MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5a Titration Buffers

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This deck focuses on 5a Titration Buffers, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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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 is the definition of a buffer in aqueous acid–base chemistry?

Answer: A solution that resists pH change upon small acid/base addition. Buffers maintain stable pH by neutralizing small amounts of added acid or base through reversible reactions involving weak acids/bases and their conjugates.

Flashcard 2: State the Henderson–Hasselbalch equation for a weak acid buffer.

Answer: $\text{pH}=\text{p}K_a+\log\left(\frac{[A^-]}{[HA]}\right)$. Derived from the acid dissociation equilibrium, this equation relates pH to the acid's pKa and the ratio of conjugate base to acid concentrations.

Flashcard 3: What is the relationship between $K_a$ and $\text{p}K_a$?

Answer: $\text{p}K_a=-\log(K_a)$. pKa quantifies acid strength as the negative logarithm of Ka, where lower pKa indicates stronger acids.

Flashcard 4: What is the pH at the equivalence point in a strong acid–strong base titration at $25\,^{\circ}\text{C}$?

Answer: $\text{pH}=7$. Neutralization of strong acid by strong base produces neutral salt solution with pH = 7 at 25°C.

Flashcard 5: Find the pH when $\text{p}K_a=4.76$ and $\frac{[A^-]}{[HA]}=1.0$.

Answer: $\text{pH}=4.76$. Using Henderson-Hasselbalch, when ratio = 1, log(1) = 0, so pH equals pKa directly.

Flashcard 6: What is the relationship between $K_a$ and $K_b$ for a conjugate pair at $25\,^{\circ}\text{C}$?

Answer: $K_aK_b=K_w=1.0\times10^{-14}$. For conjugate acid-base pairs, the product of Ka and Kb equals the water dissociation constant Kw at 25°C.

Flashcard 7: Find the pH when $\text{p}K_a=6.30$ and $\frac{[A^-]}{[HA]}=10$.

Answer: $\text{pH}=7.30$. Henderson-Hasselbalch gives pH = pKa + log(10) = pKa + 1, yielding 7.30.

Flashcard 8: At what condition does a buffer have maximum buffering capacity?

Answer: When $\text{pH}=\text{p}K_a$ and $[A^-]=[HA]$. Maximum capacity occurs when the buffer can equally resist acid or base additions, which is at equal concentrations and pH = pKa.

Flashcard 9: Find the equivalence volume: $25.0\,\text{mL}$ of $0.100\,\text{M}$ HCl titrated by $0.200\,\text{M}$ NaOH.

Answer: $V_{\text{NaOH}}=12.5\,\text{mL}$. Using Ma Va = Mb Vb for 1:1 titration, Vb = (0.100 * 25.0) / 0.200 = 12.5 mL.

Flashcard 10: State the equivalence condition for any titration in terms of moles of titrant and analyte.

Answer: Moles titrant added = stoichiometric moles required for analyte. Equivalence occurs when the titrant provides exactly the moles needed to react completely with the analyte per the stoichiometry.

Flashcard 11: State the formula used to find moles from molarity and volume (with volume in liters).

Answer: $n=MV$. Molarity times volume in liters gives moles, essential for calculating stoichiometry in titrations.

Flashcard 12: Identify the pH relative to $\text{p}K_a$ at the half-equivalence point of a weak acid titration.

Answer: $\text{pH}=\text{p}K_a$. At half-equivalence, [HA] = [A-], so Henderson-Hasselbalch simplifies to pH = pKa.

Flashcard 13: Find $\frac{[A^-]}{[HA]}$ when $\text{pH}=5.00$ and $\text{p}K_a=4.00$.

Answer: $\frac{[A^-]}{[HA]}=10$. From rearranged Henderson-Hasselbalch, ratio = 10^(pH - pKa) = 10^(1) = 10.

Flashcard 14: What is the relationship between $K_b$ and $\text{p}K_b$?

Answer: $\text{p}K_b=-\log(K_b)$. pKb measures base strength as the negative logarithm of Kb, with lower pKb indicating stronger bases.

Flashcard 15: What is the relationship between $\text{p}K_a$ and $\text{p}K_b$ at $25\,^{\circ}\text{C}$?

Answer: $\text{p}K_a+\text{p}K_b=14$. Taking negative logs of the Ka Kb = Kw relation yields pKa + pKb = 14 at 25°C for conjugate pairs.

Flashcard 16: Identify the primary purpose of an indicator in an acid–base titration.

Answer: To signal the endpoint by a visible color change near equivalence. Indicators change color at a specific pH near the equivalence point to visually indicate when to stop the titration.

Flashcard 17: Identify the dominant species when $\text{pH}<\text{p}K_a$ for a weak acid/conjugate base system.

Answer: The protonated form $HA$ dominates. At pH below pKa, the equilibrium favors the undissociated acid form according to Le Chatelier's principle.

Flashcard 18: In a weak acid–strong base titration, is the pH at equivalence point less than, equal to, or greater than $7$?

Answer: Greater than $7$. The equivalence point forms the conjugate base of the weak acid, which hydrolyzes to produce a basic solution with pH > 7.

Flashcard 19: What two components must be present in an effective buffer solution?

Answer: A weak acid and its conjugate base (or weak base and conjugate acid). Effective buffers require both components to shift equilibrium and absorb added H+ or OH- ions without significant pH change.

Flashcard 20: In a weak base–strong acid titration, is the pH at equivalence point less than, equal to, or greater than $7$?

Answer: Less than $7$. The equivalence point forms the conjugate acid of the weak base, which hydrolyzes to produce an acidic solution with pH < 7.

Flashcard 21: What is the typical effective buffering range for a weak acid buffer relative to $\text{p}K_a$?

Answer: Approximately $\text{p}K_a\pm^1$ pH unit. Buffers are effective within about one pH unit of pKa where the ratio of components allows significant buffering against additions.

Flashcard 22: What is the formula for the equivalence-point volume in a $1{:}1$ acid–base titration?

Answer: $M_aV_a=M_bV_b$. For 1:1 stoichiometry, the product of molarity and volume is equal for acid and base at equivalence.

Flashcard 23: Identify the dominant species when $\text{pH}>\text{p}K_a$ for a weak acid/conjugate base system.

Answer: The deprotonated form $A^-$ dominates. At pH above pKa, the equilibrium shifts to favor the dissociated conjugate base form.

Flashcard 24: What is the pH at the half-equivalence point in a weak acid–strong base titration?

Answer: $\text{pH}=\text{p}K_a$. At half-equivalence, half the acid is neutralized, making [HA] = [A-], so pH = pKa from Henderson-Hasselbalch.

Flashcard 25: Which expression gives the buffer ratio $\frac{[A^-]}{[HA]}$ in terms of pH and $\text{p}K_a$?

Answer: $\frac{[A^-]}{[HA]}=10^{\text{pH}-\text{p}K_a}$. Rearranging the Henderson-Hasselbalch equation isolates the ratio as an exponential function of the pH-pKa difference.