AP CHEMISTRY

Acid-Base Equilibria

BRILLIANT MINDS ACADEMY – BRILLIANTMINDSTESTPREP.COM

1. Introduction to Acid-Base Equilibria

Acid-base equilibria is one of the most important topics in chemistry, forming the foundation for understanding countless chemical and biological processes. From the pH of blood to industrial manufacturing, acid-base reactions govern the behavior of solutions around us.

Learning Objectives

  • Understand and apply Brønsted-Lowry acid-base theory
  • Calculate pH and pOH of strong and weak acids/bases
  • Use equilibrium constants (Ka and Kb) in calculations
  • Apply the Henderson-Hasselbalch equation to buffer solutions
  • Analyze acid-base titrations and titration curves
  • Understand buffer capacity and the common ion effect

2. Acid-Base Theories

2.1 Arrhenius Theory (Historical)

The Arrhenius theory, while limited, provides a basic understanding:

  • Arrhenius Acid: A substance that produces H⁺ ions when dissolved in water
  • Arrhenius Base: A substance that produces OH⁻ ions when dissolved in water

2.2 Brønsted-Lowry Theory (Primary Theory for AP Chemistry)

This more comprehensive theory is the foundation of AP Chemistry acid-base chemistry:

Brønsted-Lowry Definitions

Acid: A proton (H⁺) donor

Base: A proton (H⁺) acceptor

In every acid-base reaction, there is an acid-base conjugate pair. When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

HA + H₂O ⇌ H₃O⁺ + A⁻
acid₁ base₂ acid₂ base₁

Example 1: Identifying Conjugate Acid-Base Pairs

Question: Identify the conjugate acid-base pairs in the following reaction:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Solution:
Pair 1: NH₃ (base) and NH₄⁺ (conjugate acid)
NH₃ accepts a proton to become NH₄⁺

Pair 2: H₂O (acid) and OH⁻ (conjugate base)
H₂O donates a proton to become OH⁻

Key Principle: The stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate acid. Equilibrium always favors the formation of the weaker acid and weaker base.

3. Autoionization of Water and the pH Scale

3.1 Autoionization of Water

Water can act as both an acid and a base (amphoteric), undergoing self-ionization:

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

The equilibrium constant for this reaction is called the ion product constant for water:

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

3.2 The pH and pOH Scales

Important Formulas

pH = -log[H₃O⁺] or pH = -log[H⁺]
pOH = -log[OH⁻]
pH + pOH = 14.00 (at 25°C)
[H₃O⁺] = 10⁻ᵖᴴ
[OH⁻] = 10⁻ᵖᴼᴴ
Solution Type pH [H₃O⁺] vs [OH⁻]
Acidic pH < 7 [H₃O⁺] > [OH⁻]
Neutral pH = 7 [H₃O⁺] = [OH⁻]
Basic pH > 7 [H₃O⁺] < [OH⁻]

4. Strong Acids and Strong Bases

4.1 Strong Acids

Strong acids dissociate completely (100%) in aqueous solution. You must memorize these seven strong acids:

Strong Acid Formula Name
1 HCl Hydrochloric acid
2 HBr Hydrobromic acid
3 HI Hydroiodic acid
4 HNO₃ Nitric acid
5 HClO₄ Perchloric acid
6 H₂SO₄ Sulfuric acid (first proton only)
7 HClO₃ Chloric acid

4.2 Strong Bases

Strong bases dissociate completely in aqueous solution. Common strong bases include:

  • Group 1A hydroxides: LiOH, NaOH, KOH, RbOH, CsOH
  • Heavy Group 2A hydroxides: Ca(OH)₂, Sr(OH)₂, Ba(OH)₂

Example 2: pH of a Strong Acid

Question: Calculate the pH of a 0.0045 M HCl solution.

