🧫 Unit 8: Acids and Bases

Dive into proton transfer chemistry — pH, Ka/Kb, buffers, and titrations.

Exam Weight: 11–15%  |  Topics 8.1–8.10

📑 In This Unit

8.1 — Introduction to Acids and Bases

There are multiple models for understanding acid-base chemistry. The AP exam primarily uses the Brønsted-Lowry model.

ModelAcid DefinitionBase DefinitionScope
ArrheniusProduces H⁺ in waterProduces OH⁻ in waterAqueous solutions only
Brønsted-LowryProton (H⁺) donorProton (H⁺) acceptorAny solvent; gas phase too
Conjugate Acid-Base Pairs:
When a Brønsted-Lowry acid donates a proton, it becomes a conjugate base. When a base accepts a proton, it becomes a conjugate acid.

Example: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
• NH₃ is the base (accepts H⁺) → NH₄⁺ is its conjugate acid
• H₂O is the acid (donates H⁺) → OH⁻ is its conjugate base
Amphoteric (Amphiprotic) Substances

Water is the most common amphiprotic substance — it can act as either an acid or a base depending on the reaction partner. Other examples include HSO₄⁻, HCO₃⁻, and H₂PO₄⁻.

Autoionization of water:
H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C

In pure water: [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M → pH = 7.00 (neutral)
Kw applies to ALL aqueous solutions, not just pure water.
pH scale illustration

Acids and bases in titration. The pH scale ranges from 0 (strongly acidic) to 14 (strongly basic). (Wikimedia Commons)

Strong vs. Weak
StrongWeak
Ionization100% ionized in waterPartially ionized (< 100%)
EquilibriumSingle arrow (→); goes to completionDouble arrow (⇌); equilibrium established
Ka or KbVery large (not usually given)Small (Ka or Kb << 1)
Strong acidsHCl, HBr, HI, HNO₃, H₂SO₄, HClO₃, HClO₄
Strong basesLiOH, NaOH, KOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂

8.2 — pH and pOH of Strong Acids and Bases

For strong acids and bases, the calculation is straightforward because they ionize completely.

Key relationships:
• pH = −log[H₃O⁺]
• pOH = −log[OH⁻]
• pH + pOH = 14.00 (at 25°C)
• [H₃O⁺] = 10−pH
• [OH⁻] = 10−pOH
• [H₃O⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴
Example — Strong acid:
What is the pH of 0.0025 M HCl?
HCl is a strong acid → [H₃O⁺] = 0.0025 M
pH = −log(0.0025) = 2.60
Example — Strong base:
What is the pH of 0.010 M Ba(OH)₂?
Ba(OH)₂ → Ba²⁺ + 2OH⁻ (strong base, 100% dissociation)
[OH⁻] = 2 × 0.010 = 0.020 M
pOH = −log(0.020) = 1.70
pH = 14.00 − 1.70 = 12.30
AP Exam tip: Watch for diprotic/triprotic strong acids. H₂SO₄ is strong for the first proton only (the second dissociation has Ka2 = 0.012, which is weak). For dilute solutions of Ba(OH)₂, Ca(OH)₂, and Sr(OH)₂, remember to multiply [OH⁻] by 2.

8.3 — Weak Acid and Base Equilibria

Weak acids and bases only partially ionize. We must use equilibrium expressions and ICE tables to calculate pH.

Weak acid equilibrium:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)

Ka = [H₃O⁺][A⁻] / [HA]

Weak base equilibrium:
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Kb = [BH⁺][OH⁻] / [B]
Key relationship: For any conjugate acid-base pair:
Ka × Kb = Kw = 1.0 × 10⁻¹⁴

This means: the stronger the acid (larger Ka), the weaker its conjugate base (smaller Kb), and vice versa.
Example — Weak acid pH calculation:
Find the pH of 0.100 M acetic acid (CH₃COOH), Ka = 1.8 × 10⁻⁵

CH₃COOHH₃O⁺CH₃COO⁻
I0.10000
C−x+x+x
E0.100 − xxx
Ka = x² / (0.100 − x) ≈ x² / 0.100 = 1.8 × 10⁻⁵
x² = 1.8 × 10⁻⁶ → x = 1.34 × 10⁻³ M = [H₃O⁺]
Check: 1.34 × 10⁻³ / 0.100 = 1.34% < 5% — approximation valid!
pH = −log(1.34 × 10⁻³) = 2.87
Percent Ionization
% ionization = ([H₃O⁺]eq / [HA]initial) × 100%

For weak acids:
• % ionization increases as initial concentration decreases (dilution effect).
• % ionization increases as Ka increases (stronger weak acid).

