unit 3 chemical pathways

8/9/2019 Unit 3 Chemical Pathways

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UNIT 3

Chemical pathways AREA OF STUDY 1 Chemical analysis

AREA OF STUDY 2 Organic chemical pathways

AREA OF STUDY 1

CHAPTER 1 Introduction — chemical analysis

CHAPTER 2 Gravimetric analysis

CHAPTER 3 Volumetric analysis

CHAPTER 4 Instrumental methods of analysis — chromatograp

CHAPTER 5 Instrumental methods of analysis — spectroscopy

AREA OF STUDY 2

CHAPTER 6 Organic chemistry — naming compounds

CHAPTER 7 Organic pathways

CHAPTER 8 Living chemistry


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CHAPTER

1Introduction — chemicalanalysis

You will examine:

• the broad range of applications for chemical analysis

• the difference between qualitative and quantitative analysis.

You will revise:

• writing chemical equations

• acid–base reactions

• redox reactions

• precipitation reactions

• the mole

• calculations based on chemical equations (stoichiometry)

• dilution of solutions.

Explosives, weapons and drugs pose a

serious threat to the security of our airports

and the community. Samples and swabs

of the clothing and skin of passengers are

taken by airport security and analysed

using instrumental chemical analysis. One

machine, Ionscan™, can trace minute

amounts of explosives and drugs. The

machine also tests for the starting materials

used in the production of these dangerous

chemicals. The qualitative analysis positively

identifies which dangerous chemicals

are discovered. The quantitative analysis

accurately determines how much of the

substances are present.


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3CHAPTER 1 Introduction — chemical analysis

Where is chemical analysisimportant?To know accurately what chemicals are in something is a right that today istaken for granted. This ranges from products that we buy to substances thatare found in the environment. Chemical analysis is used to provide us with

this information. It should be remembered, however, that these applicationsof analytical chemistry are not exhaustive — new areas and methods ofanalysis are continually being discovered and developed.

Analysing consumer productsToday’s consumers are much better educated and much more demandingabout the products that they buy. Consumers want high-quality products thatmeet their specific wants and needs. For example, there is a wide range ofsports drinks and smoothies, each one designed to meet a particular dietaryrequirement. The exact contents of each drink should be carefully analysedand monitored for purity and quality.

Research and development in many areas continues to have an

impact. New prescription drugs, for example, are constantly beingevaluated and released for public consumption. Older ones are beingmonitored as drug companies strive to obtain a competitive edge overtheir rivals. Medical research may often lead to the realisation thatsmaller amounts of a drug can perform the same function, usually

with fewer side effects.Quality itself, or at least the perception of quality, has always been

a powerful selling point. Products such as Dynamo® laundry powderand Cussons Morning Fresh® dishwashing liquid compete in theirrespective markets on the basis of their ‘quality’. The consumer isasked to evaluate such claims and decide between these and the mul-titude of other brands available in supermarkets. The competitivenature of the marketplace means that a product must live up to the

claims made for it, and do so consistently, for it to become popular with consumers. At the other end of the scale, the past few years have seen therise of the so-called ‘budget’ brands, competing on a ‘value for money’ basis.

In either case, both the manufacturer and the consumer have an obviousinterest in the chemical composition of the product, and would expect controlmeasures to be adequate and appropriate for maintaining its composition.Control measures include process control (monitoring the steps in a chemicalmanufacturing process) and quality control (monitoring the final product).

The environmentThe word ‘quality’ has also come to have an environmental meaning asconsumers become increasingly aware of the effects that their actions and

their numbers have on their surroundings. Governments have respondedto this pressure by enacting legislation and setting up agencies such as theEnvironment Protection Authority (EPA). This means that many products nowhave to meet stringent standards.

For example, manufacturers of detergents must now ensure that their prod-ucts are biodegradable. The history of detergents shows that earlier products

were almost too good at producing a lather. Unused detergents in waste water were often responsible for fouling rivers and sewage treatment plants withlarge amounts of froth. This was because they could not be broken down bythe enzymes in bacteria, fungi and other organisms. Biodegradability is nowalso a big advertising point as well as being a legal requirement.

The Ionscan™ machine analyses

samples from clothing and skin

for illicit substances. It signals an

alarm if these substances are

found in sufficient quantities.


