Buffer Capacity - AP Chemistry

SECTION 1

Historical Context & Motivation

The concept of chemical buffering arose from a practical problem: biological and industrial processes often require pH stability, yet adding even small quantities of strong acid or base to pure water causes enormous pH swings. Early physiologists noticed that blood resists pH change far more effectively than simple salt solutions, prompting systematic investigation into why certain mixtures stabilize hydrogen-ion concentration. The quantitative treatment of this resistance—what we now call buffer capacity—developed over roughly a century of work linking equilibrium theory with real-world applications in medicine, environmental science, and manufacturing.

1884 • Arrhenius Acid–Base Theory

Svante Arrhenius defined acids as H⁺ producers and bases as OH⁻ producers in water, providing the first framework to quantify hydrogen-ion concentration in solution.

1908 • Henderson's Buffer Equation

Lawrence Joseph Henderson derived the relationship between pH, pKₐ, and the conjugate acid–base ratio, giving chemists the algebraic tool to predict buffer pH.

1916 • Hasselbalch Logarithmic Form

Karl Albert Hasselbalch rewrote Henderson's equation in logarithmic form, producing the Henderson–Hasselbalch equation used universally in buffer calculations today.

1922 • Van Slyke's Buffer Index

Donald Van Slyke introduced β = dCb/dpH, the first rigorous quantitative measure of buffer capacity, enabling comparison of different buffer systems under controlled conditions.

1966 • Buffer Design in Biochemistry

Norman Good and colleagues developed a family of synthetic biological buffers (Good's buffers), optimizing buffer capacity and biocompatibility for enzymatic research.

Knowing that a buffer resists pH change is only part of the story. The deeper question—and the one the AP exam expects you to answer—is how much strong acid or base can a particular buffer absorb before it effectively fails? That question drives the entire discussion of buffer capacity.

SECTION 2

Core Principles & Definitions

Before quantifying buffer capacity, it is essential to distinguish it from the mere existence of a buffer. A buffer solution contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in appreciable concentrations. The Henderson–Hasselbalch equation tells us the pH of such a mixture, but it says nothing about how much perturbation the system can withstand. Buffer capacity fills that gap: it measures the quantitative resistance of a buffer to pH change upon addition of strong acid or base.

Buffer Capacity (β)

The number of moles of strong acid or base required to change the pH of one liter of buffer by one unit. Higher β means a more resistant buffer.

Dependence on Concentration

Doubling both [HA] and [A⁻] doubles buffer capacity because there are twice as many moles of weak acid and conjugate base available to neutralize added strong acid or base.

Dependence on Ratio

Buffer capacity is maximized when [A⁻]/[HA] = 1, i.e., pH = pKₐ. As the ratio deviates from unity, one component is depleted faster, reducing overall capacity.

Effective Buffer Range

A buffer functions effectively within approximately ±1 pH unit of its pKₐ. Outside this window the buffer capacity drops below a practical threshold.

SECTION 3

Visual Explanation — Buffer Capacity Curve

The bell-shaped curve shows how buffer capacity (β) varies with pH for an acetic acid / acetate buffer. The peak occurs at pH = pKₐ, where [HA] = [A⁻]. The shaded region marks the effective buffer range of pKₐ ± 1.

The diagram above reveals the central insight of buffer capacity: it is not a fixed property of a solution but rather a function of pH. At the peak (pH = pKₐ), the buffer contains equal moles of weak acid and conjugate base, so it can neutralize added strong acid and strong base with equal efficiency. As the pH drifts away from pKₐ, one component becomes scarce relative to the other, and the capacity drops. Beyond roughly ±1 pH unit from pKₐ, the minor component is so depleted that the buffer provides negligible resistance and effectively fails.

SECTION 4

Mathematical Framework

Two equations are central to buffer-capacity problems on the AP exam. The first is the Henderson–Hasselbalch equation, which predicts buffer pH; the second is the Van Slyke expression for buffer capacity itself.

