General Chemistry 2 • Acids and Bases

Buffer Components & Behavior — Identify buffer components and predict buffer behavior

Understanding how conjugate acid–base pairs resist pH changes in chemical and biological systems.

An atom with laboratory flask and electron orbit for General Chemistry 2
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The concept of buffering arose from a practical frustration: chemists and physiologists in the late nineteenth century repeatedly observed that certain biological fluids and carefully prepared laboratory solutions stubbornly resisted changes in acidity, even when small amounts of strong acid or base were introduced. Blood, for instance, maintained a remarkably stable pH near 7.4 despite the continuous metabolic production of carbon dioxide and lactic acid—a phenomenon that purely aqueous solutions could never replicate. The quest to explain and harness this pH-stabilizing behavior drove decades of research linking equilibrium chemistry to real-world applications in medicine, biochemistry, and industrial processes.

1884 • Arrhenius Acid–Base Theory

Svante Arrhenius proposed that acids produce H⁺ ions and bases produce OH⁻ ions in water, providing the first quantitative framework for understanding acid–base chemistry and laying the groundwork for pH-related phenomena.

1909 • Sørensen Introduces pH

Søren Sørensen, working at the Carlsberg Laboratory in Copenhagen, defined the pH scale as −log[H⁺], giving scientists a convenient logarithmic metric to quantify acidity and compare the behavior of different solutions.

1916 • Henderson–Hasselbalch Equation

Lawrence Henderson and Karl Hasselbalch independently derived the relationship pH = pKₐ + log([A⁻]/[HA]), providing a practical tool to calculate the pH of buffer solutions and predict their response to added acid or base.

1923 • Brønsted–Lowry Theory

Johannes Brønsted and Thomas Lowry generalized acid–base definitions to proton donors and acceptors, respectively, expanding buffer theory beyond aqueous solutions and enabling the analysis of conjugate acid–base pairs in diverse solvents.

1966 • Good's Buffers for Biochemistry

Norman Good and colleagues published a set of zwitterionic buffers specifically designed for biological research, optimizing pKₐ values near physiological pH and minimizing interference with enzymes—a milestone in applied buffer chemistry.

The central question that emerges from this historical trajectory is deceptively simple: what specific chemical components must be present in a solution to give it buffering capacity, and how can we predict—quantitatively—the extent to which a buffer will resist pH changes? Answering this question requires a firm understanding of weak acid–base equilibria, conjugate pairs, and the interplay between Le Chatelier's principle and the Henderson–Hasselbalch equation.

SECTION 2

Core Principles & Definitions

A buffer solution is an aqueous system that resists significant changes in pH upon the addition of moderate amounts of strong acid or strong base. Buffers achieve this resistance not through inertness but through an active chemical equilibrium involving two complementary species—a conjugate acid–base pair. When H⁺ is added, the basic component of the pair neutralizes it; when OH⁻ is added, the acidic component neutralizes it. The result is a dynamic equilibrium that shifts in response to perturbation, as predicted by Le Chatelier's principle, thereby stabilizing the hydrogen ion concentration.

1

Weak Acid + Conjugate Base Buffer

Composed of a weak acid (HA) and its conjugate base (A⁻), typically supplied as a soluble salt. Example: acetic acid (CH₃COOH) with sodium acetate (CH₃COONa). The weak acid neutralizes added OH⁻, while the conjugate base neutralizes added H⁺.
2

Weak Base + Conjugate Acid Buffer

Composed of a weak base (B) and its conjugate acid (BH⁺), again supplied as a salt. Example: ammonia (NH₃) with ammonium chloride (NH₄Cl). The weak base neutralizes added H⁺, while the conjugate acid neutralizes added OH⁻.
3

Buffer Capacity

The amount of strong acid or base a buffer can absorb before its pH changes significantly (usually defined as > 1 pH unit). Buffer capacity depends on the total concentration of the conjugate pair and is maximized when [HA] ≈ [A⁻].
4

