BK2 ans

HKDSE CHEMISTRY – A Modern View

(Chemistry)

Coursebook 2

Suggested answers

Chapter 14 Introduction to acids and alkalis Page

Number

Class Practice 1

Chapter Exercise 3

Chapter 15 Concentration of solutions

Class Practice 6

Chapter Exercise 9

Chapter 16 Indicators and pH

Class Practice 12

Chapter Exercise 13

Chapter 17 Strength of acids and alkalis

Class Practice 15

Chapter Exercise 16

Chapter 18 Salts and neutralization

Class Practice 18

Chapter Exercise 21

Chapter 19 Volumetric analysis involving acids and alkalis

Class Practice 26

Chapter Exercise 30

© Aristo Educational Press Ltd. 2009

Part Exercise 36

Chapter 20 Hydrocarbons from fossil fuels

Class Practice 39

Chapter Exercise 40

Chapter 21 Consequences of using fossil fuels

Class Practice 41

Chapter Exercise 42

Chapter 22 Homologous series, structural formulae and naming of

carbon compounds

Class Practice 44

Chapter Exercise 46

Chapter 23 Alkanes and alkenes

Class Practice 48

Chapter Exercise 49

Chapter 24 Addition polymers

Class Practice 51

Chapter Exercise 53

Part Exercise 55

© Aristo Educational Press Ltd. 2009

HKDSE CHEMISTRY A Modern View (Chemistry) Coursebook 2

Chapter 14 Introduction to acids and alkalis

Class Practice

A14.11. Zinc granules sink to the bottom.2. Zinc granules dissolve to give a colourless solution.3. Effervescence occurs./Colourless gas bubbles evolve.4. The test tube becomes warm as heat is given out.5. Hissing sound is heard.

A14.21. (a) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

(b) ZnO(s) + H2SO4(aq) ZnSO4(aq) + H2O(l)(c) Fe(OH)2(s) + 2HCl(aq) FeCl2(aq) + 2H2O(l)(d) CuCO3(s) + H2SO4(aq) CuSO4(aq) + CO2(g) + H2O(l)(e) Ca(HCO3)2(s) + 2HCl(aq) CaCl2(aq) + 2CO2(g) + 2H2O(l)

2. (a) Magnesium dissolves./Effervescence occurs. (Colourless gas bubbles evolve.)/The solution becomes warm./The resultant solution is colourless.

(b) Zinc oxide dissolves./The resultant solution is colourless.(c) Iron(II) hydroxide dissolves./The resultant solution is green.(d) Copper(II) carbonate dissolves./Effervescence occurs. (Colourless gas

bubbles evolve.)/The resultant solution is blue.(e) Calcium hydrogencarbonate dissolves. /Effervescence occurs. (Colourless

gas bubbles evolve.)/The resultant solution is colourless.

A14.3(a) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)(b) ZnO(s) + H2SO4(aq) ZnSO4(aq) + H2O(l)

ZnO(s) + 2H+(aq) Zn2+(aq) + H2O(l)(c) 2KHCO3(aq) + H2SO4(aq) K2SO4(aq) + 2CO2(g) + 2H2O(l)

HCO3(aq) + H+(aq) CO2(g) + H2O(l)

A14.41. Magnesium reacts with the hydrogen ions (from ascorbic acid) to give hydrogen

gas.Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

2. (a) There is no colour change to the litmus paper. When there is no water, liquid ethanoic acid (pure) cannot ionize to give H+

(aq) and thus cannot exhibit acidic properties. (b) The blue litmus paper turns red.

In the presence of water, ethanoic acid ionizes to give H+(aq) and thus can exhibit acidic properties.

3. Fizzy drink tablets or powder should be stored in a cool and dry place.

© Aristo Educational Press Ltd. 2009 1


A14.5(a) 1 (monobasic), HCOOH ⇌ HCOO + H+

(b) 2 (dibasic), (COOH)2 ⇌ (COO)2 + 2H+

A14.6(a) 2KOH(aq) + CO2(g) K2CO3(aq) + H2O(l)(b) (NH4)2SO4(aq) + Ca(OH)2(s) CaSO4(s) + 2NH3(g) + 2H2O(l)

A14.7Magnesium hydroxide, aluminium hydroxide, zinc hydroxide and lead(II) hydroxide

A14.81. (a) (i) Potassium hydroxide (ii) OH(aq) + H+(aq) H2O(l)

(b) (i) Magnesium oxide, zinc hydroxide and copper(II) oxide(ii) MgO(s) + 2H+(aq) Mg2+(aq) + H2O(l)

Zn(OH)2(s) + 2H+(aq) Zn2+(aq) + 2H2O(l)CuO(s) + 2H+(aq) Cu2+(aq) + H2O(l)

2. (a) (i) Pb(NO3)2(aq) + Ca(OH)2(aq) Pb(OH)2(s) + Ca(NO3)2(aq) (ii) Pb2+(aq) + 2OH(aq) Pb(OH)2(s)

(b) (i) CuSO4(aq) + 2KOH(aq) Cu(OH)2(s) + K2SO4(aq) (ii) Cu2+(aq) + 2OH(aq) Cu(OH)2(s)



Chapter 14 Introduction to acids and alkalis

Chapter Exercise

1. sour, red, hydrogen, carbon dioxide, salts, water2. H+(aq)3. water, water, hydrogen, H+

4. dibasic, monobasic5. salt, water, metal oxides, ammonia, alkali6. limewater, carbon dioxide7. bitter, blue, salts, water, carbonates, water, salts, ammonia, water, metal

hydroxides8. hydroxide, OH

9. H+(aq), oxidizing, oxidizing, dehydrating10. Concentrated11. B12. B13. C14. C15. D16. B17. D18. D19. B20. B21. B22. C23. B24. D25. D26. C27. A28. B29. C30. B

31. (a) Hydrochloric acid(b) Citric acid/ascorbic acid(c) Hydrochloric acid(d) Sulphuric acid

32. (a) Ethanoic acid(b) 1 (Monobasic)(c) (i) Effervescence occurs./Magnesium ribbon dissolves./A colourless

solution is formed./The solution becomes warm./A hissing sound is heard. (Any FOUR)

(ii) 2CH3COOH(aq) + Mg(s) (CH3COO)2Mg(aq) + H2(g)(d) (i) Carbon dioxide is formed.

(ii) Na2CO3(aq) + 2CH3COOH(aq) 2CH3COONa(aq) + CO2(g) + H2O(l)



33. (a) Carbon dioxide(b) Gas Y can turn limewater milky. Carbon dioxide reacts with calcium

hydroxide to form the insoluble white solid of calcium carbonate.Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)

(c) In the presence of water, the solid acid ionizes to give H+(aq) which reacts with sodium hydrogencarbonate to give carbon dioxide.

(d) H+(aq) + HCO3(aq) H2O(l) + CO2(g)

(e) 2NaHCO3 Na2CO3 + CO2 + H2O

34. (a) (i) Fe(s) + H2SO4(aq) FeSO4(aq) + H2(g) (ii) Fe(s) + 2H+(aq) Fe2+(aq) + H2(g)(b) (i) No reaction

(ii) No reaction(c) (i) ZnO(s) + 2HCl(aq) ZnCl2(aq) + H2O(l)

(ii) ZnO(s) + 2H+(aq) Zn2+(aq) + H2O(l)(d) (i) CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

(ii) CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l)(e) (i) KHCO3(aq) + HNO3(aq) KNO3(aq) + CO2(g) + H2O(l)

(ii) HCO3(aq) + H+(aq) CO2(g) + H2O(l)

35. (a) Monobasic acid is the acid that gives one hydrogen ion per molecule in aqueous solution. Dibasic acid is the acid that gives two hydrogen ions per molecule in aqueous solution.

(b) (i) 2HA(aq) + Zn(s) ZnA2(aq) + H2(g)(ii) H2B(aq) + Zn(s) ZnB(aq) + H2(g)(iii) 2HA(aq) + CuO(s) CuA2(aq) + H2O(l)(iv) H2B(aq) + CaCO3(s) CaB(aq) + CO2(g) + H2O(l)(v) HA(aq) + NaHCO3(aq) NaA(aq) + CO2(g) + H2O(l)

36. (a) Fe2+(aq) and Cu2+(aq)Iron(II) hydroxide – dirty greenCopper(II) hydroxide – pale blue

(b) Mg2+(aq)Mg2+(aq) + 2OH(aq) Mg(OH)2(s)

(c) OH(aq)NH3(aq) + H2O(l) ⇌ NH4

+(aq) + OH(aq)(d) As an active ingredient in window cleaners.

37. (a) Stir solid calcium hydroxide with water for a long time, then filter the suspension to get the clear filtrate.

(b) Hydroxide ion(c) 2NH4NO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2NH3(g) + 2H2O(l)(d) (i) A white precipitate is formed.

Ca2+(aq) + CO32(aq) CaCO3(s)

(ii) A pale blue precipitate is formed.Cu2+(aq) + 2OH(aq) Cu(OH)2(s)



38. (a) A: Concentrated hydrochloric acidB: Concentrated sulphuric acidC: Concentrated nitric acid

(b) Corrosive(c) Light speeds up the decomposition of concentrated nitric acid to give the

brown gas NO2. The gas dissolves in water to form a yellow solution. 4HNO3(aq) 2H2O(l) + 4NO2(g) + O2(g)

(d) Wash the affected area with plenty of water.

39. Dissolve some pure vitamin C in distilled/deionized water to get a dilute solution.

Add some zinc granules to the vitamin C solution. Colourless gas bubbles are evolved from the surface of the granules. When the gas is collected and tested with a burning splint, it will give a ‘pop’ sound, indicating that the gas is hydrogen.

