Solutions Manual for chemist 2

5/22/2018 Solutions Manual for chemist 2

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SECOND EDITION

HSC

Contexts

Solutions Manual


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Solutions Manual: Module 1

Production of materials

Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2006.This page from Chemistry Contexts 2 Teachers Resource second editionmay be reproduced for classroom use.

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+ Cl2C3H6Cl2 1,2-dichloropropane

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

practice only 2-bromoheptane forms

(Markovnikovs rule).

c C6H12+ H2O C6H13OH 3-hexanol

4 a Addition of hydrogen chloride to 1-butene


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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 800900oC) 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

(Markovnikovs 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 Br2across the double bond and

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

cyclohexane.

9 Investigation


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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

b i


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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

b formation of polystyrene


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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 greaterhardness 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 durablef 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 afeedstock. 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 -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

the formation of cellulose is an example of condensation polymerisation.


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5 Cellulose is made from repeating units of -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 amylaseand amylopectin) is made from long-chain repeating units of -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 tomake 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

bacteriums cell walls and separating this from the cell debris.b The bacterium used isAlcaligenes eutrophus.

c The plastic produced is of a type known as PHAspolyhydroxyalkanoateswhich

have similar properties to polypropene.


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.


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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-propanolb 4-chloro-2-pentanol


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c 1,4-butanediol

d 2-methyl-2-propanol

2 a

b

c

3

2-methyl-propanol 83C


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2-butanol 100C

2-methyl-1-propanol 108C

1-butanol 118C

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-tastingchemicals. 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.


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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 = 5C

Specific heat capacity water = 4.18 J K1g1

Specific heat capacity copper = 0.387 J K1g1

Heat released per 1.10 g methanol = (mCT(copper) + mCT(water))

= (200 0.387 5 + 100 4.18 5)

= 2477 J released by 1.10 g methanol

Heat released per mole of methanol = 2477 10.1

32

= 72 058 J mol1

Molar heat of combustion = 72 kJ mol1

Review exercise 3.3

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

ethylene.

2

3 a carbohydrates, e.g. glucose, sucrose, starch, plus water and yeast


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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 1215%.

4 Fermentation of glucose occurs best in the absence of oxygen (anaerobic conditions) and

at a temperature of 3540oC.

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


1 a

b


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c

2 a 1-propanol

b 1,4-butanediol

31-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 Investigation8 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 = 18.4C

Specific heat water = 4.18 J K1g1

Methanol

Heat released by 2 g = mCT


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= (500 4.18 18.4)

= 38 456 J

i Heat of combustion/g = 38 456 2

= 19 228 J g1= 19.2 k J g1

ii Heat of combustion/mole =38 456

32

2

= 61 5296 J mol1

1-propanol

= 615 kJ mol1


= (500 4.18 28.5)

= 59 565 J


= 29 782.5 J g1= 29.8 kJ g1

ii Heat of combustion/mole =

= 1 789 928 J mol1

= 1790 kJ mol1

1-butanol


= (500 4.18 31.5)

= 65 835 J


= 32 917.5 J

= 32.92 kJ g1

ii Heat of combustion/mole =

= 2 439 845.1 J

= 2440 kJ mol1


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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.


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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: Cu

2+

(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; Cu2+(aq) + 2eCu(s); copper would coat the tin

c Sn(s) Sn2+(aq) + 2e; Ag+(aq) + eAg(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


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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 Brand HBr both have bromine in 1 oxidation state.

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

Review exercise 4.3

1 The salt bridge allows ions to move between each half-cell.


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2 a

b See diagram in part a.

c anode: oxidation Mg(s) Mg2+

(aq) + 2e

cathode: reduction Cu2+(aq) + 2eCu(s)

3 a 3Pb2+(aq) + 6e3Pb(s) cathode

2Cr(s) 2Cr3+(aq) + 6e anode

b Cu(s) Cu2+(aq) + 2e anode

2Ag+(aq) + 2e2Ag(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, Cr2O7

2

b Mg, Zn, H2, H2S

3 a Cl2

b Sn4+

c Au3+


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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) + 2SO4

2(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) + SO4

2(aq) + 2ePbSO4(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) + 5eMn2+(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) + 2e2Cl(aq) + S(s) + 2H+(aq) + 2e

Cl2(g) + 2e 2Cl(aq) E= +1.36 V

H2S(aq) S(s) + 2H+(aq) + 2e E= 0.14 V


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ii cell potential = +1.36 0.14

= +1.22 V


c i 2Cu+(aq) + eCu2+(aq) + Cu(s) + e

Cu+(aq) Cu2+(aq) + e E= 0.16 V

Cu+(aq) + eCu(s) E= +0.52 V

ii cell potential = 0.16 + 0.52

= +0.36 V


4 a H2O2(aq) + 2H+(aq) + 2e2H2O(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 H2O2and H2C2O4at 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.


