Acid–Base Equilibria (5A)
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MCAT Chemical and Physical Foundations of Biological Systems › Acid–Base Equilibria (5A)
A blood-gas analyzer is calibrated using a bicarbonate buffer where $pK_a = 6.1$ for the $\mathrm{H_2CO_3}/\mathrm{HCO_3^-}$ pair. Two calibration solutions are prepared at 37°C:
Solution 1: $\mathrm{HCO_3^-} = 20\ \mathrm{mM}$, $\mathrm{CO_2(aq)} = 1.0\ \mathrm{mM}$
Solution 2: $\mathrm{HCO_3^-} = 10\ \mathrm{mM}$, $\mathrm{CO_2(aq)} = 1.0\ \mathrm{mM}$
Assume $\mathrm{H_2CO_3}$ tracks with $\mathrm{CO_2(aq)}$ and the Henderson–Hasselbalch relationship applies qualitatively.
Which statement best describes the system’s response when moving from Solution 1 to Solution 2?
pH is unchanged because $pK_a$ is constant and therefore fixes pH for any mixture.
pH increases because lowering $[\mathrm{HCO_3^-}]$ reduces buffering, allowing $[\mathrm{H^+}]$ to fall.
pH increases because CO$_2$ is the conjugate base of $\mathrm{HCO_3^-}$ and remains constant.
pH decreases because the base-to-acid ratio decreases when $[\mathrm{HCO_3^-}]$ is halved at constant CO$_2$.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically the bicarbonate buffer system's pH dependence on component ratios. The Henderson-Hasselbalch equation for this system is: pH = pKa + log([HCO₃⁻]/[CO₂]), where pKa = 6.1. In Solution 1, the ratio is 20/1 = 20, while in Solution 2, it's 10/1 = 10, representing a decrease in the base-to-acid ratio. Choice B correctly identifies that pH decreases because the base-to-acid ratio decreases when [HCO₃⁻] is halved at constant CO₂, resulting in a smaller log term in the Henderson-Hasselbalch equation. Choice A incorrectly suggests that lowering [HCO₃⁻] would cause pH to increase, which contradicts the mathematical relationship. In future problems, remember that decreasing the numerator in the Henderson-Hasselbalch ratio always decreases pH.
A clinician models blood buffering by preparing 1.0 L of a weak acid buffer containing 20 mM lactic acid (HLac) and 20 mM lactate (Lac$^-$) at 37°C. For $\mathrm{HLac} \rightleftharpoons \mathrm{H^+} + \mathrm{Lac^-}$, $pK_a = 3.9$. A small bolus of strong acid is added, increasing $\mathrm{H^+}$ transiently, and the mixture is allowed to re-equilibrate without changing volume.
Based on the buffer principle, which conclusion about the buffer system is most consistent?
The added $\mathrm{H^+}$ is primarily consumed by Lac$^-$ to form HLac, so the pH decreases only modestly.
The pH is unchanged because weak acid buffers maintain pH exactly at $pK_a$ regardless of added acid.
The pH change is large because buffering is ineffective when $[\mathrm{HLac}] = [\mathrm{Lac^-}]$.
The added $\mathrm{H^+}$ is primarily consumed by HLac to form Lac$^-$, so the pH increases.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically buffer action when strong acid is added. Buffer systems resist pH changes by converting added H⁺ or OH⁻ to weak acid or conjugate base forms. In this lactate buffer system with equal concentrations of HLac and Lac⁻, added H⁺ ions will react with the conjugate base (Lac⁻) to form more weak acid (HLac), following the equilibrium HLac ⇌ H⁺ + Lac⁻. Choice A correctly states that added H⁺ is consumed by Lac⁻ to form HLac, resulting in only a modest pH decrease due to the buffer's resistance. Choice B incorrectly reverses the reaction direction - HLac cannot consume H⁺ as it would violate the equilibrium principle. In future buffer problems, identify which species (weak acid or conjugate base) will react with the added strong acid or base.
In a study of metabolic acidosis, investigators compare two 100 mL buffer preparations at 37°C using the same weak acid/conjugate base pair with $pK_a = 6.8$.
Buffer X: 50 mM HA and 50 mM A$^-$
Buffer Y: 5 mM HA and 5 mM A$^-$
A small, identical amount of strong acid is added to each buffer (same moles of $\mathrm{H^+}$), and volume change is negligible.
Which statement best describes the system’s response?
Neither buffer changes pH because weak acid buffers neutralize all added strong acid until equivalence is reached.