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UNIT 3 Chemical pathways4

Sport A product is not always something that we buy in a bottle or a tube. The

International Olympic Federation (IOF) ‘sells’ the Olympic Games as itsproduct, the Australian Football League (AFL) ‘sells’ football, the Victorian

Racing Club (VRC) ‘sells’ horseracing, and so on. For their own credibility, andto gain public confidence, as well as to attract sponsors, it is in the interest of

these bodies to market a quality product. In recent years this has also come tomean a drug-free product.

TABLE 1.1 Caffeine concentrations of doping samples from a recentOlympic Games

Concentration (mg L–1) Distribution (%)


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Law and governmentGovernment agencies employ chemists who check the quality of productsand the environment to make sure that the products conform to legislativerequirements that ultimately protect the community. The work of these chem-ists consists of analysis — determining what is in a product and, in manycases, how much. Just as importantly, they may also have to verify that certainsubstances are not present in a product, or, if they are, that they do not exceed

prescribed maximum limits.Chemistry and other sciences have many forensic applications (that is,

applications in resolving legal problems). In recent times, with advancessuch as DNA analysis and the perfection of techniques requiring very smallsamples, the area of forensic analysis has made spectacular headlines relatingto paternity cases and criminal investigations.

Types of chemical analysisThe analytical chemist uses methods that range from sophisticated tovery simple. Techniques such as gravimetric analysis (analysing bymass) and volumetric analysis (analysing with accurate concentra-

tions and volumes of solution) may be used. These techniques mayalready be familiar to you through your laboratory work. More likely,however, the analyst will use advanced instrumental techniques suchas atomic absorption spectroscopy (AAS) and various types of chro-matography. Due to advances in computer-chip technology,instruments for these techniques and many others have becomecheaper, more powerful and more user-friendly. For example, acommon instrumental set-up today is to have a gas chromatograph(GC) connected to a mass spectrometer (MS) and a computer. Oncethe instrument has been configured and calibrated, a sample can beinjected, its components identified and measured, and accurateresults printed out quickly and reliably. Previously, chemists would

have had to manually measure and interpret gas chromatographs from a

paper printout one peak at a time, then run a separate mass spectrum.

Qualitative and quantitative analysis are important in forensic science.

Chemical analysis is important

in many areas including the

food and drugs industries,

environmental protection, sport

and forensics.

A technician injects a sample into a

gas chromatograph. This sensitiveinstrument can analyse samples as

small as a microlitre (10 –6 L).


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Two types of analysis are performed in an analytical laboratory: qualitativeanalysis and quantitative analysis.

Qualitative analysisIn qualitative analysis, the chemist is merely interested in what is present.In testing a urine sample from a racehorse, for example, the mere presenceof a banned drug (or its metabolised products) is all that is necessary. In the

confectionery industry, a sample of imported food dye might be tested to see whether or not it contains chemicals that are banned in this country.

Quantitative analysisOn the other hand, the question of how much may need to be answered — thisis quantitative analysis. A brewer of a low alcohol beer, for example, will needto know whether its alcohol content is below a certain limit. Health authori-ties might need to know whether the level of mercury in samples of fish isbelow the allowed level.

A logical sequence often used in many analytical procedures is to performa qualitative analysis first to find out what is present, then to perform aquantitative analysis to find the various amounts of substances present.

Deciding on an analytical method When deciding on a method of analysis, the properties of the substance underinvestigation must be considered. A chemist will therefore give careful thoughtto such physical properties as melting and boiling temperature, colour andsolubility. The way the substance reacts chemically will also be considered.

Acid–base reactions will often be important, while in other cases, redoxreactions might be chosen as the basis for analysis. For example, to analysea sample of oven cleaner containing the base sodium hydroxide, an acidmight be used. However, a sample of bleach containing the oxidant sodiumhypochlorite would most likely be analysed by reaction against a suitable

reductant. In other cases, the most appropriate method for identifying anddetermining a substance may be through adding something that will form aprecipitate, or alternatively, something that will produce a colour that can bematched against a set of standards (colorimetric analysis).