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2

3 a If the Zn and MnO2were 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) + SO4

2(aq) + 2ePbSO4(s) + 2H2O(l)

(+4) (+2) reduction

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

7 Leadacid batteries are very heavy and weight is an important issue for solar-powered

cars in the World Solar Challenge.


The following reactions will occur:

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

Zn(s) + NiSO4(aq) ZnSO

4(aq) + Ni(s)

2 a i Yes. Fe2+ions will not oxidise Ni.


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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 galliums 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; O2reduced

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; O2reduced

6 a All theEvalues would be 2.93 V higher.

b Calculated e.m.f. values for cells are based on the difference inEbetween two

half-cell reactions. These would be unchanged.


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7 a i

ii Cu2+(aq) + 2eCu(s)E= +0.16 V

Cr(s) Cr3+(aq) + 3eE= +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) + 2e2Cl(aq)

ii Cl2(g) + 2e2Cl(aq)E= +1.36 V

Zn(s) Zn2+(aq) + 2eE= +0.76 V

Zn(s) + Cl2(g) Zn2+(aq) + 2Cl(aq)

iii e.m.f = 1.36 + 0.76

= 2.12 V


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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) + 2eCu(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 O2would 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+ Fe

3++ e E = 0.77 V

MnO4+ 8H++ 5e Mn

2++ 4H2O E = +1.51 V

E(cell) = +0.74 V

ii 2Cl Cl2+ 2e E = 1.36 V

MnO4+ 8H++ 5e Mn

2++ 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 2Cr

3++ 7H2O E = +1.33 V

E(cell) = 0.03 V

Thus a redox reaction should occur in i, iiand iii.

b ivabove shows that Cr2O72will not oxidise the Clion of HCl, but iishows that

MnO4

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

permanganate titrations.


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13 a An anodeis the electrode at which oxidation takes place. A cathodeis the electrode

at which reduction takes place.

b Asalt bridgeallows a flow of ions between two half-cells. An external circuit

allows a flow of electrons from anode to cathode.

c Asecondary cellcan be recharged, whereas aprimary cellcannot.

14 a i anode: Pb(s) + SO42(aq) PbSO4(s) + 2e

cathode: PbO2(s) + 4H+(aq) + SO4

2(aq) + 2ePbSO4(s) + 2H2O(l)

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

cathode: PbSO4(s) + 2ePb(s) + SO4

2(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) + 4e4OH(aq)

b Al(s) Al3+(aq) + 3e E= +1.66 V

O2(g) + 2H2O(l) + 4e4OH(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 O2from the air as a cathode.1618 Investigation

CHAPTER 5: NUCLEAR CHEMISTRY

Review exercise 5.1

1 a Atoms are not conserved in nuclear reactions.

b Chemical reactionsinvolve changes in electrons between atoms.Nuclear reactions

involve the particles in the nucleus.

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


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d Far more energy is obtained from nuclear reactions.

2 a i 17O : 8 protons; 9 neutrons

ii52Fe : 26 protons; 26 neutrons

iii137I : 53 protons; 84 neutrons

iv258Md : 101 protons; 157 neutrons

b i

ii Nb

93

41 iii Pb

208

82

3 a large mass; very small mass; no mass

b 2+ charge; 1 charge; no charge

c least penetrating, about 5 cm through air

about 1 m through air, stopped by aluminium

can pass through several cm of lead

4 a He4

2

b e01

c n1

0

d p11

e e01

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 andelectrons. 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 H31 He3

2 + e0

1

b


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c

2 a

b Pu23994 U235

92 + He4

2

c U23492 Th230

90 + He4

2

3 a N127 C12

6 + e0

1

b B85 Be8

4 + e0

1

c

4 a Be7

4 + e0

1 Li7

3

b Np23293 + e0

1 U232

92

c

5 a beta emission

b positron emission

c alpha emission

d positron emission

6 a positron or electron capture

b decay

c 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


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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 N157 + p11 C

126 + He

42

b Fe5626 + H21 He42 + Mn5425

c Li63 + n10 H

31 + He

42

d Co5927 + H21 Co

6027 + p

11

e Cr5324 + He42 Mn

5725 + n

01

4 The nuclear transformations in a, b, dand ewould all take place in a cyclotron, as the

charged particles are used to bombard the nucleus.

