Acid A

1. Answer: C

2. Answer: C

3. Answer: A

4. Answer: D

5. Answer: D

6. Answer: B Phosphoric acid is a tribasic acid.

7. Answer: C

8. Answer: C

9. Answer: C Sulphuric acid is a strong acid and a dibasic acid, therefore it has more ions and has higher conductivity. 10. Answer: D

11. Answer: C Concentrated KOH is a strong alkali.

12. Answer: A Lemon juice is the most acidic one amongst the four.

13. Answer: B

14. Answer: D Solid sodium hydroxide shows no alkaline properties.

15. Answer: D Oxalic acid is a weak acid.

16. Answer: A Nitric acid is a strong acid and contains no acid molecules.

17. Answer: D All of them are slightly acidic substances. 18. Answer: D CaCO3 and CaSO4 are insoluble in water.

19. Answer: D CaSO4 forms a protective layer on CaCO3 and prevents further reaction.

20. Answer: D Iron(II) hydroxide is a green precipitate.

21. Answer: C Phenolphthalein changes colourless in acidic and neutral solution. Vinegar is acidic; shampoo is alkaline while salt solution is neutral.

22. Answer: C Soda lime is a mixture of sodium hydroxide and calcium oxide which are strongly alkaline.

23. Answer: A Strong acid usually has a low pH value. However, its pH value can be higher if the acid solution is dilute. A dilute solution can still be a strong acid since it is completely ionized in water.

24. Answer: C Blue litmus paper cannot be used to compare the strength of acids as it can change to red colour only.

25. Solution: (a) (i) A, B, C, D (ii) E, F (b) (i) CuO (s) + H2SO4 (aq) →CuSO4 (aq) + H2O (l) (ii) Zn (s) + 2HCl (aq) →ZnCl2 (aq) + H2 (g) (iii) HCl (aq) + NaOH (aq) →NaCl (aq) + H2O (l) (c) C>B>D Both B and C are stronger acid than D. B give more ions when in solution than A. 26. Solution: (a) (i) Weakly acidic (ii) Because the water and alkali solutions of the perm can change the shape of the hair by breaking some of the bonds in the protein molecules of hair. (iii) The strongly alkaline solution of the perm can make the hair weaker and easy to break. (b) (i) The baking powder makes the breads or cakes ‘rise’ and spongy. (ii) The baking powder dissolves in water to form hydrogen ions. The hydrogen ions − + react with sodium hydrogencarbonate to form carbon dioxide gas. H (aq) + HCO3 (aq) → CO2(g) + H2O(l) Moreover, on heating, solid sodium hydrogencarbonate decomposes to form sodium carbonate, carbon dioxide gas and water vapour. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)

27.

Solution: (a) A and B (b) D (c) E (d) A (e) A can be ammonia solution / window cleaner. B can be sodium hydroxide / oven cleaner. C can be ethanoic acid / vinegar. D can be hydrochloric acid / stomach acid. E can be sodium chloride / pure water. F can be carbonic acid / unpolluted rain water. 28. Solution: (a) PH decreases (the solution becomes less alkaline) and the color change of the indicator depend on the amount of water added. If only a small amount of water is added, the solution is still alkaline and the indicator remains red in color. If a large amount of water is added, phenolphthalein indicator changes from red to colorless (when pH less than 8.3). (b) Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g) being added pH ∼ 1 pH = 7 pH ∼ 5-6 pH increase and the methyl orange indicator changes from red to yellow. (c) H2SO4(aq) + ZnO(s) → ZnSO4(aq) + H2O(l) pH ∼ 1 being added nearly neutral (pH ≈ 7) pH increase and the phenolphthalein indicator remains red in colour. (d) CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq) being added pH ∼ 13 neutral (pH = 7) pH decrease and the litmus indicator remains blue in colour. 29. Solution: (a) pH paper / pH meter / universal indicator (b) limewater > ‘Milk of magnesia’ > blood > distilled water > rain water > orange juice (c) The chemical present in ‘Milk of magnesia’ is magnesium hydroxide. The chemical present in limewater is calcium hydroxide. (d) Carbon dioxide is dissolved in unpolluted rain water. Sulphur dioxide and nitrogen dioxide are dissolved in polluted rain water. (e) Limewater is alkaline which can turn pH paper (universal indicator) to blue. pH meter shows pH = 12. Distilled water is neutral which can turn pH paper (universal indicator) to green. pH meter shows pH = 7. Rainwater is acidic which can turn pH paper (universal indicator) to yellow / orange / red. pH meter shows pH = 5-6(unpolluted) or pH = 3-4(polluted).

