Acid-Base Reactions and Buffers

Learning Objective 8.4.A Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.

Quick Notes

Strong acid + strong base gives complete neutralization.
Weak acid + strong base forms a buffer if the weak acid is in excess.
Weak base + strong acid forms a buffer if the weak base is in excess.
A buffer is a solution that resists changes in pH when small amounts of acid or base are added.
The Henderson–Hasselbalch equation is used to calculate buffer pH: pH = pK_a + log([A⁻]/[HA]).

Full Notes

Strong Acid + Strong Base

When a strong acid is mixed with a strong base, they neutralize each other in a complete reaction. This is represented by the net ionic equation:

AP Chemistry net ionic equation H+ plus OH− goes to water showing complete neutralization

The acid donates a proton (H⁺), and the base provides a hydroxide ion (OH⁻), forming water. A salt also forms from the conjugate ions of the acid and base, but these ions are spectators and don’t affect pH directly.

Weak Acid + Strong Base

When a weak acid reacts with a strong base, the reaction can be represented by the equation:

AP Chemistry reaction of weak acid with strong base forming conjugate base and water HA(aq) + OH−(aq) → A−(aq) + H2O(l)

This reaction also goes to completion.

Where HA(aq) is the weak acid, OH⁻(aq) comes from the base and A⁻(aq) is the conjugate base of the weak acid.

Weak acid in excess: A buffer forms (see below). Use the Henderson–Hasselbalch equation to calculate pH (see 8.9).
Strong base in excess: pH is determined from [OH⁻] (see examples below).
Equimolar amounts: Only A⁻ remains, and it hydrolyzes slightly to form OH⁻. Use an ICE table and K_b to calculate pH.

Weak Base + Strong Acid

Also a complete reaction:

AP Chemistry reaction of weak base with hydronium forming conjugate acid and water B(aq) + H3O+(aq) → HB+(aq) + H2O(l)

Where B(aq) is the base, H₃O⁺(aq) comes from the strong acid reacting with water and HB⁺(aq) is the conjugate acid of the base.

Weak base in excess: A buffer forms. Use the Henderson–Hasselbalch equation (see 8.9).
Strong acid in excess: pH is found from [H₃O⁺].
Equimolar amounts: Only HB⁺ remains and slightly ionizes to form H₃O⁺. Use K_a to calculate pH from an ICE table.

Weak Acid + Weak Base

This reaction does not go to completion. Instead, it establishes an equilibrium:

AP Chemistry equilibrium of weak acid and weak base forming conjugates HA(aq) + B(aq) ⇌ A−(aq) + HB+(aq)

The position of equilibrium — and the final pH — depends on the relative strengths of the acid and base (i.e., their K_a and K_b values).

What Is a Buffer?

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers work by having both a weak acid and its conjugate base (or a weak base and its conjugate acid) in equilibrium.

Types of Buffer Systems

1. Acidic Buffers: Made from a weak acid and its salt (that contains the acid’s conjugate base).

For example the weak acid ethanoic acid (CH₃COOH) and its salt sodium ethanoate (CH₃COONa).

AP Chemistry acidic buffer made by mixing acetic acid and sodium acetate

When added to a solution of the ethanoic acid, the CH₃COONa would dissociate and release CH₃COO⁻ ions, which is the conjugate base (A⁻) of the ethanoic acid.

You can also prepare a buffer by reacting a weak acid with a limited amount of strong base:
CH₃COOH + NaOH → CH₃COONa + H₂O
This produces both HA and A⁻ in the same solution.

2. Basic Buffers: Made from a weak base and its salt (that contains the conjugate acid of the base).

For example the weak base ammonia (NH₃) and its salt ammonium chloride (NH₄Cl).

AP Chemistry basic buffer made by mixing ammonia solution with ammonium chloride

When added to a solution of ammonia, the NH₄Cl would dissociate and release NH₄⁺ ions, which are the conjugate acid ions of the ammonia.

How Do Acidic Buffers Work?

An equilibrium is established in the buffer system between HA, A⁻ and H⁺.

AP Chemistry buffer equilibrium between HA, H+ and A−

The concentration of HA and A⁻ in the mixture must be much greater than the concentration of H⁺. This ensures the position of equilibrium is sensitive to changes in H⁺ concentration more than changes to HA and A⁻ concentration. Equilibrium position can shift to keep H⁺ ion concentration nearly constant.

