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  1. AP Chemistry

  3. Buffer Capacity

AP CHEMISTRY • ACIDS AND BASES

Buffer Capacity

Quantifying how much acid or base a buffer can neutralize before its pH changes significantly.

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.

1

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.
2

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.
3

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.
4

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.
✦ KEY TAKEAWAY
KEY TAKEAWAY
SECTION 3

Visual Explanation — Buffer Capacity Curve

Buffer Capacity (β) vs. pHpHβ (mol L⁻¹ pH⁻¹)02468101214β_max at pH = pKₐpKₐ ≈ 4.75 (acetic acid)effectiverangeeffectiverange← pKₐ − 1pKₐ + 1 →
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.
AP Exam Tip
SECTION 5

Factors Affecting Buffer Capacity

Comparing Buffer Capacity: Concentration vs. RatioEffect of Total Concentration(pH = pKₐ in all cases)β (mol/L/pH)0.140.25 M0.290.50 M0.581.00 MEffect of [A⁻]/[HA] Ratio(C = 1.00 M in all cases)β (mol/L/pH)0.251:100.581:10.2510:1Higher total concentration and a 1:1 ratio both maximize β
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.

Summary of factors affecting buffer capacity
FactorEffect on βExam Strategy
↑ Total concentration Cβ increases proportionally (β ∝ C)Compare moles of HA + A⁻ available in each buffer
[A⁻]/[HA] → 1β maximized (β_max = 0.576 C)Check if pH ≈ pKₐ; if yes, capacity is at its peak
[A⁻]/[HA] → 10 or 0.1β reduced to ~43% of maxBuffer near edge of effective range; less capacity on one side
Volume of bufferMore volume → more moles at same concentrationCompare moles, not just molarity, when volumes differ
SECTION 6

Worked Example

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.

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

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.

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.

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

Strengths and limitations of buffer systems
AspectStrengthLimitation
pH StabilityMaintains nearly constant pH against moderate additions of strong acid or baseCapacity is finite; once one component is consumed, pH changes abruptly
PredictabilityHenderson–Hasselbalch allows accurate pH prediction given concentrationsAssumes dilute ideal-solution behavior; fails at high ionic strength without activity corrections
TunabilityChoosing a weak acid whose pKₐ matches the target pH yields maximum capacityNo single buffer covers the full pH range; different acids needed for different ranges
Biological UseBlood bicarbonate buffer maintains pH 7.35–7.45 critical for enzyme functionTemperature changes shift Kₐ and therefore pH, requiring recalibration of buffer recipes
✦ KEY TAKEAWAY
KEY TAKEAWAY
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.

Buffer capacity at the AP level versus advanced treatments
ConceptAP-Level TreatmentAdvanced / College Treatment
Buffer CapacityQualitative: 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 EffectsAssumed negligible; ideal dilute solutions.Debye–Hückel corrections for ionic strength alter effective Kₐ and shift the optimal buffer pH.
Polyprotic BuffersTreated 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 BloodCO₂/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.

In a general or analytical chemistry course, you will encounter the formal derivative β = dCb/dpH, where Cb 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/dCb), 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

PROBLEM 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
PROBLEM 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
PROBLEM 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
PROBLEM 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.
PROBLEM 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.
SUMMARY

Buffer Capacity — Key Concepts Review

Varsity Tutors • AP Chemistry • Buffer Capacity