Step 1: Recognize that HCl is a strong acid and dissociates completely.
HCl → H⁺ + Cl⁻
Step 2: Since HCl dissociates completely:
[H⁺] = 0.0045 M
Step 3: Calculate pH:
pH = -log[H⁺]
pH = -log(0.0045)
pH = 2.35

Example 3: pH of a Strong Base

Question: Calculate the pH of a 0.025 M NaOH solution.

Step 1: NaOH is a strong base and dissociates completely.
NaOH → Na⁺ + OH⁻
Step 2: Since NaOH dissociates completely:
[OH⁻] = 0.025 M
Step 3: Calculate pOH:
pOH = -log[OH⁻]
pOH = -log(0.025)
pOH = 1.60
Step 4: Calculate pH:
pH = 14.00 – pOH
pH = 14.00 – 1.60
pH = 12.40

5. Weak Acids and Weak Bases

5.1 Weak Acid Equilibria

Weak acids only partially ionize in solution. Any acid not on the strong acid list is considered weak. The most common weak acid is acetic acid (CH₃COOH).

HA + H₂O ⇌ H₃O⁺ + A⁻

The acid dissociation constant (Ka) measures the strength of a weak acid:

Ka = [H₃O⁺][A⁻] / [HA]

Important Relationships

pKa = -log(Ka)

Ka = 10⁻ᵖᴷᵃ

The smaller the pKa, the stronger the acid.

The larger the Ka, the stronger the acid.

5.2 ICE Tables for Weak Acids

To solve weak acid equilibrium problems, we use ICE (Initial, Change, Equilibrium) tables:

Example 4: pH of a Weak Acid Solution

Question: Calculate the pH of a 0.100 M solution of acetic acid (CH₃COOH). Ka = 1.8 × 10⁻⁵

Step 1: Write the equilibrium reaction:
CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻
Step 2: Set up an ICE table:

  CH₃COOH H₃O⁺ CH₃COO⁻
Initial (M) 0.100 ≈0 0
Change (M) -x +x +x
Equilibrium (M) 0.100 – x x x
Step 3: Write the Ka expression and substitute:
Ka = [H₃O⁺][CH₃COO⁻] / [CH₃COOH]
1.8 × 10⁻⁵ = (x)(x) / (0.100 – x)
Step 4: Apply the 5% approximation rule. If Ka is much smaller than the initial concentration (Ka/C < 0.05), we can assume x is negligible compared to 0.100:
1.8 × 10⁻⁵ ≈ x² / 0.100
x² = 1.8 × 10⁻⁶
x = 1.34 × 10⁻³ M
Step 5: Check the approximation: (1.34 × 10⁻³ / 0.100) × 100% = 1.34% < 5% ✓
The approximation is valid.
Step 6: Calculate pH:
[H₃O⁺] = x = 1.34 × 10⁻³ M
pH = -log(1.34 × 10⁻³)
pH = 2.87

5.3 Weak Base Equilibria

Weak bases partially accept protons in solution. The most common weak base is ammonia (NH₃).

B + H₂O ⇌ BH⁺ + OH⁻

The base dissociation constant (Kb) measures the strength of a weak base:

Kb = [BH⁺][OH⁻] / [B]
Critical Relationship: For a conjugate acid-base pair:
Ka × Kb = Kw = 1.0 × 10⁻¹⁴
pKa + pKb = 14.00

Example 5: pH of a Weak Base Solution

Question: Calculate the pH of a 0.150 M solution of ammonia (NH₃). Kb = 1.8 × 10⁻⁵

Step 1: Write the equilibrium reaction:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Step 2: Set up an ICE table:

  NH₃ NH₄⁺ OH⁻
Initial (M) 0.150 0 ≈0
Change (M) -x +x +x
Equilibrium (M) 0.150 – x x x
Step 3: Write the Kb expression:
Kb = [NH₄⁺][OH⁻] / [NH₃]
1.8 × 10⁻⁵ = (x)(x) / (0.150 – x)
Applying the 5% rule:
1.8 × 10⁻⁵ ≈ x² / 0.150
x² = 2.7 × 10⁻⁶
x = 1.64 × 10⁻³ M
Step 4: Calculate pOH and pH:
[OH⁻] = 1.64 × 10⁻³ M
pOH = -log(1.64 × 10⁻³) = 2.79
pH = 14.00 – 2.79
pH = 11.21