Similarly, if instrumental analysis is being considered, the properties of thesubstance under test will also be critical to the final choice. Many metals, forexample, lend themselves to atomic absorption spectroscopy. The boilingtemperature of a substance will be important in distinguishing between gaschromatography and high performance liquid chromatography if chromato-graphic instruments are being considered.

Another important consideration for the analyst will be the degree ofaccuracy required. This could mean the difference between choosing atedious but accurate method in preference to a faster but less accurate one. Is

a qualitative determination all that is needed, or is a quantitative one calledfor? An appreciation of the strengths and weaknesses of common analyticaltechniques will therefore be necessary, so that the most appropriate methodis chosen.

Standard tests to identify substances will also be important. In the analysisof an unknown compound, for example, it would be important to establishthat an evolved gas is carbon dioxide rather than oxygen or hydrogen.

All the above considerations are important in designing a method ofanalysis. If no particular standard method suits, the chemical analyst may haveto display initiative and creativity in designing a new, and maybe innovative,method to get the job done.

Qualitative analysis finds

what substances are present.

Quantitative analysis measuresthe quantity of substance

present.

The type of analysis used

will be determined by the

properties of the substanceto be analysed. For example,

atomic absorption spectroscopy

can be used for metal ions.

Gas chromatography is used

to analyse volatile organic

molecules.


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Revision questions

1. Classify the following analyses as either qualitative or quantitative.(a) A pregnancy test in which a chemical is added to a sample of urine and

a colour change is sought(b) A chlorine test in which the colour of a chemical is compared against

reference standards to estimate the chlorine level in a home swimmingpool(c) Placement of a detector in the exhaust pipe of a car during a tune-up

procedure to measure the level of carbon monoxide emissions(d) Testing for monosodium glutamate, MSG, in a sample of food claimed to

be ‘MSG free’

2. Oil tankers need to clean residual oil from their tanks before they load withnew cargo. The washings are supposed to be stored on board but unscrupu-lous captains sometimes dump these wastes at sea to reduce costs. In Victoria, there have been numerous cases of such wastes washingashore and polluting beaches. Sometimes this has occurred in the vicinity ofPhillip Island and has affected its colony of fairy penguins. There have been instances where the analysis of such material has led to

the prosecution of offending tanker captains.(a) Describe how qualitative analysis might assist in such cases.(b) Given that such oil is usually a complicated mixture, describe how

quantitative analysis could lead to the identification of the ship causingsuch pollution.

The landing of astronauts on the

Moon is still regarded as one

of the greatest technologicalachievements of all time.

In order for them to get to

the Moon and home again,

vital chemical calculations were

performed to determine the amount

of fuel required. Also, to keep the

air in their spacecraft breathable,

it was critical that the amount of

lithium hydroxide to carry in order

to remove the exhaled carbon

dioxide was calculated correctly.


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Chemical equations revisited An analytical chemist must have a thorough knowledge of chemical reactiontypes, including a sound understanding of acid–base, redox and precipitationreactions. Skills such as equation writing and the ability to do calculationsfrom equations (stoichiometry) are also essential.

The remainder of this chapter will outline some of these important ideas. As

you have already encountered some of them in your earlier work, a revision orreview is presented, rather than a more in-depth discussion.

As in most other areas of chemistry, chemical equations are essential tothe work of the chemical analyst. In many cases an ionic equation is appro-priate when we want to know exactly what is reacting. In others a full formulaequation (also called a molecular equation) is useful.

Writing chemical equationsThe law of conservation of mass states that matter can neither be creatednor destroyed in a chemical reaction. In effect, this means that all chemicalreactions are merely rearrangements of existing atoms, and that chemicalequations must therefore be ‘balanced’.

To write a chemical equation, we must know the names and formulae of theoriginal substances (the reactants) and the new substances (the products).

The chemical equation is then derived by using the following steps.• Write out the word equation (names of reactants on the left, products on

the right, and an arrow between them).• Replace the name of each substance with its correct chemical formula.• Insert coefficients to balance the equation and thus indicate the rela-

tive proportions of the substances involved. (This step reflects the law ofconservation of mass.)

• Add the correct symbol of state for each substance.

Sample problem: 1.1 Writing chemical equations

To analyse the sodium hydroxide content of some drain cleaner, an analyticalchemist chose to react it with sulfuric acid. Determine the equation for theresulting reaction, given that sodium sulfate was the salt formed under theconditions of the analysis.

