Solutions Manual for chemist 2

HSc

CHAPTER 1: ETHYLENE Review exercise 1.1

1 The word ‘petrochemical’ means any chemical that has been derived from petroleum, a liquid fossil fuel.

2 Fossil fuels are non-renewable resources with a huge demand. They are wanted both as a source of energy and as a source of petrochemical substances.

Review exercise 1.2

1 Ethylene (ethene) is the most widely used feedstock material derived from petroleum.

2 a Catalytic cracking produces shorter chain-length, lower-mass hydrocarbons from high mass petroleum fractions. Thermal steam cracking converts ethane and propane to ethylene.

b The proportions of products obtained from fractional distillation do not match the demands for the different products. Catalytic cracking allows for specific control of the types and amounts of products and hence more efficient use of the petroleum feedstock and its products.

c C9H20(g) → C7H16(g) + C2H4(g)

3 The catalysts used in cracking are zeolites. These solid crystalline substances adsorb the gaseous reactants, weakening their bonds and hence lowering the activation energies.

Review exercise 1.3

1 Alkenes are more reactive than alkanes because of the presence of the double bond, a centre of high electron density, in alkenes.

2 Place hexane and 1-hexene into separate test tubes. Add bromine water to a depth of about 1 cm, shake and allow to settle. If the bromine water is decolourised then the test tube contained 1-hexene and the product of this addition reaction is 1,2-dibromoethane.

However, there may also be some 2-bromo-1-ethanol and hydrogen bromide as a result of the water present. If there is no change to the colour of the bromine water then the test tube contained hexane.

3 a C3H6 + Cl2 →C3H6Cl2 1,2-dichloropropane

b C7H14 + HBr →C7H15Br 2-bromoheptane, 1-bromoheptane. In

practice only 2-bromoheptane forms (Markovnikov’s rule).

c C6H12 + H2O → C6H13OH 3-hexanol

b Chlorination of 2-butene

c Addition using chlorine water to 2-butene

d Hydration of 1-butene

e Oxidation of 2-pentene by reacting with cold dilute potassium permanganate

Chapter 1 — Application and investigation 1 Investigation

2 a Cracking is used to produce hydrocarbons with lower molecular mass which have greater market demand, e.g. petrol, branch-chained alkanes to improve the

performance of petrol; and to produce ethene which can be used as a starting material for many organic compounds.

b In catalytic cracking the material to be cracked is passed over a zeolite catalyst at a temperature of around 500oC. The reactants adsorb to the surface of the catalyst, which weakens their bonds, lowering the activation energy.

c Catalytic cracking uses zeolite crystals as catalysts to lower the activation energy in the cracking of high molecular mass hydrocarbons. Thermal cracking uses very high temperatures (around 800–900oC) to crack hydrocarbons such as ethane and

propane, to produce needed alkenes.

3 C7H16(g) → C5H12(g) + C2H4(g)

4 Investigation

5 Alkenes contain a reactive double bond, which readily undergoes addition reactions in order to gain a more stable single bond.

6 a 1,2-dichloropropane

b 1-chlorobutane; 2-chlorobutane. In practice only 2-chlorobutane forms (Markovnikov’s rule).

c ethanol

7 Investigation and class experimental work:

a 2-pentene with HCl

b 2-pentene with Cl2

c propene with H2O in the presence of a catalyst

8 Add bromine water to samples of each of cyclohexene and cyclohexane, shake and allow to settle. Cyclohexene undergoes an addition reaction with Br2 across the double bond and

thus the bromine water is decolourised. There is no rapid visible reaction with cyclohexane.

CHAPTER 2: POLYMERS Review exercise 2.1

1 Poymerisation is the chemical reaction whereby monomers link together to form polymers.

2 Addition polymerisation: all atoms in the monomer are present in the polymer chain. Unsaturated monomers join together via breaking of a C = C double bond.

Condensation polymerisation: a reaction between two monomers, which can be different, during which a small molecule, such as water, is eliminated.

3 a Monomers have C = C double bond.

b Monomers each contain a functional group which may be different. Common groups are –COOH (carboxylic acid), –OH (alcohol) and –NH2 (amine) group.

4 a i

ii

ii

5

6

7 a Increased branching decreases hardness by making the polymer less compact and rigid.

b Cross-linking between polymer chains increases hardness because cross-linking is produced by covalent bonds linking chains together.

c Decreasing the length of polymer chains decreases hardness because there are less dispersion forces between the chains.

d Increasing the orderly arrangement of polymer chains increases hardness by making the polymer more dense and less flexible.

e Adding a plasticiser to a polymer decreases hardness because plasticisers are generally chosen to soften a polymer.

Review exercise 2.2

1 a formation of polyethylene

c formation of polyvinylchloride

2 a chloroethylene

b phenylethylene or ethenylbenzene

3 LDPE is softer and more flexible due to greater branching of chains causing reduced dispersion forces.

HDPE has fewer branched chains, causing greater dispersion forces and hence greater hardness and strength.

4 1 monomer unit has molar mass of 62.5 g, so 15 000 monomer units have molar mass of 937 500 g.

5 a polystyrene because it is a cheap, lightweight insulating polymer

b low-density polyethene (LDPE) because it is mouldable, impermeable and flexible

c polypropene/nylon because it is lightweight and strong and easily formed into fibres

d LDPE because it is lightweight, cheap and flexible

e HDPE because it is tough and durable

f HDPE because it is rigid, tough and durable

g Teflon because it is tough, frictionless and resistant to heat and chemicals

h polyethene terephthalate (PET) because it is lightweight, tough and easily blown into a shapeable film

Review exercise 2.3

1 Most of the synthetic polymers made today are derived from the petrochemical industry, which relies on the non-renewable fossil fuel petroleum. However, these supplies will be rapidly used up, especially if we continue to demand their use as both a fuel and a

feedstock. An alternative, renewable source of raw materials is needed.

2 Biomass is organic material derived from living organisms: plant material such as sugar and cellulose; animal material such as dung; and domestic and industrial organic waste.

3 Cellulose is a long-chain polymer of β-glucose units, C6H12O6, where every second

glucose unit is inverted to produce straightened chains.

4 A glucose unit loses an H from an –OH group and joins to a carbon on a second glucose unit which has lost its –OH group; this links the glucose units with the loss of H2O and so

5 Cellulose is made from repeating units of β-glucose with inversion of every second unit. This produces long, straight chains of cellulose which are linked to each other by

hydrogen bonding. In plants, cellulose acts as a structural material. Starch (both amylase and amylopectin) is made from long-chain repeating units of α-glucose, which are also highly branched. This results in tightly coiled, compact, insoluble starch molecules, which are used as energy stores in plants.

6 Existing plant cellulose can be modified to produce biopolymers, such as celluloid and rayon, or it can be broken down into smaller units that can be used to build new polymers such as corn-starch polymers.

Review exercise 2.4

1 Silk is a biopolymer used in its natural form; rayon uses cellulose in a chemically

modified form; lactic acid can be produced from the breakdown of starch and then used to make polylactic acid (PLA).

2 Advantages are that biopolymers come from plant material, which is a renewable

resource, and that plastics made from biopolymers are easily broken down by bacteria and fungi.

Disadvantages include the cost of production and that they are easily biodegradable, which is not useful in some applications.

3 a Plastic-producing bacteria are grown in fermentation vats and fed on molasses or methanol. Extraction of plastic from the bacterium involves breaking down the bacterium’s cell walls and separating this from the cell debris.

b The bacterium used is Alcaligenes eutrophus.

c The plastic produced is of a type known as PHAs—polyhydroxyalkanoates—which have similar properties to polypropene.

Chapter 2 — Application and investigation

1 Monomer: small repeating units that join by covalent bonds to form a polymer, e.g. ethylene.

Polymer: large-chain molecule consisting of small repeating units called monomers, e.g. polyethylene.

Polymerisation: the chemical reaction by which monomers link together to form polymers, e.g. formation of polyethylene by addition reaction.

2 Investigation

3 Polyethylene can have many different molecular chain lengths. Those with around 40 000 atoms per molecule are used for food wrap films; 60 000 atoms per molecule make milk containers; 80 000 atoms per molecule make bleach containers and 800 000 atoms per molecule can be used in artificial ice rinks. As chain length increases, density, hardness and melting point increase. Branching is possible in polyethylene, and as branching increases, density, hardness and melting point decrease, increasing the uses that can be made of polyethylene.