Buffer X shows a smaller pH change because it has greater buffer capacity (higher total concentration of conjugate pair).
Both buffers show the same pH change because they have the same $[\mathrm{A^-}]/[\mathrm{HA}]$ ratio.
Buffer Y shows a smaller pH change because dilute buffers resist pH change more strongly.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically the concept of buffer capacity. Buffer capacity depends on the total concentration of the conjugate acid-base pair - higher concentrations provide more molecules to neutralize added acid or base. Both buffers have the same initial pH since they have identical [A⁻]/[HA] ratios, but Buffer X has 10-fold higher total concentration (100 mM vs 10 mM total). Choice B correctly identifies that Buffer X shows a smaller pH change because it has greater buffer capacity due to higher total concentration of the conjugate pair. Choice C incorrectly suggests dilute buffers resist pH change more strongly, when the opposite is true - concentrated buffers have greater capacity. In future problems, remember that buffer capacity increases with total buffer concentration, not just the ratio of components.
A researcher titrates 50.0 mL of 0.10 M benzoic acid (HBz) with 0.10 M NaOH at 25°C to assess buffer range for a topical formulation. For $\mathrm{HBz} \rightleftharpoons \mathrm{H^+} + \mathrm{Bz^-}$, $pK_a = 4.2$. The formulation is intended to keep pH within a range where buffering is effective.
Based on the passage, which conclusion about the buffer system is most consistent with effective buffering during titration?
Buffering is most effective at very high pH because Bz$^-$ is maximally deprotonated and can absorb added base.
Buffering is most effective when pH is within about $\pm 1$ unit of $pK_a$, where both HBz and Bz$^-$ are present in comparable amounts.
Buffering is most effective at the equivalence point because all HBz has been converted to Bz$^-$.
Buffering is most effective at very low pH because HBz is maximally protonated and can absorb added acid.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically the effective buffering range of weak acid-base pairs. Buffer capacity is maximized when both the weak acid and conjugate base are present in significant, comparable amounts, which occurs when pH is near pKa. The Henderson-Hasselbalch equation shows that when pH = pKa ± 1, the ratio [A⁻]/[HA] ranges from 0.1 to 10, providing substantial amounts of both species. Choice A correctly identifies that buffering is most effective when pH is within about ±1 unit of pKa, where both HBz and Bz⁻ are present in comparable amounts. Choice B incorrectly identifies the equivalence point, where essentially all HBz is converted to Bz⁻, leaving no weak acid to neutralize added base. In future buffer problems, remember the practical buffering range is pKa ± 1 pH unit.
In an ex vivo study of acid–base balance, researchers prepared a bicarbonate buffer intended to approximate plasma. The solution contained $\mathrm{HCO_3^-} = 24\ \mathrm{mM}$ and dissolved CO$_2$ at $\mathrm{CO_2(aq)} = 1.2\ \mathrm{mM}$ at 37°C. For the equilibrium $\mathrm{CO_2(aq)} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2CO_3} \rightleftharpoons \mathrm{H^+} + \mathrm{HCO_3^-}$, use $pK_a = 6.1$ for the $\mathrm{H_2CO_3}/\mathrm{HCO_3^-}$ pair and assume $\mathrm{H_2CO_3}$ is proportional to $\mathrm{CO_2(aq)}$ under these conditions. The system is then exposed to a sudden increase in CO$_2$ (simulating hypoventilation) that raises $\mathrm{CO_2(aq)}$ by 50% while total buffer volume remains constant.
Which statement best describes the system’s response?
pH decreases because increased $[\mathrm{CO_2(aq)}]$ drives formation of $\mathrm{H^+}$, lowering the $[\mathrm{HCO_3^-}]/[\mathrm{CO_2}]$ ratio.
pH increases because $[\mathrm{HCO_3^-}]$ rises proportionally with $[\mathrm{CO_2(aq)}]$ in an open system.
pH increases because added CO$_2$ consumes $\mathrm{H^+}$ as equilibrium shifts left toward $\mathrm{H_2CO_3}$.
pH remains unchanged because buffers completely resist pH change until $[\mathrm{HCO_3^-}]$ is exhausted.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically the bicarbonate buffer system's response to CO₂ changes. The bicarbonate buffer system maintains pH through the equilibrium CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻, where increased CO₂ drives the equilibrium rightward. When CO₂(aq) increases by 50%, the system responds by producing more H⁺ ions, which decreases pH according to the Henderson-Hasselbalch equation: pH = pKa + log([HCO₃⁻]/[CO₂]). Choice B correctly identifies that increased [CO₂(aq)] drives H⁺ formation and lowers the [HCO₃⁻]/[CO₂] ratio, resulting in decreased pH. Choice A incorrectly suggests CO₂ consumes H⁺, when actually CO₂ produces H⁺ through carbonic acid formation. In future questions, remember that increased CO₂ always leads to decreased pH in aqueous systems due to carbonic acid formation.
