Department of Molecular Sciences

HSC Enrichment Program 13th July 2018

Welcome to ...

Macquarie University’s HSC Enrichment Program 2018

CHEMISTRY 13 July 2018

8.30 am – 9.00 am Registration

9.00 am – 9.10 am Welcome A/Prof Joanne Jamie

9.10 am – 11.20 am The Production of Materials Ethene, polymers and ethanol

A/Prof Joanne Jamie Electrochemistry

Dr Damian Moran

11.20 am – 11.40 am Morning Tea

11.45 am – 1.15 pm The Acidic Environment Ms Artchaki Chandrasegar and Mr Alexander

Comerford

1.15 pm – 2.00 pm Lunch

2.00 pm – 4.10 pm Chemical Monitoring and Management Monitoring and Management in the

Chemical Industry Mr Shane Sandot

Chemistry and the Atmosphere and Monitoring Water Quality Mr Alexander Comerford

4.30 pm – 5.30 pm Top Marks Education Study Skills Workshop

Mr Tim Mason

Contents

The Production of Materials Ethene, Polymers and Ethanol Electrochemistry The Acidic Environment Chemical Monitoring and Management Monitoring and Management in the Chemical Industry Chemistry and the Atmosphere Monitoring Water Quality Appendix Nuclear Chemistry Top Marks Education Study Skills Workshop

Ethene, Polymers and Ethanol For access to past exams, answers and many helpful hints see http://www.boardofstudies.nsw.edu.au/. Ethene

Ethene (also known as ethylene) is one of the simplest organic molecules and belongs to the family of compounds known as alkenes. These are unsaturated compounds, meaning that they do not contain the maximum number of hydrogen atoms per molecule. Ethene contains just two carbon atoms, connected by a double bond.

C CH

H

H

HH2C CH2 CH2CH2 C2H4

Ethene is an extremely valuable ‘building’ block in chemistry, but does not occur naturally in great amounts. It is used in the production of polymers, solvents and detergents. It is a colourless, sweet-smelling and flammable gas. It can be produced by treating ethanol with concentrated sulfuric acid. This results in the loss of water and is therefore known as a dehydration reaction.

C CH

H

H

HC CH

H

OH

HH H

Conc. H2SO4

(Cat.)

Ethanol Ethene

H2O

Industrially, ethene is produced from fractions obtained by the distillation of crude oil/petroleum. The various fractions contain long-chained alkanes (saturated hydrocarbons) and alkenes, which are broken down into small molecules in a process known as ‘cracking’. Cracking can occur by different means, and the two major methods used are catalytic cracking and steam (or thermal) cracking. Catalytic cracking takes place at a lower temperature than steam cracking (500 C versus 700-900 C necessary for steam cracking) with a considerable saving in energy. The catalyst used is known as a zeolite, which is an inorganic compound composed mainly of aluminium, silicon and oxygen, with some metal ions such as Na+ (sodium ions). The zeolites contain cavities and tunnels through their interiors, providing a large surface area for the fragmentation of the long-chained alkanes and alkenes. By-products of the breakdown of the long-chained hydrocarbons are propene (a three-carbon alkene, H2C=CH-CH3) and ethene, and these are recovered and used as starting materials in production of other compounds.

Steam cracking (also known as thermal cracking) is not catalytic, and in this process a mixture of the long-chained alkanes and alkenes is passed through very hot metal tubes (heated at 750–900 C with steam). As mentioned previously, ethene is the simplest alkene. Shown below are the structures and IUPAC names of some other simple alkenes. It should be noted that 1-butene (but-1-ene), 2-butene (but-2-ene) and 2-methylpropene are all isomers of one another (they contain the same number and types of atoms, but they are connected differently to each other), as can be seen by looking at their molecular formula, which is C4H8. Recall also that the general formula for an alkene is CnH2n (where ‘n’ is the number of carbon atoms). The corresponding alkanes (CnH2n+2) are named as ethane, propane, butane, pentane and 2-methylpropane.

Naming of alkanes, alkenes (& other simple organic compounds), isomers, production of ethene by dehydration & catalytic cracking to produce ethene and other low molecular weight hydrocarbons are often examined (e.g. see 2016 Q11; 2014 Q4; 2013 Q18; 2011 Q1 and 14; 2008 Q16). The double bond of alkenes is a region of high electron density, and this reacts with numerous small molecules in a process that ultimately leads to an ‘opening out’ of the double bond to form two single bonds, as illustrated below.

X YC C

H

H

H

H

C C

X

H

Y

H

H H

addition product

This general type of reaction is known as an addition reaction, because the small molecule X-Y adds to the alkene. Several examples of addition reactions are given below. Whilst ethene is the alkene in each case, it should be noted that all alkenes undergo these reactions. You should be able to write balanced equations for addition

reactions, name the starting materials and products and be able to recognise the reagents used when the starting alkene and addition product are provided. Table 1. Some Addition reactions with Ethene

Representative Alkene

X-Y Addition Product

Reaction Type

C C

H

H H

H

H H (H2)

(with Ni catalyst) C C

H H

HH

H H

Hydrogenation

C C

H

H H

H

H Cl (HCl)

C C

H H

ClH

H H

Hydrogen halide

addition (HCl, HBr or HI)

C C

H

H H

H

Br Br (Br2) C C

H H

BrBr

H H

Halogenation

(Bromination or chlorination with

Cl2)

C C

H

H H

H

H OH (H2O)

(with H2SO4 catalyst) C C

H H

OHH

H H

Hydration

(H3PO4 also commonly used as

catalyst)

C C

H

H H

H

(HOBr, from H2O and Br2)HO Br C C

H H

BrOH

H H

Halohydrin formation

(Bromohydrin, with Cl2 and H2O

known as chlorohydrin)

Production of ethanol (for its use as a solvent, amongst other things) from ethene is an industrially important process. It is often conducted under high pressure at 300 oC. While H2SO4 can be used, as noted above, H3PO4 is more commonly used industrially. Alkanes, due to the lack of the double bond, DO NOT undergo addition reactions. This allows alkenes and alkanes to be readily distinguished from each other. For example, the addition of the orange-red solution of Br2 in a solvent (e.g. H2O or inert solvent such as CHCl3) leads to almost instantaneous decolourisation of the Br2. Distinguishing between alkanes and alkenes and addition reactions with alkenes are often examined (see e.g. 2017 Q7; 2016 Q15; 2014 Q5; 2013 Q18 and 36; 2011 Q11; 2010 Q1 and Q24; 2008 Q16).

Given an addition product of any alkene, you should be able to identify the starting alkene. Practice by answering the questions below.

(a) Draw and name the starting alkene and name the product indicated in each case:

C C CH2CH3

CH3 CH3

H2O and H2SO4 (Cat)

CH3

Br OHBr2 and H2O

CHCH3 CH3

OH

(b) Name the starting alkene and draw and name the expected products from treatment of the alkene with the reagent indicated:

CH CH CH3CH3

H2 and Ni

C C CH3CH3

CH3 CH3

C CH2CH3

CH3

Br2

HCl

Polymers

Apart from undergoing the typical addition reactions of alkenes as shown above, ethene also undergoes some other industrially important reactions. The most important of these is its conversion to polyethylene (named systematically as polyethene) in a process known as polymerisation. In this process, the small building-blocks (monomers) combine together to form a large product molecule, the polymer, which sometimes is also called a macromolecule. When ethene molecules combine together to form polyethylene, no atoms are lost. This type of polymer is therefore known as an addition polymer, because the small ethene monomers simply add together to give the polyethylene polymer.

CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2

CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 continuescontinues

The polymer of repeating CH2 units can be written in abbreviated form as:

CH2 CH2n

This indicates that the starting material was a C2 unit, rather than a C1 unit, which may seem to be the case upon examination of the polymer alone, without an understanding of the chemistry. The addition polymerisation of ethene and other alkenes is initiated by a catalyst/initiator (shown as I below). The initiator is often an organic peroxide and the reaction proceeds through a radical process (see below for polyethylene formation). The Initiator (I) (organic peroxide) attacks the alkene, breaking the double bond, and producing an unpaired reactive electron (radical) on the end of the growing chain (initiation).

This will attack another alkene, increasing the length of the growing chain (propagation). Chain length continues to grow in this fashion until two growing chains combine (termination).

Table 2. Common Addition Polymers, their Starting Monomers and their Uses Polymer Polymer Structure Monomer

(Building Block) Examples of Use

Polyethylene

(LDPE and HDPE)

CH2 CH2n

C CH

H

H

H

ethene

LDPE: wrapping materials, carry bags, lining milk cartons, squeeze bottles HDPE: bowls, kitchen utensils, buckets, milk crates, freezer bags

Polypropylene (Polypropene)

CH2 CH

nCH3

H2C CH CH3

propene

Moulded chairs, rope and twine, carpets, car bumpers

Polyvinyl chloride

[Poly(chloroethene)]

CH2 CHnCl

C CH

H

Cl

H

vinyl chloride

Electrical insulation, drainage pipes, guttering, garden hoses

Polyacrylonitrile

[Poly(cyanoethene)]

CH2 CH

nC N

H2C CH CN

acrylonitrile

Clothing, furnishings, carpet

Polystyrene

[Poly(ethylbenzene)]

CH2 CHn

CH CH2

styrene(ethenylbenzene)

Foam (drink cups, and packaging), tool handles, containers, insulation

Polytetrafluoroethylene

PTFE, or Teflon [Poly(tetrafluoroethene)]

CF2 CF2n

C CF

F

F

F

tetrafluoroethylene(tetrafluoroethene)

Non-stick cookware surfaces, electrical insulation, pipe thread sealant

Note the general pattern in all of the addition polymers in Table 2, and the requirement that a double bond is present in the monomeric starting material. You should be able to look at any addition polymer and recognise the starting monomer and vice-versa. Past exams have asked questions on addition polymers, including how ethene can be used to make new materials; identifying the monomer or polymer in a polymerisation reaction (when given one of the two) (e.g. 2017 Q12; 2016 Q17; 2012, Q2; 2005, Q4) and being able to name the monomer (e.g. Q6 2010). In 2009 Q19, the chemical and physical process involved in making polyethylene, polyvinyl chloride or polystyrene needed to be discussed with relevant chemical equations (no structures were given). In 2011 Q27, the properties and uses of polystyrene and a biopolymer needed to be discussed. In 2012, students needed to recognise polymerisation of ethene in the presence of a catalyst (a

peroxide) shown by using models. In 2013, students needed to show the steps for forming polyethylene, along with a biopolymer. In 2015 Q25, students needed to describe the steps involved in the process of addition polymerisation (as noted on previous page). In 2016 Q4, students needed to identify the application and reason of suitability for that application of polystyrene.

Polyethylene is typically produced from two methods: a high-pressure method, which produces branched polymers and a low-pressure method that produces linear (unbranched) polymers. The branched polymers are less dense than the linear polymers because the branches need to occupy a region of space, and hence the polymer is not able to ‘fold-up’ as efficiently as it could if there were no branches present. Hence, the polymer with branching is known as low-density polyethylene (LDPE). LDPE has a typical mp of 80 C and is quite flexible due to weaker attractive forces between polymer chains. These polymers tend to be more transparent due to the fact that there is less scattering and refraction of light as it passes through the solid. On the other hand, the main chains of linear polymers are able to pack closely together and as a result the attractive forces holding the chains together are stronger. The linear polymer derived from treating ethene under the low-pressure method (the Ziegler-Natta process) is referred to as high-density polyethylene (HDPE). HDPE appears white (in the absence of dyes) and has a typical mp of 135 C (so high density polyethylene has the higher melting point).

Melting/Softening Point: Several factors affect the melting point (mp) of polymers. A high degree of order within the polymer (minimal or no branching from the main chain) will lead to an increased mp. For a given polymer, an increase in the average chain length will increase the mp.

Chemical Stability: Teflon, PTFE, is highly resistant to chemical attack because it is composed solely of C-C and C-F bonds, having C-F bonds in place of the C-H bonds present in most other polymers, and C-F bonds are stronger than C-H bonds.

Flexibility or Rigidity: The flexibility is determined by the size and shape of the side chains on the addition monomers. As the size of the side chain increases, so too does the stiffness in the resulting polymer. Hence polypropylene is stiffer than polyethylene, whilst polyvinyl chloride is stiffer again, and polystyrene is even more inflexible.

When it comes to manufacturing a material, one very important consideration that should not be overlooked is that of cost – generally, the more elaborate the starting monomer, the more expensive the final polymer.

Past exams have asked questions relating the properties exhibited by a polymer to the structure of the polymer and ideally relating the properties to the uses of that polymer (e.g. 2016 Q4; 2015 Q25; 2011 Q27; 2007 Q26).

Condensation polymers are another class of polymers. These are formed when the monomer units react together to eliminate (or ‘kick-out’, or ‘remove, or ‘lose’) a smaller molecule, which is often water. Before looking at some examples of condensation polymers, it is useful to consider some simple reactions. Illustrated below is a reaction between a carboxylic acid and an amine, to produce an amide.

CH3CH2CH2CH2CH2 C

O

OH

CH3CH2CH2CH2CH2 C

O

N CH2CH2CH2CH2CH2CH3

H

N CH2CH2CH2CH2CH2CH3

H

H

1-aminohexanehexanoic acid

Hexyl hexanamidenew bond

Another example is esterification, where we have a reaction between an alcohol and a carboxylic acid to produce an ester. This is often catalysed by an acid (e.g. H2SO4). Note simple esters are often sweet smelling and are used in perfumes and as flavouring agents because of this property.

Given that this is the way that a carboxylic acid and an amine/alcohol react together when they are on the different molecules, it becomes apparent that if we can make a compound in which the amine/alcohol and carboxylic acid groups are in the same molecule, we can make a polymer. Indeed, for the reaction of an amine and carboxylic acid, this is the way that nature uses amino acids to make proteins.

propyl acetate1-propanol acetic acid(propyl ethanoate)(ethanoic acid)

CH3 CH2 CH2 O H HO CO

CH3H2SO4 CH3 CH2 CH2 O C

OCH3+

conc.

N CH C

R O

OHH

H

N CH C

R O

H

HN CH C

R O

HN CH C

R O

HN CH C

R O

OHH

amino acid building blocks R = an attached group,for example, if R = CH3, the amino acid is Alanine

a polypeptide

N CH C

R O

H n

the repeat unit of a polypeptide

N CH C

R O

OHH

H

N CH C

R O

OHH

H

N CH C

R O

OHH

H

denotes a new bond

It should also be apparent that we could construct a polymer by condensing two different monomers that have two reactive groups on them. For example, a diamine (contains two amines in the same molecule) and a diacid (contains two carboxylic acid groups in the same molecule) react to produce a polyamide, as illustrated below. For each amide bond that gets made, we also make a molecule of water.

N (CH2)6 NH

H

H

HC (CH2)4 C

HO

O

OH

OC (CH2)4 C

HO

O

OH

ON (CH2)6 N

H

H

H

H

C (CH2)4 CHO

O O

N (CH2)6 N

H H

C (CH2)4 C

O O

N (CH2)6 N

H H

H

N (CH2)6 N

H H

C (CH2)4 C

O O

1,6-hexanedioic acid 1,6-diaminohexane

n

the repeat unit in nylon 6,6

a polymer

Another example of two different monomers being used to form a polymer is the reaction of a dialcohol (a diol, contains two hydroxy (OH) groups in the same molecule) with a dicarboxylic acid. A diol reacts with a diacid to form a polyester. For each ester bond that gets made, we also make one molecule of water.

O CH2 CH2 O HH C C

HO

O O

OHO CH2 CH2 O HH C C

HO

O O

OH

O CH2 CH2 OH C C

O O

O CH2 CH2 O C C

OO

OH

C C

O O

O CH2 CH2 On

the repeat unit in polyethylene tetraphthalate, PET

a polyester

1,2-ethanediolterephthalic acid

denotes a new bond

conc. H2SO4 (cat) /heat

As well as polymerisation, questions on alcohols, carboxylic acids and esters are often asked in exams, including names, structures and reactions (including set up), e.g. 2017 Q4; 2016 Q22; 2015 Q9 and Q11; 2014 Q18 and Q26; 2013 Q18 and Q21; 2012 Q21; 2011 Q8; 2010 Q22.

Cellulose is the major component of plant material, which is the major mass of living material. It is a biopolymer and is a condensation polymer from the monomer glucose. The condensation leads to loss of water between two glucose molecules.

Cellulose

Cellulose contains the basic building blocks for making starting molecules for petrochemicals, such as ethene (2 C atoms), propene (3 C atoms) and butene (4 C atoms, a starting point for synthetic rubber). Considerable scientific effort is in place to look at cellulose as an alternative source of chemicals we now obtain from oil. Industries have begun production of ethanol from cellulose. This is being regarded as a viable source of ethanol for use as a biofuel (see below for more details).

OO

OHOH

HOH2C

OOH

OHOH2C

OH

OOO

OHOH

HOH2C

n

OOH

H

CH2OH

HH

OH OH

H

H

OH

123

4

5

6

Glucose

OH

HO

H

HO H

CH2OHH

OHH

OH

Glucose in its puckered ring (chair) arrangement

12

3

45

6

Recognising the structure of cellulose, that it is a condensation polymer and loses water during the condensation, is a major component of biomass, and is of interest as a source of chemicals we now obtain from oil is a consistent theme in past exams. In 2016 Q26, students had to explain why cellulose is classified as a condensation polymer as well as discuss the need for research into biopolymers.

The take home message: Condensation polymers need two reactive groups on the same molecule, or polymer growth will end. We can have a self-polymerising unit, as in the case of amino acids, or two units that will form bonds with each other specifically, as in 1,2-ethanediol and terephthalic acid to give polyester, or a diamine and a diacid in the case of nylon synthesis.

Ethanol

As seen earlier, ethanol (via dehydration) is a potential source of ethene (an alternative to the cracking of hydrocarbons).

C CH

H

H

HC CH

H

OH

HH H

Conc. H2SO4

(Cat.)

Ethanol Ethene

H2O

Ethanol is also commonly used as a solvent because it is a very polar molecule: the C-O and O-H bonds are polar due to the O atom being more electronegative than the C and H. It can therefore hydrogen-bond with other molecules.