Solution:

Following the above steps, we get:

sulfuric acid + sodium hydroxide sodium sulfate + water

H2SO4 + NaOH Na2SO4 + H2O

H2SO4 + 2NaOH Na2SO4 + 2H2O

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

Revision questions

3. Under different conditions, the reaction in the above example can producesodium hydrogen sulfate (instead of sodium sulfate). Write the equation forthis reaction.

4. When methane, CH4, is burnt in air, carbon dioxide and water are produced. Write the equation for this reaction.

WEBLINKChemical equations

Chemical equations summarise

chemical reactions by

indicating the substances

present before the reaction

(reactants) and the newsubstances produced by the

reaction (products). They also

indicate the relative amounts

of these substances, as well as

their physical states.


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TABLE 1.2 Table of ions for use in writing ionic formulae

Cations

+1 +2 +3

lithium Li+

sodium Na+

potassium K +

caesium Cs+

silver Ag +

copper (I) Cu+

ammonium NH4+

magnesium Mg 2+

calcium Ca2+

barium Ba2+

iron (II) Fe2+

nickel Ni2+

copper (II) Cu2+

zinc Zn2+

tin (II) Sn2+

lead (II) Pb2+

manganese (II) Mn2+

mercury (II) Hg 2+

strontium Sr2+

aluminium Al3+

chromium (III) Cr3+

iron (III) Fe3+

Anions

–1 –2 –3

hydride H−

fluoride F−

chloride Cl−

bromide Br−

iodide I−

hydroxide OH−

nitrate NO3−

hydrogen carbonate HCO3−

hydrogen sulfate HSO4−

chlorate ClO3−

hydrogen sulfite HSO3−

nitrite NO2−

permanganate MnO4−

oxide O2−

sulfide S2−

sulfate SO42−

carbonate CO32−

sulfite SO32−

dichromate Cr2O72−

chromate CrO42−

thiosulfate S2O32−

nitride N3−

phosphate PO43−

An important note When balancing an equation, the numbers in the formulae must not bealtered. To do so would change the substance to a new one, perhaps one thatdoes not even exist. Instead, numbers (or coefficients) are inserted in front ofthe whole formula. These coefficients then tell us the relative proportions ofmoles (or molecules) of the substances that are involved in the reaction.

VICTORIA

Once assigned, a registration number uniquely identifies a car. In a similar way, once

determined, the symbols and numbers in a formula uniquely identify a particular chem-

ical. In either case they cannot be changed without referring to something different. For

example, H 2 O is safe to drink, but H 2 O 2 is not. (H 2 O 2 is a strong oxidant.)


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Ionic equations While full formula equations are very useful and in all respects correct, theysometimes make a reaction appear more complicated than it actually is. This isespecially true of reactions occurring in water, where the reaction may be onlybetween a few different ions. In such situations it may be more informative to

write the ionic equation, rather than the full equation. To write an ionic equation, we usually begin with the full equation. We

then use our knowledge of dissociation and ionisation to write down the ionsthat are present. Dissociation is the process in which ions from an ionic solidseparate as the solid dissolves in water. The end result is that mobile ions areproduced from ions that were originally static. Ionisation is a process that pro-duces ions. A common way that this occurs is when an acid reacts with water.The end result is that mobile ions are produced from a substance which didnot originally contain ions.

Any ions that remain unchanged (called ‘spectator ions’) may then becancelled out, leaving behind the ionic equation.

Sample problem: 1.2 Deriving an ionic equation (1)

Silver bromide is an important chemical in photography. It can be made inthe laboratory by mixing silver nitrate and potassium bromide solutions. Thesilver bromide forms as a precipitate.

Derive the ionic equation for this reaction, given that the full equation forthis reaction is:

AgNO3(aq) + KBr(aq) AgBr(s) + KNO3(aq)

Solution:

As the three aqueous-phase substances will dissociate into their constituentions upon dissolving, the above equation may be rewritten:

Ag +(aq) + NO3

−(aq) + K +(aq) + Br−(aq) AgBr(s) + K +(aq) + NO3

−(aq)

As the potassium and the nitrate ions remain unchanged (they are spectatorions), they may be cancelled out to leave the ionic equation:

Ag +(aq) + Br−(aq) AgBr(s)

Sample problem: 1.3 Deriving an ionic equation (2)

Derive the ionic equation for the acid–base reaction shown in sampleproblem 1.1 on page 8.

