The problems are on the acids and bases worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Acids are substances that release hydrogen ions, H⁺, in water. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on acids and bases, from sort the household liquids to pure potassium chloride: each solution shows the formula, the substitution and the final answer with its unit.
The answers show the key chemistry behind each result. For equations, check the formulae first and then the balancing; for explanations, make sure you link your point to H⁺ ions or to ionisation.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4, IGCSE 7
Problem 1 Sort the household liquids
1. Acids: lemon juice, milk, stomach acid. Neutral: pure water. Alkalis: bleach, soapy water.2. The lowest pH is 1.5, so stomach acid is the most acidic.
3. At pH 2 universal indicator is red.
Problem 2 Name the salt
1. Sodium chloride.2. Copper(II) sulfate.
3. Potassium nitrate.
4. Zinc chloride (and hydrogen gas).
Problem 3 Complete the word equations
1. salt + water.2. salt + water + carbon dioxide.
3. salt + hydrogen.
4. salt + water.
Problem 4 Ions in acids and alkalis
1. The hydrogen ion, H⁺.2. The hydroxide ion, OH⁻.
3. H⁺(aq) + OH⁻(aq) → H₂O(l).
4. pH 7.
Problem 5 Reading indicator colours
1. Alkaline: all three colours are the alkaline colours.2. Blue or violet.
3. The hydroxide ion, OH⁻.
Problem 6 An ant sting
1. The alkali neutralises the acid in the sting, so the pain eases.2. Strong alkalis are corrosive and would burn the skin.
3. Neutralisation.
Problem 7 Balance the equations
1. Already balanced: CuO + H₂SO₄ → CuSO₄ + H₂O.2. Mg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O.
3. Already balanced: Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂.
4. Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (2 Al, 6 Cl, 6 H and 3 O on each side).
Problem 8 Ionic equations
1. Full ionic: Na⁺ + OH⁻ + H⁺ + Cl⁻ → Na⁺ + Cl⁻ + H₂O. Cancelling gives H⁺(aq) + OH⁻(aq) → H₂O(l).2. Na⁺ and Cl⁻.
3. CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g); the Cl⁻ ions are spectators.
Problem 9 Order the method
1. B, D, A, E, C.2. The excess guarantees that all the acid has reacted, so the crystals are not contaminated with acid.
3. Some zinc oxide stays undissolved however long you stir.
Problem 10 Choose the method
1. (i): zinc oxide is an insoluble base, so the excess can be filtered off.2. (ii): potassium hydroxide is soluble, so exact volumes are needed.
3. (iii): the insoluble barium sulfate forms at once and is filtered, washed and dried.
Problem 11 Mass of zinc sulfate
1. M(ZnO) = 81; n = 4.05 ÷ 81 = 0.0500 mol.2. Ratio 1 : 1, so 0.0500 mol.
3. M(ZnSO₄) = 65 + 32 + 64 = 161; m = 0.0500 × 161 = 8.05 g.
Problem 12 Fertiliser from ammonia
1. M(NH₃) = 17 g/mol; 8500 ÷ 17 = 500 mol.2. M(NH₄NO₃) = 80 g/mol; 500 × 80 = 40 000 g = 40 kg.
3. 0.92 × 40 = 36.8 kg.
Problem 13 Comparing pH values
1. A, with the lowest pH.2. 4 − 2 = 2 units, so 10 × 10 = 100 times.
3. 6 − 2 = 4 units, so 10⁴ = 10 000 times.
4. The volume increases tenfold, so [H⁺] falls tenfold and the pH rises by 1 to pH 3 (for a strong acid).
Problem 14 Strong or weak?
1. Acid X is strong: it is fully split into ions and has no whole molecules.2. Acid X has the lower pH because it has more H⁺ ions.
3. Acid X reacts faster, because more H⁺ ions are present to collide with the magnesium.
Problem 15 Dilution data
1. Each tenfold dilution raises the pH by 1 unit.2. pH 4.0.
3. Ethanoic acid is a weak acid: only a small fraction of molecules ionise, so [H⁺] is lower and the pH is higher.
Problem 16 Acid rain on a statue
1. n(H⁺) = 2400 × 1.0 × 10⁻⁴ = 0.24 mol.2. The ratio CaCO₃ : H⁺ is 1 : 2, so n = 0.12 mol.
3. m = 0.12 × 100 = 12 g.
4. pH 3.0 has ten times more H⁺, so the mass lost is ten times larger (120 g).
Problem 17 Treating a fish pond
1. 50 × 1000 = 50 000 dm³.2. n(H⁺) = 50 000 × 1.0 × 10⁻⁵ = 0.50 mol.
3. n(Ca(OH)₂) = 0.50 ÷ 2 = 0.25 mol; M = 74; m = 18.5 g.
4. Calcium hydroxide is only slightly soluble and a weak excess leaves a mildly alkaline pond, but a large excess makes the water too alkaline for fish.
Problem 18 Spot the errors
1. Strength is confused with concentration, in both statements: strong means fully ionised, weak means partly ionised.2. 1.0 mol/dm³ ethanoic acid (weak) has pH 2.4, lower than 0.001 mol/dm³ hydrochloric acid (strong) at pH 3.0. So a concentrated weak acid can be more acidic than a dilute strong acid.
3. "Hydrochloric acid is strong because it ionises completely. Ethanoic acid is weak because it ionises only partly."
Problem 19 Copper sulfate crystals
1. n = 0.500 × 0.0500 = 0.0250 mol.2. M(CuO) = 79.5; 0.0250 × 79.5 = 1.99 g.
3. M(CuSO₄·5H₂O) = 159.5 + 90 = 249.5; 0.0250 × 249.5 = 6.24 g.
4. 5.10 ÷ 6.2375 × 100 = 81.8%.
Problem 20 Pure potassium chloride
1. Potassium hydroxide is soluble, so the excess would dissolve and there would be nothing to filter off.2. Find the exact volume of acid that neutralises the alkali using an indicator, then repeat without the indicator using that volume. Evaporate part of the water and crystallise.
3. n(KOH) = 0.200 × 0.0250 = 0.00500 mol; n(KCl) = 0.00500 mol; M = 74.5; m = 0.373 g.