5.4 Percent Ionization

Percent ionization measures the degree to which a weak acid or base dissociates:

% Ionization = ([H₃O⁺] at equilibrium / [HA] initial) × 100%

For the acetic acid example above:

% Ionization = (1.34 × 10⁻³ / 0.100) × 100% = 1.34%

Key Trend: Percent ionization increases as the solution becomes more dilute. Weak acids ionize more in dilute solutions.

6. Buffer Solutions

6.1 What is a Buffer?

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers consist of:

  • A weak acid and its conjugate base (e.g., CH₃COOH and CH₃COO⁻)
  • OR a weak base and its conjugate acid (e.g., NH₃ and NH₄⁺)

How Buffers Work

When acid (H⁺) is added: The conjugate base neutralizes it

When base (OH⁻) is added: The weak acid neutralizes it

This dual action maintains a relatively constant pH.

6.2 The Henderson-Hasselbalch Equation

This equation is the key to solving buffer problems efficiently:

pH = pKa + log([A⁻] / [HA])

For basic buffers:

pOH = pKb + log([BH⁺] / [B])
Remember: “Ion on top!” The conjugate (ionic species) always goes in the numerator of the log ratio.

Example 6: pH of a Buffer Solution

Question: Calculate the pH of a buffer solution containing 0.50 M acetic acid (CH₃COOH) and 0.75 M sodium acetate (CH₃COONa). Ka = 1.8 × 10⁻⁵

Step 1: Calculate pKa:
pKa = -log(1.8 × 10⁻⁵)
pKa = 4.74
Step 2: Identify the acid and conjugate base:
[HA] = [CH₃COOH] = 0.50 M
[A⁻] = [CH₃COO⁻] = 0.75 M
Step 3: Apply the Henderson-Hasselbalch equation:
pH = pKa + log([A⁻] / [HA])
pH = 4.74 + log(0.75 / 0.50)
pH = 4.74 + log(1.5)
pH = 4.74 + 0.18
pH = 4.92

Example 7: Buffer After Adding Strong Acid

Question: A buffer contains 0.40 M NH₃ and 0.30 M NH₄Cl. Calculate the pH after adding 0.050 mol of HCl to 1.0 L of this buffer. Kb for NH₃ = 1.8 × 10⁻⁵

Step 1: Calculate initial moles:
mol NH₃ = 0.40 mol (in 1.0 L)
mol NH₄⁺ = 0.30 mol (in 1.0 L)
mol HCl added = 0.050 mol
Step 2: HCl reacts with NH₃ (the base component):
NH₃ + HCl → NH₄⁺ + Cl⁻

mol NH₃ after = 0.40 – 0.050 = 0.35 mol
mol NH₄⁺ after = 0.30 + 0.050 = 0.35 mol

Step 3: Calculate pKb and pKa:
pKb = -log(1.8 × 10⁻⁵) = 4.74
pKa = 14.00 – pKb = 14.00 – 4.74 = 9.26
(We need pKa because Henderson-Hasselbalch uses pH)
Step 4: Apply Henderson-Hasselbalch:
We can use moles directly since volume is constant:
pH = pKa + log([NH₃] / [NH₄⁺])
pH = 9.26 + log(0.35 / 0.35)
pH = 9.26 + log(1)
pH = 9.26 + 0
pH = 9.26

6.3 Buffer Capacity

Buffer capacity is the amount of acid or base a buffer can neutralize before a significant pH change occurs. Buffer capacity depends on:

  • Concentration: Higher concentrations = greater capacity
  • Ratio: Maximum capacity when [A⁻]/[HA] = 1 (pH = pKa)
Buffer Range: Buffers are most effective within ±1 pH unit of the pKa
Effective buffer range: pKa – 1 < pH < pKa + 1

7. Polyprotic Acids

Polyprotic acids can donate more than one proton. Each proton removal has its own Ka value.