Solution:

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

Rewriting to show the actual ions present, we get:

2H+(aq) + SO42−(aq) + 2Na+(aq) + 2OH−(aq)

2Na+(aq) + SO42−(aq) + 2H2O(l)

After cancelling out the spectator ions, we get:

2H+(aq) + 2OH−(aq) 2H2O(l)or

H+(aq) + OH−(aq) H2O(l)

Do moreIonic equations

Unit: 3

AOS: 1

Topic: 1

Concept: 8

To write an ionic equation,rewrite the formula equation

to show all the aqueous ions

that are present. Take care to

preserve their relative amounts.

Then simply cancel out all

ions that occur on both sides

of the equation (the so-called

‘spectator ions’). The ionic

equation is what remains.


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Revision question

5. The equation for the reaction between silver nitrate and calcium chlorideis:

2AgNO3(aq) + CaCl2(aq) 2AgCl(s) + Ca(NO3)2(aq)

Derive the ionic equation for this reaction.

Reaction patternsIt can be dangerous to generalise too much in chemistry, but there aremany obvious patterns that present themselves when a large range of dif-ferent reactions are examined. Knowledge of these patterns can often makethe writing of equations much easier. However, it should always be remem-bered that exceptions can exist and that all reactants and products should beidentified by suitable analysis to avoid any possible uncertainty.

Here is a list of patterns that you will find useful.

1. acid + base salt + water

Example: Nitric acid is added to calcium hydroxide solution. The saltformed will be calcium nitrate. The equation is

2HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2H2O(l)

This pattern also applies to an acid reacting with a basic oxide, as well asto an acidic oxide reacting with a base. (Remember that a salt is any ioniccompound, not only NaCl.)

2. acid + reactive metal salt + hydrogen

Example: Zinc metal is added to hydrochloric acid. This will produce zincchloride as the salt. The equation is

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

3. acid + metal carbonate salt + carbon dioxide + water

Example: Some marble chips containing calcium carbonate are analysedby adding hydrochloric acid. The salt produced will be calcium chloride.The equation is

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

This pattern also applies to an acid reacting with a metal hydrogencarbonate.

4. metal carbonate metal oxide + carbon dioxide

Example: Some calcium carbonate is heated strongly to decompose it.Calcium oxide and carbon dioxide will be formed. The equation is

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

5. hydrocarbon + oxygen carbon dioxide + water

Example: Some propane from a camp stove is burnt. The chemicalproducts will be carbon dioxide and water. The equation is

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

Special note: A hydrocarbon is a compound that contains only carbonand hydrogen. The following general equation is very useful for balancingreactions of this type.

C x H y + ( x + y

4) O2 x CO2 +

y

2 H2O

If this formula leads to any ‘halves’, the coefficients can be doubled.

Do moreReaction types

Unit: 3

AOS: 1

Topic: 1

Concept: 7


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Revision question

6. Write chemical equations for the following reactions.(a) Hydrochloric acid reacting with potassium hydroxide(b) Carbon dioxide (an acidic oxide) reacting with lithium hydroxide(c) Zinc carbonate being decomposed by heating

(d) Sodium carbonate solution reacting with nitric acid(e) Ethyne, C2H2 (also known as acetylene), being burnt during some oxy-acetylene welding

Some common types of chemicalreactions

We will now look at some of the more common types of chemical reactionsused in the analytical laboratory. As mentioned previously in this chapter,these are:1. acid–base reactions

2. redox reactions3. precipitation reactions. You have probably studied acid–base reactions in some detail already as

part of your work in unit 2. The revision box on the opposite page contains asummary of the important points relevant to these types of reactions.

Redox reactionsOxidation and reduction reactions (or redox reactions as they are oftencalled) form another very large group of important reactions. Since thebeginnings of human history, these reactions have played an important rolein daily life. The use of fire is a simple example of an oxidation process. Theproduction of metals from their ores is an example of an important reductionprocess.

