The common path
Acid–base or redox, every titration calculation follows the same path:
- Equivalence point计量点: where the reactants have reacted exactly according to the equation.
- Endpoint终点: where the color change is seen and the titration stops. A good indicator puts the endpoint as close as possible to the equivalence point.
Acid–base titrations
- Choosing an indicator: its color-change range must fall within the steep jump in pH around the equivalence point. A weak acid titrated with a strong base has an equivalence pH above 7, so use phenolphthalein; a weak base titrated with a strong acid has an equivalence pH below 7, so use methyl red.
- Standardizing: NaOH absorbs CO₂ from the air, so its concentration can’t be set just by weighing; it’s standardized against a primary standard基准物质 such as potassium hydrogen phthalate (KHP, KHC₈H₄O₄).
Redox titrations with KMnO₄
- MnO₄⁻ is purple and is reduced to nearly colorless Mn²⁺. Once MnO₄⁻ is in slight excess, the solution turns pale pinkish purple; that’s the endpoint, so no separate indicator is needed.
- The solution is acidified with sulfuric acid (Cl⁻ from hydrochloric acid would be oxidized by MnO₄⁻).
- The mole ratio comes from the balanced equation; common ones are n(Fe²⁺) : n(MnO₄⁻) = 5 : 1, n(H₂C₂O₄) : n(MnO₄⁻) = 5 : 2 and n(H₂O₂) : n(MnO₄⁻) = 5 : 2.
From the balanced equation, n(analyte) : n(titrant) = ?
Example 1: standardizing NaOH with KHP
0.5105 g of KHP (204.22 g/mol) is dissolved in water and titrated with NaOH solution to the phenolphthalein endpoint, using 24.35 mL. KHP is a monoprotic acid. Find the NaOH concentration.
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Moles of KHP.
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Mole ratio. KHP is monoprotic and reacts 1 : 1 with NaOH, so n(NaOH) = 2.500 × 10⁻³ mol.
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NaOH concentration.
Example 2: hydrogen peroxide by KMnO₄ titration
10.00 mL of drugstore hydrogen peroxide is diluted to 100.0 mL. A 10.00 mL portion of the diluted solution is acidified and titrated with 0.02000 M KMnO₄, using 17.60 mL. The reaction is 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O. Find [H₂O₂] in the drugstore solution.
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Moles of MnO₄⁻.
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Moles of H₂O₂ from the mole ratio. n(H₂O₂) : n(MnO₄⁻) = 5 : 2:
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Concentration of the diluted solution. 8.800 × 10⁻⁴ mol ÷ 0.01000 L = 0.08800 M.
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Concentration of the drugstore solution. Diluting 10.00 mL to 100.0 mL is a 10-fold dilution:
That’s 0.8800 × 34.02 ≈ 29.9 g of H₂O₂ per liter, about 3%, matching the label.
- CH₃COOH titrated with NaOH, equivalence pH ≈ 8.7
The equivalence point is basic, near phenolphthalein’s range.
- NH₃ titrated with HCl, equivalence pH ≈ 5.3
The equivalence point is acidic, so use methyl red.
- Acidified Fe²⁺ titrated with KMnO₄
A slight excess of MnO₄⁻ turns the solution pale pinkish purple, so it indicates its own endpoint.
- HCOOH titrated with KOH, equivalence pH ≈ 8.2
A weak acid titrated with a strong base has a basic equivalence point.
- CH₃NH₂ titrated with HCl, equivalence pH ≈ 5.9
A weak base titrated with a strong acid has an acidic equivalence point.
- Acidified oxalic acid titrated with KMnO₄
The color of MnO₄⁻ again marks the endpoint.
- Benzoic acid titrated with NaOH, equivalence pH ≈ 8.6
A basic equivalence point, so phenolphthalein.
- Acidified hydrogen peroxide titrated with KMnO₄
No separate indicator is needed.
Common mistakes
Mistake 1: “The equivalence point is where the solution is neutral.” Only in a strong acid–strong base titration is the equivalence pH 7. When a weak acid is titrated with a strong base, the solution at equivalence contains the conjugate base, so its pH is above 7.
