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6.1.17Additional lessons

Buffer Calculations (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 131 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 5The Equation

This equation makes it simple to find the pH of an acidic buffer.

Form
  • Form pH = pKa + log₁₀([A⁻] / [HA]), where A⁻ is the conjugate base and HA is the weak acid.
Origin
  • Origin It comes from rearranging Ka = [H⁺][A⁻] / [HA] and taking the negative logarithm.
Equal amounts
  • Equal amounts When [A⁻] = [HA] the logarithm is zero, so pH = pKa.
Ratio only
  • Ratio only Only the ratio matters, so moles can be used in place of concentrations when both are in the same volume.

Part 2 of 5Using the Equation

Using the Equation
Design
  • Design To make a buffer of a chosen pH, select an acid whose pKa is within about one unit of the pH required.
Ratio of ten
  • Ratio of ten A ratio [A⁻]/[HA] of 10 gives pH = pKa + 1, and a ratio of 0.1 gives pH = pKa − 1.
After adding acid
  • After adding acid Subtract the moles of H⁺ from A⁻ and add the same number to HA, then recalculate the ratio.
After adding base
  • After adding base Subtract the moles of OH⁻ from HA and add it to A⁻, then recalculate.

Part 3 of 5Method

1
Find the amounts, in moles, of the weak acid and the conjugate base present.
2
If acid or alkali has been added, adjust the amounts according to the neutralisation reaction.
3
Insert the new ratio into the equation with the pKa of the acid.
4
Calculate the logarithm, then add it to the pKa.
5
Compare with the pH before the addition to show how small the change is.

Part 4 of 5Worked Example

Question

  • A buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COONa in 1.00 dm³ (pKa = 4.76).
  • Find its pH, and the pH after adding 0.010 mol NaOH (no volume change).

  1. Initially [A⁻]/[HA] = 1, so pH = 4.76 + log₁₀(1) = 4.76.
  2. NaOH reactsHA falls to 0.090 mol and A⁻ rises to 0.110 mol.
  3. pH = 4.76 + log₁₀(0.110 / 0.090) = 4.76 + 0.087 = 4.85.

Answer

  • pH 4.76 before;
    • 4.85 after, a change of only 0.09

Part 5 of 5Exam Checklist & Tips

6.1.17 – Key Points for the Exam

  • Henderson–Hasselbalch equation pH = pKa + log₁₀([A⁻]/[HA]).
  • Buffer ratio The ratio of conjugate base to weak acid.
  • pKa The −log₁₀ of the acid dissociation constant.
  • Neutralisation of buffer The reaction of added H⁺ with A⁻, or added OH⁻ with HA.

Common Mistakes

  • Inverting the ratio to [HA]/[A⁻] in the logarithm.
  • Forgetting to adjust the amounts after adding acid or base.
  • Using the equation for a mixture of a strong acid and its salt.
  • Using the concentrations of added solutions without allowing for the increase in volume.

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