7.1 Diprotic Acid Example: H₂SO₃

First ionization: H₂SO₃ ⇌ H⁺ + HSO₃⁻ Ka1 = 1.3 × 10⁻²
Second ionization: HSO₃⁻ ⇌ H⁺ + SO₃²⁻ Ka2 = 6.3 × 10⁻⁸
Key Pattern: Ka1 > Ka2 > Ka3 (if present)
Each successive proton is harder to remove because you’re removing a positive ion from an increasingly negative species.

Example 8: pH of a Polyprotic Acid

Question: Calculate the pH of a 0.10 M H₂SO₃ solution. Ka1 = 1.3 × 10⁻² and Ka2 = 6.3 × 10⁻⁸

Step 1: Since Ka1 >> Ka2, essentially all H⁺ comes from the first ionization. We can ignore the second ionization for pH calculation.
Step 2: Set up ICE table for first ionization:
H₂SO₃ ⇌ H⁺ + HSO₃⁻

  H₂SO₃ H⁺ HSO₃⁻
Initial (M) 0.10 0 0
Change (M) -x +x +x
Equilibrium (M) 0.10 – x x x
Step 3: Since Ka1 is relatively large (1.3 × 10⁻²), we cannot use the approximation. Use the quadratic equation:
1.3 × 10⁻² = x² / (0.10 – x)
x² + 0.013x – 0.0013 = 0
Using the quadratic formula:
x = 0.031 M
Step 4: Calculate pH:
[H⁺] = 0.031 M
pH = -log(0.031)
pH = 1.51

8. Acid-Base Properties of Salts

When salts dissolve in water, they can create acidic, basic, or neutral solutions depending on their component ions.

8.1 Rules for Predicting Salt Solution pH

Cation Anion Result Example
From strong base From strong acid Neutral (pH = 7) NaCl, KNO₃
From strong base From weak acid Basic (pH > 7) NaCH₃COO, K₂CO₃
From weak base From strong acid Acidic (pH < 7) NH₄Cl, NH₄NO₃
From weak base From weak acid Compare Ka and Kb NH₄CH₃COO
Metal Cation Acidity: Small, highly charged metal cations (Al³⁺, Fe³⁺, Cr³⁺) act as weak acids in water due to polarization of surrounding water molecules.

Example 9: pH of a Salt Solution

Question: Calculate the pH of a 0.20 M solution of sodium acetate (NaCH₃COO). Ka for acetic acid = 1.8 × 10⁻⁵

Step 1: Identify the acid-base properties:
Na⁺ comes from strong base (NaOH) → neutral
CH₃COO⁻ comes from weak acid (CH₃COOH) → acts as a base
Solution will be basic
Step 2: Calculate Kb for acetate ion:
Ka × Kb = Kw
Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
Kb = 5.6 × 10⁻¹⁰
Step 3: Write equilibrium and ICE table:
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

  CH₃COO⁻ CH₃COOH OH⁻
I 0.20 0 ≈0
C -x +x +x
E 0.20 – x x x
Step 4: Solve for x:
5.6 × 10⁻¹⁰ = x² / 0.20
x² = 1.12 × 10⁻¹⁰
x = 1.06 × 10⁻⁵ M = [OH⁻]
Step 5: Calculate pH:
pOH = -log(1.06 × 10⁻⁵) = 4.97
pH = 14.00 – 4.97
pH = 9.03

9. Acid-Base Titrations

9.1 Types of Titrations

Titrations involve the gradual addition of a solution of known concentration (titrant) to a solution of unknown concentration (analyte) until the equivalence point is reached.