Add a few sodium hydroxide pellets to the vitamin C solution and stir. The colourless mixture becomes warm because neutralization reaction is exothermic.

Add one spatula measure of potassium carbonate solid to the vitamin C solution. Colourless gas bubbles are given off. The gas can turn limewater milky, indicating that it is carbon dioxide.

40. Concentrated hydrochloric acid, nitric acid and sulphuric acid are three common concentrated acids found in the school laboratory. They would ‘eat away’ other substances such as metals, clothes, paper and the skin.

Concentrated hydrochloric acid is highly corrosive because of its high concentration of hydrogen ions.

The corrosive nature of concentrated nitric acid is mainly due to its oxidizing property.

Concentrated sulphuric acid is also highly corrosive because of its dehydrating and oxidizing properties. It is even more corrosive than concentrated hydrochloric acid and nitric acid. It quickly dehydrates the skin, causing severe burns.

Concentrated sodium hydroxide and potassium hydroxide solutions are common concentrated alkalis found in the school laboratory. They are very corrosive especially when they are hot.

They attack the skin readily and stain it yellow or even black.



Chapter 15 Concentrations of solutions

Class Practice

A15.1(a) No. of moles of K2SO4 = 0.15 mol

Molarity of the solution = = 0.30 mol dm3 (or M)

(b) Molar mass of K2SO4 = 39.1 2 + 32.1 + 16.0 4 g mol1 = 174.3 g mol1

Concentration of the solution = 0.30 mol dm3 174.3 g mol1 = 52.3 g dm3

A15.2

(a) (i) Concentration in g dm3 = = 20 g dm3

(ii) Molar mass of KOH = 39.1 + 16.0 + 1.0 g mol1 = 56.1 g mol1

No. of moles of KOH = = 0.178 mol

Molarity of the solution (mol dm3) = = 0.356 mol dm3 (or M)

(b) (i) Volume of the solution in dm3 = dm3 = 1.0 dm3

Concentration in g dm3 = = 10 g dm3

(ii) Molar mass of KOH = 39.1 + 16.0 + 1.0 g mol1 = 56.1 g mol1

No. of moles of KOH = =0.178mol

Molarity of the solution (mol dm3) = =0.178 mol dm3 (or M)

(c) (i) Volume of the solution in dm3 = dm3 = 0.25 dm3

Concentration in g dm3 = = 12 g dm3

(ii) Molar mass of NaNO3 = 23.0 + 14.0 + 16.0 3 g mol1 = 85.0 g mol1

No. of moles of NaNO3 = = 0.0353 mol


(d) (i) Concentration in g dm3 = = 2.65 g dm3

(ii) Molar mass of Na2CO3 = 23.0 2 + 12.0 + 16.0 3 g mol1 = 106 g mol1

No. of moles of Na2CO3 = = 0.0500 mol




(e) (i) Volume of the solution in dm3 = dm3 = 0.100 dm3

Concentration in g dm3 = = 98.0 g dm3

(ii) Molar mass of H2SO4 = 1.0 2 + 32.1 + 16.0 4 g mol1 = 98.1 g mol1

No. of moles of H2SO4 = = 0.100 mol


A15.3

No. of moles of Na2SO4 = 0.40 mol dm3 dm3 = 0.0200 mol

A15.4

Volume of NaOH solution = = 1.6 dm3

A15.5(a) No. of moles of Na2SO4 in 500.0 cm3 of solution

= 0.50 mol dm3 dm3 = 0.250 mol

Mass of Na2SO4 = 0.250 mol (23.0 2 + 32.1 + 16.0 4) g mol1 = 35.5 g(b) No. of moles of Na2CO3 in 0.25 dm3 of solution

= 0.15 mol dm3 0.25 dm3 = 0.0375 molMass of Na2CO3 = 0.0375 mol (23.0 2 + 12.0 + 16.0 3) g mol1 = 3.98 g

A15.61. 1 mole of Al2(SO4)3 contains 2 moles of Al3+ ions and 3 moles of SO4

2 ions. molarity of Al3+ ions = 0.50 M 2 = 1.0 M (or 1.0 mol dm3)

Molarity of SO42 ions = 0.50 M 3 = 1.5 M (or 1.5 mol dm3)

2. No. of moles of K2CO3 = 0.20 mol dm3 dm3 = 0.0050 mol

1 mole of K2CO3 contains 2 moles of K+ ions and 1 mole of CO32 ions.

number of moles of K+ ions = 0.0050 mol 2 =0.010 mol Number of moles of CO3

2 ions = 0.0050 mol

A15.71. (MV)before dilution = (MV)after dilution

0.1 = M

M = 0.01Molar concentration of diluted NaOH(aq) = 0.01 M



2. After diluting ten times, the molarity of the diluted solution will be M = 0.05

M(MV)before dilution = (MV)after dilution

0.5 = 0.05

V = 250Volume of distilled water to be added = (250 – 25) cm3 = 225 cm3



Chapter 15 Concentrations of solutions

Chapter Exercise

1. moles, per dm3

2. moles, volume3. molarity, volume4. A5. C6. D7. C8. D9. D10. D11. A12. C13. D14. A15. C16. B17. C18. D19. C20. B21. B22. B23. B24. B

25. mol dm3 or M

26. (a) Molarity of K2CO3 = = 1.40 M

(b) Molar mass of K2CO3 = 39.1 2 + 12.0 + 16.0 3 g mol1= 138.2 g mol1

Concentration of K2CO3 in g dm3 = 1.40 mol dm3 138.2 g mol1

= 193 g dm3

27. Molar mass of (COOH)22H2O = (12.0 + 16.0 2 + 1.0) 2 + 2 (1.0 2 + 16.0) g mol1 = 126 g mol1

No. of moles of (COOH)22H2O used = = 0.115 mol

Molarity of the solution = = 0.230 M

28. (a) No. of moles of Mg(NO3)2 present = 1.5 mol dm3 dm3

= 0.1875 mol



(b) Molar mass of Mg(NO3)2 = 24.3 + (14.0 + 16.0 3) 2 g mol1

= 148.3 g mol1

Mass of Mg(NO3)2 used = 0.1875 mol 148.3 g mol1 = 27.81 g

29. Molar mass of NaOH = 23.0 + 16.0 + 1.0 g mol1 = 40.0 g mol1

No. of moles of NaOH used = = 0.100 mol

Volume of solution prepared = = 0.833 dm3 (or 833 cm3)

30. Molar mass of Na2CO310H2O = 23.0 2 + 12.0 + 16.0 3 + 10 (1.0 2 + 16.0) g mol1 = 286 g mol1

Molarity of Na2CO3 = = 0.350 M

1 mol of Na2CO3 contains 2 moles of Na+ ions and 1 mole of CO32 ions.

molarity of Na+ ions = 0.350 M 2 = 0.700 MMolarity of CO3

2 ions = 0.350 M

31. (a) (MV)before dilution = (MV)after dilution

11.0 = M

M = 5.50Molarity of diluted HCl(aq) = 5.50 M

(b) (MV)before dilution = (MV)after dilution

5.50 = M

M = 2.20Molarity of diluted HCl(aq) = 2.20 M

(c) (MV)before dilution = (MV)after dilution

2.20 = M

M = 1.22 Molarity of diluted HCl(aq) = 1.22 M

32. (a) (MV)before dilution = (MV)after dilution

0.65 = 0.45

V = 65.0Volume of distilled water needed = 65.0 – 45.0 cm3 = 20.0 cm3

(b) (MV)before evaporation = (MV)after evaporation

0.45 = M

M = 0.975Molarity of HNO3 after evaporation = 0.975 M



33. (a) Mass of NH3 in 1 cm3 of the solution = 0.68 g cm3 1 cm3 = 0.231

g Mass of NH3 in 1 dm3 of the solution is 0.231 g 1000 = 231 g

(b) Molarity of NH3 solution = = 13.6 M


13.6 1 = 1 V V = 13.6

Volume of distilled water = 13.6 – 1 dm3 = 12.6 dm3

34. (a) No. of moles of oxalic acid

= = 0.0143

mol

(b) Molarity of oxalic acid = = 0.572 M


0.572 = M

M = 0.0572Molarity of the diluted oxalic acid = 0.0572 M

35.

Formula of solute

Molar mass

(g mol1)

Mass of solute (g)

Volume of solution

MolarityConcentration of

solution (in g dm3)solution cation anion

(a) NaCl 58.5 117 2.00 dm3 1.00 M 1.00 M 1.00 M 58.5

(b) AgNO3 170 17.0 1.00 dm3 0.100 M 0.100 M 0.100 M 17.0

(c) Na2CO3 106 2.65 500 cm3 0.050 M 0.100 M 0.050 M 5.30

(d) MgCl2 95.3 29.4 1.50 dm3 0.206 M 0.206 M 0.412 M 19.6

(e) C6H12O6 180 0.36 12.0 cm3 0.167 M 30.0




Class Practice

A16.1(a) pH = log [H+(aq)] = log (1.000) = 0(b) pH = log [H+(aq)] = log (0.100) = 1(c) pH = log [H+(aq)] = log (0.010) = 2(d) pH = log [H+(aq)] = log (0.001) = 3

A16.21. A: weakly acidic

B: strongly acidicC: weakly alkalineD: neutral

2. (a) The pH of the solution would increase. Magnesium reacts with the hydrogen ions in aqueous solution to give hydrogen gas.Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)When H+(aq) ions are removed, [H+(aq)] decreases and the pH of the solution increases.

(b) The pH of the solution would decrease. Hydrogen chloride gas dissolves in water and ionizes completely to give H+(aq) ions. When H+(aq) ions are formed, [H+(aq)] increases and the pH of the solution decreases.