HENDERSON–HASSELBALCH EQUATION

pH = pKₐ + log([A⁻] / [HA])

pKₐ = −log Kₐ of the weak acid; [A⁻] = concentration (or moles) of conjugate base; [HA] = concentration (or moles) of weak acid. This equation is valid when the x-is-small approximation holds, i.e., Kₐ is small relative to initial concentrations.

VAN SLYKE BUFFER CAPACITY

β = 2.303 × C × Kₐ[H⁺] / (Kₐ + [H⁺])²

β = buffer capacity (mol L⁻¹ pH⁻¹); C = total buffer concentration ([HA] + [A⁻]); Kₐ = acid dissociation constant; [H⁺] = hydrogen-ion concentration. At pH = pKₐ, [H⁺] = Kₐ, and this simplifies to β_max = 2.303 × C / 4 ≈ 0.576 C.

MAXIMUM BUFFER CAPACITY

β_max = 2.303 × C / 4 ≈ 0.576 C

This maximum occurs when pH = pKₐ. For a 1.00 M total buffer, β_max ≈ 0.576 mol L⁻¹ pH⁻¹. Practically, this means about 0.58 mol of strong acid or base per liter is needed to shift the pH by one unit.

ℹ

AP Exam Tip

SECTION 5

Factors Affecting Buffer Capacity

The left panel shows that β scales linearly with total buffer concentration C when pH = pKₐ. The right panel shows that for fixed C, β peaks at a 1:1 conjugate pair ratio and drops symmetrically as the ratio deviates from unity.

The bar charts crystallize the two independent levers that control buffer capacity. First, total buffer concentration (C) acts as a simple multiplier: doubling C doubles β because there are twice as many moles available to react with added strong acid or base. Second, the [A⁻]/[HA] ratio governs how symmetrically the buffer can respond. At a 1:1 ratio, both the acid and base reservoirs are equally stocked; at a 10:1 ratio the acid reservoir is nearly empty, so even a small addition of strong base can consume the remaining HA and crash the buffer. The AP exam frequently presents scenarios where you must identify which of two buffers has greater capacity, and the answer almost always hinges on these two factors.

Factor	Effect on β	Exam Strategy
↑ Total concentration C	β increases proportionally (β ∝ C)	Compare moles of HA + A⁻ available in each buffer
[A⁻]/[HA] → 1	β maximized (β_max = 0.576 C)	Check if pH ≈ pKₐ; if yes, capacity is at its peak
[A⁻]/[HA] → 10 or 0.1	β reduced to ~43% of max	Buffer near edge of effective range; less capacity on one side
Volume of buffer	More volume → more moles at same concentration	Compare moles, not just molarity, when volumes differ

Summary of factors affecting buffer capacity

Two colorless stirred samples, separate pH probes and matched dispensing pipettes accompany exact inventory bars and calculated pH changes. Compare two 100 mL acetic-acid buffers with pKa = 4.76 and a tenfold difference in total concentration. Adding 0.50 mmol HCl changes their pH from about 4.76 to 4.72 and 4.29, respectively, when the added volume is negligible.

Equal pH is not equal reserve. Compare two 100 mL acetic-acid buffers with pKa = 4.76 and a tenfold difference in total concentration. Adding 0.50 mmol HCl changes their pH from about 4.76 to 4.72 and 4.29, respectively, when the added volume is negligible. Compare two 100 mL acetic-acid buffers with pKa = 4.76 and a tenfold difference in total concentration. Adding 0.50 mmol HCl changes their pH from about 4.76 to 4.72 and 4.29, respectively, when the added volume is negligible.

SECTION 6

Worked Example

Worked example

1
Step 1 — Identify the Two Buffers

Buffer A: 500.0 mL of 0.200 M CH₃COOH and 0.200 M CH₃COO⁻ (total C = 0.400 M). Buffer B: 500.0 mL of 0.100 M CH₃COOH and 0.300 M CH₃COO⁻ (total C = 0.400 M). Both use acetic acid, pKₐ = 4.74.

