Chemistry · Lesson 21

Practical 3: percentage purity, water of crystallisation and redox titrations

WAECNECO

What the syllabus asks

WAEC: percentage purity, water of crystallisation, relative molecular mass; redox titrations with KMnO₄ (iron(II), ethanedioate); redox reagents as tests. Links c5 (redox) and cp2.

NECO: percentage purity, water of crystallisation, relative molecular mass; redox titrations with KMnO₄ (iron(II), ethanedioate); redox reagents as tests. Links c5 (redox) and cp2.

What you will learn

Calculate percentage purity from titration data.
Find water of crystallisation in hydrates.
Determine relative molecular mass.
Explain the conditions of permanganate titrations.
Use redox ratios correctly.
Calculate percentage of iron in a sample.
Interpret redox colour changes in tests.
Diagnose errors in redox titrations.

Key points

Purity ≤ 100%.
106 + 18x for washing soda.
KMnO₄: no indicator, H₂SO₄ only.
1 : 5 for iron, 2 : 5 for oxalate.
Warm oxalate; do not heat iron(II).
Top of meniscus for KMnO₄.
K₂Cr₂O₇ orange → green.

Start here: Percentage purity

A sample from the market or a store room is rarely pure. Caustic soda sold for soap-making in Aba picks up moisture and carbon dioxide; washing soda loses some of its water of crystallisation. Titration measures how much of the reacting substance is really present.

Percentage purity = (mass concentration of pure substance found ÷ mass concentration of sample dissolved) × 100

Worked example 21.1: impure caustic soda

B was made by dissolving 5.00 g of impure NaOH in distilled water and making up to 1 dm³. 20.0 cm³ of B required 23.00 cm³ of 0.100 mol dm⁻³ HCl. Find the percentage purity of the NaOH. [Na = 23, O = 16, H = 1]

Solution. HCl + NaOH → NaCl + H₂O, 1 : 1. C_B = 0.100 × 23.00 ÷ 20.0 = 0.115 mol dm⁻³. Mass concentration of pure NaOH = 0.115 × 40 = 4.60 g dm⁻³. Percentage purity = 4.60 ÷ 5.00 × 100 = 92.0%. Percentage impurity = 8.00%.

Check your answer is possible

A purity above 100% means something has gone wrong: an inverted ratio, the wrong molar mass, or a titre from the wrong solution. Examiners give no credit for an impossible answer even if the method is mostly right, so recheck before moving on.

What the impurity is, and what it is not

The impurity must be something that does not react with the titrant. In caustic soda that has absorbed CO₂, part of the “impurity” is Na₂CO₃, which does react with acid. At school level the question will state or imply that the impurity is inert; accept that assumption and say so if asked to comment.

Worked example 21.1b: purity of washing soda as anhydrous carbonate

5.30 g of a commercial “soda ash” sample (impure anhydrous Na₂CO₃) was dissolved and made up to 1 dm³. Titration with HCl showed the solution to be 0.0480 mol dm⁻³ in Na₂CO₃. Find the percentage purity.

Solution. Pure Na₂CO₃ = 0.0480 × 106 = 5.088 g dm⁻³. Purity = 5.088 ÷ 5.30 × 100 = 96.0%.

Purity when the volume is not 1 dm³

Always bring the sample mass to the same volume basis as your titration result. “2.65 g made up to 250 cm³” is 2.65 × 4 = 10.6 g dm⁻³. Comparing 10.6 g dm⁻³ with a result in g dm⁻³ is correct; comparing 2.65 g directly with it is the commonest error in purity questions.

Worked example 21.12: purity of washing soda from a market

1.40 g of anhydrous sodium trioxocarbonate(IV) bought at a market was dissolved and made up to 250 cm³. 25.0 cm³ of this solution needed 24.00 cm³ of 0.100 mol dm⁻³ HCl. Calculate the percentage purity. [Na = 23, C = 12, O = 16] (Illustrative data.)

Solution.

Moles HCl = 0.100 × 24.00 ÷ 1000 = 0.00240 mol.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂: moles Na₂CO₃ in 25.0 cm³ = 0.00120 mol.
In 250 cm³ (ten times as much) = 0.0120 mol; mass = 0.0120 × 106 = 1.272 g.
Percentage purity = 1.272 ÷ 1.40 × 100 = 90.9%.

A purity above 100% means an arithmetic slip, usually forgetting the ×10 for the volumetric flask.

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Practice questions

1. Percentage purity =

mass of pure substance ÷ mass of sample × 100
mass of sample ÷ mass of pure substance × 100
moles ÷ mass × 100
molar mass ÷ 100

Show answer

A: mass of pure substance ÷ mass of sample × 100
Pure mass found by titration over sample mass.

2. 1.40 g of impure Na₂CO₃ contains 0.0120 mol Na₂CO₃. Its purity is [Na₂CO₃ = 106]

85.0%
100%
45.4%
90.9%

Show answer

D: 90.9%
0.0120 × 106 = 1.272 g; 1.272 ÷ 1.40 × 100.

3. Na₂CO₃·xH₂O has molar mass 286 g mol⁻¹. x is [H = 1, C = 12, O = 16, Na = 23]

Show answer

C: 10
106 + 18x = 286.

4. H₂C₂O₄·xH₂O has molar mass 126 g mol⁻¹. x is

Show answer

B: 2
90 + 18x = 126.

5. In acidified KMnO₄ titrations, the end point is

the first permanent pale pink colour
colourless to yellow
red to orange
blue-black

Show answer

A: the first permanent pale pink colour
KMnO₄ is self-indicating.

Key words

Percentage purity: Percentage by mass of the pure substance in a sample.
Water of crystallisation: Water chemically bound in a crystal, e.g. 10H₂O in washing soda.
Self-indicator: A titrant whose own colour shows the end point (KMnO₄).
Autocatalysis: Catalysis of a reaction by one of its products (Mn²⁺).
Constant mass: Repeated heating until two weighings agree.
Redox titration: Titration based on electron transfer.

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