A16.3B: Incorrect conclusion. When the litmus is blue, the pH of solution must be greater than 8 which is alkaline.C: Incorrect conclusion. Phenolphthalein is colourless when the pH of the solution is below 8.3. This solution can be acidic, neutral or weakly alkaline.




Chapter Exercise

1. (a) higher(b) equal(c) lower

2. H+(aq), pH, pH3. [H+(aq)]4. higher, [H+(aq)]5. colour, ethyl orange, itmus, henolphthalein6. Universal, Universal, pH7. pH meter8. B9. A10. C11. A12. A13. A14. A15. D16. B17. A18. C19. A20. A21. C22. B23. A

24. (a) pH = log [H+(aq)] = log (1.8 105) = (4.7) = 4.7 Since the pH < 7, this solution is acidic. (b) pH = log [H+(aq)] = log (7.2 109) = (8.1) = 8.1 Since the pH > 7, this solution is alkaline.

(c) pH = log [H+(aq)] = log (0.02) = (1.7) = 1.7 Since the pH < 7, this solution is acidic.

25. (a) Using the equation, pH = log [H+(aq)] 5.2 = log [H+(aq)] [H+(aq)] = 6.3 106 M (b) 8.4 = log [H+(aq)] [H+(aq)] = 4.0 109 M

(c) 3.68 = log [H+(aq)] [H+(aq)] = 2.09 104 M

26. Refer to Coursebook 2 Page 54 Figure 16.3.

27. (a) Use pH paper/universal indicator/pH meter/data-logger with pH electrode. (Any TWO)

(b) Oven cleaner, distilled water, orange juice(c) Orange juice



28. (a) pH meter(b) Lightweight/compact/waterproof/extremely easy to use/quick to obtain

accurate readings (Any TWO)

29. One way to determine the pH of an aqueous solution is to use acid-base indicators. An acid-base indicator is a special dye added to the solution to be tested. Its colour changes with different pH values of the solution.

Some common acid-base indicators are litmus, methyl orange and phenolphthalein.

However, these indicators cannot tell accurately the pH of a solution. They can only give a rough idea of the acidity or alkalinity.

Another way to measure pH value of a solution is to use universal indicator. It is a mixture of several indicators which gives different colours in different pH ranges. Universal indicator paper is also called pH paper.

The most accurate way to tell the pH of a solution is to use pH meter or data-logger. Both involve the dipping of a pH electrode into the testing solution.

The pH of a solution is simply read from the display of the meter.




Class Practice

A17.1(a) OH(aq)/hydrated hydroxide ion(b) NH3(aq) + H2O(l) ⇌ NH4

+(aq) + OH(aq)(c) NaOH(aq) is a stronger alkali than NH3(aq). (Being a stronger alkali, NaOH(aq)

can dissociate to a greater extent to give a high concentration of mobile ions (Na+

(aq) and OH(aq)) in aqueous solution. This explains the higher electrical conductivity of 0.1 M NaOH(aq).)

(d) As NaOH(aq) is a stronger alkali than NH3(aq), 0.1 M NaOH(aq) is more alkaline than 0.1 M NH3(aq). Thus, the former has a higher pH.

A17.2(a) Sulphuric acid molecules(b) Water molecules, hydrogen ions, nitrate ions, hydroxide ions(c) Water molecules, citric acid molecules, hydrogen ions, citrate ions, hydroxide

ions(d) Water molecules, ammonia molecules, ammonium ions, hydroxide ions,

hydrogen ions

A17.3Put equal volumes (about 80.0 cm3) of 0.1 M HCl(aq) and 0.1 M CH3COOH(aq) into two 100 cm3 beakers separately. Use the electronic balance to weigh equal masses (about 0.4 g) of lumps of calcium carbonate and put them into the two acids. Observe and compare the rates of evolution of gas bubbles from the two reaction mixtures.The acid that has a higher rate of evolution of gas should be the stronger acid ― HCl(aq) in this case. It can be explained that the stronger acid can ionize more completely to give a higher concentration of hydrogen ions and its reaction with calcium carbonate will be faster.

A17.41. (a) concentrated/weak

(b) dilute/weak(c) dilute/strong(d) concentrated/strong

2. (a) H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + CO2(g) + H2O(l)2H+(aq) + CO3

2(aq) CO2(g) + H2O(l)(b) CH3COOH(aq) + KHCO3(aq) CH3COOK(aq) + CO2(g) + H2O(l)

CH3COOH(aq) + HCO3(aq) CH3COO(aq) + CO2(g) + H2O(l)

or H+(aq) + HCO3(aq) CO2(g) + H2O(l)

A17.5(a) The statement is wrong. Hydrochloric acid is a strong acid which can ionize in

water completely. It has such a high pH (5) because it is only a dilute solution. It should be noted that pH is a measure of [H+(aq)] but not a measure of strength of acid or alkali.

(b) The statement is wrong. Ethanoic acid is only a weak acid whatever its concentration is.




Chapter Exercise

1. (a) completely, highly, ydrochloric(b) slightly, thanoic

2. (a) completetly, odium hydroxide(b) slightly, mmonia

3. (a) higher(b) lower(c) higher

4. D5. C6. B7. A8. C9. B10. B11. A12. A

13. In the presence of water, NH3(aq) reacts with water to give OH(aq) ions which make the solution alkaline.NH3(g) + H2O(l) ⇌ NH4

+(aq) + OH(aq)

Red litmus turns blue in the alkaline solution.

14. (a) Ethanoic acid is a weak acid while hydrochloric acid is a strong acid.(b) A strong acid is one which completely/highly ionizes in water.

A weak acid is one which only slightly/incompletely/partially ionizes in water.

(c) Since hydrochloric acid is a stronger acid than ethanoic acid, the former ionizes more completely than the latter. For the same molar concentration of the two acids, hydrochloric acid has a higher molar concentration of hydrogen ions. Thus, it has a lower pH.

(d) By using pH meter(e) 1. Put a known volume of 1.1 M hydrochloric acid into the container and

record the brightness of the light bulb.2. Repeat step 1 using the same volume of 1.1 M ethanoic acid.3. Compare the brightness of the light bulb in steps 1 and 2.4. The brightness of the light bulb indicates the strength of the acid. The

brighter the light bulb, the stronger the acid is.

15. (a) HA(aq) ⇌ H+(aq) + A (aq)(b) H+(aq), OH(aq), HA(aq), A(aq)

16. (a) HCl(aq) is a strong acid. 0.1 M HCl will ionize completely to give 0.1 M [H+(aq)].

(b) pH = log [H+(aq)] = log (0.1) = 1



(c) CH3COOH(aq) is a weak acid, so it ionizes only slightly to give a small amount of H+(aq).0.1 M CH3COOH(aq) gives H+(aq) ions with concentration lower than 0.1 M. Hence, the pH of 0.1 M CH3COOH is not 1 but higher than 1.

17. The statement of the S4 student is wrong. The pH value is a measure of hydrogen ion concentration. The higher the hydrogen ion concentration, the lower is the pH value. A weak acid ionizes only very slightly in water giving a low concentration

of hydrogen ions. The resultant acid solution can have a pH of 5. However, this can also occur for a strong acid. Although a strong acid

ionizes completely in water, if the solution is a very dilute one, the concentration of hydrogen ions will not be too high. Thus, a dilute solution of a strong acid can also have a pH of 5.

18. When a piece of blue litmus paper is dipped in some solid citric acid, there is no colour change of the litmus paper. When some solid citric acid is dissolved in some deionized water, the aqueous solution turns the blue litmus paper red.

When a magnesium ribbon is put in some solid citric acid, there is no gas formed. When a magnesium ribbon is put into an aqueous solution of citric acid, colourless gas bubbles (H2(g)) are evolved.

When sodium carbonate powder is mixed with solid citric acid, there is no sign of gas formed. When some sodium carbonate powder is dissolved in aqueous solution of citric acid, colourless gas bubbles (CO2(g)) are evolved.




Class Practice

A18.1(a) 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)

OH(aq) + H+(aq) H2O(l)(b) CH3COOH(aq) + KOH(aq) CH3COOK(aq) + H2O(l)

CH3COOH(aq) + OH(aq) CH3COO(aq) + H2O(l) or H+(aq) + OH(aq) H2O(l)

(c) Fe2O3(s) + 6HCl(aq) 2FeCl3(aq) + 3H2O(l)Fe2O3(s) + 6H+(aq) 2Fe3+(aq) + 3H2O(l)

A18.2(a) 20 cm3

(b) (i) Alkaline(ii) For equal concentrations of NaOH(aq) and HCl(aq), equal volumes would

neutralize each other. As there is 30 cm3 alkali added, it is in excess. Thus, the resultant solution is alkaline.

(c) Neutralization is an exothermic reaction. No more heat is given out when the neutralization is complete. Addition of excess alkali will cool down the mixture. Thus, the temperature drops.

(d) The temperature change will be more or less the same as the change when 20.0 cm3 of 2 M HCl is used. Although the volumes of the two solutions are doubled, the numbers of moles of H+(aq) and OH(aq) ions reacting are also doubled. Twice as much heat is given out, but this is used to heat up twice the volume of the solution. For this reason, both experiments have the same rise in temperature.