To illustrate titration behavior in a biologically relevant context, a researcher titrates 25.0 mL of 0.10 M formic acid (HCOOH) with 0.10 M NaOH at 25°C. For $\mathrm{HCOOH} \rightleftharpoons \mathrm{H^+} + \mathrm{HCOO^-}$, $K_a = 1.8 \times 10^{-4}$ (so $pK_a \approx 3.74$). Assume volumes are additive.
Which statement best describes the system’s response at the half-equivalence point?
pH is 7.0 because the solution is buffered and therefore neutral.
pH is greater than 7.0 because NaOH is a strong base and dominates before equivalence.
pH is approximately equal to $\tfrac{1}{2}pK_a$ because only half the acid remains.
pH is approximately equal to $pK_a$ because $[\mathrm{HCOOH}] \approx[\mathrm{HCOO^-}]$.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically the pH at the half-equivalence point of a weak acid titration. At the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base, creating equal concentrations of the weak acid (HCOOH) and its conjugate base (HCOO⁻). According to the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]), when [A⁻] = [HA], the log term equals zero, making pH = pKa. Choice A correctly identifies that pH ≈ pKa because [HCOOH] ≈ [HCOO⁻] at the half-equivalence point. Choice D incorrectly suggests pH = ½pKa, which has no theoretical basis in acid-base chemistry. In future titration problems, remember that the half-equivalence point always occurs when pH = pKa for monoprotic weak acids.
A physiology group studies intracellular pH regulation during transient ischemia, where anaerobic metabolism increases lactic acid production. In a simplified cytosolic model buffer at 37°C, the dominant pair is $\mathrm{H_2PO_4^-}/\mathrm{HPO_4^{2-}}$ with $pK_a = 6.8$. Initially, $\mathrm{H_2PO_4^-} = 30\ \mathrm{mM}$ and $\mathrm{HPO_4^{2-}} = 15\ \mathrm{mM}$. A pulse of strong acid is introduced, consuming some $\mathrm{HPO_4^{2-}}$ and forming additional $\mathrm{H_2PO_4^-}$, with negligible volume change.
Based on the passage, which conclusion about the buffer system is most consistent?
pH increases because converting $\mathrm{HPO_4^{2-}}$ to $\mathrm{H_2PO_4^-}$ removes an acid from solution.
pH is unchanged because phosphate buffers only work near pH 7.4, not near their $pK_a$.
pH decreases because the ratio $[\mathrm{HPO_4^{2-}}]/[\mathrm{H_2PO_4^-}]$ decreases after acid addition.
pH decreases because added acid shifts the equilibrium toward insoluble phosphate salts governed by $K_{sp}$.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically how phosphate buffers respond to acid addition. The phosphate buffer system H₂PO₄⁻/HPO₄²⁻ maintains pH through the Henderson-Hasselbalch equation: pH = pKa + log([HPO₄²⁻]/[H₂PO₄⁻]). When strong acid is added, it reacts with the conjugate base HPO₄²⁻ to form more H₂PO₄⁻, decreasing the [HPO₄²⁻]/[H₂PO₄⁻] ratio. Choice B correctly identifies that pH decreases because this ratio decreases after acid addition, as predicted by the Henderson-Hasselbalch equation. Choice A incorrectly suggests that converting HPO₄²⁻ to H₂PO₄⁻ removes acid, when actually H₂PO₄⁻ is the acidic form in this conjugate pair. In future buffer problems, remember that decreasing the base-to-acid ratio always decreases pH.
A pharmacology team evaluates a weakly acidic drug (HA) formulated with its conjugate base (A$^-$) to resist pH changes after injection. At 25°C, the drug has $pK_a = 5.0$. The formulation is adjusted so that initially $\mathrm{A^-} = 10\times\mathrm{HA}$ in the syringe. After injection into a compartment where a small amount of strong base is present, some HA is converted to A$^-$ without significant dilution.
