8.4 Acid-Base Reactions and Buffers

When you mix an acid and a base, the leftover species set the pH. Strong acid plus strong base reacts completely to form water, so you use the excess reagent to find pH after neutralization. For AP Chemistry, start with moles or millimoles before deciding whether the mixture is a buffer.

AP Chem Unit 8.4 Acid-Base Reactions and Buffers

In AP Chem Unit 8.4, acid-base mixture problems start with stoichiometry. Write the reaction, convert to moles or millimoles, identify the limiting reagent, and then use the major species left after reaction to decide how to find pH.

Strong acid-strong base mixtures use excess H+ or OH-. Weak acid-strong base and weak base-strong acid mixtures can form a buffer, leave excess strong reagent, or create a slightly basic or acidic equimolar solution. When a buffer is present, use the Henderson-Hasselbalch relationship with the conjugate acid-base pair.

Why This Matters for the AP Chemistry Exam

This topic ties chemical equilibrium directly to acid-base mixtures, which shows up across Unit 8. You will use these ideas to predict major species in a mixture, calculate pH from excess reagent or from the Henderson-Hasselbalch equation, and reason through what happens at each stage of an acid-base reaction. This thinking sets up titration curves in the next topic, where you identify equivalence points, half-equivalence points, and buffer regions. Expect to do dimensional analysis carefully, track moles and total volume, and pay attention to significant figures.

Key Takeaways

• Start most acid-base mixture problems by writing the net ionic equation, then convert to moles and find the limiting reagent.
• Strong acid plus strong base reacts as H+(aq) + OH-(aq) → H2O(l) and goes to completion; find pH from the excess reagent and total volume.
• A buffer contains both members of a conjugate acid-base pair and resists pH change because each member neutralizes added acid or base.
• For a weak acid with strong base (or weak base with strong acid), check three cases: weak reagent in excess (buffer), strong reagent in excess (use excess moles), or equimolar (salt hydrolysis).
• Use the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), when a buffer is present.
• When two weak species react, the position of equilibrium depends on the equilibrium constant for the proton transfer, so all four species can coexist.

Understanding Concentration Relationships in Acid-Base Mixtures

When acids and bases react, the concentrations of all species present are connected through several relationships:

• Stoichiometry: the mole ratios in the balanced equation
• Equilibrium considerations: most acid-base reactions are equilibrium processes
• Equilibrium constants: Ka, Kb, and Kw
• Mass balance: the total amount of acid and base added equals the sum of all forms present
• Charge balance: the solution stays electrically neutral

Together these let you predict and calculate the concentrations of major species in any acid-base mixture.

Defining Buffers

A buffer is a mixture of a weak acid (HA) and its conjugate base (A-) in the same solution. What makes a buffer useful is that it resists changes in pH. When you add a small amount of acid or base, the pH barely moves. If you add acid, the conjugate base reacts with it; if you add base, the weak acid reacts with it.

Buffers matter in both chemistry and biology. They help organisms maintain stable internal conditions. As an example, your blood is buffered to stay in a narrow pH range. 🩸

Reactions with Acids and Bases

Strong Acid Strong Base Reactions

A strong acid strong base reaction is probably the type you saw first when acids and bases were introduced, before equilibrium entered the picture.

Strong acid-strong base reactions go to completion quantitatively. The reaction proceeds nearly 100% to products, and the limiting reactant is completely consumed. An example is the reaction between HCl and NaOH to form NaCl and H2O. When you look at the ions, Na+ and Cl- are spectators, so the net ionic equation is just:

H+ + OH- → H2O

Use a single arrow (→) here because the reaction goes nearly to completion.

Here is an example with a strong acid strong base reaction:

What is the pH after the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl?

