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

What this quiz covers

This quiz focuses on 5a Titration Buffers, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. At equivalence, excess $\mathrm{OH^-}$ from the titrant remains unreacted, forcing the pH above 7.
2. At equivalence, acetate acts as a weak base and generates $\mathrm{OH^-}$ by hydrolysis, increasing pH above 7. (correct answer)
3. At equivalence, $pH$ must equal $pK_a$ because $[\mathrm{HA}]=[\mathrm{A^-}]$ at that point.
4. At equivalence, the solution is neutral because the moles of acid and base added are equal.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). At the equivalence point of a weak acid-strong base titration, all the weak acid (HA) has been converted to its conjugate base (A⁻), creating a solution of sodium acetate. Acetate ion is a weak base that undergoes hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻, producing hydroxide ions and increasing the pH above 7. The basicity of acetate can be quantified using Kb = Kw/Ka = (1.0×10⁻¹⁴)/(10⁻⁴·⁷⁶) ≈ 5.8×10⁻¹⁰, confirming it acts as a weak base. Choice D incorrectly assumes that equal moles of acid and base always produce a neutral solution, which is only true for strong acid-strong base titrations. For weak acid-strong base titrations, remember that the equivalence point pH > 7 due to hydrolysis of the conjugate base formed.

Question 2

1. The pH is approximately 8.30 and the amounts of B and $\mathrm{BH^+}$ are comparable. (correct answer)
2. The pH is approximately 7.00 because neutrality indicates maximal resistance to added acid.
3. The pH is far below 8.30 because strong acid addition shifts equilibrium completely to $\mathrm{BH^+}$ in the buffer region.
4. The pH is highest at the equivalence point because $\mathrm{BH^+}$ concentration is maximal there.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Maximum buffering occurs when both the weak base B and its conjugate acid BH⁺ are present in comparable amounts, which happens when pH ≈ pKa of the conjugate acid. Since pKa(BH⁺) = 8.30, the buffer region centers around pH 8.30. At this pH, the Henderson-Hasselbalch equation gives log([B]/[BH⁺]) ≈ 0, meaning [B] ≈ [BH⁺]. This equal distribution provides optimal resistance to pH change from added acid or base. A common misconception is that pH 7.00 (neutrality) indicates optimal buffering, but buffer effectiveness depends on the specific pKa, not on neutrality. To identify buffer regions in weak base titrations, look for pH values near the pKa of the conjugate acid, where both base and conjugate acid forms coexist.

Question 3

1. Decrease $[\mathrm{HCO_3^-}]$ so that more $\mathrm{CO_2}$ can dissolve and neutralize added $\mathrm{H^+}$.
2. Increase $[\mathrm{HCO_3^-}]$ to provide more conjugate base to consume added $\mathrm{H^+}$. (correct answer)
3. Increase temperature to raise $pK_a$ so that pH remains fixed near 7.4 regardless of acid load.
4. Remove dissolved $\mathrm{CO_2}$ while keeping $[\mathrm{HCO_3^-}]$ constant, because eliminating the acid form strengthens buffering against added acid.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity against added acid depends on having sufficient conjugate base (HCO₃⁻) to neutralize H⁺ ions by the reaction: HCO₃⁻ + H⁺ → H₂CO₃. Since total dissolved CO₂ is held constant by the gas reservoir, increasing [HCO₃⁻] by adding NaHCO₃ provides more conjugate base to consume added acid while maintaining the CO₂ equilibrium. The system will resist pH decrease more effectively with higher [HCO₃⁻]. A common error is thinking that decreasing [HCO₃⁻] would help by allowing more CO₂ to dissolve, but CO₂ is already at equilibrium with the gas phase. To maximize buffer capacity against acid, always ensure adequate conjugate base is present to neutralize the expected acid load.

