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4.5.9Unit 4. Rates, Equilibria and Further Organic Chemistry

Buffer Calculations — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 41 of 56 Unit quiz

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Part 1 of 5The Method

4.5.9 Buffer Calculations

The Ka expression is rearranged to give the hydrogen ion concentration.

1
Write the expression Ka = [H⁺][A⁻] / [HA] and rearrange it to give [H⁺] = Ka × [HA] / [A⁻].
2
Take the concentration of A⁻ to be the concentration of the salt, because the salt dissociates fully.
3
Take the concentration of HA to be that of the acid, because its own dissociation is suppressed by the common ion.
4
Substitute values and calculate pH = −log₁₀[H⁺].

Part 2 of 5Useful Shortcuts

Same volume
  • If acid and salt are in one volume, the ratio of moles can replace the ratio of concentrations.
Equal amounts
  • When [HA] = [A⁻], [H⁺] = Ka and the pH equals the pKa of the acid.
Partial neutralisation
  • For acid reacted with alkali, subtract moles of alkali from HA and add the same moles to A⁻.
Buffer range
  • The most effective buffering occurs within about one pH unit of the pKa.

Part 3 of 5Adding Acid or Alkali

Adding Acid or Alkali
Adding acid
  • Added H⁺ converts A⁻ to HA, so moles of A⁻ fall and moles of HA rise by the same amount.
Adding alkali
  • Added OH⁻ converts HA to A⁻, so moles of HA fall and moles of A⁻ rise by the same amount.
Total volume
  • After mixing, the volume changes, but if both species are in the same volume it cancels in the ratio.
Checking
  • The new pH should change only slightly if the buffer is working within its capacity.

Part 4 of 5Worked Example

Question

  • A buffer contains 0.50 mol dm⁻³ ethanoic acid and 0.25 mol dm⁻³ sodium ethanoate (Ka = 1.74 × 10⁻⁵ mol dm⁻³).
  • Calculate its pH, then the pH after mixing 50.0 cm³ of 0.20 mol dm⁻³ ethanoic acid with 25.0 cm³ of 0.10 mol dm⁻³ NaOH.

  1. Buffer 1: [H⁺] = 1.74 × 10⁻⁵ × 0.50 / 0.25 = 3.48 × 10⁻⁵ mol dm⁻³, so pH = 4.46.
  2. Mixturemoles HA = 0.0100, moles NaOH = 0.00250, so HA = 0.00750 mol and A⁻ = 0.00250 mol after reaction.
  3. [H⁺] = 1.74 × 10⁻⁵ × 0.00750 / 0.00250 = 5.22 × 10⁻⁵ mol dm⁻³, so pH = 4.28.

Answer pH = 4.46 for the first buffer and pH = 4.28 for the mixture.

Part 5 of 5Exam Checklist & Tips

4.5.9 – Key Points for the Exam

  • Common ion An ion present in the solution from two sources.
  • pKa −log₁₀ Ka.
  • Mole ratio Ratio of amounts of two species in the mixture.
  • Buffer range Region of pH where a buffer is effective.

Common Mistakes

  • Using the concentrations of H⁺ instead of the acid concentration in the ratio.
  • Dividing the wrong way round, putting salt over acid.
  • Forgetting to account for neutralisation when alkali is added.
  • Rounding intermediate values too early.

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