30. Solution: (a) They all contain hydrogen atoms. (b) When acids dissolve in water, hydrogen ions are produced. The ions are responsible for the conduction of electricity and the colour change of the litmus paper. (c) (i) Magnesium (or any metal above Cu in E.C.S.) reacts with acids to give hydrogen gas. Or, sodium hydrogencarbonate reacts with acids to give carbon dioxide gas. (ii) Mg (s) + 2H+ (aq) → Mg2+(aq) + H2(g) or, − HCO3 (aq) + 2H+(aq) → H2O(l) + CO2(g)

31. Solution: (a) X is sodium hydrogencarbonate / potassium hydrogencarbonate / sodium carbonate / potassium carbonate / calcium carbonate (or any metal carbonate / hydrogencarbonate) (b) The solid acid ionizes in water to give hydrogen ions. The hydrogen ions react with hydrogencarbonate / carbonate to give carbon dioxide gas. − (c) HCO3 (aq) + H+(aq) → H2O(l) + CO2(g) or, − CO32 (aq) + 2H+(aq) → H2O(l) + CO2(g) or, CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g) (d) The fizzy powder is insoluble in methylbenzene. There is no observable change. The acid will not ionize when there is no water. There are no H+ ions and the chemical in fizzy powder will not react. (e) Pass the gas into limewater. Carbon dioxide turns limewater milky. (f) Citric acid

32. Solution: (a) (i) Iron nail dissolves and colourless gas evolves. (ii) Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) + (iii) Fe(s) + 2H (aq) → Fe2+(aq) + H2(g) (b) (i) Colourless gas evolves. (ii) KHCO3(aq) + HCl(aq) → KCl(aq) + H2O(l) + CO2(g) − + (iii) HCO3 (aq) + H (aq) → H2O(l) + CO2(g) (c) (i) Lime water turns milky (white solid forms) at first and finally changes to colourless solution. (ii) CO2 (g) + Ca(OH)2 (aq) → CaCO3(s) + H2O (l) CO2 (g) + CaCO3(s) + H2O (l) → Ca(HCO3)2 (aq) − 2+ (iii) CO2(g) + Ca (aq) + 2OH (aq) → CaCO3(s) + H2O(l) (no ionic equation) CO2(g) + CaCO3(s) + H2O(l) → Ca(HCO3)2(aq) (no ionic equation) (d) (i) Brown iron(III) oxide dissolves and colourless gas evolves. (ii) 3H2SO4(aq) + Fe2O3(s) → Fe2(SO4)3(aq) + 3H2O(l) + (iii) Fe2O3(s) + 6H (aq) → 2Fe3+(aq) + 3H2O(l)

33. Solution: (a) During reaction, sodium carbonate completely dissolves and large amount of colourless gas evolves. Sodium carbonate reacts with dilute sulphuric acid and gives out carbon dioxide gas.In case of calcium carbonate, calcium carbonate will not dissolve and the evolution of gas stops after a short moment. An insoluble layer of CaSO4(s) forms on the surface of CaCO3(s) and prevents further reaction. Hence, no more gas evolves. (b) Na2CO3(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + CO2(g) CaCO3(s) + H2SO4(aq) → CaSO4(s) + H2O(l) + CO2(g) − (c) CO32 (aq) + 2H+(aq) → H2O(l) + CO2(g) − + CaCO3(s) + 2H (aq) + SO42 (aq) →CaSO4(s) + H2O(l) + CO2(g) (no ionic equation)

