Acids, Bases, and pH

pH Scale

$\text{pH} = -\log[H^+]$ $\text{pOH} = -\log[OH^-]$ $\text{pH} + \text{pOH} = 14 \quad (\text{at 25°C})$

Key Relationship: [H⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴ at 25°C

Strong vs. Weak Acids and Bases

Strong Acids (completely dissociate):

HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄

For a strong acid: [H⁺] = acid concentration Example: 0.01 M HCl → [H⁺] = 0.01 M → pH = -log(0.01) = 2.0

Strong Bases (completely dissociate):

NaOH, KOH, Ca(OH)₂, Ba(OH)₂

For a strong base: [OH⁻] = base concentration (×2 for Ca(OH)₂) Example: 0.001 M NaOH → [OH⁻] = 0.001 M → pOH = 3.0 → pH = 11.0

Weak Acids and Bases

Only partially dissociate. Characterized by Ka (acid dissociation constant) or Kb (base dissociation constant).

For weak acid HA ⇌ H⁺ + A⁻: $K_a = \frac{[H^+][A^-]}{[HA]}$

Relationship: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (for conjugate acid-base pairs)

Buffer Solutions

A buffer resists pH changes when small amounts of acid or base are added. It 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⁺)

Henderson-Hasselbalch Equation

$\text{pH} = \text{p}K_a + \log\frac{[A^-]}{[HA]}$

Example: A buffer contains 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵) and 0.15 M sodium acetate.

pH = -log(1.8 × 10⁻⁵) + log(0.15/0.1) = 4.74 + 0.18 = 4.92

Buffer capacity is greatest when [A⁻] = [HA], making pH = pKa.

Neutralization Reactions

Acid + Base → Salt + Water

Example: HCl + NaOH → NaCl + H₂O

At the equivalence point, moles of H⁺ = moles of OH⁻: $M_a \times V_a = M_b \times V_b$

Example: How many mL of 0.5 M NaOH are needed to neutralize 25 mL of 0.3 M HCl?

V_b = (0.3 × 25) / 0.5 = 15 mL