Effective Buffer Range

A buffer is most effective when pH lies within approximately ±1 unit of the pKₐ of the weak acid (or pKb of the weak base). Outside this range, one component is depleted relative to the other, and buffering capacity drops sharply.
5

Identifying Non-Buffers

A solution of a strong acid and its conjugate base (e.g., HCl/Cl⁻) is NOT a buffer because the conjugate base of a strong acid has negligible basicity. Similarly, mixing a strong base with its conjugate acid does not produce a buffer.
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KEY TAKEAWAY

Think of a buffer as a chemical shock absorber in a car's suspension system. Just as the shock absorber contains a piston (which pushes fluid one way) and a spring (which pushes back the other way), a buffer contains two reactive partners—one that absorbs protons and one that donates protons. Together, they smooth out the 'bumps' (additions of acid or base) that would otherwise cause dramatic pH swings. Remove either partner, and the system loses its ability to absorb the impact.

SECTION 3

Visual Explanation — How Buffers Respond

The diagram below illustrates the two key scenarios a buffer faces: the addition of a strong acid and the addition of a strong base. In each case, one member of the conjugate pair is consumed while the other is produced, keeping [H⁺] relatively constant. The equilibrium shifts according to Le Chatelier's principle, and the pH changes only modestly as long as neither component is fully consumed.

Buffer Response Mechanism: Weak Acid (HA) / Conjugate Base (A⁻) SystemSCENARIO 1 — Strong Acid (H⁺) AddedH⁺added acidreacts withA⁻producesHANet: A⁻ + H⁺ → HA[A⁻] decreases, [HA] increases✓ pH decreases only slightlySCENARIO 2 — Strong Base (OH⁻) AddedOH⁻added basereacts withHAproducesA⁻+H₂ONet: HA + OH⁻ → A⁻ + H₂O[HA] decreases, [A⁻] increases✓ pH increases only slightlyBoth components must be present in appreciable concentration for buffering to occur.
In Scenario 1, the conjugate base A⁻ (cyan) neutralizes added H⁺, converting it to HA (amber). In Scenario 2, the weak acid HA neutralizes added OH⁻, converting it to A⁻ and water. In both cases, the ratio [A⁻]/[HA] shifts modestly rather than the overall [H⁺] changing dramatically.

Notice the symmetry: each perturbation is absorbed by the component whose chemistry directly opposes the change. The conjugate base A⁻ acts as a proton sponge when acid is added, while the weak acid HA donates protons to neutralize hydroxide when base is added. This dual-response mechanism is what distinguishes a true buffer from a simple solution of a weak acid or a simple solution of a salt.

SECTION 4

Mathematical Framework

The quantitative behavior of buffers is governed by the equilibrium expression for the weak acid dissociation and its logarithmic rearrangement. Below are the essential equations, presented in order of derivation so that the mathematical logic is transparent.

WEAK ACID EQUILIBRIUM

Ka=[H+][A−]/[HA]Kₐ = [H⁺][A⁻] / [HA]

Kₐ = acid dissociation constant; [H⁺] = hydrogen ion concentration; [A⁻] = conjugate base concentration; [HA] = undissociated weak acid concentration. All concentrations are equilibrium values.

HENDERSON–HASSELBALCH EQUATION

pH=pKa+log([A−]/[HA])pH = pKₐ + log([A⁻] / [HA])

Derived by taking −log of both sides of the Kₐ expression. pKₐ = −log(Kₐ). When [A⁻] = [HA], the log term equals zero and pH = pKₐ. This is the point of maximum buffer capacity.

ANALOGOUS FORM FOR BASIC BUFFERS

pOH=pKb+log([BH+]/[B])pOH = pK_b + log([BH⁺] / [B])

For a weak base B and its conjugate acid BH⁺. pKb = −log(Kb). Convert to pH via pH = 14.00 − pOH (at 25 °C). Alternatively, use pH = pKₐ(conjugate acid) + log([B]/[BH⁺]).