Mistake 2: “Redox titrations are 1 : 1, just like acid–base ones.” The mole ratio comes from the balanced equation (electrons gained = electrons lost). MnO₄⁻ gains 5 electrons and Fe²⁺ loses 1, so each MnO₄⁻ reacts with 5 Fe²⁺.
Mistake 3: “A deeper endpoint color is safer.” Stop at the first pale color that lasts 30 seconds. A deep color means you’ve gone past the endpoint, so the volume is too large and the result too high.
Quick check
Before a buret is filled with KMnO₄ solution, what should it be rinsed with last?
Rinsing with the titrant keeps leftover water from diluting it.
How are the equivalence point and the endpoint related?
With a well-chosen indicator, the endpoint is close to the equivalence point.
Why does a KMnO₄ titration need no separate indicator?
The color of MnO₄⁻ itself marks the endpoint.
Which acid is usually used to acidify a KMnO₄ titration?
Cl⁻ would be oxidized by MnO₄⁻, nitric acid is itself an oxidizing agent, and acetic acid is too weak; sulfuric acid doesn’t interfere.
How many moles of Fe²⁺ react completely with 20.00 mL of 0.02000 M KMnO₄?
n(MnO₄⁻) = 4.000 × 10⁻⁴ mol, × 5 = 2.000 × 10⁻³ mol.
KHP is a good primary standard for NaOH mainly because it is:
That makes the weighing accurate and keeps weighing errors small.
AP-style practice
To find the mass percent of iron in an iron ore, a student dissolves 0.2500 g of ore, reduces all the iron to Fe²⁺, acidifies with sulfuric acid, and titrates with 0.01750 M KMnO₄ to the endpoint, using 22.80 mL. The reaction is MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
(a) Give the oxidation number of manganese before and after the reaction. How many electrons does each MnO₄⁻ gain?
(b) How does the student know the endpoint has been reached?
(c) Calculate the moles of MnO₄⁻ used.
(d) Calculate the mass percent of iron in the ore (Fe 55.85 g/mol).
(e) The buret was rinsed with distilled water but not with the KMnO₄ solution. Is the measured mass percent of iron too high or too low? Explain.
Model answer and scoring
(a) Mn goes from +7 to +2, so each MnO₄⁻ gains 5 electrons.
(b) At first the purple of each drop of KMnO₄ disappears immediately; when the solution turns pale pinkish purple and stays that way for 30 seconds after swirling, MnO₄⁻ is in slight excess and the endpoint has been reached.
(c) n(MnO₄⁻) = 0.02280 L × 0.01750 M = 3.990 × 10⁻⁴ mol.
(d) n(Fe) = 5 × 3.990 × 10⁻⁴ = 1.995 × 10⁻³ mol; m(Fe) = 1.995 × 10⁻³ × 55.85 = 0.1114 g; mass percent = 0.1114 ÷ 0.2500 × 100% = 44.57%.
(e) Too high. Leftover water dilutes the KMnO₄, so its real concentration is below 0.01750 M and more volume is needed to reach the endpoint; the calculation still multiplies 0.01750 M by this larger volume, so the calculated n(MnO₄⁻) and mass of iron are too large.
Scoring: (a) +7 → +2, gaining 5 electrons; (b) a lasting pale pinkish purple; (c) the correct moles; (d) the 5 : 1 mole ratio and the correct result; (e) too high, with the chain “diluted → volume too large → result too high.”
Summary
- Titration calculations: V × c → n(titrant) → mole ratio from the balanced equation → n(analyte) → concentration or mass (multiply by any dilution factor).
- For acid–base titrations, choose the indicator by the equivalence pH; standardize NaOH against a primary standard such as KHP.
- KMnO₄ titrations indicate their own endpoint and are acidified with sulfuric acid; the first lasting pale color is the endpoint, and overshooting makes the result too high.
AP tip: Redox titration questions often ask first for the balanced equation or the number of electrons, then use it for the mole ratio. Write the equation and the ratio on your answer sheet first, and the later calculations go more smoothly.
Sources: atomic masses from the periodic table supplied with the AP Chemistry exam (giving KHP 204.22 g/mol and H₂O₂ 34.02 g/mol); the experimental data are illustrative.