9.2 Strong Acid – Strong Base Titration

Key Characteristics

  • Sharp pH change at equivalence point
  • Equivalence point pH = 7.00
  • Any indicator with transition range 4-10 will work

Example 10: Strong Acid – Strong Base Titration

Question: 25.0 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the pH after adding:

(a) 0.0 mL of NaOH
(b) 12.5 mL of NaOH
(c) 25.0 mL of NaOH (equivalence point)
(d) 30.0 mL of NaOH

(a) pH at 0.0 mL NaOH:
This is just the pH of 0.100 M HCl
pH = -log(0.100) = 1.00
(b) pH at 12.5 mL NaOH (halfway to equivalence):
mol HCl initially = 0.0250 L × 0.100 M = 0.00250 mol
mol NaOH added = 0.0125 L × 0.100 M = 0.00125 mol
mol HCl remaining = 0.00250 – 0.00125 = 0.00125 mol
Total volume = 25.0 + 12.5 = 37.5 mL = 0.0375 L
[H⁺] = 0.00125 mol / 0.0375 L = 0.0333 M
pH = -log(0.0333) = 1.48
(c) pH at 25.0 mL NaOH (equivalence point):
At equivalence point, moles of acid = moles of base
They completely neutralize each other
Only H₂O and spectator ions (Na⁺, Cl⁻) remain
pH = 7.00
(d) pH at 30.0 mL NaOH (past equivalence):
mol NaOH added = 0.0300 L × 0.100 M = 0.00300 mol
mol HCl = 0.00250 mol
Excess NaOH = 0.00300 – 0.00250 = 0.00050 mol
Total volume = 25.0 + 30.0 = 55.0 mL = 0.0550 L
[OH⁻] = 0.00050 mol / 0.0550 L = 0.00909 M
pOH = -log(0.00909) = 2.04
pH = 14.00 – 2.04 = 11.96

9.3 Weak Acid – Strong Base Titration

Key Characteristics

  • pH > 7 at equivalence point (weak conjugate base present)
  • Buffer region before equivalence point
  • At halfway point: pH = pKa
  • Less sharp pH change than strong acid-strong base

Example 11: Weak Acid – Strong Base Titration

Question: 50.0 mL of 0.100 M acetic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵) is titrated with 0.100 M NaOH. Calculate the pH:

(a) Initially
(b) At halfway point
(c) At equivalence point

(a) Initial pH:
This is a weak acid problem (Example 4 approach)
Using ICE table: [H⁺] = 1.34 × 10⁻³ M
pH = 2.87
(b) pH at halfway point (25.0 mL NaOH added):
At halfway point, half the acid is neutralized
[CH₃COOH] = [CH₃COO⁻]
Using Henderson-Hasselbalch:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = pKa + log(1) = pKa
pKa = -log(1.8 × 10⁻⁵) = 4.74
pH = 4.74
(c) pH at equivalence point (50.0 mL NaOH added):
mol CH₃COOH initially = 0.0500 L × 0.100 M = 0.00500 mol
mol NaOH added = 0.0500 L × 0.100 M = 0.00500 mol
All acetic acid converts to acetate ion
Total volume = 100.0 mL = 0.100 L
[CH₃COO⁻] = 0.00500 mol / 0.100 L = 0.0500 M

Now solve as a weak base problem:
Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰
Using approximation: x² / 0.0500 = 5.6 × 10⁻¹⁰
x = [OH⁻] = 5.3 × 10⁻⁶ M
pOH = 5.28
pH = 8.72

9.4 Selecting Indicators

Indicator pH Transition Range Color Change Best For
Methyl orange 3.1 – 4.4 Red to yellow Strong acid – strong base
Bromocresol green 3.8 – 5.4 Yellow to blue Weak base – strong acid
Methyl red 4.2 – 6.3 Red to yellow Strong acid – strong base
Bromothymol blue 6.0 – 7.6 Yellow to blue Strong acid – strong base
Phenolphthalein 8.3 – 10.0 Colorless to pink Weak acid – strong base