BUFFER CAPACITY (β)

β=ΔCb/ΔpHβ = ΔC_b / ΔpH

β = buffer capacity (mol·L⁻¹·pH⁻¹); ΔCb = moles of strong base (or acid) added per liter; ΔpH = resulting change in pH. Higher β means greater resistance. β is maximized when [HA] = [A⁻] and when total buffer concentration is large.

⚠

Important Assumptions

The Henderson–Hasselbalch equation assumes: (1) the weak acid or base has a well-defined Kₐ or Kb (i.e., it is not strong); (2) the solution is sufficiently dilute that activity coefficients are approximately 1; and (3) the concentrations of HA and A⁻ are large enough relative to the amount of acid or base added that the ratio [A⁻]/[HA] does not change drastically. Violating assumption (3) means the buffer has been overwhelmed.

SECTION 5

Effective Buffer Range & Titration Curves

A buffer's effectiveness is not unlimited. As the ratio [A⁻]/[HA] departs from unity, the log term in the Henderson–Hasselbalch equation grows in magnitude, and the solution's ability to resist further change diminishes. The conventional effective buffer range is pKₐ ± 1, corresponding to [A⁻]/[HA] ratios between 0.1 and 10. Outside this window, one component is present at less than 9% of the total, and the solution behaves essentially as either a dilute weak acid or a dilute salt of the conjugate base. The titration curve of a weak acid visually reveals the buffer zone as the relatively flat region surrounding the half-equivalence point, where exactly half the acid has been converted to its conjugate base.

Titration Curve of a Weak Acid (HA) with Strong Base (NaOH)pHVolume of NaOH Added (mL)2468101214012.52537.550BUFFER ZONE (pKₐ ± 1)Half-equiv. pt.pH = pKₐEquivalence pt.The flat region around the half-equivalence point is where buffering is most effective.
The pink curve shows pH versus volume of NaOH added during titration of a weak acid. The amber-shaded buffer zone (pKₐ ± 1) is the region of minimal slope where pH changes only slightly per unit volume of titrant. At the half-equivalence point (cyan dot), exactly half the HA has been converted to A⁻, so [HA] = [A⁻] and pH = pKₐ. The equivalence point (green dot) marks the complete conversion of HA to A⁻.
Ratio [A⁻]/[HA]log([A⁻]/[HA])pH relative to pKₐBuffer Quality
0.1−1.00pKₐ − 1.00At lower limit of effective range
0.5−0.30pKₐ − 0.30Good
1.00.00pKₐMaximum capacity
2.0+0.30pKₐ + 0.30Good
10+1.00pKₐ + 1.00At upper limit of effective range

Relationship between conjugate pair ratio and buffer effectiveness

SECTION 6

Worked Example — Designing and Evaluating an Acetate Buffer

Consider the following problem: You prepare a buffer by dissolving 0.400 mol of acetic acid (CH₃COOH) and 0.300 mol of sodium acetate (CH₃COONa) in enough water to make 1.00 L of solution. The Kₐ of acetic acid is 1.8 × 10⁻⁵ (pKₐ = 4.74). (a) Calculate the initial pH. (b) Calculate the pH after adding 0.050 mol of NaOH to this buffer.

Part (a): Initial pH of the Acetate Buffer

  1. 1

    Step 1 — Identify the Buffer Components

    The weak acid is CH₃COOH (HA) with an initial concentration of 0.400 M. The conjugate base is CH₃COO⁻ (A⁻), supplied by sodium acetate, with an initial concentration of 0.300 M. Since both a weak acid and its conjugate base are present in appreciable amounts, this is a valid buffer system.
  2. 2

    Step 2 — Apply the Henderson–Hasselbalch Equation

    pH = pKₐ + log([A⁻]/[HA]) = 4.74 + log(0.300/0.400). The ratio is 0.300/0.400 = 0.750.
  3. 3

    Step 3 — Evaluate the Logarithm

    log(0.750) = −0.125. Therefore, pH = 4.74 + (−0.125) = 4.74 − 0.125.
    pH = 4.62

Part (b): pH After Adding 0.050 mol NaOH

  1. 1

    Step 1 — Write the Neutralization Reaction

    OH⁻ reacts with the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The 0.050 mol of NaOH will convert 0.050 mol of HA into A⁻.
  2. 2

