AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/4/4-6)
Unit 4 · Topic 4.6
4.6 Introduction to Titration
In a titration, you add a solution of known concentration (the titrant) to a sample (the analyte) until exactly enough has been added to react with all of it. That moment is the equivalence point, and the volume of titrant used tells you how much analyte was there.
Key terms
- titrant
- analyte
- equivalence point
- endpoint
- indicator
The setup
The titrant is in a buret, a long graduated tube with a valve at the bottom that lets you add precise volumes. Its concentration is known exactly. The analyte, the substance you want to measure, is in a flask below, usually with a few drops of indicator.
The titrant must react with the analyte quickly, completely and in a known ratio. Acid-base reactions and some redox reactions work well.
Equivalence point and endpoint
The equivalence point is when the moles of titrant added are exactly enough to react with all the analyte, according to the balanced equation. For HCl + NaOH, that's when moles of NaOH added = moles of HCl present.
You can't see the equivalence point directly, so you watch for a change that signals it, such as an indicator changing color. The observed change is the endpoint. A well-chosen indicator makes the endpoint happen at, or extremely close to, the equivalence point.
Some titrations don't need an added indicator. Permanganate ion, MnO₄⁻, is deep purple, and its product Mn²⁺ is nearly colorless. When titrating with permanganate, the first lasting pale pink color in the flask means all the analyte is used up.
The calculation
Every titration calculation follows the same three steps. The mole-ratio step matters whenever the coefficients aren't 1 : 1. Sulfuric acid, H₂SO₄, gives up two H⁺ ions, so it needs 2 mol NaOH per mole of acid.
- Moles of titrant used = molarity of titrant × volume added (in L).
- Moles of analyte = moles of titrant × (analyte coefficient ÷ titrant coefficient) from the balanced equation.
- Concentration of analyte = moles of analyte ÷ volume of analyte (in L). Or multiply moles by molar mass to get its mass.
Errors and their effects
Adding water to the analyte flask doesn't change the moles of analyte, so it doesn't change the titrant volume needed. Students often get this one wrong. Titration curves of pH against volume come later, in topic 8.5.
| Mistake | Effect on titrant volume recorded | Calculated amount of analyte |
|---|---|---|
| Adding titrant past the endpoint (overshooting) | too large | too high |
| Rinsing the buret with water instead of titrant, diluting it | too large | too high |
| Adding extra water to the analyte flask | no change | no change |
| Some analyte spilled before titrating | too small | too low |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A 1 : 1 acid-base titration
25.00 mL of HCl is titrated with 0.1000 M NaOH. The endpoint is reached after 18.40 mL of NaOH. What is the concentration of the HCl?
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- Step 1: Moles NaOH = 0.1000 M × 0.01840 L = 1.840 × 10⁻³ mol.
- Step 2: HCl + NaOH → NaCl + H₂O is 1 : 1, so moles HCl = 1.840 × 10⁻³ mol.
- Step 3: [HCl] = 1.840 × 10⁻³ mol ÷ 0.02500 L = 0.07360 M.
Answer: 0.07360 M HCl
- Example 2Calculator allowed
A 1 : 2 ratio (classic trap)
20.00 mL of H₂SO₄ requires 31.20 mL of 0.1500 M NaOH to reach the equivalence point: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is the concentration of H₂SO₄?
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- Step 1: Moles NaOH = 0.1500 M × 0.03120 L = 4.680 × 10⁻³ mol.
- Step 2: Ratio: 1 mol H₂SO₄ per 2 mol NaOH, so moles H₂SO₄ = 4.680 × 10⁻³ ÷ 2 = 2.340 × 10⁻³ mol.
- Step 3: [H₂SO₄] = 2.340 × 10⁻³ ÷ 0.02000 L = 0.1170 M.
- Step 4: Using M₁V₁ = M₂V₂ here would give 0.2340 M, twice the right answer, because that shortcut assumes a 1 : 1 ratio.
Answer: 0.1170 M H₂SO₄
- Example 3Calculator allowed
A redox titration
A 25.00 mL sample of Fe²⁺ solution is titrated with 0.02000 M KMnO₄ in acid. The first permanent pink color appears after 15.00 mL. The reaction is MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Find [Fe²⁺].
Show the solutionHide the solution
- Step 1: Moles MnO₄⁻ = 0.02000 M × 0.01500 L = 3.000 × 10⁻⁴ mol.
- Step 2: Ratio: 5 Fe²⁺ per 1 MnO₄⁻, so moles Fe²⁺ = 5 × 3.000 × 10⁻⁴ = 1.500 × 10⁻³ mol.
- Step 3: [Fe²⁺] = 1.500 × 10⁻³ ÷ 0.02500 L = 0.06000 M.
Answer: 0.06000 M Fe²⁺
Common mistakes
- Using M₁V₁ = M₂V₂ for a titration that isn't 1 : 1. Always go through moles and the mole ratio.
- Treating the endpoint and the equivalence point as the same thing by definition. The endpoint is what you observe; the equivalence point is the true stoichiometric point.
- Thinking that adding water to the analyte flask changes the result.
- Using the total volume in the flask, rather than the original analyte volume, to find the analyte's concentration.
On the exam
- Titration calculations are a lab staple on the free-response section. Show moles of titrant, the mole ratio and the final concentration, each with units.
- Expect error-analysis questions: decide whether the titrant volume recorded goes up or down, then follow that through to the calculated result.
Connected topics
Videos
Check yourself
4 questions on 4.6 Introduction to Titration. Pick an answer to see if you got it, and why.
| Trial | Initial buret reading (mL) | Final buret reading (mL) | Volume of NaOH added (mL) |
|---|---|---|---|
| 1 | 0.50 | 19.05 | 18.55 |
| 2 | 19.05 | 37.50 | 18.45 |
| 3 | 0.20 | 18.70 | 18.50 |
Hypothetical data. A student titrates three 25.00 mL samples of an HCl solution with 0.100 M NaOH, using phenolphthalein as the indicator. The reaction is HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l).
Using the average volume of NaOH from all three trials, what is the concentration of the HCl solution?
At the equivalence point of each trial, which of the following is true?
Suppose the student had rinsed the buret with distilled water but not with the NaOH solution before filling it. How would this mistake affect the calculated concentration of HCl?
A 20.0 mL sample of H₂SO₄(aq) requires 30.0 mL of 0.200 M NaOH to reach the equivalence point: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l). What is the concentration of the H₂SO₄?
0 of 4 answered