Titration and Buffers (5A)
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MCAT Chemical and Physical Foundations of Biological Systems › Titration and Buffers (5A)
A 20.0 mL sample of 0.200 M formic acid is titrated with 0.200 M NaOH at 25°C:
$\mathrm{HCOOH + OH^- \rightarrow HCOO^- + H_2O}$
Given $pK_a(\mathrm{HCOOH}) = 3.75$. Consider two points on the titration: Point P occurs when 5.0 mL of NaOH has been added; Point Q occurs when 19.0 mL of NaOH has been added. Assume volume changes are not negligible for pH but focus on conceptual species dominance.
Which statement best explains why the solution at one of these points has greater buffer capacity against added base?
Point Q has greater buffer capacity because adding base decreases pH less when the solution is more basic due to the logarithmic pH scale.
Point Q has greater buffer capacity because it is closer to equivalence, where the conjugate base concentration is highest.
Point P has greater buffer capacity because both HCOOH and HCOO− are present in appreciable amounts, enabling neutralization of added $\mathrm{OH^-}$.
Point P has lower buffer capacity because $pH$ must equal $pK_a$ only at the equivalence point, not earlier in the titration.
Explanation
This question tests understanding of titration and buffer systems (Foundational Concept 5A). Buffer capacity depends on having both acid and conjugate base present in appreciable amounts. At Point P (5.0 mL NaOH added), 25% of the formic acid has been neutralized, creating a mixture with significant amounts of both HCOOH and HCOO⁻. At Point Q (19.0 mL added), 95% has been neutralized, leaving very little HCOOH remaining. When base is added, it reacts with HCOOH: HCOOH + OH⁻ → HCOO⁻ + H₂O. Point P can neutralize much more added base due to its higher HCOOH concentration. A common error is thinking that proximity to the equivalence point improves buffering, but near equivalence, one component is nearly depleted. To assess buffer capacity, check that both components are present in reasonable amounts (typically 10:1 to 1:10 ratios work well).
A laboratory technician must prepare 500 mL of a buffer at pH 7.40 using the $\mathrm{H_2PO_4^- / HPO_4^{2-}}$ system at 25°C:
$\mathrm{H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}}$, $pK_a = 7.21$.
They can choose between two stock solutions: (i) 0.10 M $\mathrm{NaH_2PO_4}$ and 0.10 M $\mathrm{Na_2HPO_4}$, or (ii) 0.010 M $\mathrm{NaH_2PO_4}$ and 0.010 M $\mathrm{Na_2HPO_4}$. The pH will be set by mixing appropriate volumes of the acid and base stocks; no other acids/bases will be added. The goal is not only to reach pH 7.40 but also to maximize buffer capacity.
Which choice is most consistent with achieving the desired pH with the highest buffer capacity, assuming both options can be mixed to the same base:acid ratio?
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.
Use the 0.010 M stocks, because lower concentration buffers maintain pH more effectively by shifting equilibrium more easily.
Use either stock set, because buffer capacity depends only on $pH - pK_a$ and not on absolute concentration.
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.
An experimentalist prepares a Tris buffer for a protein purification step at 25°C using:
$\mathrm{TrisH^+ \rightleftharpoons H^+ + Tris}$ with $pK_a = 8.06$.
They initially dissolve Tris base to make 1.0 L of 0.10 M Tris (no TrisH+ added). They then adjust pH by adding HCl, forming some TrisH+. Buffer capacity is evaluated by adding a small additional amount of HCl and observing the pH change.
Which pH target during adjustment would most effectively maximize resistance to additional acid addition while still remaining a buffer (both forms present)?
Adjust to pH 7.00 because buffers are most effective at physiological pH regardless of $pK_a$.
Adjust to pH 8.06 so that $[\mathrm{Tris}] \approx[\mathrm{TrisH^+}]$, maximizing buffer capacity near $pK_a$.
Adjust to pH 6.06 so that $[\mathrm{Tris}]$ is 100× $[\mathrm{TrisH^+}]$, maximizing capacity against added acid.
Adjust to pH 10.06 so that $[\mathrm{TrisH^+}]$ is 100× $[\mathrm{Tris}]$, maximizing capacity against added acid.
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.
A weak base B is titrated with strong acid HCl at 25°C. The reaction is:
$\mathrm{B + H^+ \rightarrow BH^+}$
A 50.0 mL sample of 0.100 M B is titrated with 0.100 M HCl. The conjugate acid has $pK_a(\mathrm{BH^+}) = 8.30$. Assume ideal behavior.
At a point during the titration where both B and $\mathrm{BH^+}$ are present in substantial amounts, the solution shows minimal pH change upon addition of a small amount of HCl.
Which observation is most consistent with the solution being in its maximum buffering region during this titration?
The pH is highest at the equivalence point because $\mathrm{BH^+}$ concentration is maximal there.
The pH is far below 8.30 because strong acid addition shifts equilibrium completely to $\mathrm{BH^+}$ in the buffer region.
The pH is approximately 8.30 and the amounts of B and $\mathrm{BH^+}$ are comparable.
The pH is approximately 7.00 because neutrality indicates maximal resistance to added acid.
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.
A researcher prepares two buffers at 25°C using the ammonia/ammonium system:
$\mathrm{NH_4^+ \rightleftharpoons H^+ + NH_3}$ with $pK_a = 9.25$.
Buffer X: $\mathrm{NH_3} = 0.090,\mathrm{M}$ and $\mathrm{NH_4^+} = 0.010,\mathrm{M}$. Buffer Y: $\mathrm{NH_3} = 0.010,\mathrm{M}$ and $\mathrm{NH_4^+} = 0.090,\mathrm{M}$. Total buffer concentration is the same in both (0.100 M). A small amount of strong base is added to each.
