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1.10.7Physical Chemistry

Buffer Action

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Section 1 of 5What a Buffer Is

1.10.7 Buffer Action

A buffer keeps the pH nearly constant when small amounts of acid or alkali are added.

Acidic buffer
  • A mixture of a weak acid and its salt, such as CH₃COOH with CH₃COONa, which holds the pH below 7.
Basic buffer
  • A mixture of a weak base and its salt, such as NH₃ with NH₄Cl, which holds the pH above 7.
Reserve
  • The weak acid HA provides a reserve to remove added alkali, and the conjugate base A⁻ provides a reserve to remove added acid.
Making a buffer
  • Partly neutralise a weak acid with alkali, so some HA remains alongside the A⁻ formed.

Section 2 of 5How the Buffer Works

The equilibrium HA ⇌ H⁺ + A⁻ moves to oppose any change.

1
The mixture contains a large reserve of both HA and A⁻ because the salt is fully ionised.
2
When a small amount of acid is added, extra H⁺ reacts with A⁻ to form HA and the equilibrium shifts left.
3
When a small amount of alkali is added, OH⁻ reacts with H⁺ to form water and HA dissociates to replace the H⁺ used.
4
Because the ratio [HA] : [A⁻] changes only slightly, [H⁺] and hence pH change very little.
5
The buffer fails if so much acid or alkali is added that one of the two components runs out.

Section 3 of 5Buffer Calculations and Uses

Rearranging the Ka expression gives the pH of the buffer directly.

Buffer Calculations and Uses
Formula
  • [H⁺] = Ka × [HA] ÷ [A⁻], so pH can be found from the ratio of acid to salt.
Equal amounts
  • If [HA] = [A⁻], then [H⁺] = Ka and pH = pKa, which is the best position for buffering both ways.
Blood
  • Blood is buffered at about pH 7.4 by the carbonic acid–hydrogencarbonate system, which protects enzymes.
Everyday uses
  • Buffers are used in shampoos, foods and biological experiments to keep pH steady.

Section 4 of 5Worked Example

Question Calculate the pH of a buffer containing 0.100 mol dm⁻³ ethanoic acid and 0.200 mol dm⁻³ sodium ethanoate, Ka = 1.74 × 10⁻⁵ mol dm⁻³.

  1. [H⁺] = Ka × [HA] ÷ [A⁻].
  2. [H⁺] = 1.74 × 10⁻⁵ × 0.100 ÷ 0.200 = 8.70 × 10⁻⁶ mol dm⁻³.
  3. pH = −log₁₀(8.70 × 10⁻⁶) = 5.06.

Answer The pH of the buffer is 5.06.

Section 5 of 5Exam Checklist & Tips

1.10.7 – Key Points for the Exam

  • Buffer solution A solution that resists changes in pH when small amounts of acid or alkali are added.
  • Acidic buffer A weak acid mixed with a salt of that acid.
  • Conjugate base The species formed when an acid has donated a proton.
  • Buffer capacity The amount of acid or alkali a buffer can absorb before its pH changes significantly.

Common Mistakes

  • Saying a buffer stops pH changing completely; it only resists change within its capacity.
  • Using the total of acid and salt as [HA] instead of the individual equilibrium concentrations.
  • Forgetting that the salt is fully ionised, so it supplies almost all the A⁻.
  • Describing the addition of acid as moving the equilibrium to the right instead of to the left.
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