34. Solution: (a) Use pH paper / universal indicator / a pH meter / red litmus paper (for testing alkalis) / blue litmus paper (for tesing acids) (b) Acidic: vinegar, lemon juice Alkaline: window cleaner, ‘milk of magnesia’, toothpaste, soap (c) The ‘milk of magnesia’ is alkaline in nature. It neutralizes the excess gastric juice in our stomach. (d) The bacteria changes the food remains in our mouths to acids. The toothpaste neutralizes the acids produced. (e) (i) 7 (ii) This indicates that water is neutral. (f) Fruit juice / tomato sauce / coca-cola (g) Drain cleaner / antiacid (h) (i) Gas bubbles are formed and the magnesium strip dissolves. (ii) Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

35. Solution: (a) A： H2SO4 sulphuric acid B： CH3COOH ethanoic acid C： KCl potassium chloride D： NH3 aqueous ammonia E： NaOH sodium hydroxide (b) (i) H2SO4 is a strong acid and CH3COOH is a weak acid. (ii) A strong acid is one which ionizes completely to give hydrogen ions. (iii) Yes. Nitric acid completely ionizes in water. (c) pH paper (or universal indicator) and pH meter (d) (i) pH = 2-3 (ii) pH = 7-8 (iii) pH = 7-9

36.Solution: (a) A weak acid only slightly ionizes in water. (b) (i) Hydrochloric acid and sulphuric acid (ii) Ethanoic acid and citric acid (c) (i) Hydrochloric acid has a lower pH and higher conductivity than ethanoic acid. (ii) The pH values and conductivity depend on the number of H+ ions in the solution. Hydrochloric acid is a strong acid. It ionizes completely in water and produces more ions. (d) Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid. Sulphuric acid + produces twice the amount of H ions than hydrochloric acid. (e) No. It does not conduct electricity and has no acidic properties because water is essential for the ionization of acid. 37. Solution: (a) Reagent in bottle A or C reacts with iron(II) sulphate solution and a pale green precipitate forms. FeSO4(aq) + 2NaOH(aq) → Fe(OH)2(s) + Na2SO4(aq) FeSO4(aq) + Ca(OH)2(aq) → Fe(OH)2(s) + FeSO4(aq) Reagent in bottle A or C may be dilute sodium hydroxide solution or calcium hydroxide solution. (b) Reagent in bottle C reacts with carbon dioxide and a white precipitate forms. CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l) Reagent in bottle C is calcium hydroxide solution. Combine this deduction with that in (a), reagent in bottle A should be dilute sodium hydroxide solution. (c) Reagent in bottle B or D reacts with potassium carbonate and carbon dioxide gas forms. K2CO3(aq) + 2HCl(aq) → 2KCl(aq) + H2O(l) + CO2(g) K2CO3(aq) + H2SO4(aq) → K2SO4(aq) + H2O(l) + CO2(g) Reagent in bottle B or D may be dilute hydrochloric acid or dilute sulphuric acid. (d) Reagent in bottle C and D are mixed together and a white precipitate forms. H2SO4(aq) + Ca(OH)2(aq) → CaSO4(s) + 2H2O(l) Reagent in bottle C is calcium hydroxide solution. It reacts with dilute sulphuric acid to form an insoluble product CaSO4(s). Hence, reagent in bottle D should be dilute sulphuric acid. Combine this deduction with that in (c), reagent in bottle B should be dilute hydrochloric acid. 38.Solution: Add copper(II) chloride solution into each of four chemicals. Only sodium hydroxide gives (blue) precipitate. CuCl2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaCl(aq) Add sodium carbonate solid into the remaining three chemicals. Only hydrochloric acid gives (colourless) gas bubbles / effervescence. Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g) Put the remaining two solutions into test tubes respectively and heat it over the Bunsen burner. Evaporate the remaining two solutions to dryness. Only sodium chloride solution will leave (white) residue. NaCl(aq) heat→ NaCl(s) The remaining one is distilled water which has nothing remains upon heating. H2O(l) heat→ H2O(g)