Question 4

1. Both titrations buffer best near pH 7 because water controls pH most strongly around neutrality.
2. HA buffers best near pH 4 and HB buffers best near pH 9 because buffering is strongest when $pH \approx pK_a$. (correct answer)
3. HA buffers best near pH 9 and HB buffers best near pH 4 because the stronger acid always buffers at higher pH.
4. Both acids buffer best at the equivalence point because the conjugate base concentration is maximal there.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer action is most effective when both the weak acid and its conjugate base are present in comparable amounts, which occurs when pH is near the pKa of the acid. For HA with pKa = 4.0, maximum buffering occurs around pH 4, while for HB with pKa = 9.0, maximum buffering occurs around pH 9. During titration with NaOH, these pH regions are reached when approximately half of each acid has been neutralized (near the half-equivalence point). A common misconception is that buffering is best at the equivalence point, but at equivalence, essentially all acid has been converted to conjugate base, leaving no acid to neutralize added base. To identify buffer regions in titrations, look for pH values within ±1 unit of the acid's pKa, where both acid and conjugate base coexist in significant amounts.

Question 5

1. At half-equivalence, $[\mathrm{HA}] = [\mathrm{A^-}]$ and $pH = pK_a$, so $pK_a \approx 4.80$. (correct answer)
2. At half-equivalence, $[\mathrm{H^+}] = [\mathrm{OH^-}]$ and $pH = 7.00$, so $pK_a \approx 7.00$.
3. At half-equivalence, all HA has been converted to A−, so $pH > pK_a$ and $pK_a$ must be less than 4.80.
4. At half-equivalence, the equivalence point has been reached, so $pH$ depends only on the strong base concentration and not on $pK_a$.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). At the half-equivalence point of a weak acid titration, exactly half of the weak acid has been converted to its conjugate base, meaning [HA] = [A⁻]. The Henderson-Hasselbalch equation shows that when [A⁻]/[HA] = 1, then pH = pKa + log(1) = pKa. Since the measured pH at half-equivalence is 4.80, the pKa of the unknown acid must be 4.80. A common error is confusing half-equivalence with the equivalence point, where all acid has been converted to conjugate base. To identify the half-equivalence point, look for when the volume of titrant added is exactly half of what's needed to reach the equivalence point, or when pH = pKa for a monoprotic acid.

Question 6

1. Buffer X, because it has higher pH initially and therefore is closer to the $pK_a$ where buffering is maximal.
2. Buffer Y, because it contains more $\mathrm{NH_4^+}$ to neutralize added $\mathrm{OH^-}$ by forming $\mathrm{NH_3}$. (correct answer)
3. Both buffers respond identically because total buffer concentration is equal, so buffer capacity is independent of component ratio.
4. Buffer X, because added $\mathrm{OH^-}$ converts $\mathrm{NH_3}$ to $\mathrm{NH_4^+}$, consuming base more effectively.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). When strong base (OH⁻) is added to an ammonia/ammonium buffer, it reacts with the acidic component NH₄⁺ according to: NH₄⁺ + OH⁻ → NH₃ + H₂O. Buffer Y contains much more NH₄⁺ (0.090 M) compared to Buffer X (0.010 M), so it can neutralize more added base before significant pH change occurs. Although Buffer X has pH closer to the pKa (better for general buffering), the specific challenge is resisting base addition, which requires the acidic buffer component. A common error is assuming that higher pH always means better buffering against base, but buffer capacity against a specific perturbation depends on having enough of the appropriate component. When evaluating buffer response to base, check the concentration of the acidic component (conjugate acid) that will neutralize the base.

Question 7

1. Use the 0.010 M stocks, because lower concentration buffers maintain pH more effectively by shifting equilibrium more easily.
2. Use either stock set, because buffer capacity depends only on $pH - pK_a$ and not on absolute concentration.
3. Use the 0.10 M stocks, because higher total buffer concentration increases the amount of acid/base that can be neutralized with minimal pH change. (correct answer)
4. Use the 0.10 M stocks, because higher concentration forces $pH$ to equal $pK_a$ regardless of mixing ratio.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity increases with the total concentration of buffer components when the pH and component ratio are held constant. Using 0.10 M stocks instead of 0.010 M stocks provides 10× more moles of both H₂PO₄⁻ and HPO₄²⁻ in the final buffer. This means the buffer can neutralize 10× more added acid or base before showing significant pH change. Both stock sets can be mixed to achieve pH 7.40 by using the same volume ratio, but the absolute concentrations differ by 10-fold. A common misconception is that lower concentrations allow easier equilibrium shifts, but buffer action depends on having sufficient material to neutralize added acid or base. When preparing buffers for maximum capacity, always use the highest practical concentration that doesn't interfere with other experimental requirements.