4 Molecular mass of 1 monomer = 28, so 84 500 ÷ 28 = 3018 monomer units

5 First-hand investigation

6 Investigation based on answer to Q5

7 a b c d 8 Investigation 9 Investigation 10 Investigation CHAPTER 3: ETHANOL Review exercise 3.1 1 a 1-propanol

c 1,4-butanediol d 2-methyl-2-propanol 2 a b c 3 2-methyl-propanol 83°C

2-butanol 100°C

2-methyl-1-propanol 108°C

1-butanol 118°C

4 Heat ethanol with an excess of concentrated sulfuric acid, or heat ethanol vapour over a catalyst at 350oC.

5 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

Review exercise 3.2

1 Methylated spirits are 95% ethanol, 5% methanol and small quantities of foul-tasting chemicals. The additives are used to discourage people from drinking it.

2 Ethanol has a short non-polar chain, which allows it to dissolve some non-polar substances. The –OH functional group makes ethanol polar, which allows it to act a solvent for polar substances.

3 The ethanol molecule contains the polar –OH end and the non-polar hydrocarbon chain. Hence it is able to be used as a solvent for non-polar substances such as perfumes and aftershaves. The low boiling point ethanol evaporates easily with body heat, leaving the heavier fragrance components on the skin.

4

5 Ethanol is described as renewable because it can be derived from the starch and sugars present in various crops such as sugar cane and corn.

6 Cars using ethanol produce fewer pollutants. Ethanol is a renewable resource. However, no commercially viable method of obtaining ethanol is available and cars will need to be significantly modified to run on pure ethanol.

7 Mass of methanol burned = 1.10 g

Mass of water = 100 g

Mass of copper = 200.0 g

Temperature change = 5ºC

Specific heat capacity water = 4.18 J K–1 g–1 Specific heat capacity copper = 0.387 J K–1 _g–1

Heat released per 1.10 g methanol = –(mC∆T(copper) + mC∆T(water))

= –(200 × 0.387 × 5 + 100 × 4.18 × 5)

= –2477 J released by 1.10 g methanol Heat released per mole of methanol = –2477 × ₁32_.10

= –72 058 J mol–1 ∴ Molar heat of combustion = 72 kJ mol–1

Review exercise 3.3

1 The acid catalyst is needed to cause the water molecule to attack the double bond in ethylene.

2

b The yeasts produce enzymes which act as catalysts in the conversion of glucose to ethanol and carbon dioxide.

c ethanol and carbon dioxide

d exothermic

e Once the ethanol concentration reaches 15% in the fermentation container, the yeast is killed and the reaction ceases. Therefore naturally fermented wines have ethanol concentrations of 12–15%.

4 Fermentation of glucose occurs best in the absence of oxygen (anaerobic conditions) and at a temperature of 35–40o_C.

5 From the equation, and using n = mass (g) ÷ molar mass (g)

mass (ethanol) = 2 × mass(glucose) ÷ molar mass(glucose) × molar mass(ethanol)

= 2 × 500 ÷180.156 × 46.068

= 255.7 g

6 From the equation, and using n = mass (g) ÷ molar mass (g)

mass (ethanol) = mass (carbon dioxide lost) ÷ molar mass (carbon dioxide) × molar mass (ethanol)

= 50.0 ÷ 44.01 × 46.068

= 52.3 g

Chapter 3 — Application and investigation 1 a

c 2 a 1-propanol b 1,4-butanediol 3 1-pentanol 2-pentanol 3-pentanol

4 Water, being polar, would not readily dissolve many esters with a non-polar component. Hexane, being non-polar, would not readily dissolve the polar esters. Ethanol, with its ability to dissolve both polar and non-polar components, is suitable for dissolving esters.

5 Investigation 6 Investigation 7 Investigation

8 Depends on responses to Questions 5, 6 and 7.

9 a Assume heat absorbed by copper is negligible.

Mass of fuel = 2.00 g

Mass of water = 500 g

Change in temp = 18.4ºC Specific heat water = 4.18 J K–1 _g–1

Methanol

= –(500 × 4.18 × 18.4) = –38 456 J i Heat of combustion/g = 38 456 ÷ 2 = 19 228 J g–1 = 19.2 k J g–1 ii Heat of combustion/mole = 38 456× 32 2 = 61 5296 J mol–1 1-propanol = 615 kJ mol–1 Heat released by 2 g = –mC∆T = –(500 × 4.18 × 28.5) = –59 565 J i Heat of combustion/g = 59 565 ÷ 2 = 29 782.5 J g–1 = 29.8 kJ g–1 ii Heat of combustion/mole = = 1 789 928 J mol–1 = 1790 kJ mol–1 1-butanol Heat released by 2 g = –mC∆T = –(500 × 4.18 × 31.5) = –65 835 J i Heat of combustion/g = 65 835 ÷ 2 = 32 917.5 J = 32.92 kJ g–1 ii Heat of combustion/mole = = 2 439 845.1 J = 2440 kJ mol–1

b

10 The dehydration of ethanol to ethylene and water is carried out by heating ethanol with concentrated sulfuric acid as a catalyst, or by heating ethanol vapour over an alumina catalyst at 350oC. The hydration of ethylene to produce ethanol requires the addition of water in the presence of a sulfuric or phosphoric acid catalyst.

CHAPTER 4: ELECTROCHEMISTRY Review exercise 4.1

1 a A pink-brown deposit of copper forms on the surface of the steel wool as the copper ions convert to elemental copper; the blue solution goes paler and slowly converts to a green solution as the copper ions decrease in concentration and the elemental iron converts to Fe2+ ions with increasing concentration.

b oxidation: Fe(s) → Fe2+_{(aq) + 2e}−

reduction: Cu2+(aq) + 2e− → Cu(s)

c Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

2 a Fe(s) + Sn2+(aq) → Fe2+_{(aq) + Sn(s); the iron would dissolve and the tin would coat}

the surface of the iron; the solution would change from colourless to green

b Pb(s) + Hg2+(aq) → Pb2+_{(aq) + Hg(l); the lead would dissolve and liquid mercury}

would form

c no reaction

d 2Cr(s) + 3Fe2+(aq) → 2Cr3+_{(aq) + 3Fe(s); the chromium would dissolve and the iron}

would coat the chromium; both ions are generally green in aqueous solution

3 The iron container would dissolve as the iron was oxidised to its ions by the nickel ions, which would convert to elemental nickel.

4 a no reaction

b Sn(s) → Sn2+_{(aq) + 2e}−_{; Cu}2+_{(aq) + 2e}− _{→ Cu(s); copper would coat the tin}

c Sn(s) → Sn2+_{(aq) + 2e}−_{; Ag}+_{(aq) + e}− _{→ Ag(s); silver would coat the tin}

Review exercise 4.2 1 a CO2 : C = +4 O = –2 O2 : elemental state ∴ 0 NH3 : N = –3 H = +1 H2S : S = –2 H = +1 HCl : H = +1 Cl = –1 b SO2 : S = +4 O = –2 SO32– : S = +4 O = –2 H2S2O7 : H = +1 S = +6 O = –2 S2O82– : S = +7 O = –2

HSO3– : H = +1 S = +4 O = –2 S : 0 c NH4+ : N = –3 H = +1 ClO4– : Cl = +7 O = –2 Cu2S : Cu = +1 S = –2 MgH2 : Mg = +2 H = –1 PO43– : P = +5 O = –2

2 a redox reaction: Zn oxidised (0 → +2), H+ _{reduced (+1 → 0)}

b not a redox reaction

c not a redox reaction

d redox reaction: Fe2+ oxidised (+2 → +3), Cr6+ _{reduced (+6 → +3)}

3 NO : +2 NO2 : +4 N2O : +1 N2O3 : +3 N2O4 : +4

N2O5 : +5 HNO3 : +5 HNO2 : +3 NO2– : +3 NO3– : +5

NH2– : –3 NH4+ : –3 N2 : 0 N2H4 : –2 NH3 : –3

NH2OH : –1 NH4Cl : –3

4 Br− _{and HBr both have bromine in –1 oxidation state.}

BrO− and Br2O and HBrO each have bromine in +1 oxidation state.