From the example above: % ionization = (1.34 × 10⁻³ / 0.100) × 100% = 1.34%

8.4 — Acid-Base Reactions and Buffers

When an acid reacts with a base, the reaction goes essentially to completion if one of them is strong. This is the foundation for understanding titrations and buffers.

Types of acid-base reactions:

1. Strong acid + Strong base → salt + water
HCl + NaOH → NaCl + H₂O   (pH = 7 at equivalence)

2. Strong acid + Weak base → conjugate acid salt
HCl + NH₃ → NH₄Cl   (pH < 7 at equivalence)

3. Weak acid + Strong base → conjugate base salt
CH₃COOH + NaOH → CH₃COONa + H₂O   (pH > 7 at equivalence)

4. Weak acid + Weak base: Direction favors the side with weaker acid and weaker base (compare Ka and Kb).
Salt Hydrolysis — Predicting the pH of Salt Solutions
Salt FromCationAnionSolution pH
Strong acid + Strong baseNeutral (Na⁺, K⁺)Neutral (Cl⁻, NO₃⁻)pH = 7
Strong acid + Weak baseAcidic (NH₄⁺)Neutral (Cl⁻)pH < 7
Weak acid + Strong baseNeutral (Na⁺)Basic (CH₃COO⁻)pH > 7
Weak acid + Weak baseAcidicBasicCompare Ka vs Kb

8.5 — Acid-Base Titrations

A titration is a technique for determining the concentration of an unknown solution by reacting it with a solution of known concentration (the titrant).

Key regions of a titration curve:

• Initial point: pH of the analyte alone before any titrant is added.
• Buffer region: The region where both the weak acid/base and its conjugate are present. pH changes gradually.
• Half-equivalence point: Exactly half the acid (or base) has been neutralized. [HA] = [A⁻], so pH = pKa.
• Equivalence point: Moles of acid = moles of base. All the analyte has been neutralized.
• Post-equivalence: Excess titrant determines the pH.
Weak acid-strong base titration curve

Titration curve for a weak acid titrated with a strong base. Note the buffer region, half-equivalence point (pH = pKa), and the equivalence point above pH 7. (Wikimedia Commons)

Equivalence Point pH
Titration TypepH at Equivalence PointReason
Strong acid + Strong basepH = 7Neutral salt formed
Weak acid + Strong basepH > 7Conjugate base of weak acid is basic
Strong acid + Weak basepH < 7Conjugate acid of weak base is acidic
Equivalence point calculation:
At the equivalence point: moles acid = moles base
Macid × Vacid = Mbase × Vbase

Example: 25.0 mL of 0.100 M HCl is titrated with 0.200 M NaOH. What volume of NaOH is needed?
(0.100)(25.0) = (0.200)(VNaOH) → VNaOH = 12.5 mL
Choosing an Indicator

Select an indicator whose color change range (pKin ± 1) includes the pH at the equivalence point. For a weak acid/strong base titration (equivalence pH > 7), use phenolphthalein (range 8.2–10). For strong acid/strong base, most indicators work since the pH jump is very steep around pH 7.

8.6 — Molecular Structure of Acids and Bases

The molecular structure of a compound determines its acid or base strength. Several structural factors affect how easily a molecule donates or accepts protons.

Binary Acids (HX)
Across a period (same row): Acid strength increases with electronegativity of X.
• CH₄ < NH₃ < H₂O < HF (but all except HF are extremely weak acids in water)

Down a group (same column): Acid strength increases with size of X (weaker H—X bond is easier to break).
• HF < HCl < HBr < HI
• Bond strength is the dominant factor down a group (larger atoms form weaker bonds with H).
Oxyacids (HO—X—O...)
Same central atom, different number of oxygens: More oxygen atoms = stronger acid (they withdraw electron density from O—H bond, making it easier to release H⁺).
• HClO < HClO₂ < HClO₃ < HClO₄

Different central atoms, same structure: More electronegative central atom = stronger acid.
• HOBr < HOCl (Cl is more electronegative than Br)
Carboxylic Acids

Electron-withdrawing groups (like −F, −Cl, −NO₂) near the −COOH group stabilize the conjugate base (carboxylate anion) through inductive effects, increasing acid strength. The closer the electron-withdrawing group is to −COOH, the stronger the effect.

