AQA · A-Level · Chemistry
Course version: 7405 · Version from Sept 2016
Students should be able to: explain qualitatively the action of acidic and…
Full specification wording
Buffer action (A-level only): Students should be able to: explain qualitatively the action of acidic and basic buffers calculate the pH of acidic buffer solutions.
Qualitatively, name the species: in an acidic buffer the conjugate base removes added H⁺ and the weak acid removes added OH⁻.
Checked against the AQA 7405 specification (opens in a new tab) on .
- Point 5 of 5
- 3 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Qualitatively, name the species: in an acidic buffer the conjugate base removes added H⁺ and the weak acid removes added OH⁻. Quantitatively, rearrange Ka to give [H⁺] = Ka × [HA]/[A⁻]. Because both species share the same solution, the volumes cancel and moles can be used directly in the ratio. Take 20.0 cm³ of 0.100 mol dm⁻³ NaOH added to 25.0 cm³ of 0.100 mol dm⁻³ butanoic acid, for which Ka = 1.51 × 10⁻⁵ mol dm⁻³. There are 2.50 × 10⁻³ mol of acid and 2.00 × 10⁻³ mol of alkali, so 5.0 × 10⁻⁴ mol of acid survives and 2.00 × 10⁻³ mol of salt is formed. Then [H⁺] = 1.51 × 10⁻⁵ × (5.0 × 10⁻⁴)/(2.00 × 10⁻³) = 3.78 × 10⁻⁶ mol dm⁻³ and pH = 5.42.
What earns the marks
- The initial moles of the weak acid and of the added alkali.
- Subtracting to find the moles of weak acid remaining and the moles of salt formed.
- Substituting into [H⁺] = Ka × [HA]/[A⁻], in which the volumes cancel.
- pH = −log₁₀[H⁺] quoted to two decimal places.
- Naming the species that removes added H⁺ and the species that removes added OH⁻.
Common mistakes and how to put them right
Converting one species to a concentration using the total volume but leaving the other as moles.
Inverting the ratio, using [A⁻]/[HA], so the pH comes out too low.
Forgetting that the moles of salt formed equal the moles of alkali added.
Treating all of the original weak acid as still present after the alkali has reacted.
Subtracting rather than adding the log term when using pH = pKa + log([A⁻]/[HA]).
Exam tips
- Set the working out as a table of initial moles, change and final moles; the marks follow that structure.
- Check the answer against pKa: a buffer with more salt than acid must have a pH above the pKa.
Try the lesson
The start of the Buffer action (A-level only): Students should be able to: explain qualitatively the action of acidic and basic buffers calculate the pH of acidic buffer solutions. lesson, made from checked teaching notes for this specification point.
Have a go first
An acidic buffer contains ethanoic acid and sodium ethanoate. Which statement explains why the pH changes only slightly when a small amount of hydrochloric acid is added?
Listen
Qualitatively, name the species: in an acidic buffer the conjugate base removes added H⁺ and the weak acid removes added OH⁻. Quantitatively, rearrange Ka to give [H⁺] = Ka × [HA]/[A⁻]. Because both species share the same solution, the volumes cancel and moles can be used directly in the ratio. Take 20.0 cm³ of 0.100 mol dm⁻³ NaOH added to 25.0 cm³ of 0.100 mol dm⁻³ butanoic acid, for which Ka = 1.51 × 10⁻⁵ mol dm⁻³. There are 2.50 × 10⁻³ mol of acid and 2.00 × 10⁻³ mol of alkali, so 5.0 × 10⁻⁴ mol of acid survives and 2.00 × 10⁻³ mol of salt is formed. Then [H⁺] = 1.51 × 10⁻⁵ × (5.0 × 10⁻⁴)/(2.00 × 10⁻³) = 3.78 × 10⁻⁶ mol dm⁻³ and pH = 5.42. A strong answer earns marks for these points. the initial moles of the weak acid and of the added alkali. subtracting to find the moles of weak acid remaining and the moles of salt formed. substituting into [H⁺] = Ka × [HA]/[A⁻], in which the volumes cancel. pH = −log₁₀[H⁺] quoted to two decimal places. Exam tip: Set the working out as a table of initial moles, change and final moles; the marks follow that structure.
Key words
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qualitative buffer action
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moles rather than concentrations
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buffer pH calculation
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See every definition
- qualitative buffer action
- The conjugate base removes added H⁺ and the weak acid removes added OH⁻.
- moles rather than concentrations
- Because both species share the same solution, the volumes cancel and moles can be used directly in the ratio.
- buffer pH calculation
- Substituting into [H⁺] = Ka × [HA]/[A⁻], in which the volumes cancel, then pH = −log₁₀[H⁺] quoted to two decimal places.
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