10. Lewis Acids and Bases

The Lewis definition is the most general acid-base theory:

Lewis Definitions

Lewis Acid: An electron pair acceptor (electrophile)

Lewis Base: An electron pair donor (nucleophile)

Common Lewis acids: Metal cations (Al³⁺, Fe³⁺), molecules with incomplete octets (BF₃, AlCl₃)

Common Lewis bases: Molecules with lone pairs (NH₃, H₂O, CN⁻)

Example 12: Identifying Lewis Acids and Bases

Question: Identify the Lewis acid and Lewis base in the following reaction:

BF₃ + NH₃ → F₃B-NH₃
Solution:
Lewis Base: NH₃ (donates a lone pair of electrons from nitrogen)
Lewis Acid: BF₃ (accepts the electron pair; boron has incomplete octet)

The nitrogen lone pair forms a coordinate covalent bond with the boron atom.

Key Formulas and Relationships Summary

Concept Formula
pH definition pH = -log[H⁺]
pOH definition pOH = -log[OH⁻]
pH and pOH relationship pH + pOH = 14.00
Ion product of water Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴
Acid dissociation constant Ka = [H⁺][A⁻] / [HA]
Base dissociation constant Kb = [BH⁺][OH⁻] / [B]
Conjugate pair relationship Ka × Kb = Kw
Henderson-Hasselbalch (acidic) pH = pKa + log([A⁻]/[HA])
Henderson-Hasselbalch (basic) pOH = pKb + log([BH⁺]/[B])
Percent ionization % = ([H⁺]eq / [HA]initial) × 100%

Problem-Solving Strategies

1. Identifying Solution Type

  • Strong acid → Use concentration directly for [H⁺]
  • Strong base → Use concentration directly for [OH⁻]
  • Weak acid → Use Ka and ICE table
  • Weak base → Use Kb and ICE table
  • Buffer → Use Henderson-Hasselbalch
  • Salt → Identify cation and anion sources

2. ICE Table Method

  1. Write balanced equilibrium equation
  2. Set up ICE table with initial concentrations
  3. Express changes in terms of x
  4. Write equilibrium expression
  5. Check if 5% approximation applies (Ka/C < 0.05)
  6. Solve for x (use quadratic if approximation fails)
  7. Calculate pH or pOH

3. Buffer Problems

  1. Identify weak acid and conjugate base (or weak base and conjugate acid)
  2. If adding acid/base, calculate new amounts using stoichiometry first
  3. Apply Henderson-Hasselbalch equation
  4. Remember: use moles if volumes are different, ratios cancel

4. Titration Problems

  1. Calculate moles of acid and base
  2. Determine which is in excess (or if at equivalence)
  3. Before equivalence: excess reactant problem
  4. At equivalence: salt hydrolysis problem
  5. After equivalence: excess titrant problem

11. Practice Problems

Practice Problem 1

Calculate the pH of a solution prepared by mixing 100 mL of 0.10 M HCl with 50 mL of 0.20 M NaOH.

Practice Problem 2

A buffer is prepared by mixing 0.25 mol of benzoic acid (C₆H₅COOH, Ka = 6.3 × 10⁻⁵) and 0.15 mol of sodium benzoate in enough water to make 1.0 L of solution. Calculate:

(a) The pH of the buffer
(b) The pH after adding 0.05 mol of HCl

Practice Problem 3

Calculate the pH at which equal concentrations of H₂PO₄⁻ and HPO₄²⁻ exist in solution. For H₂PO₄⁻: Ka = 6.2 × 10⁻⁸

Practice Problem 4

A 25.0 mL sample of 0.150 M HNO₂ (Ka = 4.0 × 10⁻⁴) is titrated with 0.200 M KOH. Calculate the pH:

(a) Before any base is added
(b) At the halfway point
(c) At the equivalence point

12. Common Mistakes to Avoid

 
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