Both buffers have the same total concentration (0.400 M) and same volume.
2
Step 2 — Calculate pH of Each Buffer

Using Henderson–Hasselbalch: Buffer A: pH = 4.74 + log(0.200/0.200) = 4.74 + 0 = 4.74. Buffer B: pH = 4.74 + log(0.300/0.100) = 4.74 + log(3.00) = 4.74 + 0.477 = 5.22.

pH_A = 4.74; pH_B = 5.22
3
Step 3 — Determine Which Has Greater Buffer Capacity

Since C is identical for both, the deciding factor is the [A⁻]/[HA] ratio. Buffer A has a ratio of 1:1 (pH = pKₐ), placing it at the peak of the capacity curve. Buffer B has a ratio of 3:1, which is 0.48 pH units from pKₐ—still within the effective range but below maximum.

Buffer A has greater overall buffer capacity because its ratio is 1:1.
4
Step 4 — Quantify: How Many Moles of NaOH Can Each Buffer Absorb Before pH Exceeds 5.74?

When NaOH is added it converts HA → A⁻. Let n = moles NaOH added. Initial moles in 0.500 L: Buffer A has 0.100 mol HA and 0.100 mol A⁻; Buffer B has 0.050 mol HA and 0.150 mol A⁻. At pH = 5.74: [A⁻]/[HA] = $10^{5.74−4.74}$ = 10. After adding n mol NaOH: Buffer A → (0.100 + n)/(0.100 − n) = 10, solving gives n = 0.0818 mol. Buffer B → (0.150 + n)/(0.050 − n) = 10, solving gives n = 0.0364 mol.

Buffer A absorbs 0.082 mol NaOH vs. Buffer B's 0.036 mol NaOH before reaching pH 5.74—more than double.
5
Step 5 — Interpret the Result

Although both buffers have the same total concentration and volume, the 1:1 ratio in Buffer A gives it nearly symmetric capacity against both acid and base. Buffer B, with a 3:1 excess of A⁻, has relatively little HA left; strong base depletes the small HA reservoir quickly. This example illustrates why buffer capacity depends on both C and the ratio, and why pH = pKₐ is optimal.

SECTION 7

Strengths & Limitations of Buffers

Aspect	Strength	Limitation
pH Stability	Maintains nearly constant pH against moderate additions of strong acid or base	Capacity is finite; once one component is consumed, pH changes abruptly
Predictability	Henderson–Hasselbalch allows accurate pH prediction given concentrations	Assumes dilute ideal-solution behavior; fails at high ionic strength without activity corrections
Tunability	Choosing a weak acid whose pKₐ matches the target pH yields maximum capacity	No single buffer covers the full pH range; different acids needed for different ranges
Biological Use	Blood bicarbonate buffer maintains pH 7.35–7.45 critical for enzyme function	Temperature changes shift Kₐ and therefore pH, requiring recalibration of buffer recipes

Strengths and limitations of buffer systems

SECTION 8

Connection to Titration Curves & Advanced Theory

Buffer capacity connects directly to the shape of a titration curve. The flat, gently sloping region near the half-equivalence point on a weak-acid/strong-base titration is precisely where buffer capacity is high. The steep rise near the equivalence point corresponds to a region of near-zero buffer capacity, because essentially all of one component has been consumed. Understanding this link allows you to read a titration curve and instantly assess buffer capacity at any point.

Concept	AP-Level Treatment	Advanced / College Treatment
Buffer Capacity	Qualitative: higher C and ratio closer to 1 → greater capacity. Moles-based stoichiometric reasoning.	Quantitative: Van Slyke equation β = dCb/dpH; integration to compute exact moles tolerable for a given ΔpH.
Activity Effects	Assumed negligible; ideal dilute solutions.	Debye–Hückel corrections for ionic strength alter effective Kₐ and shift the optimal buffer pH.
Polyprotic Buffers	Treated as separate equilibria; pick the pKₐ closest to target pH.	Overlapping equilibria contribute additive buffer capacity, modeled as sum of individual β curves.
Buffer in Blood	CO₂/HCO₃⁻ system with respiratory regulation.	Open system: CO₂ partial pressure is regulated by lungs, creating a non-equilibrium but steady-state buffer with anomalously high capacity.