Fe2+ solutionZn2+ solution

zinc iron

saltbridge

–

+

voltmeter

The generation of an electric current between half-cells shows that redox reactions

involve electron transfer. In this case the electrons flow from the more reactive zinc

electrode to the less reactive iron electrode.

Redox reactions are involved in many common corrosion reactions.Protection of metals from corrosion also often involves redox reactions. Forexample, the protection of iron by a thin coating of zinc (galvanising) involvesredox reactions between dissolved oxygen, water and zinc metal and relies onthe fact that zinc is easier to oxidise than iron. Today, we recognise that

Acid–base reactions involve

the transfer of H+ ions. The

substance that donates the

H+ ion is the acid, and the

substance that accepts the H+ ion is the base.

1.1 Iron content ofsteel wool

Redox reactions involve

the transfer of electrons. In

particular, adding electrons

to a substance is the process

of reduction, and removing

electrons is the process of

oxidation.

These processes may be

represented by ionic equations.

Remember that, for

reduction, the electrons will

be on the left side of the half-

equation (just like ‘red-’ beingon the left of ‘redox’). For

oxidation, the electrons will be

on the right side of the half-

equation (just like ‘-ox’ being on

the right of ‘redox’).


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all redox reactions involve the transfer of electrons from one substance toanother. The reactions occurring on the galvanised iron may be representedby ionic half-equations that demonstrate this transfer.

Zn(s) Zn2+(aq) + 2e−

and

O2(aq) + H2O(l) + 4e− 4OH−(aq)

Revision — a summary of acid–base reactions

Brønsted–Lowry theory • Acids are substances that donate hydrogen ions (protons) to another substance.• Bases are proton acceptors.

Conjugate pairs• A conjugate pair is the original acid/base and the consequent base/acid that it turns into.

For example, HNO3(aq) + H2O(l) NO3−(aq) + H3O

+(aq).The conjugate pairs are HNO3/NO3

− and H3O+/H2O.

Note the convention of stating the acid first then the base.

Amphiprotic or amphoteric substancesThese can act as acids in some reactions and as bases in others. Water is an example of such a substance.For example, CH3COOH(aq) + H2O(l) H3O

+(aq) + CH3COO−(aq)

NH3(g) + H2O(l) NH4+(aq) + OH−(aq)

Polyprotic acidsThese are acids that can donate more than one proton (per molecule). Orthophosphoric acid, for example,is triprotic. H3PO4(l) + H2O(l) H3O

+(aq) + H2PO4−(aq)

H2PO4−(aq) + H2O(l) H3O

+(aq) + HPO42−(aq) HPO4

2−(aq) + H2O(l) H3O+(aq) + PO43−(aq)

Acid/base strengthThe strength of an acid or base is a measure of how readily it donates or accepts H + ions. In water, forexample, a strong acid/base will react extensively with the water to produce a large number of ions. A

weak acid/base in the same situation will react only slightly and hence only a few ions will be formed.Experimentally, this will be reflected in the different conductivities of solutions that have the sameconcentration.

The pH scaleThis is used to measure how acidic or how basic a solution is.

0 1413121110987654321

vinegar

soft drink

hair conditioner

rain

water tooth-

paste

pure water hair

shampoo

machine

dishwashing

powder

cloudy

ammonia

detergent

• Values less than 7 are acidic. • Values higher than 7 are basic. • A value of 7 is neutral.pH is discussed in more detail in chapter 10.

See moreBrønsted–Lowryacids and bases

Unit: 3

AOS: 1

Topic: 2

Concept: 1


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Such equations allow us to make the following definitions:• oxidation is the process of losing electrons.• reduction is the process of gaining electrons

Note that there are also some older definitions of redox reactions thatinvolve the elements oxygen and hydrogen. These older definitions of reduc-tion and oxidation are as follows:• Oxidation involves the addition of oxygen or the removal of hydrogen from

a substance.• Reduction involves the removal of oxygen or the addition of hydrogen to asubstance.

In some circumstances it might be more appropriate to use these instead ofthe modern definitions.

As the above half-equations show, the processes of oxidation and reductionare complementary. One process cannot occur without the other. This factleads to two further definitions.• An oxidant is a substance that causes something else to be oxidised. In so

doing, the substance itself is reduced. Oxidants are also called oxidisingagents.