Nuclear transformation cwould take place in a nuclear reactor, as it involves

bombardment by neutrons.

5 Fe5626 + H21 Mn

5425 + He

42


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.


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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 decay

b decay, positron emission, electron capture

5 a Pu24194 U23792 + He

42

b Rn21086 Fr

21087 + e

01

c V4823 Ti4822 + e

01

d Cd10748 + e

0

1

Ag10747

e U23492 Th23090 + He

42

f Na2611 Mg

2612 + e

01

g N127 C126 + e

01

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 -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; decay

b unstable; positron emission or electron capture

c unstable; positron emission or electron capture

d unstable; decay

e stable


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f unstable; decay

9 a 3 half-lives mg1.02

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 Co5927 + n1

0 Co60

27

b60Co 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 Cu6329 + H2

1 Zn65

30 +

b N147 + He4

2 O17

8 + H1

1

c Ca4020 + H2

1 K38

19 + He4

2

d Pu23994 + He4

2 Cm242

96 + n1

0

17 Investigation

18 particles can cause serious damage to cells.

19 Investigation

20 Investigation

21 Research based on information gathered in Questions 19 and 20


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Module 1

REVIEW

1 D

2 A

3 D

4 D

5 D

6 A

7 D

8 B

9 B

10 A

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).


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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 (ZieglerNatta 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 -glucose

c Cellulose is a formed as a condensation polymer when -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 3540oC

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.

Ethanol is a renewable fuel and petroleum is not.


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18 a

b Cell potential = +0.76 0.13 = 0.63 V

c Zinc

19 Uranium-238 is bombarded with neutrons to form einsteinium.U23892 + n15

1

0 Es253

99 + e70

1

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.


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The acidic environment


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) Zn

2+(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; Br2O7covalent

b K2O basic; Ga2O amphoteric; Br2O7acidic

2 CO2(g) + H2O(l) H2CO3(aq)

3 Natural sources of sulfur dioxide:

bacteria decomposing organic matter to make H2S, which is then oxidisedatmospherically.

volcanic gases

bushfire smoke.


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Sources due to human activities:

burning of fossil fuels

smelting of sulfide ores such as CuFeS2and ZnS.

4 SO2H2SO3; SO3H2SO4; NO2HNO3; CO2H2CO3

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 carbonemissions 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


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)

b Fe2O3(s) + 6HCl(aq) 2FeCl3(aq) + 3H2O(l)


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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)3reacts 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 Investigation8 A useful indicator is one whose colour change occurs over a selected pH range. Methyl

orange changes colour too far outside the narrow 7.27.6 range for swimming pools.

Litmus changes colour over this range, but the colour difference in the 7.27.6 range

may be too subtle to detect easily. Universal indicator undergoes many colour changes

and the change over the 7.27.6 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 sampleof 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


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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 formammonium 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

H2remains, 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 L1

b i and ii

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


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4 In an open container, the CO2would 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, H2will 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, H2will decrease.

No change to [C(s)].

ii Will increase rate of reverse reaction. [CO], [H2]and [H2O]untilequilibrium 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


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

decreased.


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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.

c [O3]; [O2]; [NO2]


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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 willincrease 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)


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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 drinksto go flat, as follows:

CO2(aq) + 19kJ CO2(g)

11 Investigation

CHAPTER 8: THE HISTORICAL DEVELOPMENT OF IDEAS OF ACIDS ANDBASES

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

1 a A BL acid is a proton donor (or has a proton pulled from it).


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b A BL base is a proton acceptor (or takes a proton).

2 a HCl, HCO3, NH4

+, HClO4, HSO4

b F, SO4

2, 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) PO4

3(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 BL acid has lost its proton

b A conjugate acid is the species that results when a BL base accepts a proton

6 a HCO3/SO4

2: HF/F

b HSO4/SO4

2: 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 NaOHii NaOH


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iii NaOH

b i Ba(OH)2

ii Ba(OH)2

iii Ba(OH)2

3 a 5 mol L1HCl

b 0.1 mol L1NaOH

c 5 mol L1NH3

d 0.1 mol L1CH3COOH

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 + H2OHClO + 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 ClOion with water

d basic due to reaction of PO43ion with water


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e acidic due to reaction of Al3+ion (actually will be aquated aluminium ion

[Al(H2O)6]3+with water)

f acidic due to reaction of H2PO4ion with water

g neutral

h basic due to reaction of CO32ion with water


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 anArrhenius 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) SO4

2(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) CO3

2(aq) + H3O+(aq)

As base: HCO3(aq) + H2O(l) H2CO3(aq) + OH

(aq)


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b The extent to which the HCO3ion reacts as a base with water exceeds the degree

to which it reacts with water as an acid. (K value is greater for HCO3as a base.)

