Week 15 · Term 1 review (Chapters 1–3) and Darslab’s own review tasks · 3 lessons
2
Review: chemical analysis
Chemistry Sec 3 Term 1, Chapters 1–3: Review
GoalRevise weeks 6–9 with new tasks: choose the tests that identify an anion or a cation, and solve titration, purity and water-of-crystallisation calculations.
New words
group reagent · one reagent that tests a whole family of ions at once, such as dilute HCl or BaCl₂ solutionend point · the moment in a titration when the indicator changes colourwater of crystallisation · the fixed number x of water molecules per formula unit in a salt such as Na₂CO₃·xH₂O
Explanation
Qualitative analysis follows the decision chart: add dilute HCl and look for a gas; if there is none, use concentrated sulphuric acid; if there is still nothing, use barium chloride solution. Then a second test confirms. For cations, the precipitate colour and what happens in extra alkali decide: white and dissolving in extra NaOH is aluminium, reddish-brown is iron(III), green-white is iron(II), and black with hydrogen sulphide is copper(II). All reactions are paper equations here. For quantitative analysis, use the mole route. Titration: MaVa ÷ na = MbVb ÷ nb, where n is the coefficient in the balanced equation, for example H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O has na = 1 and nb = 2. Percentage of a substance = mass found ÷ mass of sample × 100. Volatilisation: the loss in mass is the mass of gas or water driven off. Precipitation: weigh the precipitate, convert to moles, then to the moles of the substance in the sample.
Observation
First test
Likely ion
Colourless gas, fizzing
dilute HCl
carbonate
Cloudy yellow liquid and gas
dilute HCl
thiosulphate
Violet vapour
concentrated H₂SO₄
iodide
White precipitate, not dissolving in HCl
BaCl₂ solution
sulphate
Worked examples
Titration. 25.0 mL of a NaOH solution is neutralised by 20.0 mL of 0.100 mol/L H₂SO₄. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so na = 1 and nb = 2. MaVa ÷ na = MbVb ÷ nb gives 0.100 × 20.0 ÷ 1 = Mb × 25.0 ÷ 2, so Mb = 0.100 × 20.0 × 2 ÷ 25.0 = 0.160 mol/L. Check the order: 2.00 mmol of acid needs 4.00 mmol of alkali, and 4.00 ÷ 25.0 = 0.160.
Purity by precipitation. A 1.20 g sample containing K₂SO₄ gives 0.932 g of BaSO₄ with excess barium chloride. n(BaSO₄) = 0.932 ÷ 233 = 0.00400 mol, the same as n(K₂SO₄) because each sulphate gives one BaSO₄. m(K₂SO₄) = 0.00400 × 174 = 0.696 g. Purity = 0.696 ÷ 1.20 × 100 = 58 %.
Water of crystallisation. 2.86 g of Na₂CO₃·xH₂O loses 1.80 g of water when dried. The salt left is 2.86 − 1.80 = 1.06 g = 1.06 ÷ 106 = 0.0100 mol. The water is 1.80 ÷ 18 = 0.100 mol. x = 0.100 ÷ 0.0100 = 10, so the salt is Na₂CO₃·10H₂O.
Class activity
Mystery cards, paper only. Write six cards, each with one set of results, for example “fizzes with dilute HCl, colourless gas” or “reddish-brown precipitate with NaOH”. On six other cards write the names of the ions. Mix them and match every result to its ion, using the decision chart. For each pair, write on the back the next test that would confirm the answer. Swap with a friend to check. Do not smell or taste anything; no real tests at home.
Practice
1
A solid gives no gas with dilute HCl, and concentrated sulphuric acid releases a violet vapour. Name the anion and one confirming test.
Iodide. Silver nitrate solution gives a yellow precipitate (AgI), which does not dissolve in concentrated ammonia.
2
10.0 mL of NaOH is neutralised by 20.0 mL of 0.050 mol/L H₂SO₄. Find the concentration of the NaOH.
na = 1, nb = 2: 0.050 × 20.0 ÷ 1 = Mb × 10.0 ÷ 2, so Mb = 0.050 × 20.0 × 2 ÷ 10.0 = 0.20 mol/L.
3
2.78 g of FeSO₄·xH₂O loses 1.26 g of water. Find x. (FeSO₄ = 152 g/mol)