A18.31. (a) Soluble

(b) Insoluble(c) Soluble(d) Soluble(e) Insoluble(f) Insoluble(g) Soluble(h) Soluble

2. (a) NaNO3

(b) KNO3

(c) CuSO4

(d) PbSO4/PbCl2

(e) MgCO3 (f) AgCl(g) BaSO4



A18.41. (a) Magnesium oxide/magnesium hydroxide/magnesium carbonate (Any

TWO)Dilute hydrochloric acid

(b) MgO(s) + 2HCl(aq) MgCl2(aq) + H2O(l) ……………………….….(1)Mg(OH)2 + 2HCl(aq) MgCl2(aq) + 2H2O(l) ……………………….(2)MgCO3(s) + 2HCl(aq) MgCl2(aq) + CO2(g) + H2O(l) ……………..(3)

(c) Comparing (1) and (2):(i) Both solids dissolve to give a colourless solution./Test tubes become a

little warm.(ii) NoneComparing (2) and (3):(i) Both solids dissolve to give a colourless solution./Test tubes become a

little warm.(ii) In (2), there are no gas bubbles evolved. In (3), effervescence (CO2(g)

is given off) occurs.Comparing (1) and (3):(i) Both solids dissolve to give a colourless solution./Test tubes become a

little warm.(ii) In (1), there are no gas bubbles evolved. In (3), effervescence (CO2(g)

is given off) occurs.2. (a) CaCO3 + H2SO4 CaSO4 + CO2 + H2O

(b) Calcium sulphate is only slightly soluble. The undissolved salt forms a protective layer on the unchanged calcium carbonate, preventing further reaction.

A18.5(a) Yes. HNO3(aq) and NaOH(aq) (b) Yes. HCl(aq) and NH3(aq)(c) No. Mg(OH)2 is insoluble in water, so no standard solution can be prepared for

titration.(d) No. Pb(OH)2 is insoluble in water, so no standard solution can be prepared for

titration.

A18.61. (a) (i) AgNO3(aq) + NaBr(aq) AgBr(s) + NaNO3(aq)

(ii) Mg(NO3)2(aq) + Na2CO3(aq) MgCO3(s) + 2NaNO3(aq)(b) (i) Ag+(aq) + Br(aq) AgBr(s)

(ii) Mg2+(aq) + CO32(aq) MgCO3(s)

2. (a) Impracticable. All salts are soluble. They cannot be separated by crystallization.

(b) Practicable.(c) Impracticable. PbSO4 is insoluble. Pb2+ ions in the solid cannot combine

with the Cl ions in the aqueous solution.

A18.71. (a) Copper(II) chloride, hydrochloric acid

(b) Iron(III) sulphate, sulphuric acid(c) Zinc nitrate, nitric acid

2. (a) (CH3COO)2Ca(b) Pb(NO3)2

(c) (NH4)2SO4



A18.81. (a) CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l)

(b) Ca(OH)2(s) + 2H+(aq) Ca2+(aq) + 2H2O(l)2. Sodium hydroxide is corrosive.A18.9(a) (i) Sodium hydrogencarbonate, magnesium hydroxide and aluminium

hydroxide(ii) NaHCO3(s) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

(b) Calcium oxide: it reacts with acid and water producing a lot of heat which causes chemical burns.Sodium hydroxide: it is irritating and corrosive, so it will cause serious burns in the mouth and oesophagus.Ammonia: it is a toxic gas with an irritating smell.

A18.10(a) Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O(b) 2NH3 + H2SO4 (NH4)2SO4




Chapter Exercise1. hydrogen, H+, hydroxide, OH, oxide, O2, water2. given out, liberated, exothermic3. soluble, insoluble4. (a) metal, base, carbonate

(a) (i) metal, hydrogen(a) (ii) base, water(a) (iii) carbonate, carbon dioxide, water, filtration, crystallization(b) alkali, carbonate, alkali, water, crystallization

5. precipitation, precipitate6. ionizable, metallic, ammonium7. (a) pH

(b) acid(c) waste(d) fertilizers

8. C9. B10. C11. A12. C13. D14. C15. A16. C17. C18. D19. D20. D21. A22. B23. B24. B25. D26. A27. C28. A29. B30. D

31. (a) Neutralization(b) H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)(c) H+(aq) + OH (aq) H2O(l)(d) The temperature of the reaction mixture would increase because

neutralization is an exothermic reaction.



32. (a)

(b) Temperature rises because the reaction between an acid and an alkali is exothermic.

(c) When all the acid has just been reacted completely, no more neutralization occurs. This means that no more heat is given out. Thus, the rise in temperature stops and a maximum point is established.

(d) As neutralization has completed, no more heat is given out. Addition of excess alkali will cool down the mixture. Thus, the temperature drops.

(e) H+(aq) + OH(aq) H2O(l)(f) (i) 40.00 cm3

(ii) Same/more or less the same. Although the amount of heat released is doubled, the volume of the solution to be heated is also doubled.

33. (a) (i) Sodium ethanoateCH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

(ii) Copper(II) nitrate2HNO3(aq) + Cu(OH)2(s) Cu(NO3)2(aq) + 2H2O(l)

(iii) Zinc sulphateH2SO4(aq) + ZnO(s) ZnSO4(aq) + H2O(l)

(iv) Calcium chloride 2HCl(aq) + CaO(s) CaCl2(aq) + H2O(l)

(b) Water/H2O

34. (a) Silver iodide/silver carbonate/lead(II) iodide/lead(II) carbonate/lead(II) sulphate/barium carbonate/barium sulphate (Any THREE)

(b) Ag+(aq) + I(aq) AgI(s) 2Ag+(aq) + CO3

2(aq) Ag2CO3(s) Pb2+(aq) + 2I(aq) PbI2(s) Pb2+(aq) + CO3

2(aq) PbCO3(s) Pb2+(aq) + SO4

2(aq) PbSO4(s) Ba2+(aq) + CO3

2(aq) BaCO3(s) Ba2+(aq) + SO4

2(aq) BaSO4(s)(Any THREE)


Volume of NaOH added/cm3

Te mp

era

tur e /o CTe mp

era

tur e /o C


35. (a) Magnesium, magnesium oxide, magnesium hydroxide and magnesium carbonate

(b) Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g) MgO(s) + H2SO4(aq) MgSO4(aq) + H2O(l) Mg(OH)2(s) + H2SO4(aq) MgSO4(aq) + 2H2O(l) MgCO3(s) + H2SO4(aq) MgSO4(aq) + H2O(l) + CO2(g)

36. (a) Zinc nitrate(b) ZnCO3(s) + 2HNO3(aq) Zn(NO3)2(aq) + CO2(g) + H2O(l)

(c) Effervescence stopped./Insoluble solid Smithsonite can be seen.(d) To make sure that all the nitric acid has been used up.(e)

(f) No. of moles of HNO3 used = 1.0 mol = 0.100

mol

No. of moles of Zn(NO3)2 produced = mol = 0.0500 mol

Formula mass of Zn(NO3)2 = 65.4 + (14.0 + 16.0 3) 2 g mol1 = 189.4 g mol1

Theoretical mass of Zn(NO3)2 = 0.0500 mol 189.4 g mol1 = 9.47 g

37. (a) Titration (b) Acid X: Sulphuric acid

Alkali Y: Sodium hydroxide(c) At this point, just enough acid X has been added to neutralize the alkali Y. (d) Add a little activated charcoal to the coloured solution. Warm and then

filter the mixture. The filtrate would be a colourless solution of sodium sulphate.OR

Repeat the experiment with exactly the same volumes of acid and alkali needed for neutralization. However, do not add methyl orange indicator at the start this time.

(e) Crystallization


filter paper

inside filter

funnel

excess powdered

Smithsonite

filtrate (Zn(NO3)2(aq))


38.

Cation

AnionK+ Na+ Ca2+ Mg2+ Zn2+ Fe2+ Pb2+ Cu2+ NH4

+ Ag+ Ba2+

NO3

HCO3

CO32

CaCO3 MgCO3 ZnCO3 FeCO3 PbCO3 CuCO3 Ag2CO3 BaCO3

Cl PbCl2 AgCl

BrPbBr2 AgBr

IPbI2 AgI

SO42

*CaSO4 PbSO4 BaSO4

*CaSO4 is only sparingly soluble.

39.Salt Parent acid Parent base/alkali

Ca(NO3)2 HNO3 CaO/Ca(OH)2

NH4Cl HCl NH3

Fe2(SO4)3 H2SO4 Fe2O3/Fe(OH)3

CH3COOK CH3COOH KOH

40. (a) Calcium hydroxide(b) CaCO3(s) CaO(s) + CO2(g)

limestone

CaO(s) + H2O(l) Ca(OH)2(s)quicklime slaked lime

(c) To neutralize acids in soil.(d) To neutralize acidic liquid waste before disposal.(e) Ca(OH)2(s) + 2H+(aq) Ca2+(aq) + 2H2O(l)

41. (a) Most plants cannot grow well in soils which are too acidic. To decrease the acidity of the soils, farmers often add slaked lime to neutralize acids in soil.Ca(OH)2(s) + 2H+(aq) Ca2+(aq) + 2H2O(l)

(b) ‘Milk of Magnesia’ neutralizes the excess acid in the stomach which causes stomach pain. Mg(OH)2(s) + 2H+(aq) Mg2+(aq) + 2H2O(l)



42. Add copper(II) oxide a little at a time to dilute sulphuric acid with stirring and warming until some excess copper(II) oxide remains undissolved. This is to ensure that all the sulphuric acid has reacted completely.

Filter off the excess copper(II) oxide using filter paper and filter funnel. The filtrate is the copper(II) sulphate solution.

Boil the solution to concentrate the filtrate. Then cool the hot concentrated solution slowly to room temperature to separate out the hydrated copper(II) sulphate crystals.

Finally, filter, wash and dry the hydrated copper(II) sulphate with some filter paper.

43. Use a pipette to transfer 25.0 cm3 of dilute nitric acid to a clean conical flask. Add two drops of phenolphthalein indicator to give a colourless solution.

Run dilute sodium hydroxide solution from a burette into the flask until the solution just turns to pink colour. The resultant solution is a sodium nitrate solution together with the phenolphthalein indicator.

