# 15.3 Quantitative Methods of Analysis

F.6 Chemistry · Topic XV Analytical Chemistry · Summer Course Lesson 3

How much of a species is there? Weigh a precipitate (gravimetric) or titrate it (volumetric), and get the mole ratio right.

## Key points

- **Gravimetric analysis ／ 重量分析** finds the mass of the species being analyzed, or of a compound chemically related to it. Precipitation method: weigh the sample → dissolve → precipitate with excess reagent → stand in the hot mother liquor → filter → wash → dry → weigh → calculate.
- **Volumetric analysis ／ 容量分析** measures the volume of a solution of known concentration (the titrant) that reacts completely with the species, in a **titration**. Acid–alkali, precipitation and redox titrations.
- A **sintered glass filtering crucible** lets you filter, dry and weigh in one vessel. Dry to **constant mass** (weigh, dry, weigh again until successive readings agree) and cool in a **desiccator ／ 乾燥器**.
- **Systematic error ／ 系統誤差** (instrument, operator or method) pushes every result the same way and cannot be removed by repeating. **Random error** scatters results either side of the true value; reduce it by repeating and taking a mean.
- **Mohr titration**: Cl^−^ against standard AgNO~3~(aq) with a chromate indicator. The first excess Ag^+^ gives a **reddish brown** Ag~2~CrO~4~ precipitate. Keep the pH at about 8 with NaHCO~3~.
- **KMnO~4~ titrations** need no indicator: colourless → pale pink at the first excess MnO~4~^−^. Acidify with dilute H~2~SO~4~. KMnO~4~ contains impurities (MnO~2~), so standardize it against sodium oxalate warmed to 80 – 90 °C.
- **Permanganate index**: excess KMnO~4~ reacts with the water sample, excess sodium oxalate is added, and the oxalate left is titrated with KMnO~4~ (a **back titration**). Reported as mg of O~2~ per litre; a larger index means dirtier water.
- **Iodine titrations**: a reducing agent (vitamin C) is titrated with I~2~ generated in situ from IO~3~^−^ + I^−^; an oxidizing agent (OCl^−^ in bleach) liberates I~2~ from excess KI, then the I~2~ is titrated with thiosulphate.

## Must know

| Titration | Mole ratio | End point ／ watch out |
|---|---|---|
| MnO~4~^−^ : Fe^2+^ | 1 : 5 | self-indicating: colourless → pale pink; dilute H~2~SO~4~ |
| MnO~4~^−^ : C~2~O~4~^2−^ | 2 : 5 | same; warm to 80 – 90 °C |
| IO~3~^−^ → I~2~ | 1 : 3 | starch in at the start: dark blue appears |
| I~2~ : S~2~O~3~^2−^ | 1 : 2 | starch near the end: dark blue → colourless |
| Ag^+^ : Cl^−^ (Mohr) | 1 : 1 | first excess Ag^+^: reddish brown Ag~2~CrO~4~ |

MnO~4~^−^ + 8H^+^ + 5e^−^ → Mn^2+^ + 4H~2~O · 2MnO~4~^−^ + 5C~2~O~4~^2−^ + 16H^+^ → 2Mn^2+^ + 10CO~2~ + 8H~2~O · IO~3~^−^ + 5I^−^ + 6H^+^ → 3I~2~ + 3H~2~O · I~2~ + 2S~2~O~3~^2−^ → 2I^−^ + S~4~O~6~^2−^

## Common mistakes

- ✗ Acidify a KMnO~4~ titration with HCl. ✓ Dilute H~2~SO~4~ only: MnO~4~^−^ oxidizes Cl^−^ (extra titre), and HNO~3~ is itself an oxidizing agent.
- ✗ n(Fe^2+^) = n(MnO~4~^−^). ✓ Write the equation before the arithmetic: 1 MnO~4~^−^ ≡ 5 Fe^2+^, 2 MnO~4~^−^ ≡ 5 C~2~O~4~^2−^, 1 IO~3~^−^ → 3 I~2~, 1 I~2~ ≡ 2 S~2~O~3~^2−^.
- ✗ Stop at the 25.0 cm^3^ portion. ✓ 25.0 cm^3^ taken from 250.0 cm^3^ means × 10, then go back to the original sample (10.0 cm^3^ of bleach, 2 tablets, 0.190 g of wire).
- ✗ Add starch at the start of an iodine / thiosulphate titration. ✓ Add it near the end, when the mixture is pale yellow; added while iodine is still concentrated, the starch–iodine complex breaks up slowly and the end point is late and unclear. Starch goes in at the start only when I~2~ is made in situ (vitamin C).
- ✗ Rinse the pipette with distilled water only. ✓ Rinse the pipette with the solution it will measure; the volumetric flask is rinsed with distilled water only.

## Quick check

1. Which acid is used to acidify a KMnO~4~ titration, and why not hydrochloric acid?
2. 25.0 cm^3^ of an iron(II) solution needs 9.20 cm^3^ of 0.00500 mol dm^−3^ KMnO~4~(aq). How many moles of Fe^2+^ are in the 25.0 cm^3^?
3. In the bleach analysis, when is starch added, and what is the end-point colour change?
4. In a Mohr titration, what signals the end point, and why is the pH kept at about 8?
5. Why is a precipitate dried to constant mass and cooled in a desiccator before weighing?

## Answers

1. Dilute H~2~SO~4~. MnO~4~^−^ oxidizes Cl^−^, so the titre is too large.
2. n(MnO~4~^−^) = 0.00500 × 9.20/1000 = 4.60 × 10^−5^ mol; × 5 → n(Fe^2+^) = 2.30 × 10^−4^ mol.
3. Near the end, when the mixture is pale yellow; dark blue → colourless.
4. The first excess Ag^+^ gives a reddish brown Ag~2~CrO~4~ precipitate. Below pH 6 chromate becomes HCrO~4~^−^ (more Ag^+^ needed); in alkaline solution silver oxide precipitates.
5. Constant mass shows the solvent and volatile impurities are gone; the desiccator stops it taking up moisture while it cools.

## Did you know?

Humans cannot make vitamin C: unlike most mammals we carry a broken copy of the gene for the last enzyme in its synthesis. That is why sailors on long voyages developed scurvy until citrus juice was issued, which earned British sailors the nickname "limeys".