Question 8

1. Adjust to pH 8.06 so that $[\mathrm{Tris}] \approx [\mathrm{TrisH^+}]$, maximizing buffer capacity near $pK_a$. (correct answer)
2. Adjust to pH 6.06 so that $[\mathrm{Tris}]$ is 100× $[\mathrm{TrisH^+}]$, maximizing capacity against added acid.
3. Adjust to pH 10.06 so that $[\mathrm{TrisH^+}]$ is 100× $[\mathrm{Tris}]$, maximizing capacity against added acid.
4. Adjust to pH 7.00 because buffers are most effective at physiological pH regardless of $pK_a$.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Maximum buffer capacity occurs when pH = pKa, where [Tris] = [TrisH⁺] according to the Henderson-Hasselbalch equation. At pH 8.06 (the pKa), the buffer has equal concentrations of both forms, providing optimal resistance to pH changes from added acid or base. While the question specifically asks about resistance to additional acid, having comparable amounts of both forms ensures the buffer can respond effectively. A common error is trying to maximize one component over the other for specific resistance, but this creates a poor buffer that quickly exhausts its capacity. For maximum buffer capacity at a specific pH, always target pH = pKa, which automatically provides the optimal 1:1 ratio of buffer components.

Question 9

1. Set $[\mathrm{CH_3COOH}] = [\mathrm{CH_3COO^-}] = 0.10\,\mathrm{M}$ because maximum capacity occurs when $pH = pK_a$.
2. Choose $[\mathrm{CH_3COO^-}]$ about 10× $[\mathrm{CH_3COOH}]$ while keeping their sum at 0.20 M. (correct answer)
3. Choose $[\mathrm{CH_3COOH}]$ about 10× $[\mathrm{CH_3COO^-}]$ while keeping their sum at 0.20 M.
4. Decrease total buffer concentration well below 0.20 M so that the buffer can more easily shift equilibrium and hold pH constant.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). The Henderson-Hasselbalch equation relates pH to pKa and the ratio of conjugate base to acid: pH = pKa + log([A⁻]/[HA]). For a desired pH of 5.76 with pKa = 4.76, we need pH - pKa = 1.00, which means log([CH₃COO⁻]/[CH₃COOH]) = 1.00, so [CH₃COO⁻]/[CH₃COOH] = 10. To maintain total concentration of 0.20 M while achieving this 10:1 ratio, we need [CH₃COO⁻] ≈ 0.182 M and [CH₃COOH] ≈ 0.018 M. A common error is thinking maximum buffer capacity always occurs at pH = pKa (equal concentrations), but the question asks for pH 5.76, not maximum capacity at any pH. When preparing buffers, always calculate the required ratio from the Henderson-Hasselbalch equation first, then adjust concentrations to meet that ratio.

Question 10

1. Buffer 1 will show the smaller pH change because lower concentrations reduce ionic strength effects.
2. Buffer 2 will show the smaller pH change because higher total buffer concentration provides greater capacity at the same ratio. (correct answer)
3. Both buffers will show the same pH change because pH depends only on the $[\mathrm{A^-}]/[\mathrm{HA}]$ ratio, not concentration.
4. Buffer 2 will show the larger pH change because more $\mathrm{A^-}$ means more $\mathrm{H^+}$ is produced by the equilibrium.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity, the ability to resist pH changes, increases with the total concentration of buffer components while maintaining the same [A⁻]/[HA] ratio. Buffer 2, with ten times higher concentrations (0.10 M each) than Buffer 1 (0.010 M each), has ten times more moles of A⁻ available to neutralize the same amount of added H⁺, resulting in a smaller pH change. Both buffers start at the same pH since they have the same [A⁻]/[HA] ratio, but Buffer 2's higher concentration provides greater resistance to perturbation. Choice C incorrectly assumes pH change depends only on the ratio, ignoring that buffer capacity depends on absolute concentrations. To assess buffer capacity, consider both the ratio (which sets the pH) and the total concentration (which determines resistance to change).