Start by writing the reaction:

HCl + NaOH → NaCl + H2O

The net ionic equation, with Cl- and Na+ as spectators, is:

H+ + OH- → H2O

Multiply mL times molarity to find millimoles of each species:

10.0 * 0.100 = 1.0 mmol OH-

25.0 * 0.100 = 2.5 mmol H+

Use stoichiometry to find how much H+ is left over. All of the OH- is consumed because H+ is in excess, leaving 2.5 - 1.0 = 1.5 mmol H+.

Divide 1.5 mmol H+ by the total volume (35 mL) to find [H+] = 4.3 * 10^-2 M, which corresponds to a pH of about 1.37.

It helps to think of three "zones" in these reactions: acid in excess, base in excess, or the two equimolar. This framing carries over to titration curves in the next topic.

Weak Acid Strong Base Reactions

A weak acid strong base reaction involves a weak acid reacting with a strong base. These reactions strongly favor products, but they are written as equilibrium processes.

For this section, use the reaction of acetic acid with sodium hydroxide:

CH3COOH + NaOH → CH3COONa + H2O

Na+ is a spectator, but CH3COO- is not, because acetic acid is a weak acid and does not fully dissociate. The net ionic equation is:

HA(aq) + OH-(aq) ⇌ A-(aq) + H2O(l)

Or specifically: CH3COOH + OH- ⇌ CH3COO- + H2O

Note the equilibrium arrows. The reaction strongly favors products but is still an equilibrium.

Start these problems by writing the net ionic equation, then converting to moles. When the weak acid is left in excess along with its conjugate base, you have a buffer. For a buffer you can find pH with the Henderson-Hasselbalch equation:

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

Here is an example:

Calculate the pH in the titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH after adding 10.00 mL of 0.100 M NaOH.

Use the same setup as the strong acid strong base problem. First find millimoles, then use stoichiometry to see what remains:

• Acetic acid: 25.0 * 0.100 = 2.5 mmol
• OH- added: 10.00 * 0.100 = 1.0 mmol
• OH- is limiting, so it converts 1.0 mmol of acetic acid into 1.0 mmol of CH3COO-
• Remaining: 1.5 mmol CH3COOH and 1.0 mmol CH3COO-

Instead of leftover H+ or OH-, you now have a mixture of CH3COOH and its conjugate base CH3COO-, which is a buffer. Plug the amounts into the Henderson-Hasselbalch equation to solve for pH.

Because both species share the same total solution volume, the volumes cancel in the ratio [A-]/[HA], so you can use the mole amounts directly in the log term.

Like the strong acid strong base case, you will pass through regions with only acid, a buffer region (unique to weak acid/base reactions), an equivalence point (where mmol acid = mmol base), and a post-equivalence region.

At the Equivalence Point

At the equivalence point, the pH is not 7 the way it is for strong acid-strong base reactions. When you mix equimolar amounts of weak acid and strong base, you are left with the conjugate base (such as CH3COO-). This conjugate base undergoes hydrolysis:

CH3COO- + H2O ⇌ CH3COOH + OH-

That makes the solution slightly basic. To find the pH, use the Kb of the conjugate base and solve an equilibrium problem.

Weak Base Strong Acid Reactions

The same strategies apply to weak base strong acid reactions, which follow the general equation:

B(aq) + H3O+(aq) ⇌ HB+(aq) + H2O(l)

Like weak acid-strong base reactions, these are equilibrium processes that strongly favor products.

For example, mixing ammonia (NH3) with HCl:

NH3 + H3O+ ⇌ NH4+ + H2O

There are three scenarios:

1. Excess weak base: forms a buffer of B and HB+. Use the Henderson-Hasselbalch equation with the pKa of the conjugate acid HB+.

2. Excess strong acid: find pH from the excess H3O+ moles and total volume.

3. Equimolar amounts: at the equimolar point, only HB+ is present, and it hydrolyzes:

   HB+ + H2O ⇌ B + H3O+

   This makes the solution slightly acidic.

