AP Chemistry · Unit 8 · Topic 8.5
Acid-Base Titrations
A titration curve is a sequence of different chemical mixtures. One pH formula cannot describe every point because the species controlling pH change as titrant is added.
Begin with the equivalence calculation
Titrate 20.0 mL of 0.100 M monoprotic HA with 0.100 M NaOH. The sample contains 2.00 mmol HA, so equivalence requires 2.00 mmol OH⁻, delivered in 20.0 mL. Equivalence means stoichiometric consumption of the starting acid, not necessarily pH 7. Assume Ka = 1.00 × 10⁻⁵ and 25 °C for this original example.
Read the changing chemical regions
| NaOH volume | Dominant pH model | Approximate pH |
|---|---|---|
| 0 mL | Weak acid alone | 3.00 |
| 10.0 mL | Equal HA and A⁻ | 5.00 |
| 20.0 mL | A⁻ hydrolysis | 8.85 |
| 25.0 mL | Excess strong OH⁻ | 12.05 |
These points form a rising curve with a buffer region followed by a steep rise near equivalence.
Understand the halfway point
At 10.0 mL, hydroxide has converted 1.00 mmol HA into A⁻, leaving equal amounts of the conjugate partners. Their concentration ratio is one, so pH ≈ pKa = 5.00. This makes half-equivalence useful for estimating Ka from a measured curve. It is not the point at which all acid molecules are gone.
Work the equivalence-point pH
At equivalence, 2.00 mmol A⁻ occupies 40.0 mL, giving 0.0500 M A⁻. Its base constant is \(K_b=K_w/K_a=1.00\times10^{-9}\). The hydrolysis \(A^-+H_2O\rightleftharpoons HA+OH^-\) gives
The ionization fraction is only 0.014%, supporting the approximation. Basic pH follows from conjugate-base hydrolysis, not leftover strong base.
Account for excess titrant
At 25.0 mL, 2.50 mmol OH⁻ has been added. After neutralization, 0.50 mmol remains in 45.0 mL, so [OH⁻] ≈ 0.0111 M and pH ≈ 12.05. The added strong base dominates; applying Henderson–Hasselbalch with a negative calculated HA amount would be physically meaningless.
Choose an endpoint that tracks equivalence
An indicator changes color over a range, not at an infinitely sharp pH. Select one whose transition falls within the steep region of the expected curve. The observed endpoint estimates the theoretical equivalence volume. Overshooting increases the apparent titrant requirement and can make the calculated unknown acid concentration too high.
Collect better evidence
Use smaller titrant additions near the steep rise, record cumulative volume, and allow mixing and electrode response after each addition. Estimate the equivalence volume from the greatest slope or an appropriate curve analysis. Multiple equivalence regions can suggest a polyprotic acid, but overlapping ionization steps may not resolve cleanly.
Practice with answers
- Why is the original equivalence pH above 7? A⁻ produces hydroxide by reacting with water.
- If the acid volume doubles at the same concentration, what happens to equivalence volume? It doubles for the same titrant.
- Does half-equivalence mean half the final total volume? No; it refers to half the required titrant amount.
Factual reference: OpenStax Chemistry 2e, 14.7. The titration values, landmark table, calculations, and questions were independently created.
Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.