Review exercise 4.3

2 a

b See diagram in part a.

c anode: oxidation Mg(s) → Mg2+_{(aq) + 2e}−

cathode: reduction Cu2+(aq) + 2e− → Cu(s)

3 a 3Pb2+(aq) + 6e– → 3Pb(s) cathode 2Cr(s) → 2Cr3+_{(aq) + 6e}– _anode

b Cu(s) → Cu2+_{(aq) + 2e}– _anode

2Ag+(aq) + 2e– → 2Ag(s) cathode

Review exercise 4.4

1 a Ca2+, Pb2+, Cu2+, Ag+

b Ag, Sn, Cr, Mg

c strongest oxidising agent: Ag+ strongest reducing agent: Mg

2 a H2O2, HClO, MnO4−, Cr2O72−

b Mg, Zn, H2, H2S

3 a Cl2

b Sn4+ c Au3+

d MnO4 Review exercise 4.5 1 a Pb2+ + 2e– → Pb(s) E° = –0.13 V cathode Zn(s) → Zn2+(aq) + 2e– E° = +0.76 V anode cell potential = –0.13 + 0.76 = 0.63 V b Cl2(g) + 2e– → 2Cl–(aq) E° = +1.36 V cathode 2I–(aq) → I2(s) + 2e– E° = –0.54 V anode cell potential = +1.36 – 0.54 = 0.82 V c 2H+(aq) + 2e– → H2(g) E° = 0 V cathode Mg(s) → Mg2+(aq) + 2e– E° = +2.37 V anode cell potential = 0 + 2.37 = 2.37 V

2 Pb(s) + PbO2(s) + 4H+(aq) + 2SO42–(aq) → 2PbSO4(s) + 2H2O(l) E°(cell) = +2.04 V

anode: Pb(s) + SO42–(aq) → 2PbSO4(s) + 2e– E° = +0.36 V

cathode: PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l)

E° = 2.04 – 0.36 = +1.68 V

3 a i 2I– (aq) → I2(s) + 2e– E° = –0.54 V

MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l) E° = +1.51 V

Add equations.

10I–(aq) + 2MnO4–(aq) + 16H+(aq) → 5I2(s) + 2Mn2+(aq) + 8H2O(l)

ii cell potential = –0.54 + 1.51 = +0.97 V

iii Cell potential is positive, and therefore the reaction can occur spontaneously.

b i Cl2(g) + H2S(aq) + 2e– → 2Cl–(aq) + S(s) + 2H+(aq) + 2e–

Cl2(g) + 2e– → 2Cl– (aq) E° = +1.36 V

ii cell potential = +1.36 – 0.14 = +1.22 V

iii Cell potential is positive, and therefore the reaction can occur spontaneously.

c i 2Cu+(aq) + e– → Cu2+_{(aq) + Cu(s) + e}–

Cu+(aq) → Cu2+_{(aq) + e}– _{E° = –0.16 V}

Cu+_{(aq) + e}– _{→ Cu(s) E° = +0.52 V}

ii cell potential = –0.16 + 0.52 = +0.36 V

iii Cell potential is positive, and therefore the reaction can occur spontaneously.

4 a H2O2(aq) + 2H+(aq) + 2e– → 2H2O(l) E° = +1.78 V

H2C2O4(aq) → 2CO2(g) + 2H+(aq) + 2e– E° = +0.48 V

E° (cell) = +1.78 + 0.48 = +2.26 V

b A high activation energy may be necessary for this reaction to occur.

c Mix the H2O2 and H2C2O4 at a higher temperature and/or add a catalyst.

Review exercise 4.6

1 The dry cell is one of the most widely used sources of portable electricity. It has led to the proliferation of the portable devices we see around us. Without dry cells we would be a much less mobile society. Devices requiring electrical power would need to be in close proximity to permanent sources such as power points. Instead of torches we could still be using candles or kerosene lamps. However, the development of dry cells has led to landfill pollution.

2

3 a If the Zn and MnO2 were in direct contact, the following overall cell reaction would

occur:

Zn(s) + 2H+(aq) + 2MnO2(s) → Zn2+(aq) + Mn2O3(s) + H2O(l)

i.e. zinc is oxidised and manganese dioxide is reduced.

b The cell would be useless because electron transfer would occur directly between the reactants and not through an external circuit.

4 Zn(s) + 2H+(aq) + 2MnO2(s) → Zn2+(aq) + Mn2O3(s) + H2O(l)

5 Pb(s) + SO42–(aq) → PbSO4(s) + 2e–

(0) (+2) oxidation

PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l)

(+4) (+2) reduction

6 A 6V lead–acid battery would have three cells, each with a cell potential of 2 V.

7 Lead–acid batteries are very heavy and weight is an important issue for solar-powered cars in the World Solar Challenge.

Chapter 4 — Application and investigation

The following reactions will occur:

1 Zn(s) + CuSO4(aq → ZnSO4(aq) + Cu(s)

Zn(s) + NiSO4(aq) → ZnSO4(aq) + Ni(s)

ii No. Fe3+ will be reduced to Fe2+ and Ni oxidised to Ni2+.

b No. The steel will dissolve and displace the copper.

3 The student should determine gallium’s position in the order of reactivity by attempting to oxidise it using the ions of the other metals in the list.

4 a HNO3(aq) → NO2(g) reduction

+5 +4

b Cl2(g) → ClO3–(aq) oxidation

0 +5 c NH3(g) → N2(g) oxidation –3 0 d Fe2O3(s) → FeO(s) reduction +3 +2 e SO2(g) → SO42–(aq) reduction +4 +2

f MnO4–(aq) → MnO2(s) reduction

+7 +4

5 a Fe(s) + SO2(g) + O2(g) → FeSO4(s)

0 +4 –2 0 +2 +6 –2

Fe oxidised; S oxidised; O2 reduced

b 4FeSO4(s) + O2(g) + 6H2O(l) → 2Fe2O3.H2O(s) + 4H2SO4(aq)

+2 +6 –2 0 +1 –2 +3 –2 +1 –2 +1 +6 –2 Fe oxidised; O2 reduced

6 a All the E° values would be 2.93 V higher.

b Calculated e.m.f. values for cells are based on the difference in E° between two

7 a i

ii Cu2+(aq) + 2e– → Cu(s) E° = +0.16 V Cr(s) → Cr3+_{(aq) + 3e}– _{E° = +0.73 V}

2Cr(s) + 3Cu2+(aq) → 3Cu(s) + 2Cr3+_(aq)

iii e.m.f. = +0.89 V b i Zn(s) → Zn2+(aq) + 2e– Cl2(g) + 2e– → 2Cl–(aq) ii Cl2(g) + 2e– → 2Cl–(aq) E° = +1.36 V Zn(s) → Zn2+_{(aq) + 2e}– _{E° = +0.76 V} Zn(s) + Cl2(g) → Zn2+(aq) + 2Cl–(aq) iii e.m.f = 1.36 + 0.76 = 2.12 V

8 a The half-cells are separated in order to generate a flow of current in the external circuit.

b A salt bridge allows ions to move between each half-cell and complete the circuit.

9 a anode: Zn(s) → Zn2+_{(aq) + 2e}– _{E° = +0.76 V}

cathode: Cu2+(aq) + 2e– → Cu(s) E° = +0.34 V

cell: Zn(s) + Cu2+(aq) → Zn2+_{(aq) + Cu(s) E°(cell) = +1.10 V}

b As the cell operates, the concentration of Cu2+ decreases and the Zn2+ concentration increases. Both changes would decrease the tendency of the forward reaction to take place and hence the cell e.m.f. would decrease.

c i Transfer of electrons cannot occur, and therefore the reaction between the two half-cells will cease.

ii Transfer of charged ions between half-cells cannot occur.

10 The reduction potential of +1.7 V for Au+ is higher than that for the reduction of O2, and

therefore O2 would not oxidise the gold. The reduction potential for [Au(CN)2]–(aq) is

lower than that for O2, and therefore in the presence of cyanide ion, gold can be oxidised.