8.7 — pH and pKa

The pKa is the negative logarithm of Ka and provides a convenient way to compare acid strengths.

pKa = −log Ka    |    pKb = −log Kb

• Smaller pKa = stronger acid (larger Ka)
• Larger pKa = weaker acid (smaller Ka)
• pKa + pKb = pKw = 14.00 (at 25°C)
Relationship between pH and pKa:
• If pH < pKa: The acid form (HA) predominates.
• If pH = pKa: [HA] = [A⁻] (equal amounts).
• If pH > pKa: The conjugate base form (A⁻) predominates.

This is critical for understanding buffer chemistry and for predicting the dominant species at any given pH.
Example: Acetic acid has pKa = 4.74. In a solution at pH 6.0:
pH > pKa → the conjugate base CH₃COO⁻ predominates. The solution has more acetate ions than acetic acid molecules.

8.8 — Properties of Buffers

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are essential in biological systems (blood pH = 7.4) and many chemical processes.

A buffer consists of:
• A weak acid and its conjugate base (e.g., CH₃COOH / CH₃COO⁻), OR
• A weak base and its conjugate acid (e.g., NH₃ / NH₄⁺)

Both components must be present in significant amounts.
How Buffers Work
When acid (H⁺) is added: The conjugate base reacts with H⁺ to neutralize it.
A⁻ + H⁺ → HA
This consumes the added acid, preventing a large pH drop.

When base (OH⁻) is added: The weak acid reacts with OH⁻ to neutralize it.
HA + OH⁻ → A⁻ + H₂O
This consumes the added base, preventing a large pH increase.

The pH changes slightly because the ratio [A⁻]/[HA] changes, but the change is much smaller than it would be in an unbuffered solution.

8.9 — Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is the most important equation for buffer calculations on the AP exam.

pH = pKa + log([A⁻] / [HA])

• [A⁻] = concentration of the conjugate base
• [HA] = concentration of the weak acid
• When [A⁻] = [HA]: pH = pKa + log(1) = pKa (the log term is 0)
• When [A⁻] > [HA]: pH > pKa (log term is positive)
• When [A⁻] < [HA]: pH < pKa (log term is negative)
Example: What is the pH of a buffer containing 0.30 M CH₃COOH and 0.50 M CH₃COONa? (Ka = 1.8 × 10⁻⁵, pKa = 4.74)

pH = 4.74 + log(0.50 / 0.30) = 4.74 + log(1.67) = 4.74 + 0.22 = 4.96
Example — Adding acid to a buffer:
To the buffer above (total volume = 1.00 L), 0.050 mol HCl is added. Find the new pH.

Before: 0.30 mol HA, 0.50 mol A⁻
HCl reacts with A⁻: A⁻ + H⁺ → HA
After: HA = 0.30 + 0.050 = 0.35 mol; A⁻ = 0.50 − 0.050 = 0.45 mol

pH = 4.74 + log(0.45 / 0.35) = 4.74 + log(1.286) = 4.74 + 0.11 = 4.85
The pH changed by only 0.11 units despite adding a strong acid!
AP Exam tip: You can use moles instead of concentrations in the Henderson-Hasselbalch equation because the volume cancels in the ratio [A⁻]/[HA] (as long as both are in the same solution volume).

8.10 — Buffer Capacity

Buffer capacity is the amount of strong acid or base a buffer can neutralize before the pH changes significantly (usually defined as a change of more than 1 pH unit).

Factors that increase buffer capacity:
1. Higher concentrations of both the weak acid and conjugate base. A buffer made with 1.0 M components can absorb more acid/base than one made with 0.10 M components.

2. Equal ratio of [HA] to [A⁻] (ratio closest to 1:1). This maximizes the buffer’s ability to resist pH change in both directions.

3. The buffer is most effective when pH is within ±1 of pKa. Outside this range, one component is largely depleted.
When a buffer is “broken”:
If you add more moles of strong acid than there are moles of A⁻ in the buffer, the buffer is overwhelmed. All A⁻ is consumed, excess H⁺ remains, and the pH drops sharply. The solution is no longer buffered.