While ethene (from catalytic cracking) is currently the main industrial source of ethanol, fermentation is another significant source, and of increasing interest due to the diminishing oil reserves. Fermentation is a process in which glucose (typically) is broken down to ethanol and carbon dioxide by the action of enzymes present in yeast. For fermentation, a suitable grain or fruit is mashed up with water; yeast is added; air is excluded (i.e. conducted anaerobically - if the organism is provided with a high concentration of oxygen, it will produce energy via a different pathway); and the mixture is kept at about 25 - 37 oC. Enzymes in the mixture convert any starch or sucrose to glucose and/or fructose and then glucose or fructose are converted to ethanol and carbon dioxide.

C6H12O6(aq) 2CH3CH2OH(aq) + 2CO2(g)

C OH

H

HH

+

-

+

Ethanol readily burns through a combustion/oxidation reaction to give carbon dioxide, water and energy.

CH3CH2OH(l) + 3O2(g) 2CO2(g) + H2O(g) + energy H = -1360 kJ/mol heat of combustion

Ethanol is increasingly being used as a fuel due to it being a renewable resource, as opposed to petroleum products from fossil fuels. It is commonly being produced from starch or sugars in a wide variety of crops, including sugar cane and corn. It is often referred to as a biofuel. While, as a renewable resource, ethanol should be able to reduce the use of the non-renewable fossil fuels, there is considerable debate about how useful this is. Concerns relate to the large amount of land required for crops, as well as the energy and pollution balance of the whole cycle of ethanol production. The recent development of industries to produce ethanol from waste cellulose-based products are likely to make the production of ethanol much more advantageous. In Australia, methods are being developed to convert ethanol from wood, bagasse (waste from sugar production), crop stubble and municipal green waste as an economically feasible and environmentally-sustainable conversion of these products to ethanol.

Production of ethanol from hydration and fermentation (2006 Q2 and Q18; 2010 Q15; 2012 Q5 and Q26), its use as a solvent (2011 Q21; 2008 Q5) and comparison to water (2009 Q17), solubility (2015 Q9), hydrogen-bonding (2016 Q4; 2010 Q14), combustion reactions of it and other alcohols and organic compounds and fuels (2017 Q9; 2015 Q10, 20; 2014 Q19, 22 and 25; 2013 Q27; 2012 Q14; 2010 Q23; 2009 Q20; 2007 Q12, Q24; 2006 Q4; 2005 Q3; 2005 Q17), dehydration of it and other alcohols (2010 Q11), its use as a fuel (2008 Q24; 2007 Q1; 2006 Q20), including the advantages and disadvantages (2017 Q28), are all commonly assessed. In 2014 Q31 asked students to discuss the underlying chemistry of ethanol with relevant equations, and assess the impacts on society of TWO uses of ethanol. In 2017, Q25, students had to outline the steps, with equations, for conversion of cellulose to polyethylene – this combines several organic chemistry concepts including hydrolysis, fermentation, dehydration and addition polymerisation.

Electrochemistry / Oxidation & Reduction Reactions

1. Why study oxidation and reduction reactions?

• Energy storage and production • Corrosion (built and natural environment) • Materials production (e.g. electrolytic refining of metals) • Testing and analysis (e.g. coulometric Karl Fischer titration)

2. Key terminology and defintions

• Oxidation is a loss of electrons (e.g. A ⇌ A+ + e−) Reduction is a gain of electrons (i.e. B+ + e− ⇌ B)

• An oxidising agent (or oxidant) is a substance which oxidises some other substance by removing its electrons. Hence, the oxidising agent is reduced.

A reducing agent (or reductant) is a substance which reducing some other substance by giving it electrons. Hence, the reducing agent is oxidised.

• Oxidation and reduction (redox) reactions involve transfer of electrons from one substance to another and always occur in pairs. That is, a redox reaction is the reaction between an oxidising agent (or oxidant) and a reducing agent (reductant).

3. Oxidation Number Redox reactions are just one type of chemical reaction and sometimes it is not clear whether a

reaction involves oxidation and reduction. Oxidation numbers, also known as oxidation states, enable us to quickly assess whether a redox reaction has occurred.

The oxidation number of an element in a molecule or ion is the charge the atom of that element would carry if the molecule or ion were completely ionic. By definition, the sum of the oxidation numbers of all the elements in a species (molecule or polyatomic ion) must equal the net charge on the species. Useful rules:

Substance in elemental state e.g. Cu(s), Ag(s), H2(g), O2(g) 0

Monatomic ions e.g. Ca2+, Cl− charge on ion

Hydrogen in compounds with non-metals e.g. HCl, CH4 +1

Fluorine in all compounds −1

Oxygen in all compounds except peroxides (−1) and OF2 (+2) −2

As an example, the oxidation number of nitrogen in these examples is −3 (K3N), +4 (N2O4) and 0 (N2). Changes in oxidation number correspond to redox reactions, as:

- an increase in oxidation number corresponds to an oxidation reaction .....(loss of electrons)

- a decrease in oxidation number corresponds to a reduction reaction ......(gain of electrons)

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

4. Oxidation and Reduction (Redox) Reactions Half reactions (or half equations) are reactions which describe the oxidation and reduction processes separately in terms of electrons lost or gained.

Oxidation half-reaction: Cu ⇌ Cu2+ + 2e− (magnesium 0 +2) Reduction half-reaction: Ag+ + e− ⇌ Ag (chlorine 0 −1)

In combining half reactions into complete reactions it is necessary to balance the number of electrons, as no electrons are allowed to be left over.

Cu ⇌ Cu2+ + 2e− 2x (Ag+ + e− ⇌ Ag)

Cu + Ag2+ ⇌ Cu2++ 2Ag

Mg has been oxidised by Cl2 to Mg2+ • Cl2 has been reduced by Mg to Cl− • Cl2 is the oxidant / oxidising agent • Mg is the reductant / reducing agent

A good approach for challenging redox equations is to use the half-reactions shown for the standard reduction potentials included with the HSC examination paper. Find the relevant half-reactions, write one as a reduction half-reaction and the other reaction reversed, so that it is the oxidation half-reaction. Balance for the number of electrons and then combine the half reactions, cancelling the electrons. Finally, cancel substances (eg H+, H2O) appearing on both sides of the complete redox reaction.

2x (MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O) 5x (H2O2 ⇌ O2 + 2H+ + 2e−)

2MnO4− + 16H+ + 5 H2O2 ⇌ 2Mn2+ + 8H2O + 5O2 + 10H+

2x (MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O) 5x (H2O2 ⇌ O2 + 2H+ + 2e−)

2MnO4− + 6H+ + 5 H2O2 ⇌ 2Mn2+ + 8H2O + 5O2

5. Redox reactions and electricity Redox reactions will make electricity if the isolated half-reactions are connected using a metal wire for the electrons to flow between the oxidation and reduction processes. This occurs in all the batteries that we use.

Example: Cu(s) ⇌ Cu2+(aq) + 2e− (ox)

& Ag+(aq) + e− ⇌ Ag(s) (red) The set-up on the right is called a galvanic (or voltaic) cell. The device requires a voltage (potential) difference between the half-cells to generate electricity. Cell diagram: Cu | Cu2+ || Ag+ | Ag | change in phase (s / l / g); || a salt bridge Hint: always place the oxidation reaction at the left of the || symbol.

Negative electrode Positive electrode

6. Standard Electrode Potentials Voltage is a relative measurement. The standard potentials found in the HSC data sheet are measured relative to the standard hydrogen electrode (SHE), which by definition has a potential of 0.00 V under standard conditions (temperature 25°C, 1 M for solutions, 1 bar for gases, and pure solids or liquids for other substances). At the SHE, H+ cations are reduced or H2 molecules are oxidized at the Pt surface according to the following equation:

2H+ + 2e− ⇌H2(g)

Using the SHE, we can define a standard electrode potential, E⦵, for the half-reaction:

Mn+ + ne− ⇌ M

as the potential of the right-hand electrode less that of the left-hand electrode in a cell:

Pt, H2 | H+ || Mn+ | M

Under standard conditions, the potential difference (voltage) between the SHE and the M electrode is the standard electrode potential (E⦵) for the half-reaction:

Mn+ + ne− ⇌ M

Example: When measured against SHE,

Mg2+ + 2e− ⇌Mg E⦵ = –2.36 V Cl2 + 2e− ⇌ 2Cl− E⦵ = 1.36 V

Note: positive E⦵ value indicates spontaneous reaction in the direction as written; negative E⦵ value indicates spontaneous reaction in the reverse direction.

7. Cell Voltage We can use the same approach as used to define standard electrode potentials to calculate a galvanic cell voltage.

For example:

Cu2+ + 2e− ⇌ Cu(s) E⦵ = 0.34 V Ag+ + e− ⇌ Ag(s) E⦵ = 0.80 V

Reverse the reaction with the least positive E⦵ value and its E⦵ sign. Cu(s) ⇌ Cu2+ + 2e− E⦵ = −0.34 V 2x ( Ag+ + e− ⇌ Ag(s) ) E⦵ = 0.80 V Cu(s) + 2 Ag+ ⇌ Cu2+ + 2 Ag(s) E⦵

cell = 0.46 V

As in this case, sometimes we need to multiply throughout by a common factor to cancel the number of electrons. But, there is NO CHANGE in E⦵ values.

Note: positive E⦵

cell indicates spontaneous redox reaction as written; negative E⦵

cell indicates spontaneous redox reaction in reverse direction i.e. cathode and anode switch.

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

8. Lead-Acid Battery Example Consider PbO2(s) + HSO4

- (aq) + 3H+(aq) + 2e− ⇌ PbSO4(s) + 2H2O(l) E⦵ = 1.69 V (I)

PbSO4(s) + H+(aq) + 2e− ⇌ Pb(s) + HSO4- (aq) E⦵ = 0.36 V (II)

⦵ of (II) is less positive than E⦵ of (I) reverse (II) Cathode: PbO2(s) + HSO4

- (aq) + 3H+(aq) + 2e− ⇌ PbSO4(s) + 2H2O(l) E⦵ = 1.69 V Anode: Pb(s) + HSO4

- (aq) ⇌ PbSO4(s) + H+(aq) + 2e− E⦵ = 0.36 V

PbO2(s) + Pb(s) + 2HSO4- (aq) + 2H+(aq) → 2PbSO4(s) + 2H2O(l) E⦵ = 2.05 V

A 12-V lead-acid car battery consists of 6 galvanic cells in series, each producing 2 V. Note that the lead-acid battery can be recharged by attaching an external power source, which will reverse the direction of the overall reaction: 2PbSO4(s) + 2H2O(l) PbO2(s) + Pb(s) + 2 HSO4

- (aq) + 2H+(aq)

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

Revision Exercises 1. Oxidation numbers - Determine the oxidation number of the bold italic element in:

(a) K4P2O7 ______________ (b) NaAuCl4 _____________ (c) ICl __________________ (d) OF2 _________________

2. Balancing half-equations (a) Cu(s) Cu2+(aq) ________________________________________________

(b) Al3+(aq) Al(s) _________________________________________________

(c) F–(aq) F2(g) ___________________________________________________

(d) H2O(l) O2(g) __________________________________________________

(e) Ag+(aq) Ag(s) _________________________________________________

3. Recognising Oxidation and Reduction reactions

Increase in oxidation number Oxidation reaction Decrease in oxidation number Reduction reaction

(a) In half-reaction 2(a), oxidation number of Cu(s) is _________; oxidation number of Cu2+(aq) is

____________. This corresponds to a(n) (increase / decrease) in oxidation number. Therefore, Cu(s) has been (oxidised / reduced) to Cu2+(aq).

(b) In half-reaction 2(b), oxidation number of Al3+(aq) is ________; oxidation number of Al(s) is

______________. This corresponds to a(n) (increase / decrease) in oxidation number. Therefore, Al3+(aq) has been (oxidised / reduced) to Al(s).

(c) Half-reaction 2(c) is an example of (oxidation / reduction) reaction. (d) Half-reaction 2(d) is an example of (oxidation / reduction) reaction. (e) Half-reaction 2(e) is an example of (oxidation / reduction) reaction.

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

4. Net Redox Reactions Given: Pb(s) Pb2+(aq) + 2e– and Cl2(g) + 2e– 2Cl–(aq) The overall reaction or the net equation is: 5. Oxidants versus Reductants Based on the net equation in 4, the oxidant is ___________ and the reductant is __________. 6. Galvanic / Voltaic cells In drawing an electrochemical cell, always position oxidation reaction in the left-hand half-cell. Consider Sn2+(aq) + 2e– Sn(s) (reduction) Zn(s) Zn2+(aq) + 2e– (oxidation)

Half-reaction: ______________ ______________

V

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

(i) In the left-hand cell, electrons are (generated / consumed). (ii) For the reaction taking place in the right-hand cell, electrons are (generated / consumed). (iii) Electrons will travel from the (left- / right-) hand cell to the (left- / right-) hand cell. (iv) When electrons move along a conducting wire, _______________ is generated along the

conducting wire. (v) When electrons move, there must be a difference in ____________________ between the left-

hand electrode and the right-hand electrode. (vi) As the reaction in the left-hand cell progresses, the electrode will (gain / lose) mass. (vii) As the reaction in the left-hand cell progresses, we expect the electrode to become increasingly

(positive / negative). (viii) At the tip of the right-hand electrode, electrons (can / cannot) continue to flow from the electrode

into the electrolyte solution in the cell. (ix) In order to complete the circuit (i.e. so that current can flow through the entire electrochemical

cell), we need to introduce a ____________________________. (x) The salt bridge placed in between the two half-cells consists of agar gel entrapped with cations

(for example, ______________) and anions (for example, _____________). Anions will move towards the (left- / right-) hand cell, while cations will move towards the (left- / right-) hand cell.

7. Standard Electrode Potential (E⦵) of a Half-cell In the following set up, under standard conditions, the voltmeter records +0.34 V compared to the standard hydrogen electrode. This means, in the galvanic cell above, we will record +0.34 V when copper is being (oxidised / reduced), under standard conditions. All tabulated E values are obtained in this manner: Pb2+(aq) + 2e– Pb(s) E⦵ = –0.13 V 2H+(aq) + 2e– H2(g) E⦵ = 0.00 V Ag+(aq) + e– Ag(s) E⦵ = 0.80 V Cl2(g) + 2e– 2Cl–(aq) E⦵ = 1.40 V

(i) What is the E⦵ value for the oxidation reaction Pb(s) Pb2+(aq) + 2e–?

Answer: ___________________.

(ii) What is the E⦵ value for the oxidation reaction 2Ag(s) 2Ag+(aq) + 2e–?

Answer: ___________________.

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

8. Setting up a galvanic cell Consider two half-reactions from the E⦵ table:

Cd2+(aq) + 2e– Cd(s) E⦵= –0.40 V

Ag+(aq) + e– Ag(s) E⦵= 0.80 V E for the Cd2+(aq)|Cd(s) couple is LESS POSITIVE than E for the Ag+(aq)|Ag(s) couple REVERSE the direction of the Cd2+(aq)|Cd(s). Write down the half-equation for the Cd2+(aq)|Cd(s) couple in reversed direction and its E value. ___________________________________________________________________________ Based on the above half-equation, the Cd2+(aq)|Cd(s) couple is undergoing (oxidation / reduction). We will therefore place this couple in the (left- / right-) hand compartment of a galvanic cell. Copy down the equation for the Cd2+(aq)|Cd(s) couple in reversed direction and its E⦵ value. Copy down the equation for the Ag+(aq)|Ag(s) couple the way it is and its E⦵ value. Balance the number of electrons in the two equations. Add the two equations together and calculate the net E⦵

cell. The E⦵

cell value indicates that the net reaction is (spontaneous / non-spontaneous).

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

Additional Practice Questions 1. A galvanic cell is constructed such that one electrode component consists of an aluminium strip

placed in a solution of Al(NO3)3, and the other has a nickel strip placed in a solution of NiSO4. The overall cell reaction is

2 Al(s) + 3 Ni2+(aq) ⇌ 2 Al3+(aq) + 3 Ni(s) Assume the Al is not coated with its oxide. (a) What is being oxidised, and what is being reduced?

__________________________________________________________________

(b) Write the half-reactions that occur in the two electrode compartments.

__________________________________________________________________ __________________________________________________________________

(c) Which electrode is the anode, and which is the cathode?

__________________________________________________________________

(d) Indicate the signs of the electrodes.

__________________________________________________________________

(e) Do electrons flow from the aluminium electrode to the nickel electrode, or from the nickel to the aluminium?

__________________________________________________________________

(f) In which directions do the cations and anions migrate through the solution?

__________________________________________________________________

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

2. Based on Ni2+(aq) + 2e− Ni(s) E⦵ = −0.24 V and Ce4+(aq) +e− Ce3+(aq) E⦵ = 1.61 V

Calculate the Ecell for the reaction Ni(s) + 2 Ce4+(aq) Ni2+(aq) + 2 Ce3+(aq).

3. Based on

Ag+(aq) + e− Ag(s) E⦵ = 0.80 V Cu2+(aq) + 2e− Cu(s) E⦵ = 0.34 V Ni2+(aq) + 2e− Ni(s) E⦵ = −0.24 V Cr3+(aq) + 3e− Cr(s) E⦵ = −0.74 V

(a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive Ecell

, and calculate the value.

(b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive Ecell

, and calculate the value.

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

Example 2017 HSC Exam Questions and Answers Q10 and Q11 (2017) 1 mark each

Substances in their elemental state → � reducing agents. Zn(s) ⇌ Cu2+(aq) + 2e

Pb(s) ⇌ Pb2+(aq) + 2e

By definition, oxidation occurs at the anode, which is the negative electrode.

Sulfur dioxide - oxidation number = +4 Sulfur in elemental state -

oxidation number = 0

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

Q23 (2017) 6 marks

Taken from solution guide: The salt bridge completes the circuit in a galvanic cell. It allows movement of ions between solutions to maintain electrical neutrality in each half-cell. (Source: NESA 2017 HSC Chemistry Marking Guidelines)

Electrochemistry Summary Macquarie University Department of Molecular Sciences 2018

Example 2016 HSC Exam Questions and Answers Q16 (2016) 1 mark An electrochemical cell has the following structure.

Q21 (2016) 5 marks A student set up the following galvanic cell.

Oxidation half-reaction: Al3+(aq) + 3e Al(s) (–1.68 V) Reduction half-reaction: Cu2+(aq) + 2e Cu(s) ( 0.34 V) Overall cell potential = 0.34 – (–1.68) = 2.02 V

Electrons flow from the right half-cell to the left half-cell (reverse convention).

