A^-] = [HA] in the Henderson-Hasselbalch equation?

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

Flashcard 64: What is the relation between pH and pKapK_a when pH > pKapK_a?

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

Flashcard 65: Determine the pH if pKa=7.8pK_a = 7.8, [A−]=0.3[A^-] = 0.3 M, [HA]=0.1[HA] = 0.1 M.

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

Flashcard 66: 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 67: Find the pH if pKa=4.75pK_a = 4.75, [A−]=0.1[A^-] = 0.1 M, [HA]=0.1[HA] = 0.1 M.

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

Flashcard 68: Determine the pH if pKa=7.8pK_a = 7.8, [A−]=0.3[A^-] = 0.3 M, [HA]=0.1[HA] = 0.1 M.

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

Flashcard 69: State the formula for the Henderson-Hasselbalch equation.

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

Flashcard 70: Calculate the pH if pKa=5.4pK_a = 5.4, [A−]=0.2[A^-] = 0.2 M, [HA]=0.2[HA] = 0.2 M.

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