Buffer capacity at the AP level versus advanced treatments

In a general or analytical chemistry course, you will encounter the formal derivative β = dC_b/dpH, where C_b is the concentration of strong base added. This derivative form makes clear that buffer capacity is the slope of the titration curve inverted: where the titration curve is flat (small dpH/dC_b), buffer capacity is large, and vice versa. This elegant connection unifies two major AP topics—buffers and titrations—into a single mathematical picture.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student prepares two buffers using the same weak acid (pKₐ = 6.00). Buffer X has pH = 6.00 and total concentration C = 0.10 M. Buffer Y has pH = 7.00 and total concentration C = 0.20 M. Which buffer has the greater capacity, and why? A. Buffer X, because its pH equals pKₐ, maximizing the ratio factor enough to outweigh the lower concentration B. Buffer Y, because its higher total concentration outweighs the non-ideal ratio C. They have equal capacity because the effects cancel D. Buffer X, because pH 7.00 is outside the effective buffer range

A. Buffer X, because its pH equals pKₐ, maximizing the ratio factor enough to outweigh the lower concentration
B. Buffer Y, because its higher total concentration outweighs the non-ideal ratio
C. They have equal capacity because the effects cancel
D. Buffer X, because pH 7.00 is outside the effective buffer range

Show answer (A)

Correct: A. Buffer capacity depends on both total concentration and how close the pH is to pKₐ. The Van Slyke equation gives β = 2.303 × C × Kₐ[H⁺] / (Kₐ + [H⁺])². For Buffer X (pH = pKₐ = 6.00, C = 0.10 M): the ratio [A⁻]/[HA] = 1:1, which maximizes the ratio factor Kₐ[H⁺]/(Kₐ + [H⁺])² = 0.25. So β_X = 2.303 × 0.10 × 0.25 ≈ 0.0576. For Buffer Y (pH = 7.00, pKₐ = 6.00, C = 0.20 M): [A⁻]/[HA] = 10:1, so [H⁺] = 10⁻⁷ and Kₐ = 10⁻⁶. The ratio factor = (10⁻⁶)(10⁻⁷)/(10⁻⁶ + 10⁻⁷)² = 10⁻¹³/(1.1 × 10⁻⁶)² = 10⁻¹³/1.21 × 10⁻¹² ≈ 0.0826. So β_Y = 2.303 × 0.20 × 0.0826 ≈ 0.0381. Since β_X ≈ 0.058 > β_Y ≈ 0.038, Buffer X has the greater capacity. The 1:1 ratio at pH = pKₐ maximizes the ratio factor at 0.25 (its theoretical maximum), and this advantage more than outweighs Buffer Y's twofold concentration advantage. B is wrong because although Buffer Y has double the total concentration, the ratio factor drops to about 33% of its maximum value at pH = pKₐ + 1, which is a larger penalty than the 2× concentration boost can overcome. C is wrong because the two effects do not perfectly cancel; the ratio penalty in Buffer Y is more severe. D is wrong because pH 7.00 is at the edge of the commonly cited effective range (pKₐ ± 1) and the buffer still functions, just with reduced capacity.

Question 2: BASIC CALCULATION

A buffer contains 0.300 mol CH₃COOH and 0.300 mol CH₃COONa in 1.00 L of solution. The pKₐ of acetic acid is 4.74. If 0.050 mol of NaOH is added (assume no volume change), what is the new pH? A. 4.58 B. 4.74 C. 4.89 D. 5.22

A. 4.58
B. 4.74
C. 4.89
D. 5.22

Show answer (C)

Correct: C. Adding 0.050 mol NaOH converts 0.050 mol CH₃COOH → CH₃COO⁻. New moles: HA = 0.300 − 0.050 = 0.250; A⁻ = 0.300 + 0.050 = 0.350. pH = 4.74 + log(0.350/0.250) = 4.74 + log(1.40) = 4.74 + 0.146 = 4.89. A (4.58) would result from mistakenly subtracting from A⁻ and adding to HA (confusing acid with base addition). B (4.74) ignores the NaOH addition entirely. D (5.22) would require a ratio of 3:1, which doesn't match the stoichiometry here.