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

Ag+(aq) + e Ag(s)

Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

0.80 V – 0.34 V = 0.46 V

The Acidic Environment Module 9.3 Acids and bases are commonly encountered in our everyday lives, as part of our body, in the foods we eat and the products we use. For example, hydrochloric acid is used in our stomach to help digest food, citric acid is in orange juice and can be bad for our teeth because it can dissolve the enamel, and phosphoric acid is the principle component in “rust-converter”.

Acids and bases are also widely used in industry. Examples of commonly encountered acids include hydrochloric acid (used to clean cement off bricks), sulfuric acid (in car batteries) and ascorbic acid (vitamin C, found in orange juice), while bases include ammonia (for cleaning floors), sodium bicarbonate (baking soda, cleaner), calcium carbonate (toothpaste, calcium supplements), magnesium hydroxide (antacid tablets) and sodium hydroxide (oven/drain cleaner). Even codeine (found in some analgesics) is a base.

Identifying a household base was asked in 2002 (Q20). In 2005 they asked about antacids (Q25) and in 2007 the uses of bases (Q25). In 2009, they asked about which grouping of common items were all acids. Then in 2011, they asked about which grouping of common items were all bases. In 2016 (Q7) was asking about indicators used to distinguish lemon juice (a common acidic substance) and potato juice.

In the HSC Curriculum the Acidic Environment module is used to teach (and examine!) a wide range of concepts. These include:

The history, nature and practice of chemistry

Applications and uses of chemistry

Implications of chemistry for society and the environment

Specific topics are:

9.3.1. Indicators: natural, common lab-based; uses of outside the lab; identification of acidic, basic, neutral; Arrhenius definition.

9.3.2. Acids in the air: non-metal oxides (e.g. CO2, NO, NO2, SO2, SO3) as acids; Periodic Table; solubility; equilibria; Le Chatelier’s Principle; natural and industrial sources, chemical equations; assessment of evidence; gas volume calculations; environmental consequences.

9.3.3. Acids in Biological Systems: Brönsted-Lowry definitions; simple organic acids; pH scale; strong/weak-dilute/concentrated distinctions; degree of ionisation – weak/strong/equilibria.

9.3.4. Definitions: History – Lavoisier/Davy/Arrhenius; Brönsted-Lowry; conjugate pairs; salts – acidic/basic/neutral; amphoteric substances; neutralisation; titrations; standard solutions; buffers.

These topics mean that the module allows examination of everything from history (of chemistry) to stoichiometry (balancing reactions) to calculations (neutralisation reactions) to nomenclature (naming).

The Evolution of the Theory of Acids and Bases

Early Concepts of Acids and Bases The English word “acid” comes from the Latin word acere, which means “sour”. All acids taste sour. Well known from ancient times were vinegar, sour milk and lemon juice. In the 700’s, Geber, an Arabic alchemist, prepared nitric acid and acetic acid. Early in the 1200s, the strong mineral acids were first isolated. Sulfuric acid was made by heating green vitriol (iron(II) sulphate) and condensing the vapour into water. Other vitriols gave the same product. Mixing a vitriol with nitre (potassium nitrate) and heating produced vapours which gave nitric acid upon condensing the vapours into water. Adding sal ammoniac (ammonium chloride) to nitric acid gave aqua regia, so named for its ability to dissolve gold. Hydrochloric acid too was known in the Middle Ages. The word “alkaline” comes from the Arabic al-qily, which means “to roast in a pan” or “the calcinated ashes of plants”. By leaching the ashes with water, one can obtain a solution of sodium or potassium carbonate (to use the modern terms). This is then mixed with slaked lime (calcium hydroxide) to provide a solution of NaOH or KOH. This technique was described in writing in the 900s, but certainly existed well before then. So we have been making and using acids and bases long before it was understood what they are and how they work.

Antoine Lavoisier Antoine Lavoisier was a French Chemist from the 1700’s. His idea was that acidity was caused by the presence of oxygen in the compound. Lavoisier (in September 1777) invented the word “oxygen”: From the Greek (oxys = sour and genes = born), to indicate “acid maker”. This idea turned out to be wrong, but it is relevant to the acid nature of some oxides. Many of the strong acids contain a relatively large proportions of oxygen. Also, the test Lavoisier used to demonstrate the presence of oxygen in the air was the “nitrous air test”. This test was devised by Joseph Priestley. Lavoisier knew that nitrous air (NO2) combined with oxygen and the resulting compound made nitric acid in water. His conclusion (published about 1776) was that oxygen was the component in a compound that was responsible for the generic property of acidity.

Humphry Davy In 1779, Lavoisier concluded that oxygen was present in muriatic acid (hydrochloric acid) because it was an acid and by his definition an acid contained oxygen. In 1809, however, Humphry Davy tried to find oxygen in muriatic acid by reacting it with all manner of substances but never obtained oxygen nor any oxygen compounds. He also noted the hydrogen sulfide (H2S) and hydrogen telluride (H2Te), were acids. Davy suggested that hydrogen was the essential component of acids, rather than oxygen, but didn’t quite formalise this as a theory. Justus von Liebig took Davy’s suggestion further, by defining an acid as a compound that contains hydrogen in a form that can be displaced by a metal. Bases, however, were understood only in terms of their ability to neutralise acids.

Svante Arrhenius Svante Arrhenius’ most important contribution to chemistry was also his first - the idea of electrolytic dissociation. This was the basis of his PhD thesis of 1884 but the subject of bitter debate for the next ten years. In fact, Arrhenius’s theory was so controversial that he was awarded a fourth class “not without merit” grade (a D in an A, B, C, D, F grading system) for his degree. Because of the controversy, he chose to speak about “active” and “inactive” parts of a molecule rather than the ionised and unionised parts of which we are familiar. However, his ideas were rewarded in 1903 with

a Nobel Prize in Chemistry. This also led to his definition of an acid being a substance that provides H+ ions in aqueous solution.

You need to be aware of the history of the acid and base concepts and should be able to identify types of acids and bases. These have turned up in past exams (e.g. see 2010 Q8; 2008 Q8; 2005 Q20).

Arrhenius Definition of Acids and Bases The Arrhenius definition of acids and bases is not the most sophisticated, but for most common acids and bases in aqueous solutions it is perfectly adequate and commonly used.

An acid is a substance that provides H+ ions in aqueous solution.

e.g. HCl(g) H+(aq) + Cl–(aq)

e.g. H2SO4(l) 2H+(aq) + SO42–(aq) *Note that we do not have free H+ ions in solution as they become attached to water (via a coordinate bond) to form the hydronium ion, H3O+ (hydrated hydrogen ion). Cations and anions in general, get surrounded by water. Therefore we generally refer to ions as aquated species (aq):

e.g. HCl(g) + H2O(l) H3O+(aq) + Cl–(aq)

For convenience we commonly write:

HCl(g) + H2O(l) H+(aq) + Cl–(aq)

You should be comfortable with using H+(aq) to represent the more correct H3O+(aq)

A base is a substance that provides OH– (hydroxide) ions in aqueous solution.

e.g. NaOH(s) + H2O(l) Na+(aq) + OH–(aq)

Many common acids and bases are Arrhenius acids and bases. These include:

Arrhenius Acid

Formula Arrhenius Base Formula

Hydrochloric Acid HCl Sodium Hydroxide NaOHNitric Acid HNO3 Potassium Hydroxide KOHSulfuric Acid H2SO4 Ammonium Hydroxide NH4OHAcetic Acid CH3CO2H Barium Hydroxide Ba(OH)2

Carbonic Acid H2CO3 Calcium Hydroxide Ca(OH)2

The Arrhenius definition explains the properties of many acids and bases in the presence of water, but not all. It also does not take into account the fact that many species act as acids or bases in solvents other than water and even in the absence of water and that their reactivity is dependent on the type of solvent and/or other species present. Q8 in the 2008 HSC exam examines the definition of Arrhenius acids and bases. Q29 in the 2011 HSC exam asked you to “Justify the continued use of the Arrhenius

OH

H+ H+ O

H

HH

+

formation of hydronium ion

These reactions are ionisation reactions, e.g. HCl ionises to form H+(aq) and Cl(aq). These are not strictly dissociation reactions.

definition of acids and bases, despite the development of the more sophisticated Brönsted–Lowry definition”.

Brønsted-Lowry (B-L) Definition of Acids and Bases Brønsted and Lowry independently refined the Arrhenius definition of acids and bases. The acid definition is essentially the same as that of Arrhenius, recognising that H+ is a proton, and that this species can be found in solutions other than aqueous solutions.

An acid is a proton (H+) donor molecule or ion.

Their definition of a base departs further from that of Arrhenius. Rather than being a specific species, OH-, their definition is broader and includes the concept of oppositeness or conjugation.

A base is a proton acceptor.

e.g. NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

Ammonium ions (proton donor) react with water molecules (proton acceptor) to produce ammonia and the hydronium ion. The ammonium ion is a B-L acid as it donates a proton and water is behaving as a B-L base in the above as it accepts a proton. (Note well, it was Lowry who introduced the concept of the hydronium ion, H3O+.)

Lewis Definition of Acids and Bases The following section is not explicit in the HSC curriculum, but relates to Lewis Structures and is included here for completeness.

There is a further definition of acids and bases. This is the Lewis Definition. A Lewis acid is a chemical substance that can accept a pair of electrons while a Lewis base, B, is an electron-pair donor. Together they form an adduct; AB. You should recognise that donating a proton is the same as accepting a pair of electrons. The symmetry of opposites! This forms the basis for the definition of Lewis acids.

A Lewis base can be a Brønsted-Lowry base as it can donate a pair of electrons to a proton; the proton is a Lewis acid as it can accept a pair of electrons. However, the Lewis definition is much more general. For example, carbon monoxide is a very weak Brønsted-Lowry base but it forms a strong adduct with BF3 (a Lewis acid), because CO is a strong electron donor and BF3 is a strong electron acceptor. Neither contain H+ or OH–.

So recapping, the evolution of the definition of acids and bases from Arrhenius (H+/OH–) to Brønsted-Lowry (H+ donor/H+ acceptor) to Lewis (e– acceptor/e– donor).

Acids and Bases in Aqueous Solutions A number of compounds do not act as acids or bases until they are in aqueous solution. For example: acetic (ethanoic) acid, CH3COOH, when diluted with water, causes dry blue litmus paper to turn red. However, pure (glacial) acetic acid does not cause a colour change with dry, blue litmus paper. Why?

Answer: The acidic hydrogen in glacial acetic acid is covalently bound (locked in the molecule) and is not ionised. Therefore no hydrogen ion (or hydronium ions) are produced and the blue litmus paper

H

acid base

H

proton transfer

does not change colour. In the presence of water, glacial acetic acid reacts (ionises) to form a small amount of H3O+ ions, which cause blue litmus paper to turn red.

i.e. CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq)

Conjugate Pairs Derived from the Brønsted-Lowry definition. An acid gives up a proton to form its conjugate base and a base accepts a proton to form its conjugate acid:

Base --------------------- Conjugate acid e.g. NH3(aq) + H2O(l) NH4+(aq) + OH(aq) Acid --------------------- Conjugate base

Acid --------------------- Conjugate base e.g. NH3(aq) + OCH3(l) NH2(aq) + CH3OH(aq) strong base --------------------- Conjugate acid The conjugate base of a strong acid is an extremely weak base, e.g. HCl (strong acid) Cl

(very weak base). The conjugate base of a weak acid is a weak base, e.g. NH4+ (weak acid) NH3 (weak base). The conjugate bases of extremely weak acids, e.g. water, methanol and ethanol are very strong

bases, e.g. HO, CH3O, CH3CH2O.

Understanding conjugate pairs was assessed in 2008 (Q26), 2009 (Q7), 2011 (Q25) and 2016 (Q10).

Amphiprotic Species Some molecules such as H2O and many ions including HO, HSO4, HCO3, H2PO4 and HPO42 can act as both proton donors (Brønsted-Lowry acids) and proton acceptors (Brønsted-Lowry bases), i.e. they are amphiprotic. (amphi means “both”).

e.g. HCO3 + OH CO32 + H2O (HCO3 as an acid) HCO3 + H3O+ H2CO3 + H2O (HCO3 as a base)

Note: the ability to write equations showing the amphiprotic nature of a species is often assessed.

Amphoteric compounds have the ability to react as an acid or a base. These include substances that do not donate or accept protons.

e.g. ZnO, zinc oxide, can react with acids and bases

Conjugate base = acid – H+ Conjugate acid = base + H+

Remember charges MUST balance on right-hand and left-hand sides of the equation.H

acid base

H

proton transfer

conjugate base conjugate acid

ZnO + 2 HCl ZnCl2 + H2O base acid salt water

ZnO + 2 NaOH Na2ZnO2 + H2O acid base salt water

Amphiprotic molecules are examples of (or a subset of) amphoteric molecules. A question on amphiprotic species was in 2002 (Q6), 2004 exam (Q22) and 2017 (Q5).

Strong and Weak Acids and Bases Strong acids and bases are completely ionised. Weak acids and bases are partially ionised.

Acids such as H2SO4, HCl, HBr, HI and HNO3 are strong acids. Most other acids (e.g. ethanoic acid (also known as acetic acid), citric acid, the ammonium ion NH4+ and carbonic acid) are weak acids. The ionisation of any acid (HA) in water can be represented by the equilibrium:

HA(aq) + H2O(l) H3O+(aq) + A(aq)

The degree of ionisation [H3O

][HA]

100% where HA is any weak acid.

The larger the degree of ionisation (separation of H+ from the remainder of the initial molecule), the stronger the acid. The concentration of the acid, [HA], is that before any ionisation occurs (it is the concentration that you make to solution up to, e.g. 1.0 mol/L), while the concentration of [H3O+] is that after ionisation of HA is complete. Note: a dilute solution of a strong acid (e.g. 0.01 molar hydrochloric acid) is still derived from a strong acid and a concentrated solution of a weak acid is still derived from a weak acid. Acid strength was examined in 2007 (Q8), 2009 (Q21), 2011 (Q15, 16) and 2013 (Q11.)

Metallic oxides and hydroxides are generally strong bases. Ammonia, ethanoate (acetate) and carbonate ions are examples of weak bases. The ionisation of any base (B) in water can be represented by the equilibrium: B(aq) + H2O(l) HB+(aq) + OH(aq)

The degree of ionisation [OH – ]

[B]100% where B is any weak base.

The pH Scale pH = -log10[H3O+] often simply written as pH = -log10[H+] Note: [H3O+] = 10pH mol L1 in calculator terminology [H3O+] = INV LOG ( pH)

In pure water:

H2O + H2O H3O+ + HO

At 25 °C:

Most metallic oxides/ hydroxides (e.g. Na2O, Mg(OH)2) are basic. Most non-metallic oxides/ hydroxides are acidic (e.g. CO2) See Q21 in 2002, Q18 in 2004 and Q2 in 2009 HSC exams.

[H3O+] = [HO] = 1.0 x 107 molar, Kw = 1.0 x 1014 [HO-] = 1.0 x 1014 * [H3O+] pOH = 14 pH (and therefore pH = 14 + log[HO]) * *These are good relationships to be familiar with. See Q12 in 2016 HSC exam. A neutral solution has a pH of 7 (equal parts of OH- and H3O+ ions) An acidic solution has a pH <7 ([H3O+] > [OH]) A basic solution has a pH >7 ([OH] > [H3O+]) pH Calculations The pH of a strong acid or base can be calculated directly from the molarity of the solution. i) For 0.10 molar hydrochloric acid the pH = (log10(0.10) = 1.0

ii) For 0.05 molar Ba(OH)2

Ba(OH)2(s) Ba2+(aq) + 2 OH-(aq)

[H+] = 1 x 1014/[HO-] = 1 x 1014/(2 ( 0.05)

(pH = -log10(1 x 1013) = 13

Alternatively, [OH-] = 2 (0.05 = 0.1, pOH = -log10(0.1) = 1; pH = 14 – pOH = 14 – 1 = 13 Calculating pH turns up frequently, see e.g. 2002 (Q22a), 2003 (Q8), 2004 (Q24), 2005 (Q8), 2007 (Q10, Q21), 2008 (Q14), 2010 (Q21), 2012 (Q28), 2016 (Q12 and Q18). To calculate the pH of a weak acid/base you need to know the degree of ionisation. For example, calculate the pH of a 0.1 molar solution of ascorbic acid (chemical name for vitamin C). Note the solution has been found to be 2.8% ionised.

i) Degree of ionisation [H3O

]ascorbic acid

100% = 2.8%

ii) 2.8 = [H3O

]0.1

100%

iii) [H3O+] = 2.8 x 103 molar iv) pH = log10[2.8 x 103] = 2.6

A solution of a strong acid (100% ionised) has a lower pH value (i.e. higher [H3O+]) than the pH of a similar concentrated solution of a weak acid (not fully ionised). Exams almost always have a question on calculation of pH or ask you to explain why solutions of equal concentration of two different acids (or bases) don’t have the same pH. In 2015, students were asked to determine whether acetic acid or hydrochloric acid had the highest pH, and if so, at what concentration.

pH of Salt Solutions Generally the reaction of an acid and base produce salt and water.

e.g. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

i) The salt formed from the reaction of a strong acid and a strong base (as above) is neutral ([H3O+] = [HO]) and the pH of the salt solution is 7.

e.g. the pH of a 0.1 molar sodium nitrate solution is 7.

ii) The salt formed from a reaction of a strong acid and weak base is itself a weak acid and the pH of the salt solution is less than 7.

e.g. the pH of a 0.1 molar ammonium nitrate solution is ~5.1.

iii) The salt formed from the complete reaction of a weak acid with a strong base is itself a weak base and the pH of the salt solution is greater than 7. e.g. the pH of a 0.1 molar sodium acetate (CH3COONa) solution is ~8.9.

Ranking of the relative acidity of common acids, bases and salts is frequently asked in HSC exams (e.g. see 2007, Q9; 2010, Q3). The pH of salt solutions was tested in 2008 (Q27). As part of Q24 in 2017 students were asked to explain why the pH of a salt solution resulting from a titration of a strong base against a weak acid was not equal to 7, including a relevant chemical equation. The reaction of an acid and base to give salt and water is often referred to as a neutralisation reaction. The reaction is exothermic, i.e. releases heat.

