How to Calculate pH & pOH — Strong/Weak Acids

The Foundation: Water, pH, and pOH

All pH calculations stem from the autoionisation of water:

H₂O ⇌ H⁺ + OH⁻

At 25°C, the ion product of water is:

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴

From this, we derive the key relationships:

• pH = -log₁₀[H⁺]
• pOH = -log₁₀[OH⁻]
• pH + pOH = 14.00 (at 25°C)

These equations allow you to interconvert between [H⁺], [OH⁻], pH, and pOH.

1. Strong Acids and Bases: Complete Dissociation

Strong acids (e.g., HCl, HNO₃) and strong bases (e.g., NaOH, KOH) dissociate 100% in water.

Strong Acid Example:

For a 0.01 M HCl solution:

• [H⁺] = 0.01 M (since HCl → H⁺ + Cl⁻)
• pH = -log(0.01) = 2.00

Strong Base Example:

For a 0.001 M NaOH solution:

• [OH⁻] = 0.001 M
• pOH = -log(0.001) = 3.00
• pH = 14.00 – 3.00 = 11.00

⚠️ Assumption: Valid for concentrations > 10⁻⁶ M. For very dilute solutions, water’s contribution to [H⁺] or [OH⁻] must be considered.

2. Weak Acids and Bases: Partial Dissociation

Weak acids (e.g., CH₃COOH) establish an equilibrium:

HA ⇌ H⁺ + A⁻

The acid dissociation constant (Ka) is:

Ka = [H⁺][A⁻] / [HA]

The ICE Table Method

For a weak acid with initial concentration C: | Species | Initial (M) | Change (M) | Equilibrium (M) | |--------|------------|-----------|----------------| | HA | C | –x | C – x | | H⁺ | 0 | +x | x | | A⁻ | 0 | +x | x |

Substitute into Ka:

Ka = x² / (C – x)

When to Use the Approximation

If C / Ka > 100, then x ≪ C, so:

Ka ≈ x² / C → x ≈ √(Ka·C)

Example: 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵)

• C / Ka = 0.10 / 1.8e-5 ≈ 5,555 > 100 → approximation valid
• [H⁺] ≈ √(1.8e-5 × 0.10) = √(1.8e-6) ≈ 1.34 × 10⁻³ M
• pH ≈ -log(1.34e-3) ≈ 2.87

When to Solve the Quadratic

If C / Ka below 100, solve:

x² + Ka·x – Ka·C = 0

Example: 1.0 × 10⁻⁴ M weak acid (Ka = 1.0 × 10⁻⁵)

• C / Ka = 10 below 100 → use quadratic
• x² + (1.0e-5)x – (1.0e-9) = 0
• x = [H⁺] ≈ 9.5 × 10⁻⁵ M (vs. 1.0 × 10⁻⁴ M from approximation—5% error)

3. Buffer Solutions: The Henderson-Hasselbalch Equation

A buffer contains a weak acid (HA) and its conjugate base (A⁻). Its pH is given by:

pH = pKa + log₁₀([A⁻] / [HA])

Example: 0.30 M HA and 0.20 M A⁻ (pKa = 4.74)

• pH = 4.74 + log(0.20 / 0.30) = 4.74 + log(0.667) ≈ 4.74 – 0.176 = 4.56

💡 Buffer capacity is highest when [A⁻] = [HA] → pH = pKa

Pro Tips & Common Mistakes

• Always identify the system first: Strong vs. weak vs. buffer.
• Check the C/Ka ratio before using the approximation.
• Use concentrations, not moles, in Henderson-Hasselbalch (unless volume is identical).
• Report pH to 2 decimal places for typical lab concentrations (e.g., 0.10 M).
• Temperature matters: Kw = 1.0 × 10⁻¹⁴ only at 25°C.
• For polyprotic acids, usually only the first dissociation matters (Ka₁ >> Ka₂).

Practical Applications

• Biological systems: Blood pH buffered at ~7.4 by H₂CO₃/HCO₃⁻
• Environmental science: Monitoring acid rain (pH below 5.6)
• Industrial processes: Controlling pH in fermentation, dyeing, wastewater treatment
• Laboratory work: Preparing buffer solutions for experiments