Buffer Solutions and pH Curves

This lesson covers buffer solutions, how they work, pH calculations using the Henderson-Hasselbalch equation, pH curves for titrations, and the selection of indicators.

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Buffer Solutions

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

Types of Buffer

1. Acidic buffer (pH < 7): a weak acid and its conjugate base (the salt of the weak acid). Example: ethanoic acid (CH₃COOH) + sodium ethanoate (CH₃COONa).

2. Basic buffer (pH > 7): a weak base and its conjugate acid. Example: ammonia (NH₃) + ammonium chloride (NH₄Cl).

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How an Acidic Buffer Works

The buffer contains a large reservoir of both HA (weak acid) and A⁻ (conjugate base).

When a small amount of acid (H⁺) is added: The added H⁺ ions react with the conjugate base: H⁺ + A⁻ → HA. The A⁻ removes the added H⁺, so pH barely changes.

When a small amount of base (OH⁻) is added: The added OH⁻ ions react with the weak acid: OH⁻ + HA → A⁻ + H₂O. The HA neutralises the added OH⁻, so pH barely changes.

The buffer works because both HA and A⁻ are present in large and roughly comparable amounts, acting as reservoirs.

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Calculating Buffer pH

From Ka = [H⁺][A⁻] / [HA], rearranging:

[H⁺] = Ka × [HA] / [A⁻]

Or equivalently, the Henderson-Hasselbalch equation:

pH = pKa + log₁₀([A⁻] / [HA])

Example 1: A buffer is made from 0.40 mol dm⁻³ ethanoic acid and 0.20 mol dm⁻³ sodium ethanoate. Ka = 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH.

[H⁺] = Ka × [HA] / [A⁻] = 1.74 × 10⁻⁵ × (0.40/0.20) = 1.74 × 10⁻⁵ × 2.0 = 3.48 × 10⁻⁵

pH = −log₁₀(3.48 × 10⁻⁵) = 4.46

Or using Henderson-Hasselbalch: pH = −log₁₀(1.74 × 10⁻⁵) + log₁₀(0.20/0.40) = 4.76 + (−0.301) = 4.46

Example 2: Calculate the pH of a buffer made by mixing 50 cm³ of 0.10 mol dm⁻³ CH₃COOH with 30 cm³ of 0.10 mol dm⁻³ NaOH. Ka = 1.74 × 10⁻⁵ mol dm⁻³.

Moles CH₃COOH = 0.050 × 0.10 = 5.0 × 10⁻³ mol Moles NaOH = 0.030 × 0.10 = 3.0 × 10⁻³ mol

NaOH reacts with CH₃COOH: CH₃COOH + NaOH → CH₃COONa + H₂O