Definition:
pH is a measure of the acidity or basicity (alkalinity) of an aqueous solution. It quantifies the concentration of hydrogen ions (H⁺) or hydronium ions (H₃O⁺) in a solution.

The term "pH":
• "p" stands for "power" (mathematical power/exponent)
• "H" stands for hydrogen ion concentration
• pH = "power of hydrogen"

Key Points:
• pH is a logarithmic scale (base 10)
• Lower pH = more acidic (more H⁺ ions)
• Higher pH = more basic/alkaline (fewer H⁺ ions)
• pH scale typically ranges from 0 to 14

\[ \text{pH} = -\log[H^+] \]

or

\[ \text{pH} = -\log[H_3O^+] \]

Where:
• pH = measure of acidity/basicity
• [H⁺] = hydrogen ion concentration (mol/L or M)
• [H₃O⁺] = hydronium ion concentration (mol/L or M)
• log = logarithm base 10
• The negative sign means higher [H⁺] gives lower pH

Note: [H⁺] and [H₃O⁺] are equivalent and interchangeable. In aqueous solutions, hydrogen ions exist as hydronium ions.

\[ [H^+] = 10^{-\text{pH}} \]

or

\[ [H_3O^+] = 10^{-\text{pH}} \]

Use this formula when:
• You know the pH value
• You need to find hydrogen ion concentration
• You're working backwards from pH to concentration

Example: If pH = 3, find [H⁺]

\[ [H^+] = 10^{-3} = 0.001 \text{ M} = 1.0 \times 10^{-3} \text{ M} \]

Acidic Solutions: pH < 7

• pH 0-3: Strong acids (HCl, H₂SO₄, HNO₃)
• pH 3-7: Weak acids (CH₃COOH, citric acid)
• High [H⁺] concentration
• Examples: stomach acid (pH ~2), lemon juice (pH ~2), vinegar (pH ~3)

Neutral Solution: pH = 7

• [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M
• Pure water at 25°C
• Neither acidic nor basic

Basic/Alkaline Solutions: pH > 7

• pH 7-11: Weak bases (NH₃, baking soda)
• pH 11-14: Strong bases (NaOH, KOH)
• Low [H⁺] concentration, high [OH⁻]
• Examples: soap (pH ~10), bleach (pH ~13), drain cleaner (pH ~14)

Important: Each unit change in pH represents a 10-fold change in [H⁺]. pH 5 is 10 times more acidic than pH 6!

\[ \text{pOH} = -\log[OH^-] \]

Where:
• pOH = measure of basicity
• [OH⁻] = hydroxide ion concentration (mol/L or M)
• Used primarily for basic solutions

Reverse formula:

\[ [OH^-] = 10^{-\text{pOH}} \]

\[ \text{pH} + \text{pOH} = 14 \]

(at 25°C)

Derived formulas:
• \(\text{pH} = 14 - \text{pOH}\)
• \(\text{pOH} = 14 - \text{pH}\)

Example 1: If pH = 3, find pOH

\[ \text{pOH} = 14 - 3 = 11 \]

Example 2: If pOH = 2.5, find pH

\[ \text{pH} = 14 - 2.5 = 11.5 \]

\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]

(at 25°C)

Where:
• Kw = ion product constant for water
• [H⁺] = hydrogen ion concentration
• [OH⁻] = hydroxide ion concentration
• This relationship is ALWAYS true in aqueous solutions

Derived formulas:
• \([H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{[OH^-]}\)
• \([OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{[H^+]}\)

Logarithmic form:
\[ pK_w = \text{pH} + \text{pOH} = 14 \]

Strong acids completely dissociate:
• HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄
• [H⁺] = concentration of acid
• Direct calculation: pH = -log[acid concentration]

Example 1: Find pH of 0.0025 M HCl

HCl is strong, so [H⁺] = 0.0025 M = 2.5 × 10⁻³ M
\[ \text{pH} = -\log(2.5 \times 10^{-3}) = 2.60 \]

Example 2: Find pH of 0.055 M HNO₃

HNO₃ is strong, so [H⁺] = 0.055 M = 5.5 × 10⁻² M
\[ \text{pH} = -\log(5.5 \times 10^{-2}) = 1.26 \]

Strong bases completely dissociate:
• NaOH, KOH, LiOH, Ba(OH)₂, Ca(OH)₂
• [OH⁻] = concentration of base
• Calculate pOH first, then pH

Steps:

1. Find [OH⁻] = concentration of base
2. Calculate pOH = -log[OH⁻]
3. Calculate pH = 14 - pOH

Example: Find pH of 0.0035 M LiOH

[OH⁻] = 0.0035 M = 3.5 × 10⁻³ M
\[ \text{pOH} = -\log(3.5 \times 10^{-3}) = 2.46 \] \[ \text{pH} = 14 - 2.46 = 11.54 \]

Weak acids partially dissociate:
• CH₃COOH, HF, HNO₂, H₃PO₄
• Use acid dissociation constant (Ka)
• Requires ICE table or equilibrium calculations

Acid Dissociation Constant:

\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

pKa Formula:

\[ pK_a = -\log K_a \]

\[ K_a = 10^{-pK_a} \]

Note: Smaller Ka (or larger pKa) = weaker acid

Weak bases partially dissociate:
• NH₃, CH₃NH₂, pyridine
• Use base dissociation constant (Kb)
• Calculate pOH first, then pH

Base Dissociation Constant:

\[ K_b = \frac{[OH^-][BH^+]}{[B]} \]

pKb Formula:

\[ pK_b = -\log K_b \]

\[ K_b = 10^{-pK_b} \]

Ka and Kb Relationship:

\[ K_a \times K_b = K_w = 1.0 \times 10^{-14} \]

\[ pK_a + pK_b = 14 \]

Example 1: Calculate pH when [H⁺] = 1.6 × 10⁻⁴ M

\[ \text{pH} = -\log(1.6 \times 10^{-4}) = 3.80 \]

Example 2: Find [H⁺] when pH = 3.1

\[ [H^+] = 10^{-3.1} = 7.9 \times 10^{-4} \text{ M} \]

Example 3: Find pH when [OH⁻] = 0.015 M

Step 1: \(\text{pOH} = -\log(0.015) = 1.82\)
Step 2: \(\text{pH} = 14 - 1.82 = 12.18\)

Example 4: Find [H⁺] when [OH⁻] = 4.2 × 10⁻³ M

\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{4.2 \times 10^{-3}} = 2.4 \times 10^{-12} \text{ M} \]

❌ Forgetting the negative sign in pH formula
✅ pH = -log[H⁺], NOT log[H⁺]

❌ Confusing pH and pOH
✅ pH measures [H⁺], pOH measures [OH⁻]

❌ Using wrong base for logarithm
✅ Always use log base 10 (common logarithm)

❌ Treating weak acids like strong acids
✅ Use Ka and equilibrium calculations for weak acids

❌ Forgetting pH + pOH = 14 at 25°C
✅ This relationship is fundamental

\[ \text{pH} = -\log[H^+] \] \[ [H^+] = 10^{-\text{pH}} \]

\[ \text{pOH} = -\log[OH^-] \] \[ [OH^-] = 10^{-\text{pOH}} \]

\[ \text{pH} + \text{pOH} = 14 \] \[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]

Key Points:
• pH < 7: Acidic
• pH = 7: Neutral
• pH > 7: Basic/Alkaline
• Each pH unit = 10× change in [H⁺]