Hence more OHions 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

b diluted solution of a strong acid


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c concentrated solution of a weak acid

d diluted solution of a weak acid

10 a CH3COO(aq) + H

2O(l) CH

3COOH(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) HPO4

2(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 OHions

in waterH++ OHH2O

BrnstedLowry proton donor proton acceptor NH3(aq) + H2ONH4+

+ OHproton transfer


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13 The reaction: HCl(g) + NH3(g)NH4Cl(s) can be viewed as a BrnstedLowry 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 acidbase reaction. When nitric acid is mixed with sulfuric acid, for the

manufacture of nitroglycerine, the reaction that occurs is

H2SO4+ HNO3HSO4+ H2NO3

+

This is a BL acidbase reaction because a proton is transferred but it is not an

Arrhenius acidbase reaction.

14 Investigation

CHAPTER 9: HYDROGEN ION CONCENTRATION AND THE pH SCALE

Review exercise 9.1

1 a KW = [H+] [OH]

1.0 1014 = [H+] 1.6 10

7

[H+] = 6.2 108mol L1

b [H+] is less than 1 107mol L1pool is slightly basic

2 a [H+] = 5.0 103mol L1

[NO3] = 5.0 10

3mol L1

[OH] = 1 10145.0 10

3

= 2 1012mol L1

b [H+] = 1.5 mol L1

[Cl] = 1.5 mol L1

[OH] = 1 10141.5

= 6.7 1015mol L1

c [K+] = 0.25 mol L1

[OH] = 0.25 mol L1

[H+] = 4 1014mol L1

d [Ba2+] = 6.0 102mol L1

[OH] = 0.12 mol L1

[H+] = 8.3 1014mol L1


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Review exercise 9.2

1 a pH = log[H+]

= log[5.0 101]

= 0.30

b [H+] = 1 10140.0065

= 1.54 1012mol L1

pH =log[1.54

10

12

]= 11.81

c pH = log[3.6 103]

= 2.44

d [H+] = 1 1014(6.5 10

42)

= 7.7 1012mol L1

pH = log[7.7 1012]

= 11.1

2 a pH = 3.50; [H+] = 3 104mol L1; [OH] = 3.16 10

11mol L1

b pH = 11.90; [H+] = 1.3 10

12mol L1; [OH] = 7.9 103mol L1

c pH = 0.80; [H+] = 0.16 mol L1; [OH] = 6.3 10

14mol L1

3 pH = 7 so [H+] = 1 10

7

pH = 5 so [H+] = 1 105

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 L1gives 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 conjugatebase (some are a mixture of weak bases and their conjugate acids).


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2 a Buffered; contains approximately equal amounts of the weak acid HSO4and its

conjugate base SO42.

b Not buffered; HNO3is 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 OHconcentration.

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 bloods 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.


1 a [H+] = 0.5 mol L1

[OH] = 1 10140.5

= 2 1014mol L1

pH = log[0.5]

= 0.30

b [OH] = 3.0 1032

= 6.0 103mol L1

[H+] = 1 10140.006

= 1.67 1012mol L1

pH = log[1.67 1012]

= 11.8

2 a [H+

] = 1.0 103

mol L1


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[OH] = 1 10140.001

= 1 1011mol L1

b [H+] = 3.9 10

3mol L1

[OH] = 2.51 1012mol L1

c [H+] = 3.2 109mol L1

[OH] = 3.2 106mol L1

d [H

+

] = 7.9 1013

mol L1

[OH] = 1.3 10

2mol L1

3 a i [H+] = 0.20 mol L1

[NO3] = 0.20 mol L1

[HNO3] = 0 mol L1

ii pH = log10[0.20]

= 0.70

b As HNO2is a weak acid, it will not dissociate completely. Therefore [H+] will be

less than 0.20 mol L1and 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)

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


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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 acidbase interaction. In BrnstedLowry 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 reactionproduces 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 exothermici.e. it releases a lot of heat.