Sodium salt can be obtained as follows:Add a little activated charcoal to the above pink solution to absorb the phenolphthalein indicator. Then remove the charcoal by filtration. The filtrate should be a colourless solution of sodium nitrate. (OR Repeat the experiment with exactly the same volumes of nitric acid and sodium hydroxide needed for neutralization (as found by the above experiment). But, this time do not add phenolphthalein indicator at the start. A solution of sodium nitrate is formed.)

Boil to concentrate the sodium nitrate solution. Then cool the hot concentrated solution slowly to room temperature to separate out the sodium nitrate crystals.

Finally, filter, wash and dry the sodium nitrate crystals with some filter paper.

44. Farmers often add powdered limestone (a natural form of calcium carbonate) or slaked lime (calcium hydroxide) to neutralize the acidic soil. This is called the liming of soil which can adjust the soil pH suitable for plant growth.

A suspension of magnesium hydroxide in water is called ‘Milk of Magnesia’ which can be used as antacid to neutralize the excess acid in the stomach and help to relieve stomach pain.

Liquid wastes from the electroplating and dyeing industries are usually acidic which will cause serious water pollution and kill water life. Slaked lime or sodium carbonate are usually used to neutralize the liquid wastes.

Many common fertilizers are made by neutralization. For example, ammonium nitrate is produced by neutralizing nitric acid with ammonia. After crystallization, ammonium nitrate crystals are packed and sold to farmers.




Class Practice

A19.11. (a) Beaker

(b) Volumetric flask(c) Pipette filler(d) Electronic balance(e) Pipette

2. (a) Pipette (25.0 cm3 type)(b) Measuring cylinder(c) Electronic balance(d) Volumetric flask (250.0 cm3 type)

A19.2

(a) No. of moles of (COOH)22H2O(s) required = 0.1 mol = 0.0500 mol

Mass of (COOH)22H2O(s) required= 0.05 (12.0 + 16.0 2 + 1.0) 2 + 2 (1.0 2 + 16.0) g = 6.30 g

(b) Molarity = M = 0.102 M

(c) Yes. This is because the accurate molarity of the solution is known.(d) No. of moles of (COOH)22H2O(s) remains unchanged on dilution. (MV)before dilution = (MV)after dilution

0.102 = 0.0150

V = 36.8 Thus, the volume of 0.102 M ethanedioic acid solution required is 36.8 cm3.

A19.3(a) After washing the burette, some distilled water (a few drops) may be left inside

the burette. When the given standard hydrochloric acid fills the burette, it is diluted and the molarity of the acid is no longer known. The given hydrochloric acid is not a standard solution any more.

(b) The burette should be washed one or two times with the given standard hydrochloric acid, i.e. the solution it is to deliver.

(c) Pipette

A19.4(a) H+(aq) + OH(aq) H2O(l)(b) Pipette(c) Yellow to orange(d) Methyl orange



A19.5(a) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

No. of moles of NaOH(aq) present = 0.100 mol = 0.0250 mol

From the equation, mole ratio of NaOH : HCl = 1 : 1. no. of moles of HCl(aq) needed = 0.0250 mol

Volume of 0.100 M HCl(aq) needed = dm3 = 0.250 dm3 = 25.0 cm3

(b) Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l)

No. of moles of Na2CO3(aq) present = 0.050 mol = 0.00100 mol

From the equation, mole ratio of Na2CO3 : HCl = 1 : 2. no. of moles of HCl(aq) needed = 0.00100 2 mol = 0.00200 mol

Volume of 0.1 M HCl(aq) needed = dm3 = 0.0200 dm3 = 20.0 cm3

A19.62HCl(aq) + Na2CO3(aq) 2NaCl(aq) + CO2(g) + H2O(l)

2.65 g250.0 cm3

26.30 cm3 (25.0 cm3 used)? MMolar mass of Na2CO3 = 23.0 2 + 12.0 + 16.0 3 g mol1 = 106 g mol1

No. of moles of Na2CO3 in 250.0 cm3 of solution = mol = 0.0250 mol

No. of moles of Na2CO3 in 25.0 cm3 of solution = 0.0250 mol = 2.50 103

molFrom the equation, mole ratio of HCl : Na2CO3 = 2 : 1. no. of moles of HCl in 26.30 cm3 = 2.50 103 2 mol = 5.00 103 mol

Molarity of HCl solution = M = 0.190 M

A19.7(a) Molar mass of sodium hydroxide = 23.0 + 16.0 + 1.0 g mol1 = 40.0 g mol1

No. of moles of sodium hydroxide present = mol = 0.500 mol

Molarity of solution = M = 2.00 M

(b) No. of moles of HnA present = mol = 0.200 mol


(c) (i) No. of moles of NaOH = 2.00 mol = 0.100 mol

(ii) No. of moles of HnA = 2.00 mol = 0.0500 mol

(iii) 2 mol



(iv) 2(v) H2A(aq) + 2NaOH(aq) Na2A(aq) + 2H2O(l)

A19.8(a) 2NaOH(aq) + H2X(aq) Na2X(aq) + 2H2O(l)

25.0 cm3 30.00 cm3

0.120 M ? M

No. of moles of NaOH present = 0.120 mol = 3.00 103 mol

From the equation, mole ratio of NaOH : H2X = 2 : 1.

no. of moles of H2X needed = mol = 1.50 103 mol

Molarity of H2X(aq) = M = 0.05 M

(b) Since 1 dm3 H2X contains 6.30 g or 0.05 mol of the solute,

molar mass of H2X = g mol-1 = 126 g mol1

A19.9H2AnH2O(s) + 2KOH(aq) K2A(aq) + (n+2)H2O(l) 1.26 g0.126 g (25.0 cm3 used) 0.100 M n = ? 20.00 cm3

Molar mass of H2AnH2O = (90.0 + 18.0n) g mol1

No. of moles of KOH needed = 0.100 mol = 2.00 103 mol

From the equation, mole ratio of H2AnH2O : KOH = 1 : 2.

no. of moles of H2AnH2O present = mol = 1.00 103 mol

No. of moles of H2AnH2O =

1.00 103 =

n = 2There are 2 molecules of water of crystallization per molecule of the hydrated acid.



A19.10K2CO3(s) + 2HCl(aq) 2KCl(aq) + CO2(g) + H2O(l) 5.00 g 2.00 M(with impurity) 34.20 cm3

? % by massNo. of moles of hydrochloric acid used to react with K2CO3

= 2.00 mol = 0.0684 mol

From the equation, mole ratio of K2CO3 : HCl = 1 : 2.

no. of moles of K2CO3 in the sample = mol = 0.0342 mol

Molar mass of K2CO3 = 39.1 2 + 12.0 + 16.0 3 g mol1 = 138.2 g mol1

Mass of K2CO3 in the sample = 0.0342 mol 138.2 g mol1 = 4.73 g

Percentage purity of the K2CO3 sample = 100% = 94.6%

A19.11(a) About 6.5 g of sodium carbonate solid was weighed accurately using an electronic balance.(b) The solution was transferred to a 250.0 cm3 volumetric flask using a filter funnel.(c) 2 or 3 drops of methyl orange indicator solution were added to the conical flask.




Chapter Exercise

1. volumes2. molarity, solid, concentrated3. volumetric4. pipette, pipette5. burette6. water, hold/deliver, conical flask, water7. burette, conical flask, end point, endpoint, indicator, conical flask8. end point9. strength

(a) methyl orange, phenolphthalein(b) methyl orange(c) phenolphthalein

10. molarity, standard11. past, third, passive, grammatical12. B13. D14. B15. D16. D17. D18. D19. A20. B21. A22. C23. A24. B25. C26. B27. B28. B29. C30. B31. B32. A33. B

34. (a) A standard solution is a solution of accurately known molar concentration/molarity.

(b) Volumetric flask(c) High purity/chemically stable/easily dissolved in water/involatile/not

hygroscopic (Any THREE)(d) (1) Dissolve all the solid in a beaker with about 100 cm3 of distilled water.

(2) Pour the solution completely (include all washings) into a 250.0 cm3

volumetric flask.(3) Fill the volumetric flask with distilled water until the bottom of the

meniscus reaches the graduation mark.



(4) Stopper the flask and invert the flask several times to mix the contents thoroughly.

(e) Molar mass of anhydrous sodium carbonate= 23.0 2 + 12.0 + 16.0 3 g mol1 = 106 g mol1

No. of moles of solute used = mol = 0.100 mol


35. (a) (i) Burette(ii) Wash the burette with distilled water and then with the sulphuric acid.

Fill the burette with sulphuric acid and make sure no bubbles are trapped in the jet.

(b) (i) Although a little water would remain in the conical flask, this will not change the amount of solute present.

(ii) Its conical shape allows it to be swirled gently without spilling out the content.

(c) Methyl orange(d) To indicate that the reaction of the two solutions (sodium carbonate and

sulphuric acid) has completed.(e) A white background allows the detection of colour change to be easier and

more accurate.(f) No. of moles of sodium carbonate solution used

= 0.0527 mol = 0.00132 mol

Na2CO3 + H2SO4 Na2SO4 + CO2 + H2OFrom the equation, mole ratio of Na2CO3 : H2SO4 = 1 : 1. no. of moles of sulphuric acid in 25.80 cm3 of solution = 0.00132 mol

Molarity of sulphuric acid = M = 0.0512 M

36. (a) CO32(aq) + 2H+(aq) CO2(g) + H2O(l)

(b) Methyl orange; yellow to orange

(c) No. of moles of Na2CO3 used = mol = 0.0250 mol

Molarity of Na2CO3 prepared = M = 0.100 M

(d) (i) No. of moles of Na2CO3 = 0.100 mol = 0.00250 mol

(OR Since 25.0 cm3 is withdrawn from 250.0 cm3 of solution i.e. one-tenth of the original solution is withdrawn.

no. of moles of Na2CO3 used = mol = 0.00250 mol)



(ii) From the equation in (a), mole ratio of CO32 : H+ = 1 : 2.

no. of moles of HCl in 20.0 cm3 of solution= 0.00250 2 mol = 0.00500 mol

Molarity of the acid = M = 0.250 M

37. (a) Volumetric flask(b) (i) Pipette

(ii) Wash with distilled water and then with the solution it is to deliver i.e. the acid solution.