11 a Ni2+(aq) or Co2+(aq) or Cd2+(aq)

b Mn(s) or Al(s) or Zn(s) 12 a i Fe2+ → Fe3+ + e– E° = –0.77 V MnO4– + 8H+ + 5e– → Mn2+ + 4H2O E° = +1.51 V E°(cell) = +0.74 V ii 2Cl– → Cl2 + 2e– E° = –1.36 V MnO4– + 8H+ + 5e– → Mn2+ + 4H2O E° = +1.51 V E°(cell) = +0.15 V iii Fe2+ → Fe3+ + e– E° = –0.77 V Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O E° = +1.33 V E°(cell) = +0.56 V iv 2Cl– → Cl2 + 2e– E° = –1.36 V Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O E° = +1.33 V E°(cell) = –0.03 V Thus a redox reaction should occur in i, ii and iii.

b iv above shows that Cr2O72– will not oxidise the Cl– ion of HCl, but ii shows that

MnO4– will. Hence, it is inappropriate to use HCl to acidify solutions involving

13 a An anode is the electrode at which oxidation takes place. A cathode is the electrode

at which reduction takes place.

b A salt bridge allows a flow of ions between two half-cells. An external circuit

allows a flow of electrons from anode to cathode.

c A secondary cell can be recharged, whereas a primary cell cannot. 14 a i anode: Pb(s) + SO42–(aq) → PbSO4(s) + 2e–

cathode: PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l)

ii anode: PbSO4(s) + 2H2O(l) → 2e– + SO42–(aq) + 4H+(aq) + PbO2(s)

cathode: PbSO4(s) + 2e– → Pb(s) + SO42–(aq)

b i On discharging, the density decreases.

ii On recharging, the density increases.

c Chemical potential energy is transformed into electrical energy and some heat energy as the battery is discharged. On charging, electrical energy is transformed into chemical potential energy and some heat energy.

15 a Anode: Al(s) → Al3+_{(aq) + 3e}–

Cathode: O2(g) + 2H2O(l) + 4e– → 4OH–(aq)

b Al(s) → Al3+(aq) + 3e– E° = +1.66 V

O2(g) + 2H2O(l) + 4e– → 4OH–(aq) E° = +0.40 V

E°(cell) = +2.06 V Cell e.m.f. would be 2.06 V.

c It would weigh much less. It avoids use of corrosive H2SO4.

It uses O2 from the air as a cathode.

16–18 Investigation

CHAPTER 5: NUCLEAR CHEMISTRY Review exercise 5.1

1 a Atoms are not conserved in nuclear reactions.

b Chemical reactions involve changes in electrons between atoms. Nuclear reactions

involve the particles in the nucleus.

c Temperature, pressure/concentration and catalysts have no effect on nuclear reactions.

d Far more energy is obtained from nuclear reactions.

2 a i 17O : 8 protons; 9 neutrons

ii 52Fe : 26 protons; 26 neutrons

iii 137I : 53 protons; 84 neutrons

iv 258Md : 101 protons; 157 neutrons b i ii Nb93 41 iii 208Pb 82

3 a α large mass; β very small mass; γ no mass

b α 2+ charge; β 1– charge; γ no charge

c α least penetrating, about 5 cm through air β about 1 m through air, stopped by aluminium γ can pass through several cm of lead

4 a He4 2 b 0e 1 − c 1n 0 d 1p 1 e 0e 1

5 Photographic film: becomes dark on exposure. Darkness increases with length of exposure and greater intensity of radiation.

Geiger counter: The argon gas inside a metal tube is ionised into positive ions and electrons. These move towards electrodes which conduct a current measured by a recording device.

Scintillation counter: Electrons in substances such as zinc sulfide are excited by radiation and emit photons of light as they return to lower energy states. Radiation is measured by counting flashes of light.

Review exercise 5.2 1 a 3H 1 → He 3 2 + e 0 1 − b

c 2 a b Pu239 94 → U23592 + He42 c 234U 92 → Th 230 90 + He 4 2 3 a 12N 7 → C126 + e10 b 8B 5 → Be 8 4 + e 0 1 c 4 a Be7 4 + e 0 1 − → Li37 b 232Np 93 + e 0 1 − → 23292 U c 5 a beta emission b positron emission c alpha emission d positron emission

6 a positron or electron capture

b β decay

c α decay

7 a No, the neutron : proton ratio is too high and outside the zone of stability.

b No, the neutron : proton ratio is too low and outside the zone of stability.

c Yes, the neutron : proton ratio is in the zone of stability.

Review exercise 5.3 1 After 1st 2.6 hours: 0.1/2 = 0.05 g After 5.2 hours: 0.025 g After 7.8 hours: 0.0125 g 2 1st half-life: 0.120/2 = 0.06 g 2nd half-life: 0.06/2 = 0.03 g

3rd half-life: 0.03/2 = 0.015 g ∴3 half-lives in 24 days I-131 has a half-life of 8 days.

3 a 2.1 g = 21%

b 164 g → 20.5 g, i.e. 12.5% left = 3 half-lives, ∴ 84 years

c Half-life is 28 years.

4 Thallium-201 and gallium-67: suitable, as have a short half-life; sulfur-35 and hydrogen-3: unsuitable as half-life is too long.

Review exercise 5.4

1 To overcome the electrostatic repulsion alpha particles and protons will encounter in the nuclide. Neutrons do not carry a charge and therefore do not need to be accelerated.

2 They are produced synthetically.

3 a 15₇ N + 1₁p → C12₆ + 4₂He b Fe56₂₆ + ₁2H → 4₂He + Mn54₂₅ c Li6₃ + n1₀ → ₁3H + 4₂He d Co59₂₇ + ₁2H → Co60 27 + 11p e 53₂₄Cr + 4₂He → Mn57₂₅ + ₁0n

4 The nuclear transformations in a, b, d and e would all take place in a cyclotron, as the charged particles are used to bombard the nucleus.

Nuclear transformation c would take place in a nuclear reactor, as it involves bombardment by neutrons.

5 Fe56₂₆ + ₁2H → Mn54

25 + 42He

Chapter 5 Application and investigation 1 a 90 protons; 140 neutrons

b 96 protons; 148 neutrons

c 77 protons; 115 neutrons

2 Chemical reactions involve changes in the arrangements of electrons around the nucleus. Nuclear reactions involve changes within the nucleus.

3 a Ionisation counter: uses a tube filled with an inert gas. As radiation enters the tube it causes the gas to be ionised, forming cations and free electrons. These particles are attracted to electrodes of the opposite charge, causing a current to flow, which is heard as an audible click and can be recorded as a digital readout.

b Scintillation counter: electrons in some substances, such as zinc sulfide, are excited by radiation and emit photons of light when the electrons return to a lower energy state. The flashes of light are counted electronically to measure the amount of radiation.

4 a β decay

b α decay, positron emission, electron capture

5 a 241₉₄ Pu → 237₉₂ U + 4₂He b 210₈₆ Rn → 210₈₇ Fr + e₋0₁ c 48₂₃V → 48₂₂Ti + ₁0e d 107₄₈ Cd + e₋0₁ → 107₄₇ Ag e 234₉₂ U → 230₉₀ Th + 4₂He f ₁₁26Na → ₁₂26Mg + e₋0₁ g 12₇ N → 12₆ C + ₁0e h

6 Uranium and thorium undergo alpha emission spontaneously. Therefore helium is found within the ore.

7 Elements with high neutron-to-proton ratios emit β-particles, thereby raising the number of protons and lowering the number of neutrons. Elements with low neutron-to-proton ratios undergo either positron emission or electron capture, thereby raising the number of neutrons and lowering the number of protons.

8 a unstable; β decay

b unstable; positron emission or electron capture

c unstable; positron emission or electron capture

d unstable; β decay

f unstable; α decay 9 a 3 half-lives ∴ 0.1mg 2 0.2 ; 2 . 0 2 0.4 ; 4 . 0 2 8 . 0 = = = b 2 half-lives ∴ 12.3 × 2 = 24.6 years 10 a 35% b 20% × 500 g = 100 g remaining c 84 days d 25 days 11 a 32 days is 4 half-lives

∴ 0.625 mg remaining after 32 days

b 5 half-lives drop to 0.312 g, so < 0.5 g in 36 days

12 13 Investigation 14 a Co59 27 + n 1 0 → Co 60 27

b 60Co would be made in a nuclear reactor as it involves the capture of a neutron. In a nuclear reactor, atoms are bombarded with neutrons. A cyclotron is used to

bombard atoms with charged particles.