Volumetric Analysis of Acids and Bases Titration: The process of determining the volume of one reactant solution (acid or base), which reacts completely with a known amount of another reactant (base or acid) is called an acid-base titration. The base is often placed in a conical flask, and the acid placed in a burette to be released in a controlled manner into the base. Sometimes the burette contains the base and the conical flask contains the acid (but as strong bases like NaOH can etch glass, we try to avoid doing this if we can).

Equivalence Point: The equivalence point occurs where the numbers of moles of the two reactants are just sufficient to cause complete consumption of both reactants. Close to the equivalence point the hydrogen ion concentration changes rapidly. Often the conical flask contains a few drops of an indicator (an organic dye, the colour of which is sensitive to pH), which changes colour at an end-point. A good indicator will have an end-point close to the equivalence point. For example: When 20 mL of 0.1 M hydrochloric acid (in a beaker) is titrated with 0.1 M sodium hydroxide (from a burette) the following conductance change is observed.

0 10 20 30mL NaOH (aq) added

Con

duct

ivity

Reaction: H+

(aq) + Cl(aq) + Na+(aq) + HO

(aq) H2O(l) + Na+(aq) + Cl(aq)

As sodium hydroxide is added the hydrogen ions (H+) get used up and effectively replaced by sodium ions (Na+). Sodium ions are much larger than hydrogen ions and have a smaller conductivity. Therefore conductance falls as the equivalence point is reached. After the equivalence point, we are essentially adding NaOH to the sodium chloride solution. This adds extra ions and therefore leads to an increase in conductance.

Volumetric acid-base analysis is frequently assessed, with questions on determination of the concentration of an unknown acid or base following titration typically asked. For example, see 2016 Q29 or 2017 Q24. In addition, students have often been asked to describe in detail the order of performing a titration, sketch the glassware used (often very poorly done), determine what indicator would be best, and interpret titration curves. Modern chemistry techniques have also been assessed, for instance for Q26(b) of the 2015 exam, students were asked to describe how computer technology could identify the equivalence point.

Example of Volumetric Acid-Base Analyses From 2008 HSC exam (Q28): “A standard solution was prepared by dissolving 1.314 g of sodium carbonate in water. The solution was made up to a final volume of 250.0 mL. (a) Calculate the concentration of the sodium carbonate solution.” The molecular formula of sodium carbonate is Na2CO3, which has a molecular mass of (2 × 22.99) + 12.01 + (3 × 16.00) = 106.0 g mol-1 (NOTE: 4 significant figures) Number of moles of sodium carbonate used is 1.314 g/106.0 g mol-1 = 0.01240 moles This amount of material is in 250.0 mL, therefore the concentration is 0.01240 moles/0.2500 L = 4.959×10-2 mol L-1 “This solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.00 mL samples of the acid were titrated with the sodium carbonate solution. The average titration volume required to reach the end point was 23.45 mL. (b) Write a balanced equation for the titration reaction”

Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l) “(c) Calculate the concentration of the hydrochloric acid solution.”

The number of moles of Na2CO3 used is 0.02345 L × 4.959 × 10-2 mol L-1 = 1.163 × 10-3 moles Therefore, the number of moles of HCl reacted is 2 × 1.163 × 10-3 = 2.326 × 10-3 moles This is in a volume of 25.00 mL (the volume used in the titration), so the concentration of HCl is 2.326 × 10-3 moles/0.02500 L = 9.303 × 10-2 mol L-1. (NOTE still 4 significant figures)

See also Q23 in 2003; Q24 in 2005; Q21 in the 2007; Q28 in 2010; Q30 in 2014; Q26 in 2015; Q29 in 2016 and Q13/Q24 in 2017 HSC exams.

Titration Curves If we plot the pH of the solution being titrated against the volume of acid or bases added (we could measure the pH using a pH meter) we get a titration curve. The equivalence point is the point where the slope of the curve is the steepest (close to vertical) and is the point where the acid or base has been neutralised. For a strong acid/strong base titration the pH of the equivalence point will be about 7; for a weak acid/strong base titration it will be above 7 (basic); and for a strong acid/weak base it will be less than 7 (acidic). Examples of titration curves are shown below.

Typically we don’t use a pH meter to measure the pH as we go along, we use an indicator to alert us to the rapidly changing pH, and therefore the volume of added acid or bases that was required to reach the equivalence point. An indicator normally has a range of 1 or 2 pH units over which it changes colour. As long as the pH range over which this change occurs corresponds to the vertical part of the titration curve, it is not absolutely required that the indicator range overlaps with the equivalence point. Because the curve is close to vertical, the difference in volume is negligible. However, it is important to use an indicator that changes colour in a pH range that is close to the equivalence point. This requires considering if the system is “strong acid-strong base”, “weak acid-strong base”, or “strong acid-weak base”.

(i) Titration of a Strong Base with a Strong Acid (Base in Conical Flask)

The products are neutral so the pH at the equivalence point is close to 7. Bromothymol blue, which changes colour between pH 7.6 and 6.2 would be ideal, but because the pH change around the equivalence point is very large, the choice of indicator is not critical.

(ii) Titration of a Weak Acid with a Strong Base (Acid in Conical Flask)

At the equivalence point the solution will be weakly basic (conjugate base of weak acid). An indicator like phenolphthalein, which changes from clear to pink over the range pH 8.3 to 10, would be suitable. Methyl orange, which changes colour between pH 3.1 and 4.4, would be unsuitable because the end-point would occur before the equivalence point.

(iii) Titration of a Weak Base with a Strong Acid (Base in Conical Flask)

At the equivalence point the solution will be weakly acidic (conjugate acid of weak base). Methyl orange, which changes colour from yellow when the pH exceeds 4.4 to red when the pH falls below 3.1, is a suitable indicator. Phenolphthalein would give an end-point before the equivalence point.

Acid/Base Indicators are substances whose solutions change colour due to changes in pH. They are usually weak acids or bases, where the conjugate base or acid forms have different colours due to differences in their absorption spectra. The indicators are themselves being neutralised in the titration, so it is important that there is not much indicator added to the solution. The actual volume of “equivalence point” will be at the sum of the volume required to neutralise the acid or base of interest AND volume required to neutralise the indicator. If there is a significant amount of indicator present then the second volume may not be negligible. Hence, for an indicator to be useful for quantitative analysis it needs to be strongly coloured so that we don’t need much of it.

There are quite a few natural indicators, such as:

Beetroot - A very basic solution will change the colour of beetroot juice from red to purple.

Black Berries - Blackberries, blackcurrants, and black raspberries change from red in an acidic environment to blue or violet in a basic environment.

Blueberries - Blueberries are blue around pH 2.8-3.2, but turn red as the solution becomes even more acidic.

Cherries - Cherries and their juice are red in an acidic solution, but turn blue to purple in a basic solution.

Curry Powder and Turmeric - Curry powder contains turmeric which itself contains the pigment curcumin. This compound changes from yellow at pH 7.4 to red at pH 8.6.

Geranium Petals - Geraniums contain the anthocyanin pelargonin, which changes from orange-red in an acidic solution to blue in a basic solution.

Grapes - Red and purple grapes contain multiple anthocyanins. Blue grapes contain a monoglucoside of malvinidin, which changes from deep red in an acidic solution to violet in a basic solution.

Onion - Onions are olfactory indicators. You don't smell onions in strongly basic solutions. Red onion also changes from pale red in an acidic solution to green in a basic solution.

Red (Purple) Cabbage - Red cabbage contains a mixture of pigments used to indicate a wide pH range.

Rose Petals - The oxonium salt of cyanin turns from red to blue in basic solution.

Tea - Tea is an acid-base indicator. Its colour changes from brown in basic solution to yellow-orange as the solution becomes acidic. This is the reason why lemon juice is added to a cup of tea – not for the taste, but to reduce the intensity of the colour (called “brightening”).

Questions often (nearly always) turn up in the HSC exams referring to indicators. For example, questions on selecting the correct indicator were in the 2002, 2004 and 2005 (Q10) exams and a question on identifying an indicator was in 2003 (Q1). In 2012, Q7 asked about colours in relation to the pH range in a mixture of indicators. A different take on indicators featured in the 2015 exam, which involved preparation techniques for making and testing natural indicators (Q21 (a) and (b), respectively). Then in 2015, Q14 featured a graph of pH changes during a titration and asked, “Which pH range should an indicator have to be used in this titration”. 2014 included a question (Q7) that involved deducing a substance based on colour changes of indicators. More recently, a question on using indicators to distinguish between substances of different pH was asked in 2016 (Q7).

Buffers Buffers are mixtures of weak acid and weak base conjugate pairs. Because both components are weak they readily accept protons added to the solution (if a base) or have protons to donate to added bases (if they are an acid). In this way a buffer solution “buffers” against changes in pH.

Explaining the action of buffers involves using of Le Chatelier’s Law. The buffer acid and base are in equilibrium with their conjugates prior to adding extra acid or base. When that acid or base is added the equilibrium is disturbed and the buffer components react to counter this disturbance.

Many reactions that chemists want to perform are affected by the acidity of the solution in which they occur, so the pH of the reaction medium must be controlled. Such control is provided by using buffers.

Biochemical reactions are especially sensitive to pH. Most biological molecules contain groups of atoms that may be charged or neutral depending on pH, and whether these groups are charged or neutral has a significant effect on the biological activity of the molecule. In all multicellular organisms (that includes us!), the fluid within the cell and the fluids surrounding the cells have a characteristic and nearly constant pH. This pH is maintained in a number of ways, and one of the most important is through buffer systems. Two important biological buffer systems are the dihydrogen phosphate system and the carbonic acid system.

The phosphate buffer system operates in the internal fluid of all cells. This buffer system consists of dihydrogen phosphate ions (H2PO4-) as hydrogen-ion donor (acid) and hydrogen phosphate ions (HPO42-) as hydrogen-ion acceptor (base) (a chemist would make up a phosphate buffer by mixing, typically, Na2HPO4 and KH2PO4 , i.e. HPO42- and H2PO4-). These two ions are in equilibrium with each other as indicated by the chemical equation below.

H2PO4-(aq) H+(aq) + HPO42-(aq)

If additional hydrogen ions enter the cellular fluid, they are consumed in the reaction with HPO42-, and the equilibrium shifts to the left. If additional hydroxide ions enter the cellular fluid, they react with H2PO4-, producing HPO42-, and shifting the equilibrium to the right.

Another biological fluid in which a buffer plays an important role in maintaining pH is blood plasma. In blood plasma, the carbonic acid and hydrogen carbonate ion equilibrium buffers the pH. In this buffer, carbonic acid (H2CO3) is the hydrogen-ion donor (acid) and hydrogen carbonate ion (HCO3-) is the hydrogen-ion acceptor (base).

H2CO3(aq) H+(aq) + HCO3-(aq)

This buffer functions in exactly the same way as the phosphate buffer. Additional H+ is consumed by HCO3- and additional OH- is consumed by H2CO3.

See Q15 in 2003 HSC exam where a small part of a titration curve is used to ask about buffers. See also Q10 in 2005 paper which gives a pH curve and asks about appropriate indicators and Q8 in 2012. See also Q24 in 2015 HSC exam asking about the acetic acid/sodium acetate buffer system. Q14 in 2017 asked about the effectiveness of a resulting buffer solution.

Acids in the Air – Non-Metal Acid Oxides Carbon dioxide (CO2), sulfur dioxide (SO2) nitrogen monoxide (NO, nitric oxide) and nitrogen dioxide (NO2) all dissolve in water forming acid solutions. Most non-metal oxides (except for CO, NO and N2O which are neutral) are acidic. To detect that a non-metal oxide gas is acidic with indicator paper, the paper must be moist. Moisture enables the gas to dissolve and form the acid.

Acid rain is an unusually acidic precipitation. It can have harmful effects on plants, aquatic animals, and structures through the process of “wet deposition”. Acid rain is caused by the gases listed above reacting with the water in the atmosphere to produce acids. Much of these gases are produced through human industrial and, to a lesser extent, agricultural activities. However, it can also be caused naturally by the splitting of nitrogen compounds by the energy produced by lightning strikes, or the release of sulfur dioxide into the atmosphere by volcanic eruptions.

Pure water has a neutral pH (i.e. pH 7 at 25 °C). “Clean” or unpolluted rain has a slightly acidic pH of about 5.2, because carbon dioxide and water in the air react to form carbonic acid.

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

Carbonic acid then can ionise in water to produce low concentrations of hydronium and carbonate ions:

2 H2O(l) + H2CO3(aq) CO32−(aq) + 2 H3O+(aq)

Other gases react in a similar manner:

Oxides of nitrogen, particularly nitrogen dioxide (NO2) react with water to form nitrous acid (HNO2) and nitric acid (HNO3):

2 NO2(g) + H2O(l) HNO2(aq) + HNO3(aq)

Sulfur dioxide (SO2) reacts with water to form sulfurous acid (H2SO3):

SO2(g) + H2O(l) H2SO3(aq)

Sulfur dioxide can be oxidised gradually to sulfur trioxide (SO3):

2 SO2(g) + O2(g) 2 SO3(g)

Sulfur trioxide (SO3) reacts with water to form sulfuric acid (H2SO4):

SO3(g) + H2O(l) H2SO4(aq)

Why rain is acidic was asked in the 2009 HSC exam (Q2) and a short answer question on the build-up of oxides of nitrogen in the air was asked in Q18. See Q5 in 2015 HSC for a question on non-metal and metal oxides and their acid/base behaviour in aqueous solution.

Common Student Misconceptions about Acids and Bases If you find yourself thinking in the following ways, stop, rethink and change your ideas.

1. “Substances containing H are acidic; substances containing OH are basic.” Many substances that contain H are not acids and many substances that contain OH are not bases. Sugar (sucrose), C6H12O6, contains five OH groups, however, when dissolved in water it dissolves as intact molecules and does not ionise to produce any H+ or OH– ions. Alcohols have a characteristic hydroxyl group (OH) but when they dissolve in water they are completely surrounded by water (aquated) and do not ionise.

2. “When a ‘proton donor’ acid reacts, the nucleus of an atom loses a proton.” When we speak of acids as proton donors we are talking about the single proton in a hydrogen ion, H+, not a proton from the nucleus of some other atom. Although chemists commonly refer to acids and bases as proton donors and proton acceptors, it is important to realise that H+ is being moved from one chemical species to another, not a proton from the nucleus of an atom.

3. “Strength and concentration mean the same thing.” or “A strong acid is always a concentrated acid.” These two terms are often confused by students. Concentration refers to the number of moles of solute (acid or base) per litre of solution. Strength refers to the percent of molecules that ionise and form ions in solution. Hydrochloric acid is considered a strong acid because in aqueous solutions nearly all the molecules ionise to produce H+ and Cl- ions. Acetic acid is considered a weak acid because only about 1% of the molecules normally ionise. We can have 1 molar solutions of either, hence their concentration is the same, but in the HCl solution there will be more H+ (or more correctly H3O+) than in the acetic acid case, so the pH will be lower (more acidic) in the HCl solution case. HCl is a strong acid.

4. “An acid is something which eats material away; an acid can burn you.” There are many acids that are not strong enough to cause any corrosion or burning. This might be a characteristic of some of the strong acids (at a significant concentration), but it is not universal. It also does not tell us anything about the chemistry of the acid.

5. “To neutralise is to break down an acid or to change from an acid.” Neutralisation is a two way process – a proton transfer between and acid and a base.

6. “All acids and bases are harmful and poisonous.” Some acids and bases can be harmful, but our own bodies and many foods contain acids and bases. The harmfulness will also depend on the concentration of the acid or base. Even strong acids or bases, if sufficiently dilute, pose no danger.

7. “All salts are neutral.” Some salts are acidic or basic. For instance, if one of the ions in a salt is the conjugate base of a weak acid, such as the ethanoate ion from ethanoic (acetic) acid, then it will accept a proton to form the un-ionised acid, resulting in a basic solution. If one of the ions in the salt is the conjugate acid of a weak base, such as NH4+ from NH3, then dissolving the salt in water will result in the ion donating its proton to H2O, giving an acid solution.

8. “Bubbles or bubbling is a sign of chemical reaction, or strength of an acid or a base.” Bubbles may arise from many different processes. It is not a definitive test. It does show that a reaction is occurring and a gas is evolving.

9. “As the value of pH increases, acidity increases.” It is fairly common to get this around the wrong way. Lower pH means the solution is more acidic.

10. “As the number of hydrogen atoms increases in the formula of an acid, its acidity becomes stronger.” The number of hydrogen atoms in a molecule is not a good indicator of acid strength. Organic molecules typically have lots of hydrogen atoms, but very few of them are acidic, and those that are acidic (e.g. acetic acid), are weak acids.

Monitoring and Management in the Chemical Industry

Theroleofchemistsinindustry

Within industry, chemists are employed in production, research and development, quality control, environmental monitoring and chemical analysis. Much of the work of chemists in industry involves monitoring the reactants and products of industrial reactions and managing reaction conditions. Chemical processes in industry require monitoring and management to optimise production and ensure quality control. Manufactured products, including food, pharmaceuticals (drugs) and household chemicals, are analysed to determine or ensure their chemical composition and/or energy content.

Whyisthereaneedtomonitor?

The three main reasons to monitor are to maximise yield of products, to confirm quality/purity of final products and to monitor and control any nasties being released to the environment. Monitoring to maximise yield is important when there is a reaction in which the products that form depend on the reaction conditions used. An example of such a reaction is the oxidation of ethene. Depending on the reaction conditions, the combination of ethene with oxygen can produce CO2 and water, ethylene oxide, or acetaldehyde. Careful monitoring of reaction conditions is necessary to optimise the yield of the desired products.

Ammonia–avaluableproductofthechemicalindustry

Millions of tonnes of ammonia are produced worldwide each year, making it one of the most important industrial chemicals produced. Ammonia is used as the prime ingredient in the manufacture of numerous economically and industrially important products such as nitric acid, which is used to make explosives and fertilisers; several textiles and plastics, such as nylon and acrylics; and most cleaning products (e.g. Ajax Spray n’ wipe etc.).

TheHaberprocess‐Ammonia,equilibriumandcompromise

The synthesis of ammonia:

N2(g) + 3H2(g) 2NH3(g) H = -92 kJ/mol

This reaction is reversible with the equilibrium position at ordinary temperatures and pressures lying well to the left. To make as much ammonia as possible and thus make the process economically viable, we want the equilibrium position to be as far to the right as possible.

