About this experiment
Add sodium hydroxide from a burette into 25 mL of acid, one drop at a time or with the stopcock open, and watch the titration curve draw itself. The pH is calculated exactly from the charge balance, so the steep jump at the equivalence point and the buffer region of a weak acid appear just as they would in a careful real experiment.
Learning goals
- Read the equivalence point from a titration curve
- Compare the curves of a strong acid and a weak acid titrated with a strong base
- Choose an indicator whose color change falls inside the steep part of the curve
- Calculate an unknown concentration from the equivalence volume
How to use it
- 1Pick the acid, its concentration and the NaOH concentration in the panel.
- 2Use +2 mL to move quickly at first, then single drops as the pH starts to rise.
- 3Watch for the moment the indicator changes color and compare it with the equivalence line.
- 4Switch to acetic acid and repeat. Note the higher starting pH and the gentle buffer slope.
The model
Assumptions
- Temperature is 25 °C, so Kw = 1.0 × 10⁻¹⁴.
- Acetic acid has Ka = 1.8 × 10⁻⁵. Activity effects are ignored.
- Volumes add exactly and one drop is 0.05 mL.
Try this in class
- Q1Why is the pH at equivalence 7 for HCl but above 7 for acetic acid?
- Q2Which indicator gives a poor result for the weak acid titration, and why?
- Q3At half the equivalence volume, read the pH of the acetic acid curve. What constant does it equal?
- Q4Double the NaOH concentration. How does the equivalence volume change?
Frequently asked questions
What is the difference between the endpoint and the equivalence point?
The equivalence point is where the moles of base equal the moles of acid. The endpoint is where the indicator changes color. A good indicator makes the two almost the same.
Why does the curve rise so steeply near equivalence?
Almost all the acid has been used up, so a single drop of base changes the hydrogen ion concentration by several powers of ten.
Is the pH calculated or approximated?
It is calculated by solving the full charge balance numerically, without the usual simplifications, so it is accurate across the whole curve.