(iii) Pipette filler

(c) A reasonable average volume = cm3 = 22.00 cm3

The volumes of 1.50 M KOH delivered in titrations 1,2,3 and 4 are 22.90 cm3, 22.10 cm3, 22.00 cm3 and 21.90 cm3 respectively. The value 22.90 cm3

obviously deviates a lot from all the others and may be discarded.(d) No. of moles of KOH required to neutralize 25.0 cm3 of the excess acid

= 1.50 mol = 0.0330 mol

No. of moles of acid in 250.0 cm3 of solution = mol = 0.110 mol

No. of moles of acid in 25.0 cm3 of solution

= 0.110 mol = 0.0110 mol

Let HnA be the formula of the acid, where n is the number of ionizable hydrogen atoms.Then equation for the neutralization is:HnA(aq) + nKOH(aq) KnA(aq) + nH2O(l)

=

=

n = 3Thus, the basicity of the acid is 3, i.e. the acid is tribasic.



38. H3X(aq) + 3NaOH(aq) Na3X(aq) + 3H2O(l)

No. of moles of NaOH used = 0.5 mol = 0.00900 mol

From the equation, mole ratio of H3X : NaOH = 1 : 3. no. of moles of H3X in 25.0 cm3 of diluted solution

= mol = 0.00300 mol

No. of moles of H3X in 250.0 cm3 of diluted solution

= 0.00300 mol = 0.0300 mol

Molar mass of H3X = g mol1 = 130 g mol1

39. X(s) + H2SO4(aq) XSO4(aq) + H2O(l)

No. of moles of H2SO4 used = 0.5 mol = 0.0225 mol

From the equation, mole ratio of X : H2SO4 = 1 : 1. no. of moles of X present = 0.0225 mol

Molar mass of X = g mol1 = 65.3 g mol1

The relative atomic mass of X is 65.3.

40. No. of moles of NaOH used = 0.11 mol = 4.40 103 mol

The equation of the reaction is:FeSO4nH2O(s) + 2NaOH(aq) Fe(OH)2(s) + Na2SO4(aq) + nH2O(l)From the equation, mole ratio of FeSO4nH2O : 2NaOH = 1 : 2.

no. of moles of FeSO4nH2O present = mol = 2.20 103 mol

No. of moles =

2.20 103 =

n = 7

41. (a) CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(l)(b) Colourless to pale pink

(c) (i) No. of moles of NaOH used = 0.40 mol = 0.00800 mol

HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)From the equation, mole ratio of HNO3 : NaOH = 1 : 1. no. of moles of excess HNO3 = 0.00800 mol



(ii) Total no. of moles of HNO3 used in the experiment

= 0.50 mol = 0.0250 mol

No. of moles of HNO3 that reacted with CaCO3

= (0.0250 – 0.00800) mol = 0.0170 molFrom the equation in (a), mole ratio of CaCO3 : HNO3 = 1 : 2. no. of moles of CaCO3 present in the sample of limestone

= mol = 0.00850 mol

(iii) Molar mass of CaCO3 = 40.1 + 12.0 + 16.0 3 g mol1 = 100.1 g mol1

Mass of CaCO3 in the sample = 0.00850 100.1 g = 0.851 g

% by mass of CaCO3 in the sample = 100% = 66.5%

42. (a) In step (1): A 25.0 cm 3 type pipette should be used to transfer 25.0 cm3 of the drain cleaner sample to the volumetric flask.In step (2): Add distilled water to volumetric flask until the bottom of the meniscus reaches the graduation mark.In step (4): The burette should be washed with the standard hydrochloric acid right before the titration.In step (5): A 25.0 cm 3 type pipette should be used to transfer 25.0 cm3 of the diluted drain cleaner sample to a conical flask but not a beaker. Also, it is not necessary to use 1–2 cm3 methyl orange indicator, 1–3 drops of methyl orange indicator will be enough to allow the end point to be detected.In step (6): The colour change of the indicator at the end point should be from yellow to orange ; otherwise, the end point may have passed.

(b) (1) 25.0 cm3 of the drain cleaner sample were transferred to a 250.0 cm3

volumetric flask by using a pipette.(2) The flask was filled with distilled water until the bottom of the

meniscus reached the graduation mark.(3) The flask was stoppered and inverted several times.

43. To prepare a standard solution, the solid should: be available in a highly pure state, so no impurities should be present in the

solution. be stable in air, that is, no oxidation occurs. be easily soluble in water which ensures the molarity calculated will be

accurate. have a high molar mass. This will reduce the percentage error during

weighing. undergo complete and rapid reaction when the solution is used in

volumetric analysis.



44. The pipette must be cleaned using deionized water and then the solution it is going to hold or deliver. The main idea is that we should never affect the molarity of any given solutions: the standard solution and the solution of unknown molarity.

Use the pipette properly to deliver a correct volume of solution to the conical flask. For example, the bottom of meniscus of the solution should be on the graduation mark.

The solution should be kept at room temperature. Too hot or too cold a solution will lead to errors of volume delivered.

The last drop of solution staying in the pipette should not be blown out. Otherwise, the volume of solution delivered is more than that stated on the pipette.

45. The burette should be cleaned using deionized water and then the solution it is going to hold or deliver. The main idea is that we should never affect the molarity of any given solutions: the standard solution and the solution of unknown molarity.

The burette cannot hold too hot or too cold solutions because the volume of the burette will change under these conditions.

When the burette is filled, make sure that the jet is filled completely, i.e. no air bubbles are trapped inside the jet.

During the titration, the burette should be clamped upright. This reduces the error when taking readings on the burette.

When we take readings on the burette, we should keep our eyes at the same level as the meniscus.



Part IV Acids and bases

Part Exercise

1. A2. C3. B4. A5. C6. C7. D8. B9. C10. A11. C

12. (a) Zinc carbonate reacted with dilute hydrochloric acid to give carbon dioxide.

ZnCO3(s) + 2HCl(aq) ZnCl2(aq) + H2O(l) + CO2(g)

(b) No. of moles of carbon dioxide formed = mol = 0.0727 mol

From the equation, mole ratio of ZnCO3 : CO2 = 1 : 1. no. of moles of ZnCO3 present = 0.0727 molMass of ZnCO3 present = 0.0727 mol (65.4 + 12.0 + 16.0 3) g mol1 = 9.12 g

Percentage by mass of ZnCO3 in the sample = 100% = 91.2%

13. (a)

(b) A weak acid is an acid which incompletely/slightly/partially ionizes in water.

(c) Molar mass of ascorbic acid = 12.0 6 + 1.0 8 + 16.0 6 g mol1 = 176 g mol1

No. of moles of ascorbic acid in 100 cm3 of solution = mol = 0.188 mol

Molarity of the acid solution = M = 1.88 M

(d) The acid decomposed before boiling.(e) pH = log [H+(aq)]

3 = log [H+(aq)]



[H+(aq)] = 0.001 M

14. (a) A worker misused a bottle of concentrated ammonia solution to clean the stairs in a building in the Western District. A large amount of ammonia gas was evolved. As a result, eight people were sent to hospital after breathing in the gas.

(b) We should take necessary safety measures when we handle chemicals.(c) Concentrated ammonia solution should be kept in a glass container, with

hazard warning labels.(d) Dilute ammonia solution can be used as a cleaner. Ammonia is commonly

used in the dyeing industry and as a refrigerant.(e) Ammonia gas attacks human mucous membranes, stinging the eyes, nose

and throat, and can cause nausea and vomiting. Breathing in a high concentration of ammonia for several hours could kill.

15. (a) Methyl orange was added as an indicator to mark the end point of the strong acid-weak alkali titration. The colour would change from yellow to orange.

(b) 3HCl(aq) + Al(OH)3(s) AlCl3(aq) + 3H2O(l)

(c) No. of moles of HCl used = 0.806 mol = 0.0216 mol

From the equation, mole ratio of HCl : Al(OH)3 = 3 : 1.

no. of moles of Al(OH)3 in one tablet of Y = mol = 0.00720 mol

Molar mass of Al(OH)3 = 27.0 + (16.0 + 1.0) 3 g mol1 = 78.0 g mol1

Mass of Al(OH)3 in one tablet of Y = 0.00720 78.0 g = 0.562 g

(d) For X, price per gram of Al(OH)3 = = $1.82

For Y, price per gram of Al(OH)3 = = $1.78

Hence, Y is the better buy.

16. (a) HSO4(aq) H+(aq) + SO4

2(aq)(b) Harpic reacts with calcium carbonate to form soluble products:

CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l) lime scale (from Harpic)

(c) When the two types of cleaners mix together, heat is released. It is because the reaction is a neutralization reaction which is exothermic.NaHSO4(aq) + NaOH(aq) Na2SO4(aq) + H2O(l)

(d) Wash the affected area with plenty of water/under running tap water.

17. (a) Calcium carbonate(b) The effervescence was due to the formation of carbon dioxide; calcium

chloride solution was colourless.CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

(c) A layer of insoluble CaSO4 was formed which prevented the CaCO3 from further reaction.

(d) On heating, the CaCO3 in eggshell gave CO2 which turned limewater milky.

CaCO3(s) CaO(s) + CO2(g)



(e)

(f) Always wear safety spectacles in doing experiments, especially during heating.Remove the glass tubing from limewater before stopping heating. This is to prevent sucking back of cold limewater which would crack the hot glass.