ApplyingLeChatelier’sprinciple:

The above reaction is exothermic, thus, we would get the best yield of ammonia if we used a low temperature. Unfortunately, lowering the temperature slows down the rate of the reaction, so a compromise temperature (400°C) and a catalyst is used to give a reasonable yield and a reasonable rate of reaction. A high pressure will also increase the yield of ammonia as production of ammonia corresponds to a decrease in the number of moles of gas (4 moles on the left hand side, 2 moles on the right hand side of the equation). Removing the product (ammonia - by condensation) will drive the reaction to the right. Increasing the quantities of either reactant will also shift the reaction to the right; however, by keeping stoichiometric amounts of reactants, any left over reactants can be readily recycled. Conditions of 400°C, a total pressure of 250 atmospheres, stoichiometric quantities of H2 and N2 and an iron/magnetite catalyst give a 40% yield of ammonia at equilibrium.

Some Advice: The Haber-Bosch ammonia synthesis is introduced to further develop your understanding of chemical equilibrium. Therefore, be prepared for questions (in the context of the Haber process) along the lines of:

Explain why iron is used in the Haber process and what effect does it have? (Hint: where is a metal used in the reaction?)

Use Le Chatelier’s principle to explain how the percentage of NH3 present at equilibrium is affected by increasing temperature and/or increasing pressure.

Explain, using Le Chatelier’s principle, the trade-offs used in the Haber process to achieve the most yield at the least cost.

Note that in recent years, questions using the concepts of Le Chatelier’s principle have been set that are not Haber process related (e.g. see 2010 Q18; 2012 Q23; 2014 Q20; 2015 Q16; 2016 Q14; 2017 Q16 & Q31).

Importantfactorsindesigningindustrialprocesses: Economical starting materials: In the Haber process, N2 is obtained from the air, H2 from the

reaction of methane (readily available from natural gas) and steam in the presence of a nickel catalyst.

Ability to reclaim unused reactants: In the Haber process, stoichiometric quantities of reactants are used, so that left-over reactants can be readily recycled back into the incoming reactant stream.

Energy management: In the Haber process, the heat of reaction is utilised to heat up incoming reactants and thus minimise energy costs. The combustion of methane, whilst providing the heat for the Haber process and production of H2, simultaneously removes the O2 from air thereby effectively producing the N2 for the synthesis.

Howdoesmonitoringfitin? For the Haber process to function on an economically industrial scale, monitoring of the process is essential: Temperature and pressure must be monitored. Excess temperature will not only damage the

catalyst, but is wasteful of fuel (methane). Incoming gas stream needs monitoring to ensure stoichiometric ratio of reactants (to avoid the

build-up of one reactant), that oxygen is absent (to avoid explosive reaction with the H2) and that the concentration of CO and S containing species is sufficiently low to avoid poisoning the catalyst.

ChemicalAnalysis–Answeringlife’slittlequestions

Monitoring is clearly important for successful/economical operation of an industrial chemical process. Monitoring does not, however, stop after the final product is produced. Quality control is a very important part of the manufacturing industry. For instance, products must meet manufacturer or regulatory specifications, or have stated amounts of active ingredients, or not contain unacceptable amounts of certain impurities, etc. To answer these questions, chemical analysis is employed. The field of chemical analysis, that is, Analytical Chemistry, is central to nearly every area of science, from simple quality control, to forensic investigations, drug testing, environmental monitoring, to name a few. Chemical analysis can be reduced to two main types of determinations; qualitative analysis, and quantitative analysis. Qualitativeanalysis–What’sthere?

Qualitative analysis is determination of the presence of something. Not how much is there, but simply whether it is there or not. Youneedtoknow

Chemical tests to identify the presence of the following:- cations: barium, calcium, lead, copper and iron; anions: phosphate, sulfate, carbonate and chloride. Cations in solution are primarily identified by precipitation reactions. For example, Pb2+ can be identified by precipitation in the presence of Cl- or I- .

Lead: Pb2+(aq) + 2Cl-(aq) PbCl2(s) (white precip.) Pb2+(aq) + 2I-(aq) PbI2(s) (yellow precip.)

Barium: Ba2+(aq) + SO42-(aq) BaSO4(s) (white precip.), no precip. of F- and OH-

Calcium: Ca2+(aq) + SO42-(aq) CaSO4(s) (white precip.), precip. with F- Copper: Cu2+(aq) + OH-(aq) Cu(OH)2(s) (blue precip. that redissolves with NH3), Cu(OH)2(s) + 4NH3(aq) Cu(NH3)42+(aq) + 2OH-(aq) Iron(II): Fe2+(aq) + 2OH-(aq) Fe(OH)2(s) (white precip., quickly turns red-brown)

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) (decolourises MnO4-) Iron(III): Fe3+(aq) + 3OH-(aq) Fe(OH)3(s) (brown precip.) Fe3+(aq) + SCN-(aq) Fe(SCN)2+(aq) (deep red solution)

Tests for anions are a little more complicated, but again are mostly precipitation reactions (except for carbonate).

Sulfate: SO42-(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NO3-(aq) (white precip. after acidification) Chloride: Cl-(aq) + AgNO3(aq) AgCl(s) + NO3-(aq) (white precip., turns brown in light) AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq) (precip. dissolves in NH3) Phosphate:2PO43-(aq) + 3Ba(NO3)2(aq) Ba3(PO4)2(s) + 6NO3-(aq) (white precip. @ pH 8-11) Carbonate: addition of HNO3 makes bubbles, CO32-(aq) + 2H+(aq) CO2(g) + H2O(l)

Some advice: It’s important with all these tests to understand the chemistry involved. It is also likely that any questions asked will require you to describe (visually) the results of these tests. Thus, it’s not enough to know which test to use, you also need to know the physical outcome of each test.

Theflametest

When compounds of some elements are vaporised in a flame, they produce a distinctive colour in the flame that can be used to identify the presence of those elements. For instance, the presence of lithium will give the flame a dull red colour, sodium an intense yellow colour, K – violet, Ca – orangey red, Sr – deep red, Ba – lime green, Cu – blue green. It might be useful to attribute these colours to everyday things; sodium – lemon yellow, calcium – brick red, etc. The flame test is particularly useful for distinguishing between barium (green) and calcium (red), sodium (yellow) and potassium (violet), but can be problematic for elements with similar coloured flames (e.g. barium and copper). When dealing with mixtures of metal ions, results can be confusing, or at best, one element’s colour will dominate the overall colour of the flame (e.g. sodium), masking any other metal colours.

QuantitativeAnalysis–Howmuchofitisthere?

Determining the presence of cations and anions is only part the job. Generally, it’s very important to know how much of each ion is present. This is because there are many ions that occur naturally in the environment and are quite harmless, even beneficial in low concentration, but harmful at high concentrations (e.g. Zn and Cu). Determining amounts or concentrations of specific substances is called quantitative analysis.

QuantitativeSulfateAnalysis You are expected to have performed at least one specific quantitative analysis, namely the determination of the sulfate content of a lawn food or other fertiliser. The chemistry involved in this analysis has been covered in previous modules, so there should be nothing new to you. The analysis is simply a practical application of the chemistry you have already covered and serves to introduce you to a particular type of analytical methodology called Gravimetry. You also need to evaluate the reliability/validity (precision/accuracy) of results obtained from a quantitative analysis. To do this, you need to have a clear understanding of the difference between the reliability of an analytical result and the validity of an analytical result. Reliability is concerned with the extent to which a test instrument, or analytical procedure (such as gravimetry), is able to produce the same data when measured at different times, or by different users (assuming, of course, that the phenomena being measured has not actually changed!). This is very important because if our analyses are not reliable then the data obtained from them is at best worthless, and at worst, misleading. But reliability on its own is not enough, the instruments and procedures used in an analysis must also be shown to have validity. Validity is concerned with the extent to which a test instrument, or analytical procedure (such as gravimetry) is able to actually measure what it is supposed to measure! For instance, does the analytical procedure we are using to measure sulfate truly give us an accurate measure of sulfate? What factors might give data above or below the true value? Reliability and validity go hand in hand, and effective chemical analyses require both. It should be noted, however, that it is possible for an analysis to be highly reliable and yet not be a valid measure of what we are analysing. For example, consider the gravimetric determination of the amount of sulfate in fertiliser by precipitation of SO42- with barium ions, i.e. Ba2+

(aq) + SO42-

(aq) ⇌ BaSO4 (s) (Ksp = 1.1 x 10-10) If we were to perform this analysis five times, using only a stoichiometric amount of barium ion, then we may get a set of results that are quite reproducible (e.g. 20.0 %, 20.1 %, 19.9 %, 20.1%, 20.0 %). However, because we used only a stoichiometric amount of barium ions, a small amount of SO42- ions will remain in solution. Thus, our gravimetric procedure would not be giving us the true value for sulfate (which in this example may be 22.0 %), therefore the validity of the procedure would be said to be poor. However, if we were to use an excess of barium thereby forcing the reaction as far to the right as possible, the validity of this analysis would be greatly improved. AtomicAbsorptionSpectroscopy(AAS) All the previous tests we’ve covered are capable of detecting the ions of interest down to around 0.1 mmol (10-4 mol). The environmental monitoring requirements of today’s regulations require detection and quantitation of some elements down to the parts per billion (ppb, μg/L, 10-9 mol). An analytical technique that is regularly used to measure concentrations of metal ions at ppm to ppb levels is atomic absorption spectroscopy (AAS). AAS is an Australian invention first developed by Alan Walsh at the CSIRO in the early 1950’s. The principle of AAS is quite simple. If the ground state atoms of an element are irradiated with a wavelength of light from the atom’s own emission spectrum, then those ground state atoms will absorb some of that light. The amount of light that is absorbed will be proportional to the number of atoms present. By measuring the amount of light absorbed it is possible

to determine the concentration of atoms present. The irradiating light is derived from a special lamp (called a hollow cathode lamp) that produces an emission spectrum of the element being studied. Typically, a different lamp is required for each element being measured.

688 nmEmission Spectra Photomultiplier

688 nm

Inte

nsity

slit

Element absent

688 nm

Inte

nsity

Element present

The amount of light absorbed is proportional to concentration

The great sensitivity of AAS comes from the fact that it is measuring the nearly 100 % of atoms that are in their ground state. This differs from the older technique of atomic emission spectroscopy (AES), which only measures the very small (<0.1 %) quantity of atoms in an excited state that generate the spectrum. AAS is widely used to monitor: Low levels of heavy metals such as lead, mercury, cadmium, chromium, copper and zinc in

applications such as Cu and Zn in drinking waters, Hg in fish, Zn in oysters, lead in blood.

Concentration of trace elements in soils and foods.

Low-level contaminants in foods, medicines and other manufactured goods.

Some advice: AAS is used to introduce you to the concept of absorption spectroscopy, a technique commonly used in chemical analysis due to its specificity and quantitative nature. So apart from being able to explain the operating principle of an AAS (and its importance), you will need to be able apply the principles of absorbance spectroscopy (i.e. prepare a calibration plot and use the plot to determine a concentration). Note that questions using the concepts of absorption spectroscopy may not necessarily be AAS related (e.g. see 2009 Q25). You should also be well-versed in making and interpolating calibration plots (similar to 2016 Q27 and 2017 Q22).

Chemistry and the Atmosphere Composition and Structure of the Atmosphere Troposphere – ~11 km above earth’s surface – deeper in tropics and shallower near poles.

Temperature decreases with altitude to a minimum of ~-15oC at an altitude of around 15 km. Good convectional mixing of air due to temperature profile.

Stratosphere - from 15-50 km above earth’s surface. Temperature increases with altitude to ~0 oC. Very little mixing of air due to temperature profile, hence very stable air composition.

Mesosphere (to about 85 km) and then the thermosphere.

Composition of the atmosphere is essentially constant, with regards to N2, O2 and noble gases, up to 90 km, it just becomes less dense with altitude. N2 (78.1% v/v), O2 (21%), H2O (0-4%), Ar (1%) other minor constituents: CO2, Ne, He, CH4, Kr, H2, N2O, CO, O3, NO, NO2, NH3, SO2, H2S.

Main Atmospheric Pollutants POLLUTANT SOURCE CO combustion (cars, cigarettes, fires, slow combustion stoves) O3 photochemical smog, lightning NOx (NO + NO2) combustion (cars, power stations) hydrocarbons vehicles, factories using solvents VOCs (volatile organic compounds) industry, commerce (dry cleaners), domestic SO2 combustion, metal extraction, chemical manufacturing CFCs - 1996 from refrigerators, air conditioners, foam plastics Ozone Ozone, O3, is naturally present in the atmosphere near ground level, in clean air, at 0.02 ppm. In the stratosphere it can reach up to 12 ppm. Ozone has a very important role of absorbing most of the biologically damaging UV-B radiation from the sun. It is, however, poisonous to humans and other living organisms. Human activity is producing ozone at ground level where we do not want it and destroying it in the stratosphere where we do want it. Ozone Structure and Properties Ozone, O3, is an allotrope of the element oxygen, with the main allotrope being oxygen gas, O2. Allotropes are different forms of elements in the same state (eg diamond and graphite are allotropes of carbon). Molecular oxygen, O2, has two oxygen atoms bonded together. Ozone, O3, consists of three O atoms joined together – where both electrons in a shared pair in a covalent bond come from one atom. This is a Coordinate Covalent Bond.

Other examples of compounds with coordinate covalent bonds include: hydronium ion ammonium ion

Properties of O2 and O3

Difference in Reactivity of O2 and O3 O2(g) 2O(g) H = +498 kJ/mol O3(g) O2(g) + O(g) H = +106 kJ/mol It requires considerably less energy to break up the ozone molecule into an excited oxygen molecule and excited oxygen atom (free radical). The excited oxygen atom readily reacts with many species. Ozone in the Atmosphere

Ozone in the Troposphere Ozone, a secondary pollutant, formed in photochemical smog, is potentially very harmful, causing respiratory irritation and changes in lung function and can affect plant growth and agricultural production. It is formed from the chemical reactions in the atmosphere between exhaust gases (primary pollutants) from motor vehicles, sunlight and oxygen. The reactions

PROPERTY O2 O3 structure and bonding

linear molecule bent molecule (117o)

colour gas, colourless liquid, pale blue, bp -183oC solid, pale blue, mp –219oC

gas, pale blue liquid, deep blue, bp-111oC solid, black-violet, mp –193oC

odour odourless strong, distinctive odour essential for living

organisms poisonous to life

density similar to air ~1.5 times air water solubility sparingly soluble significantly more soluble reactivity strong oxidising

agent; forms oxides with many elements

much stronger oxidising agent; much more reactive; easily decomposes to O2

uses breathing aid making steel combustion fuel for space shuttles

sterilisation water purification bleaching agent

H3O+

HO

HH

+

N H 4+

H

N H

HH

+

depend on high motor vehicle numbers, weather conditions and sunlight. Ozone concentration is the accepted indicator of smog level.

Photochemical smog producing reactions: Ozone forms in the lower atmosphere when sunlight is very intense and concentrations of NO2 are well above clean air levels.

NO2 + sunlight NO + O O + O2 O3

NO can destroy ozone. NO + O3 NO2 + O2

NO and NO2 are predominantly formed from motor car exhausts. While levels of NO produced from the car exhaust are greater than NO2, other exhaust products including carbon compounds known as volatile organic compounds (VOCs) can convert NO back to NO2, leading to high levels of ozone.

NO + O2 + VOCs + sunlight NO2

Additionally, peroxyacyl nitrates (PANs), which can cause people’s eyes to sting during photochemical smog, can also be formed.

NO2 + VOC peroxyacyl nitrates (PAN)

Ozone in the Stratosphere

Maximum concentration (~12 ppm) of ozone occurs at about 25 km above the earth’s surface. Ozone in the stratosphere absorbs most of the solar UV-B (320-280 nm) radiation, which, if it reaches ground level, causes skin cancer, eye cataracts, decreased immune response, plant damage and severe sunburn. These reactions are occurring all the time in the stratosphere:

O2 + UV light 2O

O + O2 O3 (exothermic) O + O3 2O2 (exothermic) O3 + UV light O + O2

The Ozone Hole

In 1985 Dr Joe Farman (British Antarctic Survey) discovered that the concentration of ozone in the stratosphere over the Antarctic had been decreasing since the early 1970s by up to 50%. The ‘hole’ appears in spring ie September – October, each year. An ozone hole has also been observed over the Arctic since the mid 1990s.

What causes the ozone hole? Synthetic chlorinated fluorocarbons (CFCs or Freons) are major contributors to the destruction of stratospheric ozone. They are extremely stable and eventually find their way up to the stratosphere. CFCs are man-made molecules and were used as propellants in spray cans, as gases in the production of foams and in refrigeration systems. Reactions (example):

CCl3F + UV light Cl + CCl2F

Because of continual formation and destruction of O3 in the stratosphere there is always O as well as O2 and O3.

Cl + O3 ClO + O2 ClO is very reactive. ClO + O Cl + O2 (O from O3) Overall: O3 + O 2O2

Overall, ozone has been converted to oxygen gas and Cl atom (free radical) has not been used up, it is still available for many more ozone destroying reactions. 1 Cl atom can destroy up to 100,000 ozone molecules. This is a chain reaction. Cl acts as a catalyst. Cl is eventually removed by reactions (example):

Cl + CH4 HCl + CH3 ClO + NO2 ClONO2

Why Over the Antarctic and Why in Spring? Once spring arrives, the ozone layer in the upper atmosphere becomes severely depleted. In winter over Antarctica the stratosphere is extremely cold. This leads to solid particles forming that catalyse the reaction: • HCl + ClONO2 Cl2 + HNO3

As soon as the sun comes up in spring, it reacts with the accumulated Cl2 and the reactive Cl atoms are formed. Cl2 + UV light 2Cl Added effect with CFCs e.g. CCl3F + UV light Cl + CCl2F Cl atoms (radicals) then react with O3 as already seen. As there is only a finite amount of Cl2 formed in the winter and it is virtually all used up by early summer, the ozone destruction returns to the ‘normal’ levels due to CFCs.

