6.1.9Additional lessons
Interpreting pH Curves (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 123 of 172
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.
Section 1 of 5Key Points on the Curve
Specific points on the curve give quantitative information about the acid.
Initial pH
- Initial pH For a weak acid this reflects its partial dissociation and is higher than for a strong acid of equal concentration.
Buffer region
- Buffer region Before equivalence the flask contains both the weak acid and its conjugate base, so the pH changes only slowly when base is added.
Half-equivalence
- Half-equivalence When half of the acid has been neutralised, [HA] equals [A⁻], so the pH equals the pKa of the acid.
Equivalence
- Equivalence Here only the conjugate base remains, which makes the solution alkaline, so the pH is above 7.
Section 2 of 5Reading Values
Reading Values
Find equivalence volume
- Find equivalence volume Take the midpoint of the steep section, then halve this volume to locate the half-equivalence point.
Read pKa
- Read pKa The pH at the half-equivalence volume equals pKa, and then Ka = 10⁻ᵖᴷᵃ.
Strong acid curves
- Strong acid curves These have no buffer region, because no conjugate base is present with the acid before equivalence.
Choosing between acids
- Choosing between acids A lower initial pH and a smaller pKa both indicate a stronger acid when comparing two curves.
| Point | Species present | pH |
|---|---|---|
| Start | HA only | Set by Ka and c |
| Half-equivalence | HA = A⁻ | pKa |
| Equivalence | A⁻ only | Above 7 (weak acid, strong base) |
| After equivalence | Excess OH⁻ | Set by the excess of base |
Section 3 of 5Analysis Steps
1
Locate the vertical section and mark the equivalence volume at its midpoint.
2
Divide this volume by two to find the half-equivalence volume.
3
Read the pH at that volume on the curve; this is the pKa of the weak acid.
4
Calculate Ka from 10⁻ᵖᴷᵃ and compare with data to identify the acid.
5
Use the equivalence volume with n = cV to find the concentration of the acid, applying the stoichiometry.
Section 4 of 5Worked Example
Question
- A weak acid is titrated with NaOH.
- The equivalence point is at 20.0 cm³ and the pH at 10.0 cm³ is 4.76.
- Find pKa and Ka.
- Half-equivalence volume = 20.0 / 2 = 10.0 cm³, which agrees with the given point.
- At this point pH = pKa, so pKa = 4.76.
- Ka = 10⁻⁴·⁷⁶ = 1.7 × 10⁻⁵ mol dm⁻³.
Answer pKa = 4.76; Ka = 1.7 × 10⁻⁵ mol dm⁻³
Section 5 of 5Exam Checklist & Tips
6.1.9 – Key Points for the Exam
- Buffer region The flat part of the curve where the pH changes little because both HA and A⁻ are present.
- Half-equivalence point The point at which half of the acid has been neutralised and pH = pKa.
- pKa −log₁₀ Ka, a measure of acid strength on a logarithmic scale.
- Equivalence volume The volume of titrant needed for exact neutralisation.
Common Mistakes
- Reading pKa at the equivalence point instead of the half-equivalence point.
- Claiming the pH at equivalence is 7 for a weak acid titrated with a strong base.
- Halving the pH rather than the volume to find the half-equivalence point.
- Saying that a strong acid curve shows a buffer region.
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