Topic 16 · page 16 of 19

Redox titration calculations

Structured questions give you the steps; unstructured ones make you plan them. Both follow the same route, from moles of titrant to a mass or percentage, with the mole ratio doing the work in the middle.

15 min read16.15

Read first: Redox titration reactions and ratios

A structured calculation: iron in tablets 16.15

Further suggested practical ii is to find the percentage of iron in iron tablets by titration with potassium manganate(VII). The route from page 13 is: titrant moles, mole ratio, scale up, mass, percentage.

Worked examplePercentage of iron in a tablet (illustrative data)

A 0.350 g iron tablet (containing iron(II) sulfate) is dissolved in dilute sulfuric acid and made up to 100.0 cm³. 25.0 cm³ portions are titrated with 0.00500 mol dm⁻³ KMnO₄. The mean titre is 11.60 cm³. Find the percentage by mass of iron in the tablet.

n(MnO₄⁻) = 0.00500 × 11.60 ÷ 1000 = 5.80 × 10⁻⁵ mol
Ratio MnO₄⁻ : Fe²⁺ = 1 : 5, so n(Fe²⁺) in 25.0 cm³ = 5 × 5.80 × 10⁻⁵ = 2.90 × 10⁻⁴ mol
Scale up: 100.0 ÷ 25.0 = 4, so n(Fe²⁺) in the tablet = 4 × 2.90 × 10⁻⁴ = 1.16 × 10⁻³ mol
Mass of Fe = 1.16 × 10⁻³ × 55.8 = 0.0647 g (64.7 mg)
% Fe = 0.0647 ÷ 0.350 × 100 = 18.5%

An unstructured calculation: copper in brass 16.15

Further suggested practical i is to find the percentage of copper in brass by iodine–thiosulfate titration. In an unstructured question you plan the steps yourself. Write the equations and the overall ratio first.

Worked examplePercentage of copper in brass (illustrative data)

1.50 g of brass is dissolved in concentrated nitric acid, neutralised and made up to 250.0 cm³. Excess KI is added to 25.0 cm³ portions and the iodine released is titrated with 0.100 mol dm⁻³ Na₂S₂O₃. The mean titre is 16.40 cm³. Find the percentage of copper in the brass.

Plan. Copper(II) releases iodine, and the iodine reacts with thiosulfate:

\ce{2Cu^2+ + 4I- -> 2CuI + I2}

\ce{I2 + 2S2O3^2- -> 2I- + S4O6^2-}

2Cu²⁺ ≡ I₂ ≡ 2S₂O₃²⁻, so Cu²⁺ : S₂O₃²⁻ = 1 : 1.

n(S₂O₃²⁻) = 0.100 × 16.40 ÷ 1000 = 1.64 × 10⁻³ mol, which equals n(Cu²⁺) in 25.0 cm³
Scale up: 250.0 ÷ 25.0 = 10, so n(Cu) in the brass = 1.64 × 10⁻² mol
Mass of Cu = 1.64 × 10⁻² × 63.5 = 1.041 g
% Cu = 1.041 ÷ 1.50 × 100 = 69.4%

Brass is usually about 60 to 70% copper, so the answer is sensible.

Practical points. Dissolve the brass in nitric acid in a fume cupboard, because toxic nitrogen dioxide is given off. In the titration a white precipitate of copper(I) iodide forms, which makes the end point harder to see. Look for blue-black to colourless, with an off-white solid left in the flask.

Purity of a solid: potassium iodate(VII) 16.15

Further suggested practical iii is to prepare crystals of potassium iodate(VII), KIO₄, and measure their purity. These notes cover the purity measurement. Use your own course sheet for the preparation of the crystals.

A weighed sample is dissolved and made up to 250.0 cm³. Excess KI and dilute sulfuric acid are added to 25.0 cm³ portions, and the iodine released is titrated with thiosulfate, adding starch near the end.

\ce{IO4- + 7I- + 8H+ -> 4I2 + 4H2O}

One IO₄⁻ makes 4 I₂, and 4 I₂ need 8 S₂O₃²⁻, so IO₄⁻ : S₂O₃²⁻ = 1 : 8.