Solutions In 1987 the Montreal Protocol was developed and since added to. Western nations have banned the production of CFCs and halons (compounds of bromine, chlorine, fluorine and carbon, eg CBrClF2) and developing countries have been phasing out production. CFCs have been replaced by hydrochlorofluorocarbons, HCFCs (contain H and C as well as F and Cl) and hydrofluorocarbons, HFCs (contain no chlorine or bromine). HCFCs and HFCs contain C-H bonds that are readily attacked and decomposed in the troposphere. Additionally, HFCs have no C-Cl bonds and have no ozone depleting effects. Monitoring Ozone Concentrations in the Stratosphere Ozone has been measured by ground based instruments since the early 1900s. These instruments measure the amount of UV radiation reaching the earth’s surface at each site. Ozone concentrations are measured in Dobson Units (DU) where 1 DU is 2.7 x 1016 molecules cm-1. Nowadays satellites carry TOMS (total ozone mapping spectrophotometers) which are able to measure atmospheric ozone distribution by season, latitude and longitude. Ozone contour maps are available for viewing on the NASA website. Nomenclature of Haloalkanes Revise the rules for naming haloalkanes including isomers (refer to your text book).

Sample Past HSC Exam Questions and Answers Q35 (1998) (ii) Once spring arrives, the ozone layer in the upper atmosphere becomes severely depleted.

1. What causes the depletion of ozone at the beginning of spring? During winter ice clouds form over the Antarctic and Cl2 forms by the reaction:

HCl + ClONO2 Cl2 + HNO3 The sun comes up and Cl atoms form (from the Cl2 that has formed in the winter):

Cl2 + UV light 2Cl and CCl3F + UV light Cl + CCl2F Chlorine radicals then react with O3.

Cl + O3 ClO + O2 ClO radicals react with O atoms, regenerating the chlorine radical.

ClO + O Cl + O2 Overall: O3 + O 2O2

2. Why is the depletion of the ozone layer a concern?

Depletion of the ozone layer allows harmful UV-B light to penetrate to the earth’s surface where it causes damage to many molecules, biological, synthetic polymers etc. Skin cancer is just one effect. (iii) In the lower atmosphere, nitrogen dioxide contributes to the production of photochemical smog.

1. Name one major source of NO2(g) in the lower atmosphere. Major sources of NO2 are combustion from power stations and motor vehicles.

2. Briefly describe some of the effects of photochemical smog on humans. Photochemical smog [relatively high concentrations of NO and NO2 (NOx) and volatile organic compounds (VOCs), peroxyacyl nitrates (PANs) and ozone] causes people’s eyes to sting and water, breathing difficulties, headaches and fatigue.

2001 Explain why ozone is considered a pollutant in the troposphere but essential in the stratosphere. Ozone is poisonous to humans and other living forms. It causes respiratory difficulties, headaches and fatigue and in the lower atmosphere is formed by reactions between NO2 and O2. However in the upper atmosphere, O3 occurs naturally in much higher concentrations and has the beneficial role of screening out the harmful UV-B rays from the sun.

Q25 (2002) 6 marks (c) Discuss how CFCs damage the ozone layer, using relevant equations. In the stratosphere short wave UV radiation breaks a chlorine atom off the CFC molecule. CCl2F2 + UV light Cl + CClF2 (Answer must mention chlorine atom or radical, not chloride ion or chlorine molecule.) The chlorine atom then takes an oxygen atom from an ozone molecule forming a ClO free radical. Cl + O3 ClO + O2 The free radical, ClO, is very reactive and reacts with oxygen atoms present reforming O2 molecules and regenerating Cl atoms. ClO + O Cl + O2 The reactive Cl atom can destroy thousands of O3 molecules. 2003 – no Section B question Q27 (2004) 7 marks) A table was given showing the ozone depletion potential (ODP) of a series of compounds. These consisted of CFCs showing high ODP, some HCFCs showing a small amount of ODP and some HFCs showing zero ODP. Students were asked to discuss ozone depletion problems associated with the use of CFCs and the suitability of substitute chemicals.

a) Briefly discuss why and how CFCs were used, how stable they are and what happens when they get into the stratosphere.

b) Include a brief description (including equations) of how ozone depletion occurs. CCl2F2(g) + UV CClF2(g) + Cl(g) O3(g) + Cl(g) O2(g) + ClO(g) Regeneration: ClO(g) + O(g) O2(g) + Cl(g) c) Explain why HCFCs (still containing some Cl) and HFC (containing only C and F) are

better alternatives with HCFCs still having some ODP (but much less than CFCs). Section 1 (multiple choice) (2004) 1 mark A labelled diagram showing the layered nature of the atmosphere, showing the temperature profile and altitude, was given. Students needed to know that the layers of the atmosphere from the earth’s surface are troposphere, stratosphere, mesosphere and thermosphere. 2005 The only atmospheric related question was a multiple choice question regarding the naming of a haloalkane. Q28 (2006) 4 marks Discuss the ozone hole over the Antarctic including how it is measured and how it has developed over time.

Ozone in the stratosphere is measured by instruments on the ground, in satellites and in balloons. The instruments, UV spectrophotometers, on the ground measure the intensity of light received at a wavelength at which ozone absorbs and then at wavelengths nearby at which ozone doesn’t absorb. A comparison of these measurements allows us to calculate the concentration of ozone per unit area of the earth’s surface where the measurement was made. The spectrophotometers aboard satellites (TOMS) measure ozone concentration as a function of altitude and geographic position. Discuss how the concentration of ozone in the stratosphere over the Antarctic has decreased dramatically over the past 20 years and include the role of CFCs. The chemistry including equations for the reaction of chlorine radicals with ozone is not required in the answer. Answered earlier. Multiple choice (2007) The question asked what was the role of the chlorine free radical on ozone in the stratosphere. Answered earlier. Q16 (2007) 5 marks This showed a diagram of the layered structure of the earth’s atmosphere and asked for the layers to be identified. It then asked the effect of ozone on each of the layers. Q18 (2008) 5 marks Comparison between oxygen, O2, and ozone, O3. A multiple choice question (3) required identifying them as allotropes. A 5 mark question required drawing the Lewis dot structures for both O2 and O3 (2 marks) and asked to account for the differences in properties of O2 and O3 on the basis of molecular structure and bonding (3 marks). O2 is a linear molecule whereas O3 is bent. Due to the larger molecular weight of ozone its intermolecular forces (dispersion forces) are stronger resulting in higher melting and boiling points. O3 is much more chemically reactive as it requires less energy to decompose an ozone molecule than an O2 molecule. Q21 (2008) 4 marks Students were given a graph of atmospheric CFC-11 concentrations measured at Cape Grim Baseline Air Pollution Station, Tasmania, over time pre 1993 and post 1993 and asked to explain the graph (2 marks). Concentration of CFCs rose until 1993 mainly due to emissions of man-made CFCs. At the Montreal Protocol in 1987 many countries agreed to ban the manufacture and use of CFCs. A decrease in atmospheric CFC concentration after 1993 reflects the effectiveness of the Protocol. Why is it important to monitor the concentration of CFCs in the atmosphere? (2 marks) CFCs were found to be the cause of destruction of much ozone in the stratosphere, causing an ‘ozone hole’. It is important to monitor CFC concentrations so as to be able to check on the effectiveness of the ban and to check the compliance of signatory countries. Q18 (2009) 5 marks “There has been an increase in the concentration of the oxides of nitrogen in the atmosphere as a result of combustion. Assess both the evidence to support this statement and the need to monitor these oxides.” Evidence of the increase in atmospheric concentrations of NOx comes from direct atmospheric measurements and monitoring by organisations such as CSIRO and NSW EPA. The major sources

of NOx are the combustion of coal in power stations and hydrocarbon fuel in motor vehicles. It is important to include relevant balanced chemical equations such as: N2(g) + O2(g) → 2NO(g) (combustion) 2NO(g) + O2(g) → 2NO2(g) (oxidation of NO in lower atmosphere in presence of sunlight) NO2(g) is involved in the production of ozone, a pollutant, in the lower atmosphere in the presence of sunlight. NO2(g) + UV → NO(g) + O(g) O(g) + O2(g) → O3(g) Catalytic converters on cars reduce the emissions of NOx by emitting nitrogen as N2. NOx also contributes to acid rain which adversely affects both living organisms (eg forests and plants, aquatic life) and man-made structures. 2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq) HSC Chemistry 2010 Paper – no section B question. Q22 (2011) 4 marks Write the chemical equations for atmospheric reactions resulting in ozone depletion. Refer to equations shown previously. “Outline one method used to monitor stratospheric ozone.” Refer to previous section labeled ‘Monitoring Ozone Concentrations in the Stratosphere’. Multiple choice (2012) Question 9 asked which of a series of compounds contained a coordinate covalent bond? See Ozone Structure and Properties section of these notes. Question 12 asked for the correct IUPAC name for a hydrochlorofluorocarbon. See HSC textbook. Q29 (2012) 5 marks Draw a labelled diagram to show the layered structure of the atmosphere. In your diagram include: the names of TWO atmospheric pollutants, positioned in the layers where the detrimental impact occurs and the names of the sources of the two pollutants identified. Sample Answer right (figure) and below: O3 – photo-oxidation or industrial production of O• → O3; CO – incomplete combustion, motor vehicles, other combustion; NOx – internal combustion engine; SOx

– fossil fuels combustion; CO2 – fossil fuels; CH4 – agriculture, anaerobic processes Q30 (2013) 6 marks Account for the changes in ozone concentration above Antarctica between 1979, 1994 and 2012. Your response should include relevant equations. Students should first show the cause of ozone depletion in the stratosphere by chlorine free radicals, and how those free radicals are created by the reaction of CFC’s with UV light. Students

TroposphereO3, NOx, SOx,CO, CO2, CH4

Stratospherechlorofluorocarbons

Mesosphere

Thermosphere

should then explain the differences between the different ozone concentrations in different time periods:

- 1979: early extensive use of CFC’s - 1994: after Montreal Protocol; CFC’s banned and being phased out but ongoing

damage by CFC’s already in the atmosphere prevents the increase of O3. - 2012: Ozone hole is beginning to improve.

Students may also want to account for why the hole is over Antarctica specifically by discussing how increasing temperatures causes the rapid production of new Cl radicals faster than O3 can be produced or reproduced. Multiple Choice (2014) Question 1 asked in which layer of the atmosphere is ozone considered a pollutant? Answer troposphere. Question 11 asked which of the following does NOT represent the formation of a coordinate covalent bond? Answer A. Multiple Choice (2015) Question 1 asked in which layer of the atmosphere does ozone absorb the most UV radiation? Answer stratosphere. Question 3 asked which of the following is the most suitable replacement for CFCs in terms of reducing their environmental impact? (A) CH4, (B) CH2F2, (C) CH2ClF, (D) CHCl2CCl2F. Answer B. Q22 (2015) 7 marks This question provided a table of data for ozone concentrations over 50 years in the upper atmosphere above Antarctica and asked for the data to be plotted and then to describe a method by which this data could have been measured. Sample answer: Could be measured by a ground-based UV spectrophotometer, which analyses UV light intensity at a wavelength at which ozone absorbs, and compares the intensity to a nearby wavelength of UV that ozone does not absorb. The amount of absorption by ozone provides information about ozone concentrations. Similar measurements over time allow ozone concentrations to be monitored.

Q30 (2016) 7 marks This question addresses the steps that have been taken to reduce the damage caused to stratospheric ozone levels by CFCs. Sample answer: First, use chemical formula to show how O3 is destroyed CFCs (pick an example CFC to illustrate your point). Second describe the phased replacement of CFCs with alternate chemicals such as HCFCs and HFCs.

Question 21 (2017) (5 marks)

(a) Outline ONE effect of ozone in the troposphere and ONE in the stratosphere.

Sample answer: In the troposphere, ozone is considered a pollutant in the troposphere as it is much more reactive than molecular oxygen. With an input of energy, the molecule may react to form an oxygen free radical, which may damage biological molecules such as protein and DNA, causing harm to biological organisms. In the stratosphere ozone absorbs harmful UV-B radiation which may cause damage including skin cancer and eye cataracts in humans. Ozone absorbs UV light generating molecular oxygen and an oxygen free radical, which may then reform more ozone molecules.

(b) Qualitatively compare TWO properties of oxygen (O2) and ozone (O3).

Sample answer: Ozone is a polar molecule as its structure is bent, with its central atom containing a lone pair which gives the molecule polarity. This is contrasted with molecular oxygen which is a non-polar molecule.

Ozone is significantly less stable than molecular oxygen, this means that it requires a less input of energy to generate an oxygen free radical than in the case of molecular oxygen. This is demonstrated by the following chemical reactions:

O2(g) -> 2O(g) ∆H = +498 kJ/mol

Compared with:

O3(g) -> O2(g) + O(g) ∆H = +106 kJ/mol

(c) Using ONE chemical equation, show how a chlorine radical (Cl•) reacts with 1 ozone.

Sample Answer: Cl + O3 -> ClO + O2

Monitoring Water Quality The quality of our water on the earth is absolutely vital to all forms of life. Drinking water, water for agriculture, for recreation and for plant and animal life must be sufficiently pure and unpolluted. Different uses of water require different standards eg water for drinking needs to be purer than water used to grow crops and water used by an analytical chemist must be even more highly purified. Aspects of water quality including microbiological, physical, inorganic and organic contamination can be tested and monitored in many different ways. Water quality indicators:

Biological: bacteria, algae Physical: temperature, turbidity, colour, salinity, suspended solids, dissolved solids Chemical: pH, dissolved oxygen, biochemical oxygen demand (BOD), P and N, organic (eg

pesticides and herbicides) and inorganic (eg metal ions) compounds Aesthetic: odours, colour, floating matter Radioactive: , and radiation emitters

Tests for detection and measurement of concentration of common ions. Precipitation (gravimetric analysis) – useful as a qualitative test for many ions and for

quantitative analysis where concentrations are high. Example add AgNO3 to precipitate Cl- and add Ba2+ (as the nitrate or chloride) to precipitate SO42-.

Atomic Absorption Spectrophotometry (AAS) for metal ions. AAS was developed in the 1950s by an Australian CSIRO scientist, Alan Walsh. It uses the unique light absorption spectrum of each metal atom to measure the concentration of the metal in the water sample and is a useful method for ppm concentrations.

Ion Selective Electrodes (ISEs) for low ion concentrations eg fluoride, chloride, nitrate, sulfate ion concentrations.

Colorimetric – used commonly for measuring low phosphate concentrations. Typically ammonium molybdate, (NH4)2MoO4, is added to the phosphate sample, followed by ascorbic acid (vitamin C). This forms an intense blue colour (molybdenum blue), which can be compared to standard phosphate solutions treated identically. This method is useful for detection of 0.01 to 3 ppm. Nitrate can also be measured by a colorimetric assay.

Total Dissolved Solids (TDS) Total dissolved solids (TDS) is the mass of solid dissolved in a unit volume of water. It is generally measured as the mass of residue by evaporation of filtered water from a known volume of water sample. The residue comprises both organic and inorganic material and units are usually mg/L or ppm. As the majority of the solids are typically salts (ionic), electrical conductivity of the water is often used to measure TDS. Units are in Siemens per cm (S cm-1 or mS cm-1).

Test for Hardness.

‘Hard’ describes water that will not lather with soap but instead forms a ‘scum’. The soap ‘scum’ is the insoluble salt of the long chain (~C16-C18) alkanoic acids of the soap. The ion causing hardness is predominantly calcium but magnesium and aluminium can also be present in significant amounts and are measured by the test if present. The metal ions in the water causing hardness originate from the rocks through which the water flows. The Sydney Basin is

predominantly sandstone so its water is soft whereas Adelaide and Brisbane have relatively hard water due to the local limestone. eg Ca2+(aq) + C17H35COO-(aq) Ca(C17H35COO)2(s) Hardness is measured by an EDTA (ethylenediamine tetraacetic acid) titration, which binds the calcium and magnesium ions. Turbidity

Turbidity is the measure of the cloudiness or opaqueness of the water and is proportional to the type and amount of suspended matter. It is measured in nephelometric turbidity units (NTUs). Turbidity restricts photosynthesis to the upper layers of water and ultimately affects the fish life. Acidity

Acidity is readily measured by a pH electrode (an Ion Selective Electrode) and should be between 6.5 and 8.5 for potable (drinkable) water. Dissolved oxygen and biochemical oxygen demand (BOD) Small amounts of oxygen gas dissolve in water (~9 ppm at 20 oC) and are essential for the existence of life in the water and the breakdown of organic matter. The biochemical oxygen demand (BOD) is a measure of the dissolved oxygen required for the complete breakdown of the organic material in the water by aerobic bacteria. The BOD in natural, unpolluted waterways is typically less than 5 ppm. There are two main methods of measuring dissolved oxygen in water: Titration (Winkler method). This involves conversion of Mn2+ to Mn(IV) in alkaline solution by dissolved oxygen, followed by titrating the acidified mixture with iodide (I-), where the Mn(IV) oxidises the iodide to I2. The I2 can then be determined by titration with sodium thiosulfate, Na2S2O3, in the presence of starch. The starch forms an intensely blue colour with the I2, which disappears at the end point. Oxygen-sensitive electrode (electrolysis method). This uses a platinum disc anode at the end of an insulating rod and a silver ring as a cathode attached to the outside of the rod. This is surrounded by buffered KCl solution with a permeable membrane that allows oxygen in the test sample to diffuse to the anode where it is electrolysed: O2(aq) + 4H+(aq) + 4e- 2H2O(l) (at the cathode Ag(s) + Cl-(aq) AgCl(s) + e- ) The current from the electrolysis is measured to determine the concentration of the dissolved oxygen. One method for measuring BOD (where BOD is expected to be low) is by taking the water sample, adding nutrient and incubating it in a sealed container at 20oC for 5 days. The difference between the dissolved oxygen levels before and after incubation is the BOD.

N:P ratio N: present mainly as nitrate, nitrite, ammonia and organic N. P: present mainly as

phosphate and organic P. Sources of N and P: fertilisers, sewage, detergents.

Recommended levels (EPA): N – 0.1-1ppm P – 0.01-0.1ppm

- a ratio of 10:1 for N:P. A lower ratio indicates conditions favourable for eutrophication (the enrichment of a water body by nutrients such as fertilisers and/or sewerage) and hence high algal growth, algal blooms. Algal blooms result in low oxygen levels in the water and the death of many aerobic organisms.