15 Investigation 16 a 63Cu 29 + H 2 1 → Zn 65 30 + γ b 14N 7 + He42 → O178 + H11 c Ca40 20 + H 2 1 → K 38 19 + He 4 2 d 239Pu 94 + He42 → Cm24296 + n10 17 Investigation

18 α particles can cause serious damage to cells.

19 Investigation

20 Investigation

Module 1

11 a Ethylene is more reactive due to the presence of the double bond.

b addition c polyethylene d HI 12 a b vinyl chloride i chloroethene ii polyvinylchloride (PVC)

iii PVC is a thermoplastic with long chains with weak dispersion forces between them. This allows the polymer to be heated and remoulded. With additives, PVC can be used as a rigid plastic (e.g. guttering, water pipes) or as a flexible plastic (e.g. garden hoses).

13 a LDPE has a greater degree of branching between its polymer chains, reducing the dispersion forces between the chains. HDPE has less branching, resulting in a stronger and more dense plastic.

b cling wrap, insulation for wires

c LDPE: Organic peroxides are used as catalysts to attack the ethene double bond and form a covalent bond with a carbon atom. This new molecule then attacks another ethene molecule, causing the chain to grow (propagate). Branches are formed when the chain curls back and removes a hydrogen from a carbon in the chain, which will in turn react with another ethene molecule. The reaction terminates when two polymer radicals react together.

HDPE: Ionic catalysts (Ziegler–Natta catalysts) are used. Ethene molecules are added to the polymer on the surface of the catalyst, which reduces the amount of branching.

d The polymer chains in HDPE are packed tightly together due to less branching of the chains. This gives HDPE greater strength and toughness but reduces its flexibility.

14 a Biomass is organic material derived from living organisms, such as: plant material (e.g. sugar and cellulose); animal material (e.g. dung); and domestic and industrial organic waste.

b β-glucose

c Cellulose is a formed as a condensation polymer when β-glucose monomers

combine with the removal of a water molecule. Polyethylene forms as an addition polymer from ethylene monomers as the double bond breaks.

15 Use spirit burners containing each fuel, weigh mass before and after use (to give mass of fuel burnt), heat same volume of water, record temperature rise. Fuels are volatile and can easily ignite if care is not taken.

16 a i C6H12O6(aq) → 2C2H5OH(l) + 2CO2(g)

ii anaerobic (no oxygen); temperature 35–40oC

b Harvest cane, extract sugars, ferment to ethanol using yeast, warmth and anaerobic conditions.

c C2H5OH(g) → C2H4(g) + H2O(g)

The reaction requires a catalyst, e.g. concentrated H2SO4

17 No commercially viable method of producing ethanol from crops is yet available. Fuel mixtures containing greater than 15% ethanol require some modification to the car before being used.

18 a

b Cell potential = +0.76 – 0.13 = 0.63 V

c Zinc

19 Uranium-238 is bombarded with neutrons to form einsteinium. U

238

92 + 1510n → Es25399 + 7−01e

20 a

b Sr is in the same periodic group as calcium and therefore is chemically similar.

c 85.3 years is three half-lives ∴ > 0.125 g remains at 85 years.

21 a Iodine-131 is used in the diagnosis of thyroid misfunction and in treatment of thyroid tumours.

b I-131 is produced by neutron bombardment in a reactor.

c Benefit: short half-life and therefore is only in the body for a short time. Problem: I-131 can destroy some healthy cells when used for radiation.

CHAPTER 6: ACIDS, BASES AND INDICATORS Review exercise 6.1

1 a acetic acid, CH3COOH

citric acid, 2-hydroxypropane-1,2,3-tricarboxylic acid

b sulfuric acid, H2SO4

nitric acid, HNO3

c magnesium hydroxide, Mg(OH)2

2 Own answers

3 Own answers

4 a Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

b CO32−(aq) + 2H+(aq) → CO2(g) + H2O(l)

c 2H+(aq) + K2O(s) → 2K+(aq) + H2O(l)

d H+(aq) + OH– (aq) → H2O(l)

5 a 2Cr(s) + 2OH–(aq) + 6H2O(l) → 2[Cr(OH)4]–(aq) + 3H2(g)

b Cr(OH)3(s) + OH–(aq) → [Cr(OH)4]–(aq)

6 Acid solutions have excess hydrogen ions and base solutions have excess hydroxide ions. These ions are free to move and carry a current.

Review exercise 6.2

1 a K2O ionic bonding; Ga2O ionic; Br2O7 covalent

b K2O basic; Ga2O amphoteric; Br2O7 acidic

2 CO2(g) + H2O(l) → H2CO3(aq)

3 Natural sources of sulfur dioxide:

• bacteria decomposing organic matter to make H2S, which is then oxidised atmospherically.

• volcanic gases • bushfire smoke.

Sources due to human activities: • burning of fossil fuels

• smelting of sulfide ores such as CuFeS2 and ZnS.

4 SO2—H2SO3; SO3—H2SO4; NO2—HNO3; CO2—H2CO3

5 Increased incidence of low pH in lakes and rivers and continued damage to metal and sandstone buildings.

6 The sulfur oxides in the atmosphere usually come from burning fossil fuels in power stations or smelting sulfide ores in smelting plants. Nitrogen oxide and carbon emissions usually come from motor vehicles. Therefore acid rain usually occurs in densely populated areas or industrial areas.

Review exercise 6.3

1 a Acidic. The solution has a pH between 4 and 6.

b The pH range is 4 to 8. This solution could be neutral, acidic or basic.

c The solution is slightly basic. The pH is between 7.5 and 8.

2 a Bromothymol blue: yellow; phenolphthalein: colourless; methyl orange: red

b Bromothymol blue: blue; phenolphthalein: pink; methyl orange: yellow

c Bromothymol blue: blue; phenolphthalein: colourless; methyl orange: yellow

Chapter 6 Application and investigation

1 Not all acids are dangerous. Acids are present in foods we eat, e.g. ethanoic acid in vinegar, lactic acid in milk. Also, acids are found in our bodies, e.g. lactic acid, hydrochloric acid. The concentration of acids should be considered.

2 a 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

b KHCO3(s) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l)

c Fe2O3(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3H2O(l)

d Ba(OH)2(aq) + 2HF(aq) → BaF2(aq) + 2H2O(l)

3 a 2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na[Al(OH)4](aq) + 3H2(g)

b KOH(aq) + Fe(OH)3(s) → no reaction

c 2KOH(aq) + Zn(OH)2(s) → K2[Zn(OH)4](aq)

d 2NaOH(aq) + SO3(g) → Na2SO4(aq) + H2O(l)

4 a KHC4H4O6 + NaHCO3 → KNaC4H4O6 + CO2(g) + H2O(l)

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

d H2SO3(aq) + CaCO3(s) → CaSO3(aq) + CO2(g) + H2O(l)

5 Al(OH)3 reacts with both acids and bases, and therefore is amphoteric.

Reacting with acid: Al(OH)3(aq) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)

Reacting with base: Al(OH)3(aq) + NaOH(aq) → Na [Al(OH)4](aq)

6 Investigation 7 Investigation

8 A useful indicator is one whose colour change occurs over a selected pH range. Methyl orange changes colour too far outside the narrow 7.2–7.6 range for swimming pools. Litmus changes colour over this range, but the colour difference in the 7.2–7.6 range may be too subtle to detect easily. Universal indicator undergoes many colour changes and the change over the 7.2–7.6 range, from yellow-green to green may also be too subtle to detect easily. Bromothymol blue could be a useful indicator. The change for phenolphthalein is above a pH of 7.6 and so this indicator would not be useful.

9 Investigation

10 Acid rain has a pH less than 5; methyl orange when yellow means pH is greater than 4.5, and litmus when red indicates a pH of less than 7. Therefore the pH of the sample of rain water lies between 4.5 and 7, and so is unlikely to be classified as acid rain.

