4.5.7Unit 4. Rates, Equilibria and Further Organic Chemistry
Titration Curves — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 5 parts Lesson 39 of 56 Unit quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Section 1 of 5Shape of a Curve
A titration curve plots pH against volume of added solution.
Starting pH
- Starting pH The initial pH depends on the acid;
- a strong acid starts near 1 and a weak acid starts at a higher value.
Rapid change
- Rapid change Near the equivalence point a tiny volume of titrant causes a rapid, almost vertical rise of several pH units.
Equivalence point
- Equivalence point This is the point at which the moles of acid and base exactly match according to the balanced equation.
Final pH
- Final pH After excess alkali is added the curve flattens at a high pH close to that of the alkali solution.
Section 2 of 5Four Combinations
Four Combinations
Strong acid, strong base
- Strong acid, strong base The vertical section covers about pH 3 to 11, and the equivalence point is at pH 7.
Weak acid, strong base
- Weak acid, strong base The vertical section covers about pH 7 to 11 and the equivalence point is above 7, about 8.7 for ethanoic acid.
Strong acid, weak base
- Strong acid, weak base The vertical section covers about pH 3 to 7 and the equivalence point is below 7.
Weak acid, weak base
- Weak acid, weak base There is no sharp vertical section, so no indicator gives a clear end point for this titration.
| Indicator | pH range of colour change | Suitable for |
|---|---|---|
| Methyl orange | 3.1 to 4.4 | Strong acid, strong base; strong acid, weak base |
| Phenolphthalein | 8.3 to 10.0 | Strong acid, strong base; weak acid, strong base |
Section 3 of 5Reading the Curve
1
Find the equivalence point at the middle of the vertical section of the curve and read off the volume.
2
Use the volume and concentration to calculate the moles of the known solution, then use the equation ratio.
3
For a weak acid, find half the equivalence volume and read the pH there, which equals the pKa.
4
Choose an indicator whose colour-change range lies entirely within the vertical section of the curve.
Section 4 of 5Worked Example
Question
- 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³) is titrated with 0.100 mol dm⁻³ NaOH.
- State the volume of NaOH at equivalence, the pH at half-equivalence, and calculate the pH at equivalence.
- Moles of acid = 0.0250 × 0.100 = 2.50 × 10⁻³ mol, so equivalence needs 25.0 cm³ of NaOH.
- At half-equivalence (12.5 cm³), [HA] = [A⁻], so pH = pKa = −log₁₀(1.74 × 10⁻⁵) = 4.76.
- At equivalence [A⁻] = 2.50 × 10⁻³ / 0.0500 dm³ = 0.0500 mol dm⁻³; Kb = Kw / Ka = 5.75 × 10⁻¹⁰,
- so [OH⁻] = √ = 5.36 × 10⁻⁶; pOH = 5.27 and pH = 14.00 − 5.27 = 8.73.
- (5.75 × 10⁻¹⁰ × 0.0500)
- so [OH⁻] = √ = 5.36 × 10⁻⁶; pOH = 5.27 and pH = 14.00 − 5.27 = 8.73.
Answer
- 25.0 cm³;
- pH 4.76 at half-equivalence;
- pH 8.73 at equivalence.
Section 5 of 5Exam Checklist & Tips
4.5.7 – Key Points for the Exam
- Equivalence point Point at which reactants are in exact stoichiometric ratio.
- End point The point at which the indicator changes colour.
- Indicator A weak acid or base whose colour differs from its conjugate.
- Titrant The solution added from the burette.
Common Mistakes
- Choosing methyl orange for a weak acid with a strong base.
- Assuming every equivalence point is at pH 7.
- Reading the equivalence point from the start of the vertical section instead of its midpoint.
- Taking pKa from the equivalence volume instead of half the volume.
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