N and P are constantly monitored in streams and rivers, especially P, as it is mostly the growth-limiting nutrient. P – measured by molybdenum blue method, a colorimetric method. (P is oxidised to phosphate and then reacted with molybdate ions in an acidic solution and then reduced by ascorbic acid to yield a blue product.) N – nitrate reduced to nitrite and then colour-producing reagent added to form pink-purple azo-dye whose concentration is measured colorimetrically.

Factors Affecting Concentrations of Ions and Other Substances in Natural Water Bodies.

Rainwater can dissolve CO2 (also SO2 and NOx) from the atmosphere, cations eg Ca2+, Mg2+, Fe3+, Mn2+, Cu2+, Zn2+, and anions eg SO42-, Cl-, CO32-, from rocks and soils.

Agricultural activities and land clearing can result in fertilisers, pesticides, herbicides and inorganic salts being added to natural water.

Industrial activities and mining – organics from petrochemical industries, also heavy metals such as Cu, Hg, Cd, Cr, Pb, Zn, As, Au.

Sewage and stormwater runoff – microorganisms, organic substances. Leaching from rubbish dumps and mine tailings eg Cd, Hg, Pb from batteries, Zn, nitrate,

phosphate, from decaying organic waste. Acid Mine Drainage. Mining of coal and minerals has left large volumes of tailings (waste). Often the tailings contain pyrite or iron(II) sulfide which, when exposed to oxygen and water, result in high levels of iron, aluminium and sulfuric acid. These metals and acid find their way into surface and ground waters and soil. Acid mine drainage can go on for thousands of years. Only in the last 20 years have the potentially catastrophic effects of acid mine drainage been known. Purifying and Sanitising Mass Water Supplies. Drinking (Potable) Water Catchment is protected with people and activity excluded from the areas immediately

surrounding water storage facilities (for Sydney). Water is treated. Ferric chloride (Sydney) or aluminium hydroxide is used as a flocculating

agent ie to cause small precipitate particles to come together to form large particles which can then settle out and the water is filtered through sand and gravel. Next, water is chlorinated (by hypochlorite ions, OCl-) to kill bacteria and some viruses. Fluoride, in the form of NaF, is sometimes added (Sydney) to strengthen tooth enamel in growing children. Water quality is monitored at every stage.

Waste Water Industrial – strict guidelines have been set by the Environment Protection Authority (EPA)

and local water authorities as to what can be discharged into sewers eg organic solvents are banned and heavy metal concentrations in waste water restricted (arsenic, max 1 ppm, mercury, max 2 ppm etc).

Domestic – there are many treatment plants for Sydney water, 10 coastal and 17 inland. Level of treatment ranges from ‘high-rate’ primary (physical removal of solids from fast flowing effluent) to ‘tertiary’ (removes dissolved organic and inorganic compounds including plant nutrients, N and P).

Membrane Filters

Membrane filters are widely used for filtering very small particles (> 0.2 m) from drinking water, treated sewage and water for industry (especially beverages including bottled water). (Dirty water in at top, clean water out the bottom.) A membrane filter is a thin film of synthetic polymer with small pores of uniform size. These can filter out much smaller particles than paper or sand as they have much smaller pore sizes, liquid flows through them quickly, they are strong and they can be cleaned and reused.

Sample Past HSC Exam Questions and Answers Q26 (2002) 5 marks (a) Describe a test you have used to determine whether a given example of water is ‘hard’ or ‘soft’. (b) A sample of hard water contains 6x10-4 mol L-1 of magnesium carbonate. Calculate the mass, in mg, of magnesium carbonate in 150 mL of this sample.

(a) Shaking a water sample with soap is a method of determining whether the water is ‘hard’ or ‘soft’. If the water is hard then a precipitate or scum will form. The scum is a precipitate of calcium and magnesium stearates.

(b) moles MgCO3 in 150 mL = molarity x volume (L) = 6 x 10-4 mol/L x 0.150 L = 9 x 10-5 mol

mass MgCO3 = mol x molar mass = 9 x 10-5 mol x 84.3 g/mol = 0.007587 g = 7.587 mg = 8 mg (1 significant figure) Q27 (2002) 5 marks Describe the physical and chemical processes needed to purify and sanitise a town water supply.

(a) Firstly, protecting the catchment area is important for town drinking water. (b) Treatment of drinking water involves precipitating small particles in a process called

coagulation or flocculation. Chemicals such as Fe(OH)3 or Al2(SO4)3 are used as flocculating precipitates.

(c) Treated water is then allowed to settle and sludge forms at bottom of tank and is removed.

(d) Water is then filtered through sand and gravel (and sometimes activated charcoal) to remove suspended particles, minerals and bacteria.

(e) Clear water is then treated with chlorine to kill bacteria and sometimes fluoride is added to strengthen tooth enamel in children.

Q26 (2003) 4 marks The question asked about eutrophication and its monitoring.

(a) Eutrophication is the enrichment of a water body by nutrients which come from sources such as fertiliser runoff from agriculture or sewage from sewage treatment works, stormwater runoff etc.

(b) High levels of eutrophication of a water body can result in an algal bloom. Eutrophication can be measured indirectly by measuring the concentrations of N and P in the water. A higher ratio of N:P than 10:1 indicates conditions favourable to eutrophication. Monitoring of phosphate concentration alone is sufficient to indicate possible high nutrient concentrations that require the water body to be further analysed.

Q21 (2004) 7 marks The question asked:

(a) to distinguish between qualitative (what species are present) and quantitative (how many or what concentration of each species) analysis of a water sample. (2 marks)

(b) to describe 2 factors that affect the concentrations of ions in natural water bodies. You could describe any 2 of farms (increased concentrations of NO3-, PO43-, from fertilisers, also pesticides and herbicides), garbage tips (increased concentrations of various ions that have leached from the refuse and soil), mines (ions and other species leached from the tailings), factories, sewage treatment works etc and state that they are upstream from where the water samples have been collected. (2 marks)

(c) state which reagents you would use to identify the ions Cl-, PO43- and SO42- and what you would observe in each case. Use AgNO3 to react with Cl- and PO43- and observe a white precipitate of AgCl or a yellow precipitate of Ag3PO4. Ba(NO3)2 will react with SO42- forming a white precipitate. (3 marks)

Q14 (2005) 1 mark The question showed a river flowing from the mountains, past a limestone quarry and then a town. Results of 3 tests were given, dissolved oxygen, biochemical oxygen demand and pH. The multiple choice answer required knowing that the river:

(a) near its source will have relatively high dissolved oxygen, low biochemical oxygen demand and close to neutral pH

(b) near the limestone quarry has slightly less dissolved oxygen and slightly higher biochemical demand and slightly alkaline pH due to the slight solubility of CaCO3 and the subsequent reaction of carbonate with water to produce HCO3- and OH-

(c) near the town will have much less dissolved oxygen, much higher biochemical oxygen demand (will have effluent from the town flowing into it) and a lower pH (industrial pollution often results in acidic effluent).

Q15 (2005) 1 mark Used the example of the Winkler Method for determining dissolved oxygen concentration for a stoichiometric problem. Q25 (2005) 6 marks This question required:

(a) calculation of the percentage of dissolved solids from a water sample taken from a creek – sample must be filtered first before it is evaporated to dryness leaving the TDS

(b) suggesting and describing a chemical test to detect the presence of lead ions such as adding aqueous iodide ions and looking for a bright yellow precipitate (give a balanced equation also)

(c) justifying the need to monitor ions used by society – use one ion example only in your answer.

Q26 (2005) 7 marks This question gave a diagram of a water catchment area for a town and asked to describe 2 possible sources of contamination and assess purification methods that could be used for the water before it reaches the town. Some sources of contamination could be:

(a) farmyard – animal excrement, fertilisers (b) vegetable patch – fertilisers and pesticides/herbicides (c) logged native forest – increased runoff taking decaying organic material, from both

plant and animal origins, soil particles (d) undisturbed pine forest – possible use of fertilisers and pesticides/herbicides? (e) human recreational lake activity – fuel pollution, discarded packaging eg empty drink

bottles, gladwrap, cigarette butts etc. The second part of the question required going through the steps of water purification (flocculation, sedimentation, filtration, chlorination and sometimes fluoridation) and assess them. Q23(b )(2006) 3 marks The equilibrium reaction of the hypochlorite ion, OCl-, with water is given. OH- ions are a product. Part (i) required explanation that the regular chlorination of swimming pool water is necessary as the hypochlorite destroys bacteria. Part (ii) asked for an explanation of how the addition of hypochlorite to the pool affects the pH. It required discussion of the equilibrium between OCl- and OH- that is when the concentration of OCl- is increased the equilibrium is pushed to the right so more OH- forms and the pH increases (the water becomes more alkaline). Q27 (2007) 8 marks This question had a diagram representing equipment used in determining chloride ion concentration in a water sample. It required a description on how this could be used and a balanced equation. It then asked for a calculation of the chloride ion in ppm given the mass of a dried precipitate and volume and finally why determining the concentration of the chloride ion in water is important. Q17 (2008) 5 marks This question asked to choose 2 examples of different water catchment areas and analyse how the features of each will determine the water treatment necessary for safe drinking water. Many different examples are possible including the following two.

(a) Farmland including both horticulture and animals. Runoff may contain fertiliser, pesticides, faecal material and loose soil.

(b) Mining. There may be acid mine waste, contamination of heavy metals, loose rock and soil.

Steps in purifying water: screening, aeration, flocculation, sedimentation, filtration, chlorination, pH adjustment, luoridation. Q11 (2009) 1 mark This question asked to put in order 3 steps in the water purification process after screening and before chlorination. Screening; flocculation; settling; pH adjustment; chlorination. Q25 (2009) 7 marks Absorbance measurements of phosphate concentrations in 3 streams were given and students were asked to determine mean phosphate concentrations (2 marks) then analyse the phosphate

concentrations in the 3 streams (3 marks). The third part of the question asked why phosphate concentration in water is a quality issue (2 marks). One of the streams contained very high levels of phosphate. Possible sources are both natural (from leaching of soils and minerals locally) and man-made (runoff from fertilised farm land; runoff and stormwater containing animal waste; effluent from sewerage treatment works). High levels of phosphate in water can cause eutrophication and algal blooms. With algal blooms oxygen levels in water become reduced affecting the life of all organisms living in the water and less sunlight penetrates to lower levels of the stream further reducing organism growth and oxygen generation. Some algae are toxic causing illness in animals drinking the water. Q25 (2010) 5 marks What is the relationship between dissolved oxygen and BOD is and why is it important to monitor both in natural waterway? Firstly define both terms, dissolved O2 and BOD (see earlier notes). Explain that BOD is directly related to the amount of bacteria consuming O2 during the decay of organic matter hence BOD is a measure of microbial activity in the water. There are naturally a large variety of microorganisms in waterways, while some are harmless many are dangerous so it is important to monitor the level of microbial activity and if high then examine the waterway in more detail. Q31 (2010) 6 marks (a) This question part is an analytical one using a water sample from a dam. (i) Percentage of total dissolved solids (TDS) needs to be calculated from mass data given. Note that the mass of TDS must come from evaporation of the water sample. (2 marks) (ii) Describe a chemical test for chloride ions. Add aqueous silver ions to water sample and look for white precipitate of AgCl. An equation must be included. Ag+(aq) + Cl-(aq) → AgCl(s) Pb2+(aq) or Hg2+(aq) can be used instead of Ag+(aq). (2 marks) (b) Name an ion (other than chloride) that commonly pollutes waterways, identify its source and its effect on water quality. Ions such as SO42-, NO3-, PO43- (from fertilisers and agricultural runoff and sewerage), Na+ from sea water), Ca2+ and Mg2+ (from rocks), Cu2+, Pb2+, Zn2+ (from mining and leaching from rubbish dumps). Effects such as salination, eutrophication, water hardness and pollution by heavy metals could be noted. (2 marks) Q28 (2011) 4 marks Choose 2 of the following tests to measure the quality of stormwater and outline the chemical or physical principle for each. Hardness; Phosphate level; BOD; Total dissolved solids; Turbidity; Nitrate level. All methods are briefly described earlier in this section on ‘Monitoring Water Quality’. Q31 (2011) 4 marks Daily (6 days, Mon –Sat)) statistics are given on water quality in local river and students asked propose possible sources of contamination identified from this statistical data. Students should point out higher levels of turbidity and TDS on Tues and Thurs-Fri (mostly on the Tues) and very large increases in BOD and faecal coliforms Thurs-Fri accompanied by reduced levels of dissolved oxygen. The source of the Tues increased turbidity and TDS and slightly increased BOD could be rainfall in the region causing increased runoff of water containing soil including minerals and organic material into the river. The source of increased turbidity, TDS, BOD and faecal coliforms on Thurs-Fri could be a discharge from a sewerage treatment plant up

river from the sampling point. Microbes in the effluent require oxygen to proliferate hence the decreased dissolve oxygen levels and increased BOD. Q25 (2012) 3 marks Describe the process of monitoring waterways for eutrophication. Eutrophication can be monitored by measuring the concentrations of nitrate and phosphate ions in samples of environmental water. When nitrate and phosphate levels increase (particularly phosphate) algal abundance increases dramatically, causing problems for the waterway. These tests would need to be carried out regularly and at various locations to appropriately monitor nutrient levels, including nitrates, phosphates and oxygen, in order to monitor eutrophication and prevent the waterways becoming marsh-like or turning into swamps. Q26 (2013) 4 marks Explain how microscopic membrane filters purify contaminated waters, in terms of their design and composition. Students should describe that microscopic membrane filters are designed:

- To filter out very small particles. - With an even finer filter than paper or sand - For the treatment of water for drinking, or use in industry where very clean water is

needed. Students should also explain how they are composed of a thin film synthetic polymer whose pores are of uniform size and that this composition makes them strong, washable and reusable as well as fast to use due to high flow rate. Multiple Choice (2013) What is the purpose of the flame in atomic absorption spectroscopy (AAS)?

(a) To ionise the samples (b) To produce a spectrum (c) To atomise the substance (d) To provide the absorption wavelength

Answer: (c) To atomise the substance. Multiple Choice (2014) Drinking water is regularly tested to ensure that it is safe for consumption. Which of the following test results indicates the highest drinking-water quality?

Answer b.

Dissolved oxygen (mg/L)

Nitrate (mg/L)

Total dissolved solids

(mg/L)

Turbidity (NTU)

(a) 2 0.1 50 50

(b) 8 0.1 50 2

(c) 2 2 200 2

(d) 8 2 200 50

Q23 (2014) 3 marks The diagram shows a town situated near agriculture and industry The town relies on the river for its water supply.

(a) Identify ONE chemical species that could be a contaminant of the water supply.

Lead ions. Other answers could also include hydrogen ions and phosphate.

(b) Explain the need to monitor the levels of a contaminant in water supplies.

Lead is a heavy metal. It can accumulate in a biological system. Ingestion of lead can cause brain damage. Multiple Choice (2015) Q15 A diagram showing a water catchment area is provided. A site near a farmland, village, and within the river but opening to the ocean is identified as the site for a sample collection of water for analysis. Students were asked to choose what turbidity, BOD, pH and total dissolved solids results were most likely. Q18-19 students were provided with the following: A sample of pond water from a contaminated site was analysed to determine the concentration of lead ions using the following procedure. • A measuring cylinder was used to collect a 50 mL sample from the pond. • The sample was placed in a clean dry beaker. • 25.0 mL of 0.200 mol L−1 sodium chloride solution was added to the sample. • The precipitate of lead(II) chloride that formed was filtered, dried and weighed. It had a mass of 0.13 g. Q18 How could the reliability of the analysis of the pond water be improved?

(A) Analyse more samples from the same pond (B) Use 50 mL of distilled water as a control sample

(C) Analyse samples from different ponds on the site (D) Remove other contaminants from the sample before the analysis Answer A.

Q19 What was the concentration of lead ions in the sample?

(A) 5.0 × 10−3 mol L−1 (B) 5.8 × 10−3 mol L−1 (C) 9.3 × 10−3 mol L−1 (D) 10.7 × 10−3 mol L−1 Answer C.

Q29 (2015) 7 marks Students were given the procedure of a first-hand investigation conducted in a school laboratory to determine the percentage of sulfate in a lawn fertiliser is shown. • 2.00 g of a sample of fertiliser was ground up and placed in a beaker. • It was dissolved in about 200 mL of 0.1 mol L−1 hydrochloric acid, stirred and filtered. • Excess barium chloride solution was quickly added to this beaker and a precipitate formed. • The precipitate was then allowed to settle, filtered using filter paper and the residue collected. • The residue was dried and weighed and had a mass of 2.23 g. a) Suggest modifications that could be made to the procedure to improve the results of this

investigation. Justify your suggestions. Barium sulfate is a very fine precipitate. After the BaCl2 was added, the mixture could be heated for some time. This allows the fine BaSO4 precipitate to flocculate into larger particles, which would be trapped by the filter. The pores of most filter paper are still too large to effectively capture the BaSO4 precipitate, and so filtering should be done using a sintered glass crucible, which has small pore sizes. The sulfate will not be evenly distributed throughout the mixture, so to achieve more reliable and valid results, several samples of the fertiliser should be used so that an average value for the sulfate content in the mixture could be determined.

b) Calculate the percentage of sulfate in the original fertiliser sample. Ba2+(aq) + SO42–(aq) → BaSO4(s) mass of fertiliser used: 2.00 g; mass of precipitate formed: 2.23 g mass of SO42− = molar mass of SO42−/ molar mass of BaSO4 x 2.23 = 96.07/233.37 x 2.23 = 0.918 g ∴ % of SO42– in the fertiliser = 0.918 g/2.00 g × 100 = 45.9%

Q24 (2016) 7 marks Part A (4 marks). Students were asked to explain how microscopic membrane filters are used to purify contaminated water, including a labelled diagram. Sample answer: the key to responding to this question is the diagram, as it enables the student to indicate direction of flow, as well as the point at which the contaminated and purified material separates. Finally, it is important to note that this is a passive purification system, hence it needs inlet water pressure to drive the process. Part B (3 marks). Students are asked to explain why dissolved oxygen levels can be used to measure the extent of eutrophication. Sample answer: the first step to answering this question is to define eutrophication, then connect this condition to the depletion of oxygen levels that occurs when an algal bloom (rapidly) decays en masse. The extent of eutrophication may be assessed by measuring dissolved oxygen levels, as the decay process depletes, possibly even exhausts, the dissolved oxygen levels in the water. In the event that all the oxygen dissolved in the water is used up, anaerobic degradation processes will proceed, leading to the production of offensive smelling compounds like hydrogen sulphide.