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 = c0.005

c = 4.50 mol L1

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

Calculated moles of HCl

n = cV

= 0.02052 0.952

= 0.019535 moles

moles of HCl = moles NaOH (from equation)


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concentration of NaOH =

V

n

= 0.019535 / 0.005

c = 3.91 mol L1NaOH

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 thesolution 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 Na2CO3is 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 103moles

n = M

m

7.5 103 =

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 thecalibrated mark

5 a n =M

m


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=07.126

3162.0

= 0.0025 moles

n = cV

0.0025 = c0.25

c = 0.010 mol L1H2C2O4.2H2O

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

n(H2C2O4) = cV

n = 0.010 0.02061

= 2.068 104

moles H2C2O4

n(OH) = 2 nH2C2O4(from equation)

n(OH) = 4.136 10

4mol

n(NaOH) = cV

4.136 104 = c0.020

c = 0.0207 mol L1NaOH

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

n(NaOH) = cV

n = 0.0207 0.020

= 4.14 104mol NaOH

volume of HCl used = 20.26 1.83

= 18.43 mL

= 0.01843 LnHCl = nNaOH from equation

n(HCl) = 4.14 104moles

nHCl = cV

4.14 104 = c0.01843

c = 2.24 102HCl mol L1


1 a i NaF(aq)


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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 lessthan pH 7.

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

iii a Bromothymol blue; weak acid/strong base titration and endpoint will begreater 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 aroundpH 3.14.4 and an equivalence point assumed, so the volume of hydroxide recorded

would be too low.

4 a Running this strong base into the weak acid.


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b Running this weak base into the strong acid.

c Running this strong base into the strong acid.


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


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n =71.22

V

1.5 =71.22

V

v = 34.07 L at 0C and 100 kPa

b n =79.24

V

1.5 =79.24

V

v = 37.19 L at 25C and 100 kPa

3 Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

1 2 1 2

a nH279.24

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


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4 CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + CO2(g) + H2O(l)

1 2 1 1 1

a nCaCO3 =M

m=

1.100

10

n = 0.10 moles

From equation nCaCO3 = nCO2

nCO2 = 0.10 moles

nCO2 =71.22

V

0.1 =71.22

V

V = 2.271 L CO2at 0C and 100 kPa

b nCO2 =79.24

V

0.1 = 79.24

V

V = 2.48 L CO2at 25C and 100 kPa

c From equation nHNO3 = 2 nCaCO3

nHNO3 = 2 0.1

= 0.2 moles

nHNO3 = cV

0.2 = 1.00 V

V = 0.2 L HNO3required

d From equation nCaCO3 = nCa(NO3)2

nCa(NO3)2 = 0.10 moles

nCa(NO3)2 =M

m

0.1 =1.164

m

m = 16.41 g of Ca(NO3)2produced


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1 a 1-propanol

b 1-butanol

c 2-butanol

2 a

b

c

3 a pentanoic acid

b 5-methylhexanoic acid

4 a


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b

5 They form more H-bonds because they have a C=O group as well as an OH group,

making two H-bonds possible between two molecules.6 a methyl butanoate

b ethyl methanoate

7 a

b

8 a

b


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c

d

e

f

9 a C3H7OH + CH3COOHconc. H2SO4 CH3COOC3H7+ H2O

The ester is propyl ethanoate.

b C5H11OH + C3H7COOH conc. H2SO4 C3H7COOC5H11+ H2O

The ester is pentyl butanoate.

10 a H2SO4acts as a catalyst.

b Refluxing prevents volatile reactants from escaping before equilibrium is reachedand allows the reaction to be carried out at temperatures lower than would

normally be possible.


1 a 2HCl(aq) + CuO(s) CuCl2(aq) + H2O(l)


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b 3H2SO4(aq) + 2Fe(OH)3(s) Fe2(SO4)3(aq) + 6H2O(l)

c 2HNO3(aq) + CaCO3(s) Ca(NO3)2(aq) + CO2(g) + H2O(l)

d HCl(aq) + NaHCO3(aq) NaCl(aq) + H2O(l) + CO2(g)

2 a i Na+(aq), SO4

2(aq), OH

(aq), H

+(aq), H2O(l)

ii H+(aq), Na

+(aq), OH

(aq) all at 2 mol L

1

b i H+(aq), SO4

2(aq), H2O(l)

ii H+(aq)

c i Na+(aq), SO42(aq), H2O(l)

ii Na+(aq), SO42(aq)

3 a i Pipette: used to place an accurate volume of solution into a conical flask.

ii Burette: used to deliver variable volumes into a conical flask.

iii Conical flasks: used to mix two solutions.

b i citric acid

ii NaOH

iii deionised water

4 Sodium hydroxide cannot be made as a primary standard and hence must bestandardised by titration with an acid of accurately known concentration.

