AP Chemistry Flashcards: Henderson Hasselbalch Equation

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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.

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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.

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 $pK_a = 4.5$, what is pH when $[A^-] = 0.2$ M and $[HA] = 0.1$ M?

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

Flashcard 3: What is the relation between pH and $pK_a$ when pH < $pK_a$?

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

Flashcard 4: 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 5: If $[A^-]/[HA] = 2$, find pH given $pK_a = 7.2$.

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

Flashcard 6: 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 7: What is the effect of temperature on $pK_a$ in the Henderson-Hasselbalch equation?

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

Flashcard 8: What is the relation between pH and $pK_a$ when pH > $pK_a$?

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

Flashcard 9: What is the impact on pH if $[A^-]$ decreases in the buffer solution?

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

Flashcard 10: How is the Henderson-Hasselbalch equation derived?

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

Flashcard 11: Find pH for $pK_a = 3.5$, $[A^-] = 0.03$ M, $[HA] = 0.01$ M.

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

Flashcard 12: 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 13: 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 14: If $[HA] = 0.1$ M and $[A^-] = 0.01$ M, how does pH compare to $pK_a$?

Answer: pH < $pK_a$. More acid than base means pH below $pK_a$.

Flashcard 15: What is the effect on pH if $[A^-]$ increases relative to $[HA]$?

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

Flashcard 16: Calculate pH if $pK_a = 9.3$, $[A^-] = 0.05$ M, $[HA] = 0.1$ M.

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

Flashcard 17: What does $pK_a$ represent in the Henderson-Hasselbalch equation?

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

Flashcard 18: State the logarithmic property used in the Henderson-Hasselbalch equation.

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

Flashcard 19: For a buffer with $pK_a = 6.3$, what is pH when $[A^-] = 0.5$ M, $[HA] = 0.25$ M?

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

Flashcard 20: What happens to pH if $[A^-] = [HA]$ in the Henderson-Hasselbalch equation?

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

Flashcard 21: In the Henderson-Hasselbalch equation, what does a higher $pK_a$ indicate?

Answer: A weaker acid. Larger $pK_a$ corresponds to smaller $K_a$ value.

Flashcard 22: If $pK_a = 6.1$ and $[A^-] = 0.25$ M, $[HA] = 0.25$ M, what is the pH?

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

Flashcard 23: What does the ratio $[A^-]/[HA]$ signify in the Henderson-Hasselbalch equation?

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

Flashcard 24: How does the Henderson-Hasselbalch equation relate to buffers?

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

Flashcard 25: 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 26: Identify the meaning of $[A^-]$ in the Henderson-Hasselbalch equation.

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

Flashcard 27: What is the logarithmic form of the Henderson-Hasselbalch equation?

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

Flashcard 28: What assumption is made about $[A^-]$ and $[HA]$ in the Henderson-Hasselbalch equation?

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

Flashcard 29: Identify the meaning of $[HA]$ in the Henderson-Hasselbalch equation.

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

Flashcard 30: Find the pH if $pK_a = 8.0$, $[A^-] = 0.1$ M, $[HA] = 0.05$ M.

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

Flashcard 31: Calculate the pH if $pK_a = 5.4$, $[A^-] = 0.2$ M, $[HA] = 0.2$ M.

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

Flashcard 32: 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 33: 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 34: Find the pH if $pK_a = 4.75$, $[A^-] = 0.1$ M, $[HA] = 0.1$ M.

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

Flashcard 35: Determine the pH if $pK_a = 7.8$, $[A^-] = 0.3$ M, $[HA] = 0.1$ M.

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

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

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