11 a Na2O, Al2O3, SiO2, P4O10, Cl2O7

b K2O, Ga2O3, GeO2, As2O3, Br2O7

12 CO2(aq) + H2O(l) ↔ H2CO3(aq) carbonic acid

4NO2(g) + 2H2O(l) + O2(g) → 4HNO3(aq) nitric acid

SO2(g) + H2O(l) → H2SO3(aq) sulfurous acid

Cl2O7(l) + H2O(l) → 2HClO4 perchloric acid

13 a SO2(g) + H2O(l) → H2SO3(aq)

b H2SO3(aq) + CaCO3(s) → CaSO3(aq) + CO2(g) + H2O(l)

14 Investigation 15 Investigation 16 Investigation

CHAPTER 7: CHEMICAL EQUILIBRIUM Review exercise 7.1

1 a NH3(g) + HCl(g) ↔ NH4Cl(s)

b 2SO2(g) + O2(g) ↔ 2SO3(g)

c Ag+(aq) + Fe2+(aq) ↔ Ag(s) + Fe3+_(aq)

d 2NH3(g) ↔ N2(g) + 3H2(g)

2 a Ammonia molecules are reacting with hydrogen chloride molecules to form ammonium chloride at the same rate at which the ammonium chloride is decomposing to ammonia and hydrogen chloride.

b Sulfur dioxide reacts with oxygen to form sulfur trioxide at the same rate at which the sulfur trioxide decomposes to produce sulfur dioxide and oxygen.

c Silver ions are reduced to silver at the same rate at which the iron(III) is reduced to iron(II).

d Ammonia decomposes to nitrogen and hydrogen at the same rate at which hydrogen and nitrogen combine to form ammonia.

3 a If 0.21 mol L–1 H2 remains, 0.79 mol L–1 reacted.

Molar ratio of H2 : HI is 1 : 2

∴ 0.79 × 2 = 1.58 moles HI formed ∴ equilibrium conc HI = 1.58 mol L–1

b i and ii

c Forward and reverse reactions will occur at equal but opposing rates. The concentrations of reactants and products remain constant.

4 In an open container, the CO2 would escape and the reverse reaction would not occur. In

a sealed container, increasing the temperature would increase the pressure until equilibrium was established.

Review exercise 7.2

1 a i ↑ [NO], ↑ [H2O], ↓ [NH3]

ii Will increase rate of forward reaction until equilibrium is re-established.

b i ↓ [NH3], ↓ [O2], ↑ [H2O]

ii Will increase rate of forward reaction until equilibrium is re-established.

2 a i Concentration of H2O will decrease, concentration of CO, H2 will increase.

No change to [C(s)].

ii Will increase rate of forward reaction. This increases concentration of CO and H2, decreases [H2O].

b i Concentration of H2O will increase, concentration of CO, H2 will decrease.

No change to [C(s)].

ii Will increase rate of reverse reaction. [CO] ↓, [H2] ↓ and [H2O] ↑ until

equilibrium is re-established.

3 a i ↑ [CO], ↑ [O2], ↓ [CO2]

ii Will increase rate of reverse reaction.

b i ↓ [CO], ↓ [O2], ↑ [CO2]

ii Will increase rate of forward reaction.

4 a The system would reach equilibrium more quickly.

b Both forward and reverse rates would increase equally, therefore no change.

Review exercise 7.3 1 CO2(g) ↔ CO2(aq)

CO2(aq) + H2O(l) ↔ H2CO3(aq)

H2CO3(aq) ↔ H+(aq) + HCO3–(aq)

HCO3–(aq) ↔ H+(aq) + CO32–(aq)

2 a When the bottles are opened, the pressure in the system is decreased, causing a shift in the equilibrium to the left, where there are more moles of gas and so the change is counteracted. This results in bubbles of CO2(g).

b Cooling the drink shifts the equilibrium to the right (exothermic direction), meaning that more carbon dioxide will be dissolved into the solution. When the

drink is served cold, the rate at which CO2(g) escapes to the atmosphere is

Chapter 7 Application and investigation 1 a Changes in concentration

b Rate of reaction

2 No further change in the colour of the mixture would be detected.

3 At equilibrium, the concentration of all species remains constant, therefore there will be no change in any observable properties. The dynamic process refers to the equal but opposite reactions occurring at the molecular level.

4 concentration, pressure, temperature

5 a ↓ [NO]; ↑ [O2]; ↑ [NO2]

b Concentration of all species will double.

6 a ↓ [HBr]; ↓ [O2]; ↑ [H2O]; ↑ [Br2]

b ↑ [HBr]; ↑ [O2]; ↓ [H2O]; ↓ [Br2]

7 a i Shift equilibrium to the right.

ii Shift equilibrium to the left.

b i Shift equilibrium to the left.

ii Shift equilibrium to the right.

8 a ↑Ca(HCO3)2(s); ↓CaO(s); ↓ [H2O]; ↓ [CO2] after initial increase of CO2

b ↓Ca(HCO3)2(s); ↑CaO(s); ↑ [CO2]; ↑ [H2O] after initial decrease of H2O

c ↓Ca(HCO3)2(s); ↑CaO(s); ↑ [CO2]; ↑ [H2O]

d no change

e ↑Ca(HCO3)2(s); ↓CaO(s); ↓ [CO2]; ↓ [H2O]

9 a i Equilibrium will shift right to counteract the change: ↓ [NO2]; ↑ [SO3];

↑ [NO].

ii The forward reaction rate will increase.

b i There are equal numbers of gaseous molecules on each side, so pressure will increase the concentration of all species, but the system will remain at equilibrium.

ii Forward and reverse reactions increase equally.

c i Concentrations of all species will decrease, but the system will remain at equilibrium.

ii Forward and reverse reaction rates decrease equally.

d i The equilibrium will shift left in an endothermic direction: ↑ [SO2],

↑ [NO2], ↓ [SO3], ↓ [NO].

ii Reverse reaction is favoured.

e i No change as a catalyst does not affect the position of equilibrium.

ii Both forward and reverse reaction rates will increase.

10 a Pressure is decreased, so equilibrium will shift to counteract the change. Therefore CO2(g) is produced and bubbles form.

b It is an exothermic reaction, so warm temperatures cause the equilibrium to shift to the left and form CO2(g).

c The H+ in the orange juice causes increased production of carbonic acid. HCO3–(aq) + H+(aq) ↔ H2CO3(aq)

The ↑ [H2CO3] causes the production of more CO2(aq):

H2CO3(aq) ↔ CO2(aq) + H2O(l)

Therefore the ↑ [CO2(aq)] results in the production of more CO2(g) causing drinks

to go flat, as follows: CO2(aq) + 19kJ ↔ CO2(g)

11 Investigation

CHAPTER 8: THE HISTORICAL DEVELOPMENT OF IDEAS OF ACIDS AND BASES

Review exercise 8.1

1 a SO2(g) + H2O(l) → H2SO3(aq)

b Na2O(s) + H2O(l) → 2NaOH(aq)

2 Hydrochloric acid, HCl; hydrobromic acid, HBr; cyanic acid, HCN

3 Arrhenius theory states that bases are substances which deliver hydroxide ions in water. NaCN is a substance that acts as a base but does not contain a hydroxide group.

4 a H2SO4(aq) → 2H+(aq) + SO42–(aq)

or H2SO4(aq) → HSO4–(aq) + H+(aq), then HSO4–(aq) ↔ SO42– + H+(aq)

b HNO3(aq) → H+(aq) + NO3–(aq)

5 a 1 b 2 c 3 d 1 e 1 f 1 6 a 1 b 3 c 2

7 HOOCCOOH(aq) ↔ HOOCCOO–(aq) + H+(aq) HOOCCOO–(aq) ↔ –OOCCOO–(aq) + H+(aq)

Review exercise 8.2

b A B–L base is a proton acceptor (or ‘takes’ a proton).

