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R3.1 - Proton transfer reactions

3.1.1 Bronsted-Lowry Acid-Base Theory 3.1.2 Conjugation Acid-Base Pairs 3.1.3 Acid-Base Behaviour and Oxides 3.1.4 pH and [H+] 3.1.5 Kw and pH of Water 3.1.6 Strong Vs. Weak Acids and Bases 3.1.7 Neutralization Reaction 3.1.8 pH Curves 3.1.9 pH and [OH-] (AHL) 3.1.10 Ka, Kb, pKa and pKb (AHL) 3.1.11 Ka x Kb = Kw (AHL) 3.1.12 pH of Salt Solutions (AHL) 3.1.13 pH Curves for Acid-Base Reaction (AHL) 3.1.14 Acid-Base Indicators (AHL) 3.1.15 Indicators and Titration Points (AHL) 3.1.16 Buffer Solutions (AHL) 3.1.17 pH of Buffer (AHL)

The pH of Buffer Solutions HL Only

Specification Reference R3.1.17

Quick Notes

  • Buffer pH depends on:
    • The pKa of the weak acid (or pKb of the weak base).
    • The ratio of [acid]/[conjugate base] or [base]/[conjugate acid].
  • Dilution affects buffer capacity but pH is largely unchanged.

Full Notes

Background theory to buffers is covered here. Make sure you are comfortable with that page before looking at this one.

Two key factors determine the pH of a buffer:

Solving Buffer pH Problems

Worked Example

You have a buffer solution made by mixing 0.200 mol of ethanoic acid (CH3COOH) with 0.100 mol of sodium ethanoate (CH3COONa) in 1.00 dm³ of solution. Find the pH of the buffer solution.

  1. The Ka of ethanoic acid is 1.74 × 10⁻⁵ mol dm⁻³.
    Use the Ka expression: Ka = [H⁺][A⁻] / [HA]
    Rearranged to find [H⁺]: [H⁺] = (Ka × [HA]) / [A⁻]
  2. Substitute the values:
    [HA] = 0.200 mol dm⁻³
    [A⁻] = 0.100 mol dm⁻³
    Ka = 1.74 × 10⁻⁵
    [H⁺] = (1.74 × 10⁻⁵ × 0.200) / 0.100 = 3.48 × 10⁻⁵ mol dm⁻³
  3. Calculate pH:
    pH = –log₁₀(3.48 × 10⁻⁵) ≈ 4.46

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Matt’s exam tip

Remember pKa values can also be used instead. To convert pKa to Ka just use Ka = 10−pKa.

Effect of Dilution

Dilution decreases both [HA] and [A⁻] equally, so the ratio remains constant.

Therefore, pH changes very little. However, the buffering capacity is reduced – it cannot neutralize as much added acid or base.

Summary

Linked Course Question

Reactivity 2.3 — Linked Course Question

How does Le Châtelier’s principle enable us to interpret the behaviour of indicators and buffer solutions?

In a buffer solution, the weak acid and its conjugate base are in equilibrium:
HA ⇌ H⁺ + A⁻

  • If acid (H⁺) is added, the equilibrium shifts left, reducing [H⁺] by forming more HA.
  • If base (OH⁻) is added, it removes H⁺, so the equilibrium shifts right to replace it.

This shift helps maintain a relatively constant pH.

For acid–base indicators, the coloured forms exist in an equilibrium:
HIn ⇌ H⁺ + In⁻ (where HIn and In⁻ are different colours)

  • Adding acid increases [H⁺], shifting equilibrium left, showing the colour of HIn.
  • Adding base decreases [H⁺], shifting equilibrium right, showing the colour of In⁻.