• A reductant is a substance that causes another substance to be reduced.In so doing, the substance itself is oxidised. Reductants are also called

reducing agents.Note that in these definitions, oxidation and reduction are both processes.

As such, they are best described by their half-equations. Oxidants and reduct-ants, however, are substances and are best represented by their formulae.

Writing half-equationsIn the example mentioned on page 13, the half-equation involving the zinc

was quite simple and easy to deduce. In many redox processes, however, thesituation is not quite so simple. A good example is the reduction of orangedichromate ions, Cr2O7

2−, to green chromium(III) ions, Cr3+.To deduce these more complicated types of half-equations, we use the

following rules.

1. Write down the substance and what it turns into (i.e. its conjugate).2. Balance all atoms, except for oxygen and hydrogen.3. Balance oxygen atoms by adding water molecules.4. Balance hydrogen atoms by adding hydrogen ions.5. Balance charge by adding electrons.

Ensure both half-equations have the same number of electrons on eitherside. (If they do not, multiply through the half-equation.) Once both half-equations have been obtained, the overall ionic equation may be producedby adding the two half-equations together so that the electrons cancel out.

With these more complicated equations, water and hydrogen ions may alsocancel out (at least partially) when this is done. The appropriate symbols ofstate should also be added.

As an example, let us consider the reaction mentioned above. In testing for

alcohol levels, dichromate ions oxidise ethanol. The resulting products arechromium(III) ions and acetic acid.

If we apply the five rules to the reduction of the dichromate, we get, step bystep:

Step 1. Cr2O72− Cr3+

Step 2. Cr2O72− 2Cr3+

Step 3. Cr2O72− 2Cr3+ + 7H2O

Step 4. Cr2O72− + 14H+ 2Cr3+ + 7H2O

Step 5. Cr2O72− + 14H+ + 6e− 2Cr3+ + 7H2O

Oxidation is the process of

losing electrons. Reduction is

the process of gaining electrons.

This may be remembered by

using the mnemonic ‘OIL RIG’.

Oxidants and reductants are

substances , not processes.

An oxidant causes another

substance to be oxidised, and

is itself reduced. In a similar

way, a reductant causes another

substance to be reduced, and is

itself oxidised.

Reactants Products

left

e– on the left e– on the right

right

R E D O X

To balance redox half-

equations: write down the

reactants and products and

balance the equation; add water

molecules; add hydrogen ions;

balance charges with electrons.


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For the oxidation of the ethanol, the sequence is:

Step 1. CH3CH2OH CH3COOH

Step 2. CH3CH2OH CH3COOH

Step 3. CH3CH2OH + H2O CH3COOH

Step 4. CH3CH

2OH + H

2O CH

3COOH + 4H+

Step 5. CH3CH2OH + H2O CH3COOH + 4H+ + 4e−

To obtain the overall ionic equation, all that is now necessary is to multiplythe first half-equation by 2 and the second one by 3. This will ensure that theelectrons cancel out when the two are added together. A number of watermolecules and H+ ions will also cancel out at this stage.

Doing this, and adding these symbols, we finally get

2Cr2O72−(aq) + 3CH3CH2OH(g) + 16H

+(aq)4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)

Oxidation numbers — a valuable tool With simple redox equations, it is often possible to deduce where the electronsare coming from and where they are going. With more complicated redoxreactions, this task is more difficult. In fact, it may be quite hard to decide

whether a reaction is a redox process or not! For such situations, chemistshave invented a very useful tool — oxidation numbers. Their use greatly sim-plifies the identification of redox reactions and the subsequent identificationof oxidants and reductants. If oxidation numbers change during a reaction, itindicates that the reaction is a redox reaction.

In particular:• if an oxidation number increases , then oxidation has occurred• if an oxidation number decreases , then reduction has occurred.

To calculate oxidation numbers, the following rules are used.1. The oxidation number of an element is zero, e.g. Zn0, N2

0, O20.

2. In monatomic ions the oxidation number is the same as the charge on theion, e.g. Mg 2+, Cl−.

3. In all compounds of hydrogen the oxidation number of hydrogen is +1,e.g. Na+1O−2H+1.Exception: In metallic hydrides, the oxidation number is −1, e.g. NaH.