2 a HCl, HCO3–, NH4+, HClO4, HSO4–

b F–, SO42–, NH3, PO43–, H2O

3 a An amphiprotic substance can either accept or donate a proton, and therefore it can act as both an acid and a base.

b HPO42–(aq) + H2O(l) ↔ H2PO4–(aq) + OH–(aq)

HPO42–(aq) + H2O(l) ↔ PO43–(aq) + H3O+(aq)

4 Acid HSO4–(aq) + OH–(aq) ↔ SO42–(aq) + H2O(l)

Base HSO4–(aq) + H3O+(aq) ↔ H2SO4(aq) + H2O(l)

5 a A conjugate base is the species that results when a B–L acid has lost its proton

b A conjugate acid is the species that results when a B–L base accepts a proton

6 a HCO3–/SO42–: HF/F– b HSO4–/SO42–: NH4+/NH3 c HF/F– : H3O+/H2O Review exercise 8.3 1 a i HCl ii HCl iii HCl b i H2SO4 ii H2SO4 iii H2SO4 2 a i NaOH ii NaOH

iii NaOH b i Ba(OH)2 ii Ba(OH)2 iii Ba(OH)2 3 a 5 mol L–1 HCl b 0.1 mol L–1 NaOH c 5 mol L–1 NH3 d 0.1 mol L–1 CH3COOH

4 a HBr(aq) → H+_{(aq) + Br}–_(aq)

b H2SO3(aq) ↔ HSO3–(aq) + H+(aq)

c RbOH(aq) → Rb+(aq) + OH–(aq)

d NaF(aq) ↔ Na+_{(aq) + F}–_(aq)

5 a small extent

b large extent

c small extent

Review exercise 8.4

1 a S2–(aq) + H2O(l) ↔ HS–(aq) + OH–(aq)

b CO32–(aq) + H2O(l) ↔ HCO3–(aq) + OH–(aq)

c NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

d [Fe(H2O)6]3+(aq) + H2O(l) ↔ [Fe(OH)(H2O)5]2+(aq) + H3O+(aq)

e F–(aq) + H2O(l) ↔ HF(aq) + OH–(aq)

f HSO4–(aq) + H2O(l) ↔SO42–(aq) + H3O+(aq)

g ClO– + H2O↔ HClO + OH–

h CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq)

2 Use Table 8.5.

a neutral

b acidic due to reaction of NH4+ ion with water

c basic due to reaction of ClO– ion with water

e acidic due to reaction of Al ion (actually will be aquated aluminium ion [Al(H2O)6]3+ with water)

f acidic due to reaction of H2PO4– ion with water

g neutral

h basic due to reaction of CO32– ion with water

Chapter 8 Application and investigation

1 According to Arrhenius, acids are substances that release hydrogen ions in solution. HCl(aq) completely ionises in water to produce H+ and Cl−, and therefore HCl(aq) is an Arrhenius acid. HCl(l) cannot release hydrogen ions and so is not an Arrhenius acid.

2 a HCl(aq) → H+(aq) + Cl–(aq)

b Ca(OH)2(aq) → Ca2+(aq) + 2OH–(aq)

c H2SO4(aq) → HSO4–(aq) + H+(aq)

HSO4–(aq) ↔ SO42–(aq) + H+(aq)

d H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

e Al(OH)3(aq) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)

3 a 4

b 2 c 2

d CH2(COOH)2(aq) + 2NaOH(aq) → CH2(COONa)2(aq) + 2H2O(l)

4 a HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

b CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO–(aq)

c NH3(aq) + H2O(l) ↔ NH4+(aq) + OH–(aq)

d H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4–(aq)

HSO4–(aq) + H2O(l) ↔ H3O+(aq) + SO42–(aq)

e HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

5 a ClO3–, S2–, NH3, OH–

b H2CO3, H2S, [Fe(H2O)6]3+, N2H5+

6 a As acid: HCO3–(aq) + H2O(l) ↔ CO32–(aq) + H3O+(aq)

b The extent to which the HCO3– ion reacts as a base with water exceeds the degree

to which it reacts with water as an acid. (K value is greater for HCO3– as a base.)

Hence more OH– ions are produced than H3O+ ions.

7 a i H2C2O4/HC2O4–, H3O+/H2O ii small extent b i H2O/OH–, HCN/CN– ii small extent c i CH3COOH/CH3COO– , HS–/S2– ii large extent d i HCl/Cl–, HF/F– ii large extent

8 a HClO4(l) + H2O(l) → ClO4–(aq) + H3O+(aq)

b HCOOH(l) + H2O(l) ↔ HCOO–(aq) + H3O+(aq)

c LiOH(s) + H2O(l) → Li+(aq) + OH–(aq)

d N2H4(l) + H2O(l) ↔ N2H5+(aq) + OH–(aq)

9 a concentrated solution of a strong acid

c concentrated solution of a weak acid

d diluted solution of a weak acid

10 a CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq)

b The hydroxide ions produced raise the pH above 7. The ethanoic acid molecule formed has only a very small degree of dissociation.

11 a neutral

b basic PO43–(aq) + H2O(l) ↔ HPO42–(aq) + OH–(aq)

c basic CO32–(aq) + H2O(l) ↔ HCO3–(aq) + OH–(aq)

d acidic NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

e acidic [Cr(H2O)6]3+(aq) + H2O(l) ↔ [Cr(OH)(H2O)5]2+(aq) + H3O+(aq)

f basic SO42–(aq) + H2O(l) ↔ HSO4–(aq) + OH–(aq)

g basic CN–(aq) + H2O(l) ↔ HCN(aq) + OH–(aq)

h acidic NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

12

Acid Base Neutralisation Arrhenius produces H+ ions

in water

produces OH– ions

in water H

+ _{+ OH}–_↔H 2O

Brønsted–Lowry proton donor proton acceptor NH3(aq) + H2O↔NH4+

13 The reaction: HCl(g) + NH3(g) → NH4Cl(s) can be viewed as a Brønsted–Lowry acid–

base reaction because a proton is transferred. However, the reaction is not happening in solution and there are no hydroxide ions released, so the reaction cannot be viewed as an Arrhenius acid–base reaction. When nitric acid is mixed with sulfuric acid, for the manufacture of nitroglycerine, the reaction that occurs is

H2SO4 + HNO3 ↔ HSO4– + H2NO3+

This is a B–L acid–base reaction because a proton is transferred but it is not an Arrhenius acid–base reaction.

14 Investigation

CHAPTER 9: HYDROGEN ION CONCENTRATION AND THE pH SCALE Review exercise 9.1

1 a KW = [H+] [OH–]

1.0 × 10–14 _{= [H}+_{] × 1.6 × 10}–7

[H+] = 6.2 × 10–8 _{mol L}–1

b [H+] is less than 1 × 10–7 _{mol L}–1 _{∴ pool is slightly basic}

2 a [H+] = 5.0 × 10–3 _{mol L}–1 [NO3–] = 5.0 × 10–3 mol L–1 [OH–] = 1 × 10–14 _{÷ 5.0 × 10}–3 = 2 × 10–12 _{mol L}–1 b [H+] = 1.5 mol L–1 [Cl–] = 1.5 mol L–1 [OH–] = 1 × 10–14 _{÷ 1.5} = 6.7 × 10–15 _{mol L}–1 c [K+_{] = 0.25 mol L}–1 [OH–] = 0.25 mol L–1 [H+] = 4 × 10–14 _{mol L}–1 d [Ba2+] = 6.0 × 10–2 _{mol L}–1 [OH–] = 0.12 mol L–1 [H+] = 8.3 × 10–14 _{mol L}–1

Review exercise 9.2 1 a pH = –log[H+] = –log[5.0 × 10–1_] = 0.30 b [H+] = 1 × 10–14 _{÷ 0.0065} = 1.54 × 10–12 _{mol L}–1 pH = –log[1.54 × 10–12_] = 11.81 c pH = –log[3.6 × 10–3_] = 2.44 d [H+] = 1 × 10–14 _{÷ (6.5 × 10}–4 _{× 2)} = 7.7 × 10–12 _{mol L}–1 pH = –log[7.7 × 10–12_] = 11.1

2 a pH = 3.50; [H+] = 3 × 10–4 _{mol L}–1_{; [OH}–_{] = 3.16 × 10}–11 _{mol L}–1

b pH = 11.90; [H+] = 1.3 × 10–12 _{mol L}–1_{; [OH}–_{] = 7.9 × 10}–3 _{mol L}–1

c pH = 0.80; [H+] = 0.16 mol L–1; [OH–] = 6.3 × 10–14 _{mol L}–1

3 pH = 7 so [H+] = 1 × 10–7 pH = 5 so [H+] = 1 × 10–5 ∴ pH changes by factor of 100 Review exercise 9.3 1 a HClO2 b HClO2 c HClO2, HF, HCN

d No. Complete ionisation of 0.1 mol L–1 gives pH = 1.

Review exercise 9.4

1 a A buffer is a species that resists changes to pH despite small additions of acid or base.

b A buffer contains approximately equal amounts of a weak acid and its conjugate base (some are a mixture of weak bases and their conjugate acids).