Worked examplePurity of potassium iodate(VII) (illustrative data)

0.250 g of crystals gives a mean titre of 16.90 cm³ of 0.0500 mol dm⁻³ Na₂S₂O₃ by the method above. Find the purity. $M_\text{r}$ (KIO₄) = 230.0.

n(S₂O₃²⁻) = 0.0500 × 16.90 ÷ 1000 = 8.45 × 10⁻⁴ mol
n(IO₄⁻) in 25.0 cm³ = 8.45 × 10⁻⁴ ÷ 8 = 1.056 × 10⁻⁴ mol
In 250.0 cm³: 1.056 × 10⁻³ mol. Mass = 1.056 × 10⁻³ × 230.0 = 0.243 g
Purity = 0.243 ÷ 0.250 × 100 = 97.2%

A purity below 100% suggests soluble impurities, or water left in the crystals. Recrystallising and drying them fully would improve it.

Check yourself

25.0 cm³ of Fe²⁺(aq) needs 18.20 cm³ of 0.0150 mol dm⁻³ KMnO₄. Calculate [Fe²⁺].
2.50 g of impure iron(II) sulfate crystals, FeSO₄·7H₂O, is dissolved and made up to 250.0 cm³. A 25.0 cm³ portion needs 8.80 cm³ of 0.0200 mol dm⁻³ KMnO₄. Calculate the percentage purity. ( $M_\text{r}$ = 277.9)
2.00 g of a copper alloy is dissolved and made up to 250.0 cm³. Excess KI is added to 25.0 cm³ portions and the iodine needs a mean titre of 20.40 cm³ of 0.0800 mol dm⁻³ Na₂S₂O₃. Calculate the percentage of copper.
25.0 cm³ of hydrogen peroxide solution needs 20.50 cm³ of 0.0200 mol dm⁻³ KMnO₄ in acid (MnO₄⁻ : H₂O₂ = 2 : 5). Calculate [H₂O₂].
A 0.300 g sample of KIO₄ is made up to 250.0 cm³. A 25.0 cm³ portion is treated with excess KI and acid and needs 19.80 cm³ of 0.0500 mol dm⁻³ Na₂S₂O₃. Find the percentage purity. ( $M_\text{r}$ = 230.0)
State one hazard and its precaution when brass is dissolved in concentrated nitric acid, and one difficulty in seeing the end point in the brass titration.

Answers cover these until you have tried the questions

n(MnO₄⁻) = 0.0150 × 18.20 ÷ 1000 = 2.73 × 10⁻⁴ mol. n(Fe²⁺) = 5 × 2.73 × 10⁻⁴ = 1.365 × 10⁻³ mol. [Fe²⁺] = 1.365 × 10⁻³ ÷ 0.0250 = 0.0546 mol dm⁻³.
n(MnO₄⁻) = 0.0200 × 8.80 ÷ 1000 = 1.76 × 10⁻⁴ mol. n(Fe²⁺) in 25.0 cm³ = 5 × 1.76 × 10⁻⁴ = 8.80 × 10⁻⁴ mol. In 250.0 cm³: 8.80 × 10⁻³ mol. Mass = 8.80 × 10⁻³ × 277.9 = 2.446 g. Purity = 2.446 ÷ 2.50 × 100 = 97.8%.
n(S₂O₃²⁻) = 0.0800 × 20.40 ÷ 1000 = 1.632 × 10⁻³ mol = n(Cu²⁺) in 25.0 cm³ (ratio 1 : 1). In 250.0 cm³: 1.632 × 10⁻² mol. Mass = 1.632 × 10⁻² × 63.5 = 1.036 g. % Cu = 1.036 ÷ 2.00 × 100 = 51.8%.
n(MnO₄⁻) = 0.0200 × 20.50 ÷ 1000 = 4.10 × 10⁻⁴ mol. n(H₂O₂) = 4.10 × 10⁻⁴ × 5 ÷ 2 = 1.025 × 10⁻³ mol. [H₂O₂] = 1.025 × 10⁻³ ÷ 0.0250 = 0.0410 mol dm⁻³.
n(S₂O₃²⁻) = 0.0500 × 19.80 ÷ 1000 = 9.90 × 10⁻⁴ mol. n(IO₄⁻) in 25.0 cm³ = 9.90 × 10⁻⁴ ÷ 8 = 1.2375 × 10⁻⁴ mol. In 250.0 cm³: 1.2375 × 10⁻³ mol. Mass = 1.2375 × 10⁻³ × 230.0 = 0.2846 g. Purity = 0.2846 ÷ 0.300 × 100 = 94.9%.
Dissolving brass in concentrated nitric acid gives off toxic nitrogen dioxide: work in a fume cupboard. In the titration a white precipitate of copper(I) iodide forms, so the end point (blue-black to colourless) is seen against an off-white solid and is harder to judge.