# Lecture 7: Buffers and the Henderson-Hasselbalch Equation
## General Chemistry II
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## Learning Objectives
By the end of this lecture, students will be able to:
1. Define a buffer and explain how it resists changes in pH
2. Identify the components needed to prepare a buffer
3. Derive and apply the Henderson-Hasselbalch equation
4. Calculate the pH of a buffer solution
5. Determine the effect of adding strong acid or base to a buffer
6. Explain buffer capacity and optimal buffer range
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## Lecture Content
### I. What Is a Buffer?
A buffer is a solution that resists significant changes in pH when small amounts of strong acid or strong base are added to it. A buffer is composed of a weak acid and its conjugate base, or equivalently, a weak base and its conjugate acid. An acidic buffer pairs a weak acid (HA) with its conjugate base (A-), the latter typically supplied as a sodium or potassium salt. For example, an acetic acid / sodium acetate (CH3COOH / CH3COONa) mixture is a common acidic buffer. A basic buffer pairs a weak base (B) with its conjugate acid (BH+), such as ammonia / ammonium chloride (NH3 / NH4Cl). Both components must be present in significant concentrations for the buffer to function effectively.
### II. How Buffers Work
When a strong acid (H3O+) is added to a buffer, the conjugate base neutralizes it through the reaction A- + H3O+ -> HA + H2O. The added H+ is consumed, converting some A- into HA, but the ratio [HA]/[A-] changes only slightly as long as the amount of added acid is small relative to the buffer components.
When a strong base (OH-) is added, the weak acid neutralizes it: HA + OH- -> A- + H2O. The added OH- is consumed by converting some HA into A-, again changing the ratio only slightly.
In both cases, the pH changes only minimally because the ratio [A-]/[HA] changes only slightly. The essential insight is that the buffer works by converting a strong acid or base into a weak acid or base, thereby preventing the dramatic pH swings that would occur in an unbuffered solution.
A diagram illustrating buffer action for a CH3COOH/CH3COO- buffer. Panel A: Addition of HCl. Shows H+ being consumed by CH3COO- to form CH3COOH. Before and after bar charts show [CH3COOH] slightly increasing and [CH3COO-] slightly decreasing, with pH changing minimally. Panel B: Addition of NaOH. Shows OH- being consumed by CH3COOH to form CH3COO- and H2O. Before and after bar charts show [CH3COOH] slightly decreasing and [CH3COO-] slightly increasing, with pH changing minimally. Panel C: For comparison, the same amount of HCl added to pure water, showing a dramatic pH drop from 7 to approximately 2.
### III. The Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is derived directly from the Ka expression for a weak acid. Starting from Ka = [H3O+][A-] / [HA], solving for [H3O+] gives [H3O+] = Ka * [HA]/[A-]. Taking the negative logarithm of both sides yields pH = pKa + log([A-]/[HA]).
This is the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[weak acid]). It is valid when the x-is-small approximation holds (that is, when the buffer concentrations are much larger than Ka) and when both components are present in significant amounts. A particularly useful feature is that when [A-] = [HA], the logarithmic term vanishes (since log(1) = 0) and pH = pKa. Because volume cancels in the ratio, the equation can equivalently be written using moles instead of concentrations: pH = pKa + log(moles of A- / moles of HA). For basic buffers, the analogous equation is pOH = pKb + log([conjugate acid]/[weak base]).
### IV. Calculating Buffer pH
There are two main approaches to finding the pH of a buffer. The quick method uses the Henderson-Hasselbalch equation directly: identify [A-] and [HA] (or their mole amounts) and substitute into the equation. The systematic method uses an ICE table with the Ka equilibrium, starting with the initial concentrations of both HA and A-. The change x will be very small due to the common-ion effect.
The common-ion effect describes how the presence of A- from the added salt suppresses the ionization of HA. By Le Chatelier's principle, adding A- shifts the equilibrium HA <=> H+ + A- to the left, resulting in less ionization of HA than would occur if HA were in solution by itself.
### V. Adding Strong Acid or Base to a Buffer
When strong acid or base is added to a buffer, a two-step approach is used. In the first step (stoichiometry), treat the neutralization reaction as going to completion. Adding strong acid converts A- to HA through the reaction A- + H+ -> HA, while adding strong base converts HA to A- through HA + OH- -> A- + H2O. Calculate the new moles of HA and A- after this stoichiometric reaction.
In the second step (equilibrium), use the Henderson-Hasselbalch equation with the updated amounts: pH = pKa + log(new moles A- / new moles HA).
If all of one component is consumed, the buffer is destroyed. If all A- is consumed by added acid, you are left with excess strong acid in solution. If all HA is consumed by added base, you have excess strong base. In either case, the pH is no longer buffered and must be calculated differently.
### VI. Buffer Capacity
Buffer capacity refers to the amount of strong acid or base a buffer can absorb before a significant pH change occurs. It depends on two factors: the total concentration of the buffer components, with higher concentration providing greater capacity, and the ratio of [A-] to [HA], with maximum capacity when this ratio is near 1:1.
A buffer is most effective when the [A-]/[HA] ratio falls between 0.1 and 10, corresponding to a pH range of pKa +/- 1. The optimal buffer pH equals the pKa of the weak acid, because at this point [A-] = [HA], the buffer has its maximum capacity, and the pH is most resistant to perturbation.
A graph showing buffer capacity curves. The x-axis is "Moles of OH- added" and the y-axis is "pH." Two buffer curves are shown for the same weak acid (pKa = 4.75): one at high concentration (1.0 M total buffer) showing a gradual, nearly flat pH change, and one at low concentration (0.1 M total buffer) showing a steeper pH change for the same amount of base added. A horizontal dashed line marks pH = pKa = 4.75. The buffering region (pKa +/- 1) is shaded. Beyond the shaded region, both curves rise steeply, indicating buffer failure.
### VII. Preparing a Buffer of Desired pH
To prepare a buffer at a specific pH, first choose a weak acid whose pKa is close to the desired pH (within +/- 1 unit). Then use the Henderson-Hasselbalch equation to calculate the required [A-]/[HA] ratio. Finally, prepare the solution with that ratio and the desired total concentration.
Several buffer systems are particularly important in biochemistry. The phosphate buffer (H2PO4-/HPO4^2-) has pKa2 = 7.20 and is widely used for solutions near physiological pH. Tris buffer has pKa = 8.07. The acetate buffer (CH3COOH/CH3COO-) has pKa = 4.76. The bicarbonate buffer (H2CO3/HCO3-), with pKa1 = 6.35, is the most critical buffer system in human blood.
The blood buffer system maintains blood pH within the narrow range of 7.35 to 7.45 using the bicarbonate equilibrium: CO2(g) + H2O <=> H2CO3 <=> H+ + HCO3-. The lungs regulate CO2, and therefore H2CO3, by adjusting the rate of breathing. The kidneys regulate HCO3- by controlling its reabsorption and excretion. The blood pH is described by pH = 6.1 + log([HCO3-]/[H2CO3]), and under normal conditions the ratio [HCO3-]/[H2CO3] is approximately 20:1.
A diagram of the bicarbonate blood buffer system. A central equation shows: CO2(g) + H2O <=> H2CO3 <=> H+ + HCO3-. On the left, lungs are illustrated with arrows showing CO2 exhalation (removes acid) and CO2 retention (adds acid). On the right, kidneys are illustrated with arrows showing HCO3- reabsorption (adds base) and HCO3- excretion (removes base). The normal blood pH range of 7.35-7.45 is highlighted, with acidosis (< 7.35) and alkalosis (> 7.45) labeled on either side.
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