Question 3: INTERMEDIATE

A researcher needs a buffer at pH 9.25. She has access to NH₃ (

K_b

= 1.8 × 10⁻⁵) and NH₄Cl. She prepares 1.00 L of buffer with 0.400 mol NH₃ and 0.400 mol NH₄Cl. How many moles of HCl can be added before the buffer pH drops below 8.25? A. 0.036 mol B. 0.182 mol C. 0.327 mol D. 0.400 mol

A. 0.036 mol
B. 0.182 mol
C. 0.327 mol
D. 0.400 mol

Show answer (C)

Correct: C. First, find pKₐ of NH₄⁺: pK_b = −log(1.8 × 10⁻⁵) = 4.74, so pKₐ = 14.00 − 4.74 = 9.26 ≈ 9.25. The buffer starts at pH ≈ pKₐ = 9.25 (1:1 ratio of NH₃ to NH₄⁺). Adding HCl converts NH₃ → NH₄⁺. We need to find how many moles of HCl can be added before the pH drops to 8.25. At pH = 8.25: pH = pKₐ + log([NH₃]/[NH₄⁺]) → 8.25 = 9.25 + log([NH₃]/[NH₄⁺]) → log([NH₃]/[NH₄⁺]) = −1.00 → [NH₃]/[NH₄⁺] = 0.10 = 1/10. Let n = mol HCl added. HCl reacts with NH₃: NH₃ + HCl → NH₄⁺ + Cl⁻. After reaction: mol NH₃ = 0.400 − n, mol NH₄⁺ = 0.400 + n. Setting up the ratio: (0.400 − n)/(0.400 + n) = 1/10. Cross-multiplying: 10(0.400 − n) = 0.400 + n → 4.000 − 10n = 0.400 + n → 3.600 = 11n → n = 0.327 mol. A (0.036 mol) is far too small and would barely shift the pH. B (0.182 mol) corresponds to a ratio of about 1:2.7, giving pH ≈ 8.82, which is still above 8.25. D (0.400 mol) would consume all the NH₃, completely destroying the buffer.

Question 4: APPLIED

A biochemist is studying an enzyme that is active only between pH 6.8 and pH 7.4. She needs to choose a buffer system from the following options and prepare 500.0 mL of 0.250 M total buffer at the optimal pH for maximum capacity. (a) Which buffer system should she choose and why? System I: H₂PO₄⁻/HPO₄²⁻ (pKₐ₂ = 7.20) System II: CH₃COOH/CH₃COO⁻ (pKₐ = 4.74) System III: NH₄⁺/NH₃ (pKₐ = 9.25) (b) Calculate the masses of NaH₂PO₄ (MW = 120.0 g/mol) and Na₂HPO₄ (MW = 142.0 g/mol) needed. (c) The enzyme reaction produces 0.0080 mol H⁺. Will the buffer remain within the active range? Show calculations. (d) If the biochemist had used System II instead at the same total concentration and volume, explain qualitatively why it would be a poor choice for this application.

Show answer

(a) System I (H₂PO₄⁻/HPO₄²⁻) is optimal. Its pKₐ₂ = 7.20 falls within the required 6.8–7.4 range, so the buffer capacity peaks right in the enzyme's active window. System II (pKₐ = 4.74) and System III (pKₐ = 9.25) are both more than 2 pH units away from 7.1, meaning virtually no buffer capacity at the target pH. [1 pt for correct choice, 1 pt for pKₐ justification] (b) At pH = pKₐ = 7.20 for maximum capacity, the ratio [HPO₄²⁻]/[H₂PO₄⁻] = 1. Total buffer = 0.250 M × 0.500 L = 0.125 mol. Each component = 0.0625 mol. Mass NaH₂PO₄ = 0.0625 × 120.0 = 7.50 g. Mass Na₂HPO₄ = 0.0625 × 142.0 = 8.88 g. [1 pt for correct moles, 1 pt for correct masses] (c) Adding 0.0080 mol H⁺ converts HPO₄²⁻ → H₂PO₄⁻. New moles: H₂PO₄⁻ = 0.0625 + 0.0080 = 0.0705; HPO₄²⁻ = 0.0625 − 0.0080 = 0.0545. pH = 7.20 + log(0.0545/0.0705) = 7.20 + log(0.773) = 7.20 − 0.112 = 7.09. Since 6.8 < 7.09 < 7.4, the buffer remains within the active range. [1 pt for correct stoichiometry, 1 pt for final pH and comparison to range] (d) System II at pH 7.1 would have [A⁻]/[HA] =