For example Ethanoic Acid/Sodium Ethanoate Buffer (CH₃COOH/CH₃COO⁻)
Equilibrium reaction: CH₃COOH ⇌ H⁺ + CH₃COO⁻

AP Chemistry diagrams showing added H+ being consumed by A− in an acidic buffer

When an acid (H⁺) is added:
CH₃COO⁻ combines with added H⁺ to form CH₃COOH.
Equilibrium shifts left, reducing the increase in H⁺.
[HA] increases and [A⁻] decreases.

When a base (OH⁻) is added:

AP Chemistry diagrams showing added OH− reacting with H+ and HA dissociating to restore H+

Added OH⁻ reacts with H⁺ to form H₂O.
CH₃COOH dissociates more to replace lost H⁺.
Equilibrium shifts right, replacing H⁺ ions that reacted with the added OH⁻, resisting pH increase.
[HA] decreases and [A⁻] increases.

Matt’s exam tip

Remember that the concentration of HA and A⁻ will change when H⁺ or OH⁻ ions are added. When H⁺ ions are added to the mixture, the moles of HA will increase by the same as the moles of H⁺ added and moles of A⁻ decrease by the same amount. When OH⁻ ions are added, the moles of HA will decrease by the same as moles of OH⁻ added and the moles of A⁻ increase by the same amount.

As long as there are significant amounts of HA and A⁻ present, this system can buffer against pH changes.

Why Weak Acids and Bases Are Needed

Weak acids and bases only partially dissociate, setting up an equilibrium with their conjugates. This equilibrium helps them resist changes in pH by reacting with added H⁺ or OH⁻. Strong acids and bases fully dissociate, so they can't maintain this balance — meaning they don’t act as buffers.

pH Calculations

Worked Example

Strong acid + strong base: What is the pH of 50.0 mL of 0.10 M HCl mixed with 60.0 mL of 0.10 M NaOH?

Find moles of reactants
HCl: 0.0500 L × 0.10 mol/L = 0.00500 mol
NaOH: 0.0600 L × 0.10 mol/L = 0.00600 mol
Determine excess species
Excess OH⁻ = 0.00600 − 0.00500 = 0.00100 mol
Compute [OH⁻]
Total volume = 110.0 mL = 0.110 L → [OH⁻] = 0.00100 / 0.110 = 0.00909 M
Calculate pH
pOH = −log(0.00909) ≈ 2.04 → pH = 14 − 2.04 = 11.96

Answer: pH ≈ 11.96.

Worked Example

Weak acid + strong base (weak acid in excess means buffer forms): Find the pH when 40.0 mL of 0.10 M CH₃COOH is mixed with 20.0 mL of 0.10 M NaOH. K_a for acetic acid = 1.8 × 10⁻⁵ → pK_a ≈ 4.74.

Stoichiometry to form buffer pair
Moles CH₃COOH = 0.00400 mol; Moles NaOH = 0.00200 mol
Remaining CH₃COOH = 0.00200 mol; Formed CH₃COO⁻ = 0.00200 mol
Convert to concentrations
V = 0.0600 L → [CH₃COOH] = 0.00200/0.0600 = 0.0333 M;
[CH₃COO⁻] = 0.0333 M
Apply Henderson–Hasselbalch
pH = 4.74 + log(1) = 4.74

Answer: pH = 4.74.

Worked Example

Weak base + strong acid (weak base in excess means buffer formed): Calculate the pH when 25.0 mL of 0.20 M NH₃ is mixed with 15.0 mL of 0.20 M HCl. K_b for NH₃ = 1.8 × 10⁻⁵ → pK_a = 14 − pK_b ≈ 9.26.

Stoichiometry to form buffer pair
Moles NH₃ = 0.00500 mol; Moles HCl = 0.00300 mol
NH₃ remaining = 0.00200 mol; NH₄⁺ formed = 0.00300 mol
Convert to concentrations
V = 0.0400 L → [NH₃] = 0.00200/0.0400 = 0.0500 M;
[NH₄⁺] = 0.00300/0.0400 = 0.0750 M
Apply Henderson–Hasselbalch
pH = 9.26 + log(0.0500/0.0750) = 9.26 + log(0.667) ≈ 9.08

Answer: pH ≈ 9.08.

Summary

Strong acids and bases fully react — just look for excess species to find pH.
Weak acid/base with strong base/acid may create buffers or lead to hydrolysis reactions.
Use Henderson–Hasselbalch when a buffer forms.
Weak acid + weak base systems require full equilibrium analysis using K_a or K_b.