18. (a) The black copper(II) oxide dissolved.The colourless solution turned blue.

(b) CuO(s) + H2SO4(aq) CuSO4(aq) + H2O(l)(c) No. of moles of sodium hydroxide used to neutralize the excess acid

= 0.54 mol = 0.0154 mol

2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l)From the equation, mole ratio of NaOH : H2SO4 = 2 : 1.

no. of moles of excess sulphuric acid = mol = 0.00770 mol

Total no. of moles of sulphuric acid used = 0.55 mol = 0.0275 mol

No. of moles of sulphuric acid reacted with copper(II) oxide = (0.0275 – 0.00770) mol = 0.0198 mol Volume of sulphuric acid reacted with copper(II) oxide

= dm3 = 0.0360 dm3 (or 36.0 cm3)

(d) From the equation in (b), mole ratio of CuO : H2SO4 = 1 : 1. no. of moles of copper(II) oxide present = 0.0198 mol Mass of copper(II) oxide present = 0.0198 mol (63.5 + 16.0) g mol1 = 1.57 g

% by mass of copper(II) oxide in the original sample = 100% =

66.8%

19. (a) Ammonia(b) Characteristic pungent smell(c) (i) A pale blue (gelatinous) precipitate was formed.

(ii) The precipitate dissolved and a deep blue solution was formed.(d) Cu2+(aq) + 2OH (aq) Cu(OH)2(s)


test tube

heat

limewatereggshell

delivery tube



Class Practice

A20.1 This is due to the movements of the Earth’s crust.

A20.2There is physical separation but no chemical decomposition.




Chapter Exercise

1. fossil, plants, animals2. plants, power stations, electricity3. carbon4. sea, uels, hemicals, fuel5. hydrocarbons6. hydrocarbons, methane7. fractions, ractional distillation, efining8. heavy, higher, otter9. efinery Fuel, raw materials for chemicals

etrol Motor car fuelaphtha Raw material for town gas and chemicalserosene Jet plane fuel, household fuelas Diesel fuel for buses, trucks and factoriesuel Fuels for big ships, burnt to generate electricity in power stationsubricating Lubricating oils, candlesitumen To cover roads and roofs

10. C11. B12. B13. A14. D15. A16. C

17. (a) Fractional distillation(b) Fraction 4 contains hydrocarbons with larger molecules while fraction 1

contains hydrocarbons with smaller molecules. The van der Waals’ forces are greater between larger molecules. Therefore, fraction 4 has a higher boiling point range.

(c) Fraction 4(d) Fraction 1(e) Fraction 1. This is because it is easier to burn and it burns with a cleaner

and less sooty flame.

18. (a) The liquid alkanes in crude oil act as a solvent to dissolve the solid alkanes.(b) The difference in boiling points among the hydrocarbons(c) Fraction 1(d) Fraction 1: Refinery gas

Fraction 2: Petrol and naphthaFraction 3: KeroseneFraction 4: Gas oil

(e) As fuel oil/lubricating oils and waxes/bitumen(f) It is a liquid. It has lower density than water and is immiscible with water.

Thus, it floats on the water surface and prevents the mosquito larvae from getting enough oxygen supply.




Class Practice

A21.1(a) No. Electricity is an important source of energy, but not a fuel. Electricity is a flow

of electrons, not a substance that can be burnt to give out heat.(b) Yes. It can be burnt to give out heat.

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l)

A21.2Carbon monoxide, formed by the incomplete combustion of fuel, will build up to dangerous levels in an enclosed space. It may cause death.

A21.31. No. This is because many pollutants (e.g. sulphur dioxide, carbon monoxide,

hydrocarbons) are colourless gases.2. Carbon monoxide, nitrogen oxides (nitrogen monoxide, nitrogen dioxide),

hydrocarbons, and suspended particulates (e.g. lead, lead(II) compounds, smoke, soot)

3. Sulphur dioxide/nitrogen oxides/carbon monoxide (Any TWO)

A21.41. Both statements are correct, but the second one does not explain the first one.

Actually, the low pH of acid rain is caused by the dissolved sulphur dioxide and nitrogen dioxide.

2. Probably at that time, many factories in the Kwun Tong industrial district used fuels of high sulphur content. A large amount of sulphur dioxide was produced which caused acid rain.

A21.51. Burning fossil fuels produces carbon dioxide. More fossil fuels are burnt in

winter to keep warm. Less fossil fuels are burnt in summer.2. (a) Sulphur dioxide and nitrogen oxides

(b) SO2(g) + H2O(l) ⇌ H2SO3(aq)4NO(g) + 2H2O(l) + O2(g) ⇌ 4HNO2(aq)2NO2(g) + H2O(l) HNO3(aq) + HNO2(aq)

(c) CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l)(d) Sulphur dioxide: burn fuels of lower sulphur content.

Nitrogen oxides: burn less fuels.

A21.6(a) No(b) Nuclear power is very clean, since it produces no air pollutants such as sulphur

dioxide, carbon monoxide and nitrogen oxides.

A21.7(c) It is a cheap, clean and renewable source of energy.(d) Many countries have neither sufficient rainfall nor high grounds to make

hydroelectricity possible.




Chapter Exercise

1. gives out2. takes in3. ossil

4. x + , x ,

5. Carbon monoxide, oxic, olourless, dourless6. leakage, as7. Safety8. Air pollution9. arbon monoxide, ulphur dioxide, itrogen oxides, ydrocarbons, particulates10. oxic, nitrogen oxides, respiratory11. 5.612. sulphur dioxide, nitrogen oxides13. lants, water, building14. unleaded, catalytic converter15. low sulphur, crubbers16. ilters, lectrostatic precipitators17. nfrared radiation, arbon dioxide, warm18. rise, lobal warming19. pollutants, spillage20. save, alternative21. olar, ydroelectric, idal, ind, nuclear, iomass, non-renewable22. C23. D24. D25. B26. D27. D28. B29. D30. D

31. (a) Sulphur dioxide and nitrogen oxides (nitrogen dioxide)(b) SO2(g) + H2O(l) ⇌ H2SO3(aq)

2NO2(g) + H2O(l) HNO3(aq) + HNO2(aq)(c) CaCO3(s) + 2H+(aq) Ca2+(aq) + CO2(g) + H2O(l)(d) Use scrubbers to remove sulphur dioxide in the waste gases before they

come out of the chimneys.Fit catalytic converters to motor vehicles which convert nitrogen monoxide to nitrogen. Therefore, nitrogen dioxide will not form.

32. (a) Nitrogen dioxide and hydrocarbons(b) Nitrogen dioxide initiates a chain of chemical reactions to form ozone. The

ozone produced then oxidizes hydrocarbons in the air to produce photochemical smog.

(c) Photochemical smog irritates our eyes and attacks our respiratory system.



33. (a) This is because carbon dioxide can trap infrared radiation (heat) in the atmosphere and keep the Earth warm.

(b) There will be a rise in the Earth’s surface temperature.(c) Melting of ice at the North Pole and South Pole/climate change/wildlife in

danger (Any TWO)(d) Reduce the use of fossil fuels/stop deforestation/plant more trees/prevent

and put out forest fires (Any TWO)

34. (a) Wind power is a renewable energy source.(b) Strong wind drives the turbines of the windmill which can then drive the

generator to produce electricity.(c) It does not cause pollution problems./The supply of wind is unlimited. (Any

ONE)(d) Wind does not always blow and sometimes blows too hard./Hundreds of

windmills have to be built. (Any ONE)(e) Nuclear power/solar power/hydroelectric power/tidal power/geothermal

power/power from biomass (Any FIVE)

35. (a) Naphtha. Alkanes with 5 to 10 carbon atoms are found in naphtha.(b) Hydrogen and methane(c) PbO(s) + CO(g) Pb(s) + CO2(g)

PbO(s) + H2(g) Pb(s) + H2O(l)(d) Carbon monoxide(e) Town gas contains hydrogen. It forms a mixture with air. The mixture is

explosive. A spark can ignite the mixture, causing an explosion.

36. Formation of acid rain: Sulphur dioxide and nitrogen dioxide are two major air pollutants

responsible for acid rain. Sulphur dioxide dissolves in rainwater to form sulphurous acid:

SO2(g) + H2O(l) ⇌ H2SO3(aq) Nitrogen dioxide dissolves in rainwater to form nitric acid and nitrous acid:

2NO2(g) + H2O(l) HNO3(aq) + HNO2(aq)Environmental impacts of acid rain: It makes soils acidic, so it damages plants and destroys forests. It makes some lakes and rivers acidic, so fish and water plants cannot

survive. It speeds up the corrosion of metals and building materials e.g. limestone,

marble, sandstone, cement and concrete.Possible ways to control acid rain: Cut down pollutants from burning of fuels. Burn less fossil fuels.



Chapter 22 Homologous series, structural formulae and naming of carbon compounds

Class Practice

A22.1CO, CO2, Na2CO3, KHCO3, H2O, NH3, KOH, HCl, HNO3, NaCl (or other acceptable answers)

A22.2

(a)

(b)

(c)

A22.3

(a) Put n = 5 in CnH2n+2. The molecular formula is C5H12.(b) Put n = 11 in CnH2n+2. The molecular formula is C11H24.

A22.4(a) A and B; alkanoic acid series.(b)

A22.5(a) Yes. This is because they have the same functional group (OH).(b) A < B < C < D. Van der Waals’ forces are greater between larger molecules.