10^{7.1−4.74}

10^{2.36}

≈ 229. Almost all of the buffer would be in the conjugate base form (CH₃COO⁻), leaving negligible CH₃COOH. The buffer would have essentially no capacity to neutralize additional base, and even small amounts of acid would quickly consume the tiny HA reservoir. The buffer capacity at a pH more than 2 units from pKₐ is vanishingly small. [1 pt for explaining ratio is extreme, 1 pt for connecting to negligible capacity]

Question 5: CRITICAL THINKING

A student collects the following data during a titration of 50.00 mL of 0.200 M HA (an unknown monoprotic weak acid) with 0.200 M NaOH. | Volume NaOH (mL) | pH | |-------------------|-------| | 0.00 | 2.87 | | 10.00 | 4.15 | | 25.00 | 4.74 | | 40.00 | 5.35 | | 50.00 | 8.72 | | 60.00 | 12.30 | (a) Determine the pKₐ of the unknown acid. Justify your reasoning. (b) At which data point is the buffer capacity the greatest? Explain using buffer capacity principles. (c) Calculate the buffer capacity in terms of moles of NaOH per pH unit between the 10.00 mL and 25.00 mL data points. (d) The student claims that at 40.00 mL NaOH added, the solution has 'almost no buffer capacity left.' Evaluate this claim using the data and buffer capacity concepts.

Show answer

(a) pKₐ = 4.74. At the half-equivalence point (25.00 mL of 0.200 M NaOH added to 50.00 mL of 0.200 M HA), exactly half the acid has been neutralized, so [HA] = [A⁻] and pH = pKₐ. Since the equivalence point is at 50.00 mL, the half-equivalence point is at 25.00 mL, which gives pH = 4.74. [1 pt for pKₐ value, 1 pt for half-equivalence reasoning] (b) Buffer capacity is greatest at 25.00 mL NaOH added (pH = 4.74 = pKₐ). At this point [HA] = [A⁻], which gives the 1:1 ratio that maximizes the buffer capacity function β. The capacity at this point is β_max = 2.303C/4 where C is the total buffer concentration. [1 pt] (c) Between 10.00 mL and 25.00 mL: ΔpH = 4.74 − 4.15 = 0.59. Moles NaOH added = (25.00 − 10.00) × 10⁻³ L × 0.200 mol/L = 0.00300 mol. Average β = Δn/ΔpH = 0.00300/0.59 = 0.0051 mol/pH unit. Note: this is a total-moles capacity, not per-liter. Per liter (using average volume ≈ 67.5 mL): β ≈ 0.0051/0.0675 = 0.075 mol L⁻¹ pH⁻¹. [1 pt for correct Δn, 1 pt for correct ratio] (d) The student's claim is partially correct but overstated. At 40.00 mL, 0.0080 mol NaOH has been added to 0.0100 mol HA, leaving 0.0020 mol HA and 0.0080 mol A⁻. The ratio [A⁻]/[HA] = 4, so pH = 4.74 + log(4) = 5.35, consistent with the data. This ratio of 4:1 means the buffer is still within the effective range (pKₐ ± 1 = 3.74 to 5.74), so it does retain some capacity. However, the HA reservoir is small (only 0.0020 mol), so the buffer has very little capacity to neutralize additional base—only 0.0020 mol more NaOH would reach equivalence. The buffer retains moderate capacity against added acid but minimal capacity against added base. The claim should be refined: the buffer has asymmetric capacity, with very little remaining capacity on the base-addition side. [1 pt for recognizing asymmetry, 1 pt for quantitative support]