    Step 2 — Update Moles After Reaction

    Moles of HA remaining: 0.400 − 0.050 = 0.350 mol. Moles of A⁻ after reaction: 0.300 + 0.050 = 0.350 mol. Since the total volume is still 1.00 L, [HA] = 0.350 M and [A⁻] = 0.350 M.
  3. 3

    Step 3 — Apply Henderson–Hasselbalch Again

    pH = 4.74 + log(0.350/0.350) = 4.74 + log(1.00) = 4.74 + 0.
    pH = 4.74
  4. 4

    Step 4 — Interpret the Result

    The pH changed from 4.62 to 4.74, an increase of only 0.12 pH units despite the addition of 0.050 mol of strong base. Without the buffer, adding 0.050 mol NaOH to 1.00 L of pure water would have raised the pH from 7.00 to 12.70—a change of 5.70 units. The buffer reduced the pH change by a factor of nearly 50.
    ΔpH = 0.12 (buffered) vs. ΔpH ≈ 5.70 (unbuffered)

SECTION 7

Strengths, Limitations, and Common Buffer Systems

Choosing an appropriate buffer for a particular application requires matching the pKₐ of the weak acid to the desired pH, ensuring sufficient concentration for the expected perturbations, and avoiding chemical interference with the system under study. The table below compares several commonly encountered buffer systems across chemistry and biochemistry.

Buffer SystempKₐ (25 °C)Effective RangeCommon Use
Phosphoric acid / dihydrogen phosphate (H₃PO₄ / H₂PO₄⁻)2.151.1 – 3.1Cola beverages, acidic lab buffers
Acetic acid / acetate (CH₃COOH / CH₃COO⁻)4.763.8 – 5.8Organic synthesis, food preservation
Dihydrogen phosphate / hydrogen phosphate (H₂PO₄⁻ / HPO₄²⁻)7.206.2 – 8.2Biological systems, PBS buffer
Tris base / Tris-HCl (HOCH₂)₃CNH₂ / (HOCH₂)₃CNH₃⁺8.077.1 – 9.1Molecular biology (gel electrophoresis, PCR)
Ammonium / ammonia (NH₄⁺ / NH₃)9.258.3 – 10.3Basic pH buffers, analytical chemistry
Bicarbonate / carbonate (HCO₃⁻ / CO₃²⁻)10.339.3 – 11.3Alkaline cleaning solutions

Selected buffer systems and their properties at 25 °C

It is worth noting that the phosphate system appears twice because phosphoric acid is triprotic—each successive deprotonation has its own pKₐ. The carbonic acid / bicarbonate buffer (pKₐ₁ = 6.35) is the principal blood buffer, even though its pKₐ is more than a full unit below physiological pH of 7.4. Its effectiveness in vivo is sustained by the body's ability to regulate CO₂ partial pressure through respiration, an open-system phenomenon that shifts the equilibrium and effectively extends the buffer range.

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DESIGN PRINCIPLE

Selecting a buffer is analogous to selecting a thermostat for a building: you choose a thermostat whose set point (pKₐ) is close to your target temperature (desired pH), and you install a system with enough thermal mass (buffer concentration) to handle the expected heat loads (acid or base additions). A thermostat set for 72 °F does a poor job stabilizing 95 °F, just as an acetate buffer (pKₐ = 4.76) cannot maintain pH 7.4.

SECTION 8

Connection to Advanced Theory — Polyprotic Systems & Biological Buffering

The simple one-component buffer model extends naturally to more complex systems encountered in biochemistry and analytical chemistry. Polyprotic acids like phosphoric acid (H₃PO₄) and citric acid provide multiple buffer zones, each centered on a different pKₐ. In living organisms, buffering is maintained by an intricate interplay of the bicarbonate buffer system, the phosphate buffer system, and protein buffering via histidine residues whose imidazole side chains have a pKₐ near 6.0. The concept of buffer capacity becomes critical in clinical medicine: conditions like metabolic acidosis and respiratory alkalosis are understood as failures or compensatory adjustments of the body's buffering network.

