1.10.5Physical Chemistry
pH Curves, Titrations and Indicators
5 min read 5 parts Lesson 55 of 63
Section 1 of 5Shapes of pH Curves
The curve shows how pH changes as titrant is added to the solution in the flask.
Shapes of pH Curves
Strong acid + strong base
- Starts at low pH, rises steeply through pH 7 with a vertical section of about pH 3 to 11, and finishes at high pH.
Weak acid + strong base
- Starts at a higher pH (about 3), shows a buffering region, and the equivalence point is above 7.
Strong acid + weak base
- Starts at low pH, has a steep section ending near pH 7, and the equivalence point is below 7.
Weak acid + weak base
- Has no vertical section, so there is no sharp end point and no simple indicator works.
| Titration | pH at equivalence | Suitable indicator |
|---|---|---|
| Strong acid + strong base | 7 | Methyl orange or phenolphthalein |
| Weak acid + strong base | above 7 | Phenolphthalein |
| Strong acid + weak base | below 7 | Methyl orange |
| Weak acid + weak base | no sharp change | None |
Section 2 of 5Indicators
An indicator is a weak acid that changes colour as the pH crosses its range.
Methyl orange
- Red in acid and yellow in alkali over about pH 3.1 to 4.4.
Phenolphthalein
- Colourless in acid and pink in alkali over about pH 8.3 to 10.0.
Choice
- The colour change range must lie entirely within the steep part of the pH curve.
Few drops
- A small amount of indicator is used so that it does not react with a significant amount of titrant.
Observation
- State the colour at the end point and say that it must persist for a few seconds.
Section 3 of 5Titration Calculation
1
Write the balanced equation, such as CH₃COOH + NaOH → CH₃COONa + H₂O, to find the mole ratio.
2
Calculate moles of the solution in the burette from concentration × volume ÷ 1000.
3
Use the equation ratio to find moles of the solution in the flask.
4
Divide by the volume in dm³ of the flask solution to find its concentration.
5
Use concordant titres within 0.10 cm³ to calculate the mean titre.
Section 4 of 5Worked Example
Question
- In a titration 25.0 cm³ of ethanoic acid required 22.50 cm³ of 0.100 mol dm⁻³ NaOH.
- Calculate the concentration of the acid and choose an indicator.
- Moles NaOH = 0.100 × 22.50 ÷ 1000 = 2.25 × 10⁻³ mol.
- CH₃COOH and NaOH react 1:1, so moles acid = 2.25 × 10⁻³ mol.
- Concentration = 2.25 × 10⁻³ ÷ 0.0250 = 0.0900 mol dm⁻³.
- A weak acid with strong base has an equivalence point above 7, so phenolphthalein is suitable.
Answer The acid concentration is 0.0900 mol dm⁻³ and the indicator is phenolphthalein.
Section 5 of 5Exam Checklist & Tips
1.10.5 – Key Points for the Exam
- Equivalence point The point at which the reactants have been added in exactly the ratio shown by the equation.
- End point The point at which the indicator changes colour.
- Concordant titres Titres that agree within 0.10 cm³.
- Buffer region The relatively flat section of a weak acid curve where pH changes slowly.
Common Mistakes
- Assuming that the equivalence point is always at pH 7.
- Choosing an indicator by its colours rather than by whether its range lies within the steep section.
- Using the mean of all titres including a rough titre instead of only concordant ones.
- Forgetting to convert cm³ to dm³ when calculating concentration.