Question 29 (2017) (4 marks)

Students were given the following diagram and table showing Rivertown sits at the junction of two rivers.

A simple water purification system has been purchased for the town water supply. It consists ONLY of a sedimentation tank, pH control, sand filters and a chlorination facility.

The system is to draw water either from Site X or Site Y. A water chemist has obtained the following data from each site.

With reference to the information provided, justify which of the sites, X or Y, would be the preferred water source for the town water supply.

Sample answer: Source Y would be a more suitable water source for use by the town for the following reasons. Turbidity is caused by matter suspended in the water which may be removed by the sedimentation tank bringing source Y to acceptable turbidity levels. pH is outside of the usual range for drinking water with source Y but may be corrected by the pH control facility in the water purification system. Source X has high hardness due to high concentration of calcium, and this may cause damage to local plumbing systems. Source X also has high concentration of phosphorous which may lead to the growth of algae. Since the purification system doesn’t have an appropriate system to remove dissolved phosphate, this cannot be corrected. Source Y is therefore the more appropriate source of water for the town.

Nuclear Chemistry

Students learn to: Students:

5. Nuclear chemistry provides a range of materials

distinguish between stable and radioactive isotopes and describe the conditions under which a nucleus is unstable

process information from secondary sources to describe recent discoveries of elements

use available evidence to analyse benefits and problems associated with the use of radioactive isotopes in identified industries and medicine

describe how transuranic elements are produced

describe how commercial radioisotopes are produced

identify instruments and processes that can be used to detect radiation

identify one use of a named radioisotope: - in industry - in medicine

describe the way in which the above named industrial and medical radioisotopes are used and explain their use in terms of their chemical properties

DEFINITIONS Nucleon A proton or neutron. Mass number A The sum of the number of neutrons and protons (n + p) Atomic Number Z The number of protons in the nucleus

MAZ An atom with mass number A and atomic number Z Isotopes Atoms or nuclei with different numbers of neutrons but the same number of

protons (Z), this means an isotope of a given element will have the same atomic number (that is what defines the element under consideration), but a different atomic mass number (remember atomic mass is the result of the number of protons + the number of neutrons (both particles are located in the nucleus)). Generally named by their mass number e.g. 235U, 238U

Nuclide Another (American) name for an isotope Unstable nuclei Ones that are radioactive Radioisotope Also known as radioactive isotope - an atom with an unstable nucleus that

decays radioactively Transuranic elements Elements with an atomic number above that of uranium (which has an atomic

number Z = 92) Ionising radiation High energy radiation that causes ionisation of atoms (potentially harmful to

living things)

CHARACTERISTICS OF RADIOACTIVE EMISSIONS

Radiation Symbol Penetrating power

Relative charge

Relative mass

Nature of emission

Alpha He42 Low

Stopped by paper +2 4 He nucleus

Beta e01

Moderate Stopped by 0.5mm thickness of lead

–1 1/2000 electron

Gamma High

Stopped by many cm of lead or concrete

0 0 electromagnetic radiation

How do you expect alpha, beta and gamma particles to be affected in an electric field?

Examples of nuclear equations for radioactive decay or emission

Alpha decay decreases the mass number (A) by 4 and decreases the atomic number (Z) by 2. αRnRa 222

86226

88 (because Heα 42 )

Beta decay has no effect on the mass number but increases the atomic number by 1 ePuNp 0

123994

23993

Gamma decay has no effect on the mass or atomic number. eg, cobalt-59 in a fission reactor Co59

27 00

01

6028

6027

10 eNiCon

Remember: the number of protons determines what element it is. Frequently, both alpha and beta emission are accompanied by emission of gamma rays which carry away excess energy from the reaction. STABLE AND UNSTABLE ISOTOPES When the number of neutrons (A–Z) is plotted against the number of protons (Z) we find that all

stable nuclei lie in a narrow band called a zone of stability that gradually increases from a neutron/proton ratio of about 1 for the lighter elements (up to Z < 20) to about 1.5 for lead (Z = 80). If the neutron to proton ratio is either too high or not high enough the isotope is unstable and decays radioactively i.e., outside the zone of stability.

Beta-emission is common for all nuclei that have a neutron to proton ratio that is too high (e.g. C14

6 ).

Elements with Z > 83 are unstable and are radioactive. Several other naturally occurring isotopes are also radioactive including 40K and 14C.

Simple electrostatics can’t account for the stability of nuclei with more than one proton and physicists assume that a strong attractive force exists between all nucleons. The energy required to decompose a nucleus into protons and neutrons is called the binding energy.

Elements with an even Z have a larger number of stable nuclei than those with an odd Z. More than half the stable isotopes have even numbers of both protons and neutrons. Nuclides with Z > 83 can exhibit decay. This has little effect on the neutron/proton ratio but

reduces proton-proton repulsions.

PRODUCTION OF RADIOISOTOPES Transuranic elements: These are elements with an atomic number greater than uranium in the periodic table (i.e., having atomic numbers greater than 92). Transuranic elements do not occur naturally. Transuranic elements with atomic numbers 93 to 95 (neptunium, plutonium, and americium) are produced in nuclear reactors by bombarding natural elements with neutrons produced in nuclear fission reactions (so they are made in a nuclear reactor). Transuranic elements with atomic numbers above 95 are produced by accelerating a relatively small nucleus in a charged particle accelerator to combine with a heavy nucleus. These elements have half-lives ranging from milliseconds to seconds. Commercial radioisotopes: Radioisotopes that are used in medicine, industry and/or scientific research. They have more favourable half-lives than the transuranic elements (see “Half-lives of radioisotopes”, below). Commercial radioisotopes are produced in two main ways:

Using neutrons produced in a nuclear fission reactor, such as the production of cobalt-60 (used in medicine) from cobalt-59:

Using particle accelerators to fire protons, deuterium atoms (deuterons) or Alpha particles at a target element, such as the production of iodine-123 by the bombardment of xenon atoms with protons:

DETECTION OF RADIOACTIVITY

1) Photographic film. First used by Becquerel (1896) to show that uranium emissions could darken photographic plates. Still used today in radiation badges.

2) Geiger-Müller (GM) counter. A tube containing gas (generally argon) at a pressure of about 10 kPa. The radiation ionises the Ar atoms forming Ar+ cations and electrons. The high voltage accelerates the electrons towards the central electrode, ionising more argon atoms in its path. A cascade of electrons soon reaches the anode creating an electrical pulse. The electric pulse is amplified and detected as clicks on an audio amplifier. The positive argon cations migrate slowly to the cathode to complete the circuit. GM counters can detect alpha, beta and gamma rays.

3) Scintillation Counters. Several substances emit flashes of light when struck by alpha, beta or gamma rays. The light is emitted by a ‘phosphor’ coated onto a photomultiplier tube that amplifies an electric signal. The electrical signal generated operates an electronic counter.

4) Cloud chambers. A cloud chamber consists of a cold supersaturated vapour of water or alcohol. The vapour condenses on the ionised track left by radioactive emissions. The path of radiation can be seen and is a straight dense path for alpha particles, a less dense zig-zag track for beta particles and faint track for gamma rays.

NATURAL DECAY 238U and 235U are two naturally occurring isotopes of uranium, being alpha emitters. 238U decays to 234Th which itself is radioactive and so decays further to a final product of the stable lead isotope 206Pb. Likewise, 235U decays to 207Pb. NUCLEAR FISSION The bombarding of certain nuclei with neutrons, which then splits the target nucleus into two roughly equal fragments - ‘splitting the atom’. The reaction can be controlled such as in a nuclear reactor or uncontrolled such as an atomic bomb. Gamma rays and large quantities of energy are released in nuclear fission reactions. For example, 235U can readily be converted to more stable isotopes with about half the mass of 235U by bombarding 235U atoms with neutrons in a controlled nuclear reactor. Many decay modes are possible and more than one neutron is given off during each disintegration:

energyn2RbCs

energyn2SrXe

energyn3KrBanU

10

9037

14455

10

9038

14454

10

9236

14156

10

23592

In a nuclear bomb the number of neutrons builds up rapidly and an uncontrolled chain reaction occurs. In a nuclear reactor, boron control rods absorb all but one neutron for each 235U disintegration. This leads to a controlled steady-state nuclear reaction. TRANSURANIC ELEMENTS Nuclear reactors produce several new elements and can be used to make new elements. 235U is fissile but 238U is not. Instead, it forms a new element called neptunium Np:

eNpUnU 01

23993

23992

10

23892

Neptunium decays to plutonium, 239Pu, which is more stable and like 235U is also fissile:

ePuNp 01

23994

23993

Uranium has the highest atomic number of the naturally occurring elements, 92. Scientists have made 23 further elements with atomic numbers up to 118. These are called transuranic elements. Transuranic elements are made by bombarding heavy nuclei with other high speed positive nuclei (e.g. helium or carbon nuclei) or neutrons. For example, Americium-241, the radioisotope used in domestic smoke detectors, is a transuranic element produced bombarding plutonium with neutrons:

eAmn2Pu 01

24195

10

23994

HALF-LIVES OF RADIOISOTOPES The half-life of a radioisotope is the time, t1/2, for half the radioactive nuclei to decay to other isotopes. It is a measure of the stability of a radioisotope. Half-life is an intrinsic property of a radioisotope and is independent of the initial amount of starting material. It is also not affected by temperature, chemical combination or any other conditions.

Half-lives of some common radioisotopes are listed below. Each radioactive isotope has a characteristic half-life.

Radioisotope Radiation Emitted Half-Life Uranium-238 4.5 109 years Radium-226 1.6 103 years Carbon-14 5.7 103 yearsCobalt-60 5.3 years

Technetium-99m 6 hours Iodine-131 8 days

One half-life corresponds to the time taken for the radioactivity to fall to half its initial value (e.g. if we start with 1 g of iodine-131, after 8 days we have 0.5 g). After two half lives the remaining radioactivity will be a quarter of its initial value (0.25 g) and after N half-lives the remaining radioactivity will be reduced to (1/2)N of its initial value. This is called exponential decay. Half-lives very important for:

Assessing how long a dangerous radioisotope must be allowed to decay before the radioactivity reaches some acceptable safety threshold.

Assessing the ages of ancient rocks by measuring the ratio of Pb-206 to U-238. Pb-206 is the stable decay product when uranium-238 decays through a series of steps to the more stable nuclide Pb-206.

Carbon-14 is a naturally occurring -emitter. By measuring the C-14/C-12 ratio, that begins to drop as soon as a living organism dies and no longer absorbs 14CO2 from the atmosphere, the ages of wooden artefacts that are less than about 60,000 years old can be estimated.

APPLICATIONS OF RADIOISOTOPES

1. In Industry

Thickness gauges can be used to monitor and control the thickness of steel sheets, aluminium foil and plastic film in factories. Thicker material absorbs more radiation. The amount of radiation at the detector falls away exponentially as the thickness of the sheet increases. An isotope with low energy emission and a long half-life is used so the material absorbs a significant amount of the radiation and the radiation source does not require continual replacement. Low energy emission is also a good safety precaution. Some examples include strontium-90 (a beta emitter with a half life of 28 years), which is used for very thin films, and the gamma rays emitted by cobalt-60 (half-life 5.3 years) that are used to gauge the thickness of steel sheets.

Leak detectors are used in water pipes or underground oil pipelines by adding a radioisotope to the liquid and scanning the soil around the pipe. Short-lived isotopes including sodium-24 (a -emitter) with a half-life of 15 hours are used so that the tracer does not permanently contaminate the liquid in the pipe.

Irradiation of food and medical supplies by gamma radiation is widely used to kill bacteria in foods and in medical supplies such as wound dressings. Irradiation of food increases its shelf life and can make the food safer to ingest provided that few harmful substances such as free radicals are produced during the irradiation process. Cobalt-60 and caesium-137 are widely used for this purpose.

2. In Scientific Research

Determination of reaction mechanisms. Radioisotopes are used in research to follow individual steps that occur in reactions, such as monitoring products of photosynthesis using 14-carbon containing carbon dioxide during the biological reaction. Other chemical and biochemical processed can be followed using tritium 3H (radioactive isotope of hydrogen), sulfur-35 or phosphorus-32.

3. In Medicine

Cancer treatment. Radiation therapy for many types of cancer uses the gamma rays emitted by cobalt-60. The gamma rays penetrate deeply into body tissue because they have zero mass and charge. Their intensity is reduced by about 10-20% after passing through the human body. Cells that are most susceptible to gamma rays are in tissues that undergo rapid cell division. Cancer cells that undergo rapid cell division are somewhat selectively attacked by gamma radiation over ‘good’ cells. Thyroid cancers are treated with another emitter, iodine-131. Iodide ions in the bloodstream tend to concentrate in the thyroid gland.

Diagnosis. Radioisotopes injected into humans must have a short half-life and must be easily excreted. Technetium-99m accounts for about 80% of all medical tracers used in diagnostic medicine. The m indicates that this radioisotope is metastable. It emits only gamma rays and, because it has a half-life of only 6 hours, it is extracted on-site at the hospital from its parent molybdenum-99 by normal saline solution. It is injected into the bloodstream in association with tin compounds where it binds to red blood cells. As blood flows through the heart, liver and other organs its distribution can be followed by gamma ray imaging and abnormalities can readily be detected including blood clots, constrictions and circulation disorders. It can also detect brain tumours and thyroid abnormalities. The short half-life causes minimal damage to the patient. Positron Emission Tomography (PET) relies on imaging of gamma rays emitted by radiopharmaceuticals such as glucose incorporating fluorine-18. The fluorine-18 is a positron (positive electron e0

1 ) emitter and each emitted positron combines rapidly with a normal electron to produce two gamma ray photons that move in opposite directions. PET is widely used to monitor glucose uptake by the metabolism of brain cells. An array of detectors around the patient’s head pinpoints the sites of gamma emission and the image is analysed by a computer.

BENEFITS AND PROBLEMS FROM USE OF RADIOISOTOPES IN IDENTIFIED INDUSTRIES AND MEDICINE We have already commented on the benefits of radioisotopes to help solve industrial and medical problems. Because radioactive emissions like alpha, beta and gamma rays ionise neutral atoms and molecules (this forms the basis of the Geiger counter) the effects of this radiation on biological macromolecules and membranes can be devastating. The ionising radiation causes electrons to be lost and the resulting charged species yield free radicals with one or more unpaired electrons. Free radicals derived from water go on to attack biomolecules such as nucleic acids and proteins that regulate cell division. Such attack can lead to the development of cancers.

Irradiation of food by cobalt-60 kills bacteria that cause food to rot but this practice has risks associated with the potentially harmful compounds resulting from the free radicals produced during irradiation. Irradiation also destroys some of the vitamin content of the food. The use of cobalt-60 for treating cancers normally causes significant damage to surrounding healthy tissues. PAST PAPER QUESTIONS HSC 2017: Question 5 (1 mark) Which of the following is a transuranic element that is most likely to have been produced in a nuclear reactor?

(A) Co-60 (B) Np-239 (C) U-238 (D) Hs-265

HSC 2016: Question 9 (1 mark) Curium is produced according to this equation.

nCmXPu 10

24296

23994

What is X in the equation? (A) A proton (B) A neutron (C) A beta particle (E) An alpha particle

HSC 2015 Question 12 (1 mark) A transuranic element can be produced in a nuclear reactor according to this equation:

HSC 2015 Question 27 (5 marks) Name a radioisotope used in a non-medical industry and discuss its use in that industry in terms of its properties.

HSC 2012: Question 6 (1 mark) Cobalt-60 is produced according to the equation:

Where would a commercial quantity of cobalt-60 be produced?

(A) Cyclotron

(B) Scintillator

(C) Nuclear reactor

(D) Particle accelerator HSC 2012: Question 27 (3 marks) Iodine-131 decays through both beta and gamma emission. Iodine-123 decays through gamma emission only.

(a) Iodine-131 is used for diagnosis and therapy whereas Iodine-123 is used only for diagnosis.

With reference to the information in the table, justify the different uses of these two radioisotopes. (2 marks)

(b) Write an equation representing the decay of Iodine-131 by beta emission. (1 mark) HSC 2010: Question 2 (1 mark) Which of the following is an example of a transuranic element?

(A) C–14 (B) Co–60 (C) U–238 (D) Cm–249

HSC 2009: Question 1 (1 mark) Which of the following is an important factor in predicting the nuclear stability of an isotope?

(A) Atomic radius (B) Nuclear radius (C) The ratio of neutrons to protons (F) The ratio of electrons to protons

HSC 2008: Question 1 (1 mark) Which of the following radiations is measured with a Geiger counter?

(A) Beta (B) Infrared (C) Microwave (D) Ultraviolet

HSC 2007: Question 19 (7 marks) There are many benefits and problems associated with the use of radioisotopes in industry and medicine. Evaluate the impact on society of the use of radioisotopes in both industry and medicine. In your answer, give examples of specific radioisotopes, making reference to their chemical properties. HSC 2006: Question 16 (3 marks) Describe how technology has enabled the transuranic elements to be produced. HSC 2006: Question 3 (1 mark) Which set contains only stable nuclei?

HSC 2005: Question 1 (1 mark) Which of the following conditions would produce a radioactive isotope?

(A) Too many atoms in the sample provided (B) Too many protons and neutrons in the atom (C) Too many electrons in the outer shell of the atom (G) Too many electrons for the number of neutrons in the atom

HSC 2004: Question 26 (4 marks) Describe the benefits and problems associated with the use of ONE radioactive isotope in industry. HSC 2003: Question 3 (1 mark) Which instrument is used to detect radiation from radioactive isotopes?

(A) pH meter (B) Geiger counter (C) Ion-selective electrode (D) Atomic absorption spectrophotometer (AAS)

HSC 2003: Question 18 (4 marks) Describe how commercial radioisotopes are produced, and how transuranic elements are produced.

HSC 2002: Question 19 (5 marks) (a) Describe the conditions under which a nucleus is unstable (2 marks).

(b) The following is a flow diagram showing the sequence of products released during the decay of uranium. Use examples from the flow diagram to describe processes by which an unstable isotope undergoes radioactive decay (3 marks).

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