Which statement best describes which buffer will show the smaller pH increase upon base addition, and why?
Both buffers respond identically because total buffer concentration is equal, so buffer capacity is independent of component ratio.
Buffer Y, because it contains more $\mathrm{NH_4^+}$ to neutralize added $\mathrm{OH^-}$ by forming $\mathrm{NH_3}$.
Buffer X, because it has higher pH initially and therefore is closer to the $pK_a$ where buffering is maximal.
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.
In a physiological model of blood buffering, a closed vessel contains an aqueous bicarbonate system at 37°C:
$\mathrm{CO_2(aq) + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-}$
For the $\mathrm{H_2CO_3/HCO_3^-}$ pair, use $pK_a = 6.10$. In the model, total dissolved $\mathrm{CO_2}$ is held constant by a gas reservoir, while $\mathrm{HCO_3^-}$ can change via addition of NaHCO3. Buffer capacity is assessed by adding a small amount of strong acid and observing the pH change.
Which change would most effectively increase the system’s ability to resist a decrease in pH upon acid addition, under the stated constraints?
Decrease $[\mathrm{HCO_3^-}]$ so that more $\mathrm{CO_2}$ can dissolve and neutralize added $\mathrm{H^+}$.
Increase $[\mathrm{HCO_3^-}]$ to provide more conjugate base to consume added $\mathrm{H^+}$.
Remove dissolved $\mathrm{CO_2}$ while keeping $[\mathrm{HCO_3^-}]$ constant, because eliminating the acid form strengthens buffering against added acid.
Increase temperature to raise $pK_a$ so that pH remains fixed near 7.4 regardless of acid load.
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.
A titration is performed to compare buffer regions for two monoprotic weak acids, HA (with $pK_a = 4.0$) and HB (with $pK_a = 9.0$), each prepared as 0.10 M solutions (25.0 mL) and titrated separately with 0.10 M NaOH. The neutralization reaction is:
$\mathrm{HX + OH^- \rightarrow X^- + H_2O}$
Assume ideal behavior and that the buffer region is most effective when both acid and conjugate base are present in comparable amounts.
Which statement best explains where each titration will exhibit its greatest buffering (smallest pH change per added base)?
Both titrations buffer best near pH 7 because water controls pH most strongly around neutrality.
HA buffers best near pH 9 and HB buffers best near pH 4 because the stronger acid always buffers at higher pH.
Both acids buffer best at the equivalence point because the conjugate base concentration is maximal there.
HA buffers best near pH 4 and HB buffers best near pH 9 because buffering is strongest when $pH \approx pK_a$.
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.
A researcher designs an acetate buffer for an enzyme assay at 25°C. The conjugate pair is:
$\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}$ with $pK_a = 4.76$.
The desired buffer pH is 5.76, and the total buffer concentration $\mathrm{CH_3COOH} + \mathrm{CH_3COO^-}$ should be 0.20 M to maximize buffer capacity without affecting ionic strength beyond the assay limit. Assume ideal behavior.
Which adjustment would most effectively achieve the desired pH while maintaining high buffer capacity, based on Henderson–Hasselbalch reasoning?
Choose $[\mathrm{CH_3COO^-}]$ about 10× $[\mathrm{CH_3COOH}]$ while keeping their sum at 0.20 M.
Choose $[\mathrm{CH_3COOH}]$ about 10× $[\mathrm{CH_3COO^-}]$ while keeping their sum at 0.20 M.
Set $[\mathrm{CH_3COOH}] = [\mathrm{CH_3COO^-}] = 0.10,\mathrm{M}$ because maximum capacity occurs when $pH = pK_a$.
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.
A weak acid titration is used to characterize an unknown monoprotic acid HA. A 25.0 mL sample of 0.100 M HA is titrated with 0.100 M NaOH at 25°C. The reaction is:
$\mathrm{HA + OH^- \rightarrow A^- + H_2O}$
At the half-equivalence point, the measured pH is 4.80. Assume activity coefficients are ~1 and that HA is the only acid-base active species initially.
Based on the titration behavior, which statement is most consistent with the system at the half-equivalence point and what it implies about $K_a$ (or $pK_a$)?
At half-equivalence, $[\mathrm{HA}] = [\mathrm{A^-}]$ and $pH = pK_a$, so $pK_a \approx 4.80$.
At half-equivalence, $[\mathrm{H^+}] = [\mathrm{OH^-}]$ and $pH = 7.00$, so $pK_a \approx 7.00$.
At half-equivalence, all HA has been converted to A−, so $pH > pK_a$ and $pK_a$ must be less than 4.80.
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.
A 50.0 mL sample of 0.100 M acetic acid ($\mathrm{HA}$) is titrated with 0.100 M NaOH. The reaction is $\mathrm{HA + OH^- \rightarrow A^- + H_2O}$, and $pK_a(\mathrm{HA})=4.76$ (25°C). At the equivalence point, essentially all initial $\mathrm{HA}$ has been converted to $\mathrm{A^-}$, and the pH is determined primarily by base hydrolysis: $\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$. For acetate, $K_b=K_w/K_a$ with $K_w=1.0\times10^{-14}$.
Which statement best explains why the pH at the equivalence point is expected to be greater than 7?
Constants provided: $pK_a=4.76$; $K_w=1.0\times10^{-14}$.
At equivalence, excess $\mathrm{OH^-}$ from the titrant remains unreacted, forcing the pH above 7.
At equivalence, the solution is neutral because the moles of acid and base added are equal.
At equivalence, $pH$ must equal $pK_a$ because $[\mathrm{HA}]=[\mathrm{A^-}]$ at that point.
At equivalence, acetate acts as a weak base and generates $\mathrm{OH^-}$ by hydrolysis, increasing pH above 7.
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.