Question 11

1. Near the start of titration, when only HA is present and no conjugate base exists.
2. Near the half-equivalence point, when both HA and A− are present in comparable amounts. (correct answer)
3. Near the equivalence point, when HA and A− are present in equal amounts by definition.
4. Far past equivalence, when excess OH− dominates and the solution is maximally buffered.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). During a weak acid-strong base titration, the pH changes minimally in the buffer region near the half-equivalence point, where both HA and A⁻ are present in comparable amounts. At the half-equivalence point, [HA] ≈ [A⁻], creating optimal buffering capacity as both species can neutralize added acid or base. The pH in this region approximately equals the pKa (5.00), and small additions of NaOH are neutralized by the HA present, converting it to A⁻ with minimal pH change. Choice C incorrectly identifies the equivalence point as having equal amounts of HA and A⁻; at equivalence, essentially all HA has been converted to A⁻. To identify buffer regions in titrations, look for where both conjugate species coexist, typically centered at the half-equivalence point where pH ≈ pKa.

Question 12

1. Added $\mathrm{H^+}$ is consumed by $\mathrm{HCO_3^-}$ to form $\mathrm{H_2CO_3}$, limiting the increase in free $\mathrm{H^+}$. (correct answer)
2. Added $\mathrm{H^+}$ shifts the equilibrium to produce more $\mathrm{H^+}$, stabilizing pH by Le Châtelier's principle.
3. Because $[\mathrm{HCO_3^-}]>[\mathrm{H_2CO_3}]$, the pH must equal 6.10 and cannot change upon acid addition.
4. The buffer works only when $[\mathrm{H_2CO_3}]\gg[\mathrm{HCO_3^-}]$, so this solution will show little resistance to added acid.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer action occurs when the conjugate base component neutralizes added acid by accepting protons, preventing a large pH change. In this H₂CO₃/HCO₃⁻ system, when strong acid (H⁺) is added, the bicarbonate ion (HCO₃⁻) acts as a base and combines with H⁺ to form H₂CO₃, thereby removing free H⁺ from solution and limiting the pH decrease. This system has good buffering capacity because [HCO₃⁻] = 0.200 M is substantial and can neutralize appreciable amounts of added acid. Choice B incorrectly applies Le Châtelier's principle, suggesting the equilibrium would produce more H⁺ when H⁺ is added, which would actually decrease pH rather than stabilize it. To understand buffer action, remember that the conjugate base consumes added H⁺ while the weak acid consumes added OH⁻.

Question 13

1. Mix equal volumes of the two 1.0 M stocks to maximize total concentration and set $pH\approx pK_a$. (correct answer)
2. Use only 1.0 M $\mathrm{NaH_2PO_4}$ because the pH of a buffer equals the acid's $pK_a$ regardless of conjugate base.
3. Dilute both stocks 100-fold before mixing so that the Henderson–Hasselbalch equation is valid.
4. Mix mostly $\mathrm{Na_2HPO_4}$ with a small amount of $\mathrm{NaH_2PO_4}$ because $pH=pK_a$ only when base dominates.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). To prepare a buffer at pH 7.20 when pKa = 7.21, the Henderson-Hasselbalch equation shows that [HPO₄²⁻]/[H₂PO₄⁻] should be approximately 1:1 since pH ≈ pKa. Mixing equal volumes of the 1.0 M stocks creates [H₂PO₄⁻] = [HPO₄²⁻] = 0.5 M, achieving both the desired pH and high buffer capacity due to the high total concentration (1.0 M total). Buffer capacity increases with total buffer concentration at a fixed ratio, making this the optimal choice. Choice D incorrectly suggests that base must dominate for pH = pKa, misunderstanding the Henderson-Hasselbalch relationship. When preparing buffers near the pKa, use approximately equal amounts of acid and conjugate base forms for both correct pH and maximum buffering capacity.