Which statement best describes the system’s response?
pH decreases because adding base shifts $\mathrm{HA \rightleftharpoons H^+ + A^-}$ to the right by Le Châtelier’s principle.
pH decreases because converting HA to A$^-$ generates $\mathrm{H^+}$ directly from the weak acid equilibrium.
pH increases slightly because base converts HA to A$^-$, increasing the already-large $[\mathrm{A^-}]/[\mathrm{HA}]$ ratio.
pH remains constant because buffers prevent any pH change regardless of added base.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically buffer response to base addition when the conjugate base already predominates. Starting with [A⁻] = 10×[HA], the buffer has pH > pKa according to Henderson-Hasselbalch: pH = 5.0 + log(10) = 6.0. When strong base is added, it converts some HA to A⁻, further increasing the already-large [A⁻]/[HA] ratio. Choice B correctly identifies that pH increases slightly because the base converts HA to A⁻, though the change is small due to the already high ratio. Choice D incorrectly invokes Le Châtelier's principle in the wrong direction - adding base consumes H⁺ and shifts equilibrium left, not right. In future buffer problems, remember that adding base always increases pH by increasing the conjugate base to weak acid ratio.
In a simplified model of renal tubular fluid, an ammonium buffer is studied at 37°C. Consider the equilibrium $\mathrm{NH_4^+} \rightleftharpoons \mathrm{H^+} + \mathrm{NH_3}$ with $pK_a = 9.25$. A sample initially contains both species. During an experiment, NH$_3$ is selectively removed by diffusion into a gas phase (simulating loss to an adjacent compartment), while $\mathrm{NH_4^+}$ remains in solution and volume is constant.
What change would most likely occur following NH$_3$ removal?
pH is unchanged because only removal of $\mathrm{H^+}$ can change pH in a buffer.
pH increases because removing NH$_3$ shifts the equilibrium left, consuming $\mathrm{H^+}$ to form more NH$_4^+$.
pH decreases because removing NH$_3$ shifts the equilibrium right to replace NH$_3$, producing more $\mathrm{H^+}$.
pH decreases because NH$_3$ is a strong acid and its removal eliminates acidic character.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically how removing one component affects equilibrium position. The ammonium equilibrium NH₄⁺ ⇌ H⁺ + NH₃ will respond to NH₃ removal according to Le Châtelier's principle. When NH₃ is selectively removed, the equilibrium shifts rightward to replace the lost NH₃, which requires NH₄⁺ to dissociate and produce more H⁺ ions. Choice A correctly identifies that pH decreases because removing NH₃ shifts equilibrium right, producing more H⁺. Choice B incorrectly suggests the equilibrium would shift left, which would occur if NH₃ were added, not removed. In future equilibrium problems, remember that removing a product always shifts equilibrium toward product formation, potentially changing pH if H⁺ is involved.
A lab investigates weak acid equilibrium relevant to gastric mucosal protection. A solution contains acetic acid (HA) at 0.10 M in water at 25°C. For $\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$, $K_a = 1.8 \times 10^{-5}$. The researchers then add sodium acetate to make $\mathrm{A^-}$ large relative to $\mathrm{HA}$ while keeping the total volume essentially constant.
What change would most likely occur following the addition of sodium acetate?
The pH decreases because added A$^-$ shifts equilibrium right, producing more $\mathrm{H^+}$.
The pH increases because added A$^-$ shifts equilibrium left, decreasing $[\mathrm{H^+}]$ (common-ion effect).
The pH is unchanged because $K_a$ fixes $[\mathrm{H^+}]$ independent of initial concentrations.
Acetic acid precipitates because added A$^-$ reduces solubility via $K_{sp}$.
Explanation
This question assesses understanding of acid-base equilibria in physiological systems (5A), specifically the common-ion effect on weak acid dissociation. The common-ion effect occurs when adding a salt containing the conjugate base of a weak acid suppresses the acid's dissociation. Initially, acetic acid establishes equilibrium: HA ⇌ H⁺ + A⁻ with Ka = 1.8 × 10⁻⁵. When sodium acetate (source of A⁻) is added, the increased [A⁻] shifts the equilibrium leftward according to Le Châtelier's principle, reducing [H⁺] and thus increasing pH. Choice B correctly identifies this as the common-ion effect, where added A⁻ shifts equilibrium left and decreases [H⁺]. Choice A incorrectly suggests added A⁻ would shift equilibrium right, which violates Le Châtelier's principle. In future problems, remember that adding conjugate base always suppresses acid dissociation and increases pH.