FeatureSimple Buffer (Gen Chem)Biological Buffering (Biochem/Physiology)
ComponentsOne conjugate acid–base pairMultiple overlapping pairs (bicarbonate, phosphate, protein)
System typeClosed (constant composition)Open (CO₂ exchanged via lungs; kidneys regulate [HCO₃⁻])
Capacity regulationFixed by initial concentrationsDynamically adjusted by respiratory rate and renal excretion
Mathematical modelHenderson–Hasselbalch (one equation)Coupled equilibria, Stewart approach, alpha plots
Failure modeOverwhelmed when one component is depletedAcidosis/alkalosis when compensatory mechanisms fail

Comparison of simple and biological buffering systems

Looking ahead, courses in biochemistry and physiology will deepen your understanding of how these systems cooperate. The alpha (α) diagram, which plots the mole fraction of each species as a function of pH for a polyprotic acid, is a powerful visual tool for predicting which species dominate at any given pH and for designing multi-component buffer cocktails. For now, the essential takeaway is that every buffer system—no matter how complex—rests on the same foundational principle you have learned here: a conjugate acid–base pair in equilibrium whose ratio determines the pH.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student prepares two solutions: (A) 0.50 M HCl mixed with 0.50 M NaCl, and (B) 0.50 M CH₃COOH mixed with 0.50 M CH₃COONa. Which solution functions as a buffer? Explain your reasoning by identifying which components can respond to added acid and base.
Show answer
Solution B is the buffer. CH₃COOH is a weak acid and CH₃COO⁻ (from CH₃COONa) is its conjugate base, forming a legitimate conjugate pair. CH₃COO⁻ can neutralize added H⁺, and CH₃COOH can neutralize added OH⁻. Solution A is not a buffer because HCl is a strong acid that dissociates completely—Cl⁻ is the conjugate base of a strong acid and has no meaningful tendency to accept protons. Therefore, there is no equilibrium to shift, and the solution cannot resist pH changes upon addition of strong acid or base.

Question 2: BASIC CALCULATION

Calculate the pH of a buffer made by mixing 0.250 mol of NH₃ (Kb = 1.8 × 10⁻⁵) and 0.150 mol of NH₄Cl in 1.00 L of solution.
Show answer
First, convert to the acid form of Henderson–Hasselbalch. The conjugate acid is NH₄⁺ with pKₐ = 14.00 − pKb = 14.00 − 4.74 = 9.26. Here, NH₃ acts as the base (A⁻ analog) and NH₄⁺ acts as the acid (HA analog). Apply pH = pKₐ + log([NH₃]/[NH₄⁺]) = 9.26 + log(0.250/0.150) = 9.26 + log(1.667) = 9.26 + 0.222 = 9.48. The pH is 9.48, which falls within the effective range of 8.26 to 10.26 (pKₐ ± 1), confirming this is a well-functioning buffer.

Question 3: INTERMEDIATE

A 500.0 mL buffer contains 0.200 M formic acid (HCOOH, pKₐ = 3.75) and 0.200 M sodium formate (HCOONa). Calculate the pH after 10.0 mL of 1.00 M HCl is added. Assume the volume change is negligible.
Show answer
Moles of HCOOH initially = 0.200 M × 0.500 L = 0.100 mol. Moles of HCOO⁻ initially = 0.200 M × 0.500 L = 0.100 mol. Moles of HCl added = 0.0100 L × 1.00 M = 0.0100 mol. HCl reacts with the conjugate base: HCOO⁻ + H⁺ → HCOOH. After reaction: moles HCOO⁻ = 0.100 − 0.0100 = 0.090 mol; moles HCOOH = 0.100 + 0.0100 = 0.110 mol. Since volume is the same for both, we can use moles directly in the ratio. pH = 3.75 + log(0.090/0.110) = 3.75 + log(0.818) = 3.75 + (−0.087) = 3.66. The pH decreased by only 0.09 units from the initial value of 3.75, demonstrating effective buffering.