Question 14

1. The buffer resists both additions similarly because equal, appreciable amounts of HA and A− are present to neutralize added base or acid, respectively. (correct answer)
2. The buffer resists added HCl but not added NaOH because the pH is below the $pK_a$ when $[\mathrm{HA}]=[\mathrm{A^-}]$.
3. The buffer cannot resist either addition because strong acids and bases fully determine pH regardless of buffer concentration.
4. The buffer resists added NaOH but not added HCl because acetate is the conjugate base and therefore dominates pH control.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). A buffer resists pH changes by having both a weak acid (HA) and its conjugate base (A⁻) present to neutralize added strong base or acid, respectively. In this system with equal concentrations of acetic acid and acetate ([HA] = [A⁻] = 0.10 M), the buffer has optimal capacity to resist both acid and base additions. When HCl is added, the acetate ion (A⁻) neutralizes the H⁺ to form more HA; when NaOH is added, the acetic acid (HA) neutralizes the OH⁻ to form more A⁻. Choice B incorrectly suggests the buffer only resists acid addition, misunderstanding that buffers work bidirectionally when both components are present. To verify buffer effectiveness, check that both HA and A⁻ are present in appreciable amounts (ideally within a 10:1 ratio) and that the working pH is within ±1 unit of the pKa.

Question 15

1. The buffer has high capacity because added $\mathrm{H^+}$ is consumed by $\mathrm{CH_3COO^-}$ to form $\mathrm{CH_3COOH}$, keeping the ratio near 1. (correct answer)
2. The buffer has low capacity because $pK_a$ equals 4.76, so the pH is fixed at 4.76 regardless of added acid.
3. The buffer has high capacity because added $\mathrm{H^+}$ converts $\mathrm{CH_3COOH}$ into $\mathrm{CH_3COO^-}$, increasing pH.
4. The buffer has low capacity because equal concentrations of acid and base neutralize each other and leave no buffering species.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffers resist changes in pH by utilizing a weak acid and its conjugate base to neutralize added acids or bases, with maximum capacity when their concentrations are equal. In this acetic acid/acetate buffer, equal concentrations of CH3COOH and CH3COO- ensure the pH is at the pKa of 4.76, providing high buffering capacity. The correct answer aligns because added H+ from HCl reacts with CH3COO- to form CH3COOH, maintaining the ratio near 1 and minimizing pH change. A common error is thinking equal concentrations neutralize each other leaving no buffer, as in choice D, but they are the active buffering species. For similar questions, verify that buffer capacity is highest when [acid] ≈ [base] and both are substantial. Always check if the added amount is small relative to buffer concentrations to ensure negligible depletion.

Question 16

1. The pH is approximately equal to $pK_a$ (about 4.76). (correct answer)
2. The pH is 7.00 because half-equivalence implies neutrality.
3. The pH is well below $pK_a$ because the acid has not been fully neutralized.
4. The pH is approximately equal to $14 - pK_a$ (about 9.24).

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). In titrating a weak acid with a strong base, the half-equivalence point occurs when half the acid is neutralized, creating equal amounts of acid and conjugate base. For acetic acid titrated with NaOH, at half-equivalence, [CH3COOH] ≈ [CH3COO-], making the solution a buffer at pH ≈ pKa. This aligns with the correct answer since the Henderson-Hasselbalch equation gives pH = pKa + log(1) = 4.76. A distractor like choice B assumes neutrality at half-equivalence, but for weak acids, it's at pKa, not 7. For similar problems, confirm the half-equivalence pH equals pKa by ensuring equal conjugate forms. Double-check volumes to identify the half-equivalence point accurately.