5 a A good primary standard has a relatively high molecular weight to minimise

weighing errors, does not react with any component of the atmosphere such as

oxygen, moisture or carbon dioxide, and is available in a very pure form.

b Investigation

6 anNaHCO3=

cV = 2 1

= 2 moles

n =M

m

2 =008.84

m

m = 168 g

b nNa2CO3= cV

n = 0.05 0.5


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= 0.025 moles

n =M

m

0.025 =

m = 2.65 g

c nBa(OH)2.8H2O = cV

n = 2 1032

= 0.004

n =M

m

m = 0.004 315.444

= 1.26 g

d nH2C2O4.2H2O = cV

n= 0.150

0.25

= 0.0375 moles

n =M

m

m = 0.0375 126.068

= 4.728 g

7 a NaOH(aq) + HNO3(aq) NaNO3(aq) + H2O(l)

nHNO3 = cV

= 0.0961 0.02

= 1.9 103

From equation nNaOH = nHNO3

= 1.9 103

nNaOH = cV

1.9 103

= 0.118 V

V = 1.62 102L


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b 2NaOH(aq) + H2SO3(aq) Na2SO3(aq) + 2H2O(l)

nH2SO3 = cV

= 0.0531 0.02

= 1.062 103

moles

nNaOH = 2 nH2SO3from equation

2 1.062 103 = 0.118 V

V = 18 103L

c 2NaOH(aq) + H2C2O4.2H2O(s) Na2C2O4(aq) + H2O(l)

nH2C2O4.2H2O=

M

m

=126

134.0

= 1 103moles

2 n(oxalic acid) = n(NaOH)

n = cV

2 103 = 0.118 V

V = 16.95 103

L

8 a H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

nNaOH = cV

= 0.5 0.02

= 0.01 moles

nH2SO4 = 0.5nNaOH

= 0.005 moles H2SO4

n = cV

V =202.0

005.0

= 2.48 102L

b H2SO4(aq) + Ba(OH)2BaSO4(aq) + 2H2O(l)

nBa(OH)2 = 0.02 0.165

= 0.0033 moles


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nBa(OH)2 = nH2SO4

nH2SO4 = 0.0033 = 0.202 V

V = 1.63 102L

c Na2CO3(s) + H2SO4(aq) Na2SO4(aq) + CO2(g) + H2O(l)

nNa2CO3=

M

m

= 106

343.0

= 3.24 103

moles

nNa2CO3 = nH2SO4

nH2SO4: 3.24 103

= 0.202 V

V = 1.60 102

L

d total nOH= nOH

in Ca(OH)2+ nOHin NaOH

nCa(OH)2 =

M

m

=1.74

0746.0

= 1 103moles

moles OH = 1 10

32

= 2 103

moles

nNaOH =M

m

=40

247.0

= 6.2 103moles OH

Total nOH = 2 10

3+ 6.2 103

= 8.18 103moles

nH2SO4 = nOH0.5

= 4.08 103moles

n = cV


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= 4.08 103= 0.202 V

VH2SO4 = 2.02 102L

9 a HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

nHCl = cV

= 0.121 0.02565

= 3.1 103moles

nHCl =

nNaOH

nNaOH = cV

3.1 103 = c 0.020

c = 0.16 mol L1

b Ba(OH)2(aq) + 2HCl(aq) BaCl2(aq) + 2H2O(l)

nHCl = cV

= 0.0503 0.01523

= 7.66 104moles

nBa(OH)2 = 0.5nHCl

= 0.5 7.66 104

= 3.83 104

moles

Ba(OH)2 = 3.83 104

0.02

= 1.92 102

mol L1

10 a H2SO4(aq) + CuO(s) CuSO4(aq) + H2O(l)

nCuO =M

m=

5.79

05.1

= 0.013 moles

nCuO = nH2SO4

nH2SO4 = 0.13 moles = 2.0 V

V = 6.6 103L

b 1 mole H2SO41 mole CuSO4.5H2O

nCuSO4.5H2O = 0.013 moles

5/22/2018 Solutions Manu

Solutions Manual for chemist 2

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