Similarly, adding more moles of strong base than moles of HA will break the buffer on the basic side.
Designing a buffer:
To make a buffer at a target pH:
1. Choose a weak acid with pKa close to the desired pH.
2. Use Henderson-Hasselbalch to determine the [A⁻]/[HA] ratio needed.
3. Use high concentrations for greater capacity.

📝 Practice Multiple-Choice Questions

Test your knowledge of Unit 8. Click “Show Answer” to reveal the correct choice and explanation.

1. According to the Brønsted-Lowry model, a base is a substance that:

Answer: (C)
A Brønsted-Lowry base is a proton (H⁺) acceptor. (A) is the Arrhenius definition. (D) is the Lewis definition.

2. What is the pH of a 0.0050 M solution of NaOH?

Answer: (D)
NaOH is a strong base: [OH⁻] = 0.0050 M. pOH = −log(0.0050) = 2.30. pH = 14.00 − 2.30 = 11.70.

3. A 0.10 M solution of a weak acid HA has a pH of 3.0. What is the Ka of the acid?

Answer: (B)
pH = 3.0 → [H₃O⁺] = 1.0 × 10⁻³ M.
Ka = x² / (0.10 − x) ≈ (1.0 × 10⁻³)² / 0.10 = 1.0 × 10⁻⁶ / 0.10 = 1.0 × 10⁻⁵.

4. A solution of NaCH₃COO (sodium acetate) dissolved in water will be:

Answer: (C)
Sodium acetate is the salt of a weak acid (CH₃COOH) and a strong base (NaOH). The acetate ion (CH₃COO⁻) acts as a base in water, hydrolyzing to produce OH⁻, making the solution basic.

5. During the titration of a weak acid with a strong base, at the half-equivalence point:

Answer: (C)
At the half-equivalence point, half of the weak acid has been neutralized, so [HA] = [A⁻]. By Henderson-Hasselbalch: pH = pKa + log(1) = pKa.

6. Which of the following acids is the strongest?

Answer: (B)
The strongest acid has the largest Ka. HClO₂ has Ka = 1.1 × 10⁻², which is the largest value among the choices.

7. If a weak acid has pKa = 5.0 and the solution pH is 6.0, which species predominates?

Answer: (B)
When pH > pKa, the conjugate base form A⁻ predominates. At pH = 6.0 and pKa = 5.0: log([A⁻]/[HA]) = 6.0 − 5.0 = 1.0 → [A⁻]/[HA] = 10, so there is 10× more A⁻ than HA.

8. A buffer is prepared with 0.40 M NH₃ and 0.40 M NH₄Cl. Kb for NH₃ = 1.8 × 10⁻⁵. What is the pH?

Answer: (B)
For this base buffer, first find pKa of NH₄⁺:
Ka = Kw/Kb = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰ → pKa = 9.26
pH = pKa + log([NH₃]/[NH₄⁺]) = 9.26 + log(0.40/0.40) = 9.26 + 0 = 9.26.

9. Which of the following would NOT make an effective buffer?

Answer: (C)
HCl is a strong acid, not a weak acid. Buffers require a weak acid/base and its conjugate. NaCl is the salt of a strong acid and strong base — neither component can neutralize added acid or base effectively.

10. Which change increases buffer capacity?

Answer: (B)
Buffer capacity increases with higher concentrations of both components. More moles of HA and A⁻ means the buffer can absorb more added acid or base before being overwhelmed. Diluting (A) reduces capacity, and removing a component (D) destroys the buffer.
💡 Study Tip for Unit 8

Acids and Bases is one of the highest-weighted units on the AP exam (11–15%). Master the Henderson-Hasselbalch equation — it’s used in nearly every buffer and titration problem. Know the 7 strong acids and 6 strong bases by heart. Practice calculating pH for strong acids/bases (direct), weak acids/bases (ICE tables), and buffers (Henderson-Hasselbalch). For titration curves, be able to identify the buffer region, half-equivalence point (pH = pKa), and equivalence point, and know why the equivalence pH is above 7 for weak acid/strong base and below 7 for strong acid/weak base. Finally, practice salt hydrolysis — predicting whether a salt solution is acidic, basic, or neutral.

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