Question 4: APPLIED

A biochemist needs to prepare 2.00 L of a phosphate buffer at pH 7.40 using NaH₂PO₄ (monobasic, providing H₂PO₄⁻) and Na₂HPO₄ (dibasic, providing HPO₄²⁻). The second dissociation pKₐ of phosphoric acid is 7.20 and the total phosphate concentration must be 0.100 M. Calculate the required concentrations of each salt.
Show answer
Using Henderson–Hasselbalch: 7.40 = 7.20 + log([HPO₄²⁻]/[H₂PO₄⁻]). Solving: log([HPO₄²⁻]/[H₂PO₄⁻]) = 0.20, so [HPO₄²⁻]/[H₂PO₄⁻] = 100.2010^{0.20} = 1.585. Let [H₂PO₄⁻] = x; then [HPO₄²⁻] = 1.585x. The constraint is x + 1.585x = 0.100 M → 2.585x = 0.100 → x = 0.0387 M. Therefore [H₂PO₄⁻] = 0.0387 M and [HPO₄²⁻] = 0.0613 M. For 2.00 L: moles NaH₂PO₄ = 0.0387 × 2.00 = 0.0774 mol (= 9.28 g) and moles Na₂HPO₄ = 0.0613 × 2.00 = 0.1226 mol (= 17.4 g of the anhydrous salt). This ratio ensures pH 7.40 at 25 °C.

Question 5: CRITICAL THINKING

Human blood is buffered at pH 7.40 primarily by the carbonic acid / bicarbonate system (H₂CO₃ / HCO₃⁻, pKₐ₁ = 6.35). The Henderson–Hasselbalch equation predicts that at pH 7.40, the ratio [HCO₃⁻]/[H₂CO₃] ≈ 11.2, which is well outside the conventional effective range of pKₐ ± 1. Explain why this buffer system remains effective in vivo despite operating so far from pKₐ.
Show answer
The blood bicarbonate system is an open buffer system, fundamentally different from the closed-system model assumed by the standard Henderson–Hasselbalch analysis. Dissolved CO₂ is in equilibrium with H₂CO₃ via the reaction CO₂(aq) + H₂O ⇌ H₂CO₃, and the partial pressure of CO₂ (pCO₂) is regulated by the lungs. When excess H⁺ is produced (e.g., by metabolism), HCO₃⁻ neutralizes it to form H₂CO₃, which rapidly decomposes to CO₂ and water; the CO₂ is then exhaled, effectively removing acid from the system. Conversely, hypoventilation raises pCO₂ and shifts the equilibrium toward H₂CO₃, compensating for alkalosis. Because the respiratory system continually adjusts the denominator [H₂CO₃] (via pCO₂), the ratio [HCO₃⁻]/[H₂CO₃] is maintained near 20:1 without depleting either component. The kidneys provide a second layer of regulation by adjusting renal reabsorption and excretion of HCO₃⁻ and H⁺. Together, these homeostatic mechanisms give the bicarbonate system an effective capacity far exceeding what a closed-system buffer at the same concentration would possess.

SUMMARY

Lesson Summary

Summary

A buffer solution consists of a conjugate acid–base pair—either a weak acid with its conjugate base or a weak base with its conjugate acid. When strong acid is added, the basic member of the pair neutralizes the incoming H⁺; when strong base is added, the acidic member neutralizes the incoming OH⁻. This dual-response mechanism, governed by Le Chatelier's principle, stabilizes the pH by shifting the equilibrium rather than allowing [H⁺] to change freely.

The Henderson–Hasselbalch equation (pH = pKₐ + log([A⁻]/[HA])) is the quantitative tool for predicting buffer pH. A buffer's effective range spans pKₐ ± 1, and its buffer capacity is maximized when [HA] = [A⁻] and when the total concentration of the pair is high. Selecting a buffer for a particular application requires matching the pKₐ of the weak acid to the desired pH—a principle that extends from simple benchtop chemistry to the sophisticated multi-component buffering systems that regulate blood pH at 7.40.

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