2 a Buffered; contains approximately equal amounts of the weak acid HSO4– and its

conjugate base SO42–.

b Not buffered; HNO3 is a strong acid.

c Buffered; contains approximately equal amounts of the weak acid HClO and its conjugate base ClO–.

3 a HPO4–(aq) + H3O+(aq) ↔ H2PO4–(aq) + H2O(l)

When excess hydrogen ions are added, the system shifts to the right and thus decreases the H+ concentration.

b H2PO4+(aq) + OH–(aq) ↔ HPO4–(aq) + H2O(l)

When excess hydroxide is added, the system shifts to the right and thus decreases the OH– concentration.

4 The concentration of the acid and conjugate base ions is limited. If too much H+ or OH– is added then the system is unable to provide sufficient acid or conjugate base particles to counteract their effect.

5 Lactic acid in the blood produces lactate ions and hydrogen ions. The blood’s buffering system H2CO3(aq) + H2O(l) ↔ HCO3–(aq) + H3O+(aq) responds by combining the extra

hydrogen ions with the hydrogencarbonate ions, thus shifting the system to the left and decreasing their concentration.

Chapter 9 Application and investigation 1 a [H+] = 0.5 mol L–1 [OH–] = 1 × 10–14 _{÷ 0.5} = 2 × 10–14 _{mol L}–1 pH = –log[0.5] = 0.30 b [OH–] = 3.0 × 10–3 _{× 2} = 6.0 × 10–3 _{mol L}–1 [H+_{] = 1 × 10}–14 _{÷ 0.006} = 1.67 × 10–12 _{mol L}–1 pH = –log[1.67 × 10–12_] = 11.8 2 a [H+] = 1.0 × 10–3 _{mol L}–1

[OH–] = 1 × 10–14 _{÷ 0.001} = 1 × 10–11 _{mol L}–1 b [H+] = 3.9 × 10–3 _{mol L}–1 [OH–] = 2.51 × 10–12 _{mol L}–1 c [H+] = 3.2 × 10–9 _{mol L}–1 [OH–] = 3.2 × 10–6 _{mol L}–1 d [H+] = 7.9 × 10–13 _{mol L}–1 [OH–] = 1.3 × 10–2 _{mol L}–1 3 a i [H+] = 0.20 mol L–1 [NO3–] = 0.20 mol L–1 [HNO3] = 0 mol L–1 ii pH = –log10[0.20] = 0.70

b As HNO2 is a weak acid, it will not dissociate completely. Therefore [H+] will be

less than 0.20 mol L–1 and pH will be higher than 0.70.

4 H2CO3, H3PO4, HClO2, H2SO3, HCl

5 Investigation 6 Investigation

CHAPTER 10: VOLUMETRIC ANALYSIS AND ESTERIFICATION Review exercise 10.1

1 a NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

b 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

c 2NaOH(aq) + H2CO3(aq) → Na2CO3(aq) + H2O(l)

d 3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + H2O(l)

2 a 2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

c 2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l)

d 2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + CO2(g) + H2O(l)

3 Neutralisation is the result of an acid–base interaction. In Brønsted–Lowry terms this refers to proton transfer from the acid to the base.

4 Neutralisation results in the formation of a new bond as the proton bonds to its base. Bond formation is an exothermic process, and thus neutralisation is an exothermic process.

5 a Add solid sodium hydrogencarbonate until the ‘fizzing’ stops. The reaction produces carbon dioxide, so when the fizzing stops, this indicates that the neutralisation is complete.

b Rinse with copious amounts of water. A base should not be used, as the neutralisation reaction is exothermic—i.e. it releases a lot of heat.

Review exercise 10.2

1 2OH–(aq) + H2SO4(aq) → 2H2O(l) + SO42–(aq)

Calculate moles of OH–

n(OH–) = cV

= 1.90 × 0.02368 = 0.045 moles

∴ moles of H2SO4 : n(H2SO4) = 0.5n(OH–) (from equation)

= 0.0225 moles ∴ concentration of H2SO4

n(H2SO4) = cV

0.0225 = c × 0.005

c = 4.50 mol L–1

2 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Calculated moles of HCl

n = cV

= 0.02052 × 0.952 = 0.019535 moles

∴ concentration of NaOH =

V n

= 0.019535 / 0.005

c = 3.91 mol L–1 NaOH

3 a Pipettes are more accurately calibrated than a measuring cylinder; this accuracy is important in volumetric analysis.

b If rinsed with water, any drops left in the pipette or burette will alter the concentration of the solution and this will affect the results. Rinsing with the solution will not have this effect.

c Volume may change slightly once the solid is dissolved; the customary unit of concentration is mol/L, where the volume refers to litres of solution, not water.

4 a No. moles Na2CO3 = 0.250 × 0.1 = 0.025 moles moles = M m so 0.25 = 99 . 105 m m = 2.65 g of Na2CO3 is dissolved in water in a 250 mL

volumetric flask with solution made up to the calibrated mark b n = cV nH2C2O4.2H2O = 0.5 × 0.015 = 7.5 × 10–3 _moles n = M m 7.5 × 10–3 ₌ 07 . 126 m

m = 0.95 g oxalic acid is dissolved in water in a 500 mL volumetric flask with the solution made up to the calibrated mark

5 a n =

M m

= 07 . 126 3162 . 0 = 0.0025 moles n = cV 0.0025 = c × 0.25 c = 0.010 mol L–1 H2C2O4.2H2O

b 2NaOH(aq) + H2C2O4(aq) → Na2C2O4(aq) + 2H2O(l)

n(H2C2O4) = cV

n = 0.010 × 0.02061 = 2.068 × 10–4 _{moles H}

2C2O4

n(OH–) = 2 × nH2C2O4 (from equation)

n(OH–) = 4.136 × 10–4 _mol

n(NaOH) = cV

4.136 × 10–4 _{= c × 0.020}

c = 0.0207 mol L–1 NaOH

c NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

n(NaOH) = cV n = 0.0207 × 0.020 = 4.14 × 10–4 _{mol NaOH} volume of HCl used = 20.26 – 1.83 = 18.43 mL = 0.01843 L

nHCl = nNaOH from equation n(HCl) = 4.14 × 10–4 _moles nHCl = cV 4.14 × 10–4 _{= c × 0.01843} c = 2.24 × 10–2 _{HCl mol L}–1 Review exercise 10.3 1 a i NaF(aq)

ii basic

iii phenolphthalein

b i KNO3(aq)

ii neutral

iii methyl orange/phenolphthalein

c i NaCl(aq)

ii acidic

iii methyl orange

2 i a Phenolphthalein; as a strong acid/strong base titration and endpoint will be around 7.

b Colourless in acid changing to pink as base added, with pale pink at equivalence point.

ii a Methyl orange; as strong acid/weak base titration and endpoint will be less than pH 7.

b Yellow in base and changing to orange at equivalence point, and to red as point is passed.

iii a Bromothymol blue; weak acid/strong base titration and endpoint will be greater than pH 7.

b Yellow in acid changing to blue as base added, with pale yellow at equivalence point.

3 This is a weak acid/strong base titration with an equivalence point above pH 7. If methyl orange was used as an indicator, a colour change would be observed at around pH 3.1–4.4 and an equivalence point assumed, so the volume of hydroxide recorded would be too low.

b Running this weak base into the strong acid.

c Running this strong base into the strong acid.

Review exercise 10.4 1 a n = 71 . 22 V = 71 . 22 5 . 2 = 0.11 mol H2 b n = M m 0.11 = 016 . 2 m = 0.22 g hydrogen gas 2 a n = M m = 32 0 . 48 = 1.5 mol

n = 71 . 22 V 1.5 = 71 . 22 V v = 34.07 L at 0°C and 100 kPa b n = 79 . 24 V 1.5 = 79 . 24 V v = 37.19 L at 25°C and 100 kPa 3 Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) 1 2 1 2 a nH2 79 . 24 V n = 79 . 24 5 . 2 = 0.1 moles H2 From equation nZn = 0.1 (nH2) = 0.1 moles nZn = M m 0.1 = 4 . 65 m = 6.54 g Zn required b From equation nHCl = 2 × nH2 = 2 × 0.1 = 0.2 moles nHCl = cV 0.2 = 1.5 × V V = 0.13 L HCl required