Question 17

1. Buffer 2 shows a smaller pH decrease than Buffer 1 because it has higher total buffer concentration. (correct answer)
2. Buffer 1 shows a smaller pH decrease because lower concentration reduces ionic strength effects.
3. Both buffers show identical pH change because pH is determined only by $pK_a$.
4. Buffer 2 shows a larger pH decrease because concentrated buffers amplify pH changes on the logarithmic scale.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity directly correlates with total concentration of the conjugate pair at a fixed ratio, leading to smaller pH changes in more concentrated buffers. Here, Buffer 2 at 0.100 M total resists pH decrease from HCl better than Buffer 1 at 0.010 M. The correct answer aligns as higher concentration in Buffer 2 minimizes the pH drop. Choice D incorrectly claims concentrated buffers amplify changes, ignoring capacity principles. In comparable scenarios, compare total concentrations after confirming same ratios. Note that near pKa, capacity is maximized but still scales with concentration.

Question 18

1. Increase both $[\mathrm{H_2PO_4^-}]$ and $[\mathrm{HPO_4^{2-}}]$ by the same factor, keeping their ratio constant. (correct answer)
2. Increase only $[\mathrm{HPO_4^{2-}}]$ to raise total concentration without changing pH.
3. Decrease both components by the same factor to reduce dilution effects and stabilize pH.
4. Adjust pH to exactly equal $pK_a2$ by adding strong acid, because capacity depends only on matching $pH = pK_a$.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity increases with total concentration of the conjugate pair while pH is set by the base-to-acid ratio via Henderson-Hasselbalch. For the phosphate buffer at pH 7.20 near pKa2 7.21, increasing both [H2PO4-] and [HPO42-] proportionally maintains the ratio and pH. This aligns with the correct answer as it boosts capacity without altering pH. Choice B is incorrect because increasing only [HPO42-] would raise pH, not maintain it. For analogous questions, calculate the required ratio using pH = pKa + log([base]/[acid]). Ensure adjustments preserve the ratio for stable pH.

Question 19

1. At equivalence, the solution contains mostly $\mathrm{A^-}$, which generates $\mathrm{OH^-}$ by reacting with water. (correct answer)
2. At equivalence, the solution contains mostly HA, which generates $\mathrm{H^+}$ by reacting with water.
3. At equivalence, $pH$ must be 7 because moles of acid equal moles of base.
4. At equivalence, the $pH$ equals the $pK_a$ of the acid regardless of titrant strength.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). In weak acid-strong base titrations, equivalence point pH >7 due to hydrolysis of the weak conjugate base producing OH-. For HA with pKa 4.0, at equivalence, A- hydrolyzes to form basic solution. The correct answer aligns as A- generates OH-. Choice C assumes pH 7, valid only for strong acids. In like titrations, estimate equivalence pH ≈ (pKa + 14 + log C)/2 for dilute solutions. Confirm no buffer region at equivalence.

Question 20

1. pH decreases, and the change is typically larger than in an equimolar buffer because the acid form is relatively scarce. (correct answer)
2. pH increases because added acid converts $\mathrm{HLac}$ to $\mathrm{Lac^-}$.
3. pH does not change because the ratio is fixed at 10 by the $pK_a$.
4. pH decreases, and the change is typically smaller than in an equimolar buffer because $pH > pK_a$.

Explanation: This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity is reduced when pH deviates from pKa, leading to larger pH changes despite high base form concentration. In this lactate buffer at pH = pKa +1 with [Lac-]/[HLac]=10, added acid causes a larger pH decrease than in an equimolar buffer due to low [HLac] making the ratio more sensitive. The correct answer aligns as pH drops more significantly. Choice D incorrectly suggests smaller change, overlooking capacity reduction away from pKa. For like questions, use ΔpH approximation involving 1/[A-] + 1/[HA]. Check that capacity peaks at [acid] = [base].