Acid-Base Equilibrium and pH Calculations

Posted on Oct 15, 2025 in Chemistry

Identifying Acid-Base Problem Types

Core Acid-Base Concepts

Acid-Base Properties of Salts

• Strong acid + Strong base → Neutral salt
• Strong base + Weak acid → Basic salt
• Strong acid + Weak base → Acidic salt
• Example: NaClO is formed from NaOH (a strong base) and HClO (a weak acid), so its solution is basic.

The pKₐ and pKₑ Relationship

• The relationship for a conjugate acid-base pair is: pKₐ + pKₑ = 14
• Use this formula to convert between pKₐ and pKₑ.

The Kₐ and Kₑ Relationship

• The relationship for a conjugate acid-base pair is: Kₐ · Kₑ = 1.0 x 10⁻¹⁴
• Use this formula to convert between Kₐ and Kₑ.

Using ICE Tables for Equilibrium

For weak acids or weak bases, set up an ICE (Initial, Change, Equilibrium) table to find equilibrium concentrations.

Generic Weak Acid (HA) Example:

HA ⇌ H⁺ + A⁻

• Initial: [HA] = initial concentration, [H⁺] ≈ 0, [A⁻] = 0
• Change: -x, +x, +x
• Equilibrium: [HA] – x, x, x

The equilibrium constant expression is: Kₐ = [H⁺][A⁻] / [HA] = x² / ([HA] – x)

Assuming x is small, this simplifies to: Kₐ ≈ x² / [HA], so x = [H⁺] = √(Kₐ · [HA])

Then, calculate the pH: pH = -log[H⁺] = -log(x)

Essential Acid-Base Formulas

Step-by-Step pH Calculation Examples

Example 1: Is NaClO Acidic, Basic, or Neutral?

Problem Type: Salt classification

Steps:

1. Identify the ions and their parent acid/base:
   • Na⁺ comes from NaOH, a strong base, so it is a neutral spectator ion.
   • ClO⁻ is the conjugate base of HClO, a weak acid.
2. Determine the effect on pH: Since ClO⁻ is a conjugate base, it will accept a proton (H⁺) from water, producing OH⁻ and making the solution basic.

Conclusion: The solution is basic.

Example 2: pH of a 0.010 M Benzoic Acid Solution

Problem Type: Weak acid equilibrium

Given:

• Initial concentration [C₆H₅CO₂H] = 0.010 M
• pKₐ = 4.8, so Kₐ = 10^-4.8 = 1.58 x 10⁻⁵

Steps:

1. Set up an ICE table for the dissociation: C₆H₅CO₂H ⇌ C₆H₅CO₂⁻ + H⁺
   • Initial: 0.010, 0, 0
   • Change: -x, +x, +x
   • Equilibrium: 0.010 – x, x, x
2. Use the Kₐ expression to solve for x:

   Kₐ = x² / (0.010 – x) ≈ x² / 0.010

   x² = (1.58 x 10⁻⁵)(0.010) = 1.58 x 10⁻⁷

   x = [H⁺] = √(1.58 x 10⁻⁷) = 3.97 x 10⁻⁴ M

3. Solve for pH:

   pH = -log(3.97 x 10⁻⁴) ≈ 3.40

Example 3: pH of a 0.010 M Sodium Benzoate Solution

Problem Type: Weak base (salt hydrolysis)

Given:

• Initial concentration [NaC₆H₅CO₂] = 0.010 M
• The active species is the conjugate base, C₆H₅CO₂⁻.
• pKₐ of benzoic acid = 4.8, so pKₑ = 14 – 4.8 = 9.2
• Kₑ = 10^-9.2 = 6.31 x 10⁻¹⁰

Steps:

1. Write the hydrolysis reaction: C₆H₅CO₂⁻ + H₂O ⇌ C₆H₅CO₂H + OH⁻
2. Set up an ICE table:
   • Initial: 0.010, –, 0, 0
   • Change: -x, –, +x, +x
   • Equilibrium: 0.010 – x, –, x, x
3. Use the Kₑ expression to solve for x:

   Kₑ = x² / (0.010 – x) ≈ x² / 0.010

   x² = (6.31 x 10⁻¹⁰)(0.010) = 6.31 x 10⁻¹²

   x = [OH⁻] = √(6.31 x 10⁻¹²) = 2.51 x 10⁻⁶ M

4. Solve for pOH and then pH:

   pOH = -log(2.51 x 10⁻⁶) ≈ 5.60

   pH = 14 – pOH = 14 – 5.60 = 8.40

Common Errors to Avoid

• Forgetting to convert pKₐ to Kₐ or pKₑ to Kₑ before calculating.
• Using the wrong equilibrium constant (e.g., using Kₐ for a base hydrolysis reaction).
• Forgetting that ions from strong acids/bases (like Na⁺ or Cl⁻) are neutral spectator ions.
• Making mistakes when setting up the equilibrium concentrations in an ICE table.