Short on time? This page is the recap. For explanations, the titration curve and worked examples, use the full study guide.
Recap
- A titration finds the amount of an analyte by reacting it with a titrant of known concentration added from a burette.
- The titration reaction must be specific (reacts only with the analyte) and quantitative (goes to completion with a known equation).
- The equivalence point is reached when the titrant has exactly used up the analyte, in the ratio of the balanced equation.
- The endpoint is the observable change (indicator colour, permanganate pink, a jump in pH or a change in conductivity). It should match the equivalence point as closely as possible.
- On a titration curve, the equivalence point is where the measured property changes most sharply (or where two straight-line sections meet).
Key relationships
| Step | Relationship | Watch out for |
|---|---|---|
| 1 | Write the balanced equation | the mole ratio comes from here |
| 2 | n(titrant) = M(titrant) × V(titrant) | V in litres |
| 3 | n(analyte) = n(titrant) × (coeff. analyte / coeff. titrant) | ratio the right way up |
| 4 | c(analyte) = n(analyte) ÷ V(sample) | the sample volume, not the total |
| Predict | V(equivalence) = n(titrant needed) ÷ M(titrant) | useful for planning |
Assumptions behind the numbers
- The reaction goes to completion and is the only reaction with the titrant.
- The endpoint coincides with the equivalence point.
- Volumes and the titrant concentration are accurate.
Mistakes to avoid
- Calling the endpoint the equivalence point. One is seen; the other is calculated.
- M₁V₁ = M₂V₂ for a non-1:1 reaction. Go through moles.
- Dividing by the total volume in the flask instead of the original sample volume.
- Thinking added water in the flask changes the titre. Moles of analyte are unchanged.
- Ignoring direction of errors. More titrant recorded than needed → analyte calculated too high.
Quick self-check
- What volume of 0.100 M HCl reaches equivalence with 25.00 mL of 0.0800 M NaOH? (20.0 mL)
- What volume of 0.100 M HCl reaches equivalence with 10.00 mL of 0.0500 M Ca(OH)₂? (10.0 mL, because the ratio is 2 HCl : 1 Ca(OH)₂)
- A student overshoots the endpoint. Is the calculated analyte concentration too high or too low? (Too high)
Next: practice questions.