A22.61. (a) Hexyl group

(b) Octyl group2. (a) CH3(CH2)3CH2

(b) CH3(CH2)5CH2

A22.7(a) 2-methylpropane(b) 2,3-dimethylbutane(c) 4-ethyl-3-methylheptane



A22.8(a) 2-methylbut-2-ene(b) Propene

A22.9(a) Butan-1-ol(b) Pentan-2-ol

A22.10 (a) 3-methylbutanoic acid(b) 3,3-dimethylbutanoic acid

A22.11

(a)

(b)

(c)



Chapter 22 Homologous series, structural formulae and naming of carbon compounds

Chapter Exercise

1. carbon2. organisms3. hydrogen4. slow, carbon dioxide; water5. saturated, unsaturated6. functional group7. general formula, CH2, physical, chemical8. alkane, alkenel alkanol, alkanoic acid9. IUPAC10. Alkyl; hydrogen11. hydroxyl; carboxylic12. C13. D14. B15. C16. A17. B

18. (a) 2, 4(b) Four(c)

19. (a) Alkanes/alkane series(b) Methane, ethane and propane(c) No. They only have similar chemical properties.(d) No, they have different physical properties. This is because their molecular

sizes are different, thus the van der Waals’ forces between molecules are also different.

20. (a)

(b)

(c)



(d)

21. (a) 2-methylpropene (or methylpropene)(b) 2-methylpropanoic acid (or methylpropanoic acid)(c) Butanoic acid(d) 2-methylbutane




Class Practice

A23.12C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l)

A23.2Chloromethane, dichloromethane, trichloromethane, tetrachloromethane and hydrogen chloride.

A23.3More and more petrol is required as fuel for the ever-increasing number of motor vehicles.

A23.4(a) An oil fraction(b) The first few cm3 of gas is mainly air expelled from inside the apparatus.(c) The delivery tube should be removed from water before stopping heating. This is

to prevent sucking back of water which may crack the hot reaction tube.

A23.5Ethene can decolorize purple acidified potassium permanganate solution, but ethane cannot. ORIn the dark, ethene can decolorize the red-orange bromine solution, but ethane cannot.




Chapter Exercise

1. CnH2n+2

2. homologous, increase, similar3. alogens, xygen4. substitution5. initiation, propagation, termination, Free radicals6. petrochemical7. cracking, alkene8. petrol, alkenes

9. , nsaturated, addition10. two, more, single11. C12. D13. B14. B15. A16. C17. B

18. (a) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)(b) The gaseous product can be collected and pumped through limewater.

Limewater turns milky if carbon dioxide is present.Ca(OH)2(aq) + CO2(g) CaCO3(s) + H2O(l)

(c) During incomplete combustion, carbon monoxide, unburnt carbon particles and hydrocarbon may be produced.Unburnt carbon particles: irritate the respiratory system. Carbon monoxide: causes breathing problem and decreases the oxygen carrying ability of blood. Unburnt hydrocarbon: causes cancer.

19. (a)

(b) Unsaturated hydrocarbon(c) Thermal cracking and catalytic cracking(d) C10H22(l) C8H18(l) + C2H4(g)

20. (a) This is to avoid the paraffin vaporizing too quickly without being cracked.(b) High temperature, absence of air and presence of catalyst.(c) This is to avoid sucking back of water, which can crack the combustion

tube. (When the burner is removed, the combustion tube is cooled down and then the air inside contracts. As the air pressure inside the tube is lower, water is sucked up. The cool water will lead to a sudden contraction of the tube that may crack the tube.)

(d) The bromine water will be decolorized. The gas may be ethene.



(e) This is because under high temperatures, air inside the combustion tube expands. The gas collected by the first test tube thus usually contains a lot of air, which will not be used for testing.

21. (a) Carbon-carbon double bond(b) (i) Bromine is decolorized.

(ii) Acidified potassium permanganate solution is decolorized.(c) (i)

(ii)

(d) (i) Carbon dioxide and water(ii) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)

22. Crude oil can be distilled into different useful fractions. Supply of petrol, kerosene and gas oil cannot meet their great demand. Heavy fractions (e.g. heavy oils) are not good fuels and are in less demand. Cracking allows the breaking down of large molecules from heavy

fractions into smaller ones. Heavy fractions can be cracked to produce extra petrol and kerosene

which are in greater demand. Alkenes are also obtained from the cracking of oil fractions, which can be

used to make many useful organic chemicals, such as plastics.




Class Practice

A24.1Computer keyboard/mouse/CD case/pen(s)/antenna shield of mobile phone (Any FOUR)

A24.21. Yes. They are made from chemicals derived from petroleum.2. (a) Yes

(b) No

A24.3

(a) (i) and (iii) can undergo addition polymerization.

(i)

(iii)

(b) Propene(c) Polypropene

A24.4(a) (i) Polyethene/polythene

(ii) Polypropene/polypropylene(iii) Polyvinyl chloride

(b) (i) Ethene(ii) Propene(iii) Chloroethene/vinyl chloride

(c) (i) Flexible and good insulator of electricity.(ii) Resistant to many chemical solvents, alkalis and acids.



A24.5

A24.6(a) Weak intermolecular forces(b) The polymer chains vibrate vigorously when heated. The intermolecular forces

are overcome, and the chains can slide over one another easily. The plastic bottle thus softens and deforms.

(c) Thermoplastic(d) (i) The bottle softens and melts.

(ii) The bottle burns.(e) (i) The melting point of polypropene is low. The milk feeding bottles may

soften when they are sterilized by hot water or steam.(ii) PVC is poisonous.




Chapter Exercise

1. Petroleum2. ethane, aphtha, as oil3. polymers4. ight, unreactive, heat, electricity5. Polyethene6. Polymerization, polymers7. addition

8. monomer, polymer, small, 9. repeating unit10. natural11. incinerators12. Reduce, Reuse, Recycle, biodegradable13. D14. A15. D16. C17. C18. C

19. (a) C Repeating unit:

(b) The monomer contains at least one carbon-carbon double bond.

20. (a) (i) Plastic C. This is because plastic C is flexible.(ii) Non-poisonous

(b) Plastic D. This is because plastic D can be used at high temperatures (up to 1350C) and does not burn.

(c) Plastic A. This is because plastic A has excellent transparency.(d) (i) Perspex

(ii) Perspex does not break as easily as glass.(iii) PVC is cheaper and more resistant to weathering than rubber.(iv) Expanded polystyrene is more rigid than waxed paper.(v) Glass(vi) Polyethene is lighter than glass.

21. (a) (i)

(ii)

(iii) Addition polymerization



(b) (i)

(ii)

(iii) Addition polymerization(c) (i)

(ii)

(iii) Addition polymerization

22. (a)

(b)

(c) Tetrafluoroethene(d) It is non-toxic, heat and chemical resistant. Besides, it has a smooth and

slippery surface.

23. Problems associated with the disposal of plastic wastes: Plastics are non-biodegradable. They remain in the environment almost

forever. Putting plastics into landfills would lead to land wastage and cause

pollution of underground water. Burning plastic waste in incinerators would lead to air pollution.Solving plastic disposal problems: Reduce the use of plastics. Reuse plastic articles. Recycle plastic waste: thermoplastic waste can be recycled. Make biodegradable plastics.



Part V Fossil fuels and carbon compounds

Part Exercise

1. B2. B3. C4. C5. A6. C7. B8. B9. D10. B

11. (a)

(b) 1. Wear safety spectacles.2. Beware of burns.

(c) Fraction C(d) Fraction B(e) Fraction B(f) B < D < A < C

12. (a) COOH group(b) CnH2n+1COOH(c) 1. Each of them differs from the next one by a –CH2– group.

2. They have similar chemical properties.3. Their boiling points increase with the number of carbon atoms in the

molecules. (Any TWO)(d) Molecular formula: C6H12O2

Name of compound: hexanoic acid

13. (a) Volume of one mole of a gas at room conditions = 24 dm3 mol1

molar mass of gas Q = 1.75 g dm3 24 dm3 mol1= 42 g mol1

relative molecular mass of gas Q = 42


thermometer (0360C)

oil fraction

water

heat

rocksil soaked with crude oil


(b) Q can decolorize Br2 (in organic solvent) and KMnO4/H+(aq), and is therefore an unsaturated hydrocarbon. Since Q has a molecular mass of 42, it would be propene, CH3CH=CH2.

(c) To provide a hot catalytic surface for reaction that produces Q to occur.(d) Two possible addition products are formed:

14. (a)

(b) No. This is because alkanes do not contain carbon-carbon double bond.(c) For ethane, there is no observable change. For ethene, the bromine water is

decolorized.(d) There is no observable change. It is because polyethene does not contain

any carbon-carbon double bond.(e) (i) Plastic bags/wrapping film for food/food boxes/flexible cold water

pipes/kitchen wares (Any ONE)(ii) Polyethene is light and flexible.

15. (a) Addition polymerization(b) (i) For polymer A:

For polymer B:

(b) (ii) For polymer A:

For polymer B:

(c) All of them are thermoplastics because they all consist of chain-like giant molecules which are attracted by weak intermolecular forces.Heat the sample gently on a hot plate. It softens or melts because it is a thermoplastic.



(d) Shake a little of each of the solutions with Br2 (in organic solvent). The solution is decolorized only by the monomer solution because the following addition reaction takes place.

16. Coal contains sulphur. Sulphur dioxide is formed when coal is burnt in a power station.

S(s) + O2(g) SO2(g) Suspended particulates are carbon particles present in the flue gas

generated in the power station. Nitrogen oxides are formed when fuels are burnt at very high temperatures

in a power station. N2(g) + O2(g) 2NO(g) and 2NO(g) + O2(g) 2NO2(g)

Possible ways to reduce the emission of air pollutants: Sulphur dioxide: installation of scrubbers/use coal of lower sulphur

content Suspended particulates: installation of electrostatic precipitator


Documents

BK2 ans