Weak Acid Weak Base Reactions

The trickiest case is when both the acid and the base are weak. These reactions also reach an equilibrium:

HA(aq) + B(aq) ⇌ A-(aq) + HB+(aq)

For example: CH3COOH + NH3 ⇌ CH3COO- + NH4+

For a weak acid weak base mixture, the position of equilibrium is set by the equilibrium constant for the proton-transfer reaction. Significant amounts of all four species (HA, B, A-, and HB+) can coexist because neither reactant fully dissociates and the reaction does not go to completion.

Here is how the four reaction types compare in how far they shift toward products:

• Strong acid-strong base: complete or nearly complete (→)
• Weak acid-strong base: strongly favors products but still equilibrium (⇌)
• Weak base-strong acid: strongly favors products but still equilibrium (⇌)
• Weak acid-weak base: equilibrium may not strongly favor either direction (⇌)

Determining the Equilibrium State

To describe the equilibrium for a weak acid weak base reaction, combine the equilibrium expressions for the acid, the base, and water. The equilibrium constant for HA + B ⇌ A- + HB+ can be written as:

K = Ka(acid) × Kb(base) / Kw

Where:

• Ka is the acid dissociation constant of HA
• Kb is the base dissociation constant of B
• Kw is the ion product of water (1.0 × 10⁻¹⁴ at 25°C)

Example: Acetic Acid + Ammonia

CH3COOH + NH3 ⇌ CH3COO- + NH4+

Given: Ka(CH3COOH) = 1.8 × 10⁻⁵, Kb(NH3) = 1.8 × 10⁻⁵

K = (1.8 × 10⁻⁵)(1.8 × 10⁻⁵)/(1.0 × 10⁻¹⁴) = 3.24 × 10⁴

This large K value tells you the reaction favors products. Because Ka equals Kb here, the resulting solution has a pH near 7. When Ka is larger than Kb the mixture tends acidic, and when Kb is larger the mixture tends basic.

How to Use This on the AP Chemistry Exam

Problem Solving

• Write the net ionic equation first, then convert volumes and concentrations into moles (or millimoles).
• Identify the limiting reagent and do the stoichiometry as if the reaction goes to completion, then reason about what is left.
• Match what remains to a case: only excess strong acid or base, a buffer, or a salt solution at the equimolar point.
• For excess strong acid or base, divide the leftover moles of H+ or OH- by the total combined volume before taking the log.
• For a buffer, apply pH = pKa + log([A-]/[HA]); the volume cancels in the ratio, so mole amounts work directly.

Common Trap

• Do not assume the equivalence point pH is 7 for weak-strong titrations. The leftover conjugate species hydrolyzes and shifts pH away from 7.
• Watch your total volume. After mixing, the new volume is the sum of both solutions, and that matters for any concentration you calculate.
• Keep track of significant figures and units through every step.

Common Misconceptions

• "All neutralizations end at pH 7." Only strong acid-strong base mixtures are neutral at the equimolar point. Weak acid with strong base ends slightly basic; weak base with strong acid ends slightly acidic.
• "A buffer can hold pH at any value forever." A buffer only resists change near its pKa and only until one component is used up. Adding too much acid or base overwhelms it.
• "Any mixture of an acid and a base is a buffer." A buffer needs meaningful amounts of both a weak acid and its conjugate base. A strong acid with a strong base does not make a buffer.
• "Weak acid-strong base reactions barely proceed because the acid is weak." These reactions still go strongly toward products; the strong base drives the proton transfer.
• "You use the weak acid's Ka in Henderson-Hasselbalch for a weak base buffer." For a weak base and its conjugate acid, use the pKa of the conjugate acid HB+, not the base.
• "Hydronium and hydrogen ion are different things here." H+(aq) and H3O+(aq) refer to the same aqueous species and are used interchangeably.

Related AP Chemistry Guides

• Unit 8 Overview: Acids and Bases
• 8.1 Introduction to Acids and Bases
• 8.2 pH and pOH of Strong Acids and Bases
• 8.3 Weak Acid and Base Equilibria
• 8.5 Acid-Base Titrations
• 8.7 pH and pKa