4. In compounds of oxygen the oxidation number of the oxygen is −2.Exception: In peroxide, the oxidation number of the oxygen is −1, e.g.H2O2.

After using these rules, the following additional rules may be required.5. In a neutral molecule the sum of all the oxidation numbers must equal

zero. For example, in K 2Cr

2O

7, the oxidation number of potassium is +1

(total +2) and the oxidation number of oxygen is −2 (total −14), so theoxidation number of chromium must be +6 (total +12).

6. In a polyatomic ion the sum of the oxidation numbers must equal thecharge on the ion. For example, the oxidation number of the chromateion, CrO7

2−, is −2.

On rare occasions the following rule may be needed.7. For compounds where there is no oxygen or hydrogen present, the more

electronegative atom is assigned an oxidation number, as if it were a mon-atomic ion. The electronegativity of an element is a measure of its abilityto attract electrons in a bond.

WEBLINKRedox half-equations

Oxidation is an increase in

oxidation number.

Reduction is a decrease in

oxidation number.

Do moreOxidationnumbers

Unit: 3

AOS: 1

Topic: 2

Concept: 5


16/28


TABLE 1.3 Electronegativity values of some common elements

H

2.1

Li

1.0

Be

1.6

B

2.0

C

2.5

N

3.0

O

3.5

F

4.0

Na0.0 Mg 1.3 Al1.6 Si1.9 P2.2 S2.6 Cl3.2

Br

2.8

I

2.5

Sample problem: 1.4 Writing a redox equation from half-equations

Permanganate ions, MnO4−, are able to oxidise Fe2+ ions to Fe3+ ions.

Manganese(II) ions are produced in this process. Write the half-equations for both the oxidation and reduction processes,and hence obtain the overall redox equation for this reaction.

Solution:

For the oxidation process we get:

Fe2+(aq) Fe3+(aq) + e− [1]

If we follow the rules on page 15, we get the following half-equation for thereduction process:

MnO4−(aq) + 8H+(aq) + 5e− Mn2+(aq) + 4H2O(l) [2]

To obtain the overall redox equation, we must multiply equation [1] by 5,

so that the electrons will cancel out when the two half-equations are addedtogether.

When this is done we obtain:

5Fe2+(aq) + MnO4−(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Sample problem: 1.5 Calculating oxidation numbers

Calculate the oxidation numbers of the atoms shown below:(a) O in O2(b) Ca in CaBr2(c) Cr in Cr2O7

2−

(d) S in SF6(e) O in H2O2

Solution:

(a) O in O2 = 0 (Rule 1)

(b) Ca in CaBr2 = +2 (Rule 2)

(c) Cr in Cr2O72− = +6 (Rule 4 followed by rule 6)

(d) S in SF6 = +6 (Rule 7, followed by rule 5. The F is‘imagined’ to be F−.)

(e) O in H2O2 = −1 (Rule 4 — an exception!)


17/28


Sample problem: 1.6 Using oxidation numbers to identify a redox reaction

Is the reaction represented by the equation

Zn(s) + 2H2SO4(aq) ZnSO4(aq) + SO2(g) + 2H2O(l)

a redox reaction, and, if so, which substance has been oxidised?

Solution:

Calculating the oxidation numbers for each atom in the above equationreveals that Zn has an oxidation number of 0, while the Zn in ZnSO4 has anoxidation number of +2.

This is enough to tell us that the reaction is a redox reaction and that it is thezinc that has been oxidised.

Note that a change also occurs with the sulfur — it is +6 in H2SO4, but only +4in the SO2. Therefore, we can also say that the sulfuric acid has been reduced.

Revision questions

7. If dichromate ions, Cr2O72−, are mixed with iodide ions, I−, iodine, I2, and

chromium(III) ions are produced. Write the overall ionic equation for this reaction.

8. Assign oxidation numbers to each atom in the following equation. Decide whether or not it is a redox reaction.

2MnO2(s) + 12H2O(l) + 3S2O82−(aq) 2MnO4

−(aq) + 6SO42−(aq) + 8H3O

+(aq)

Precipitation reactionsIn precipitation reactions, soluble substances in solution are mixed and

an insoluble product is formed. This appears as a precipitate and is oftencollected by filtration.Such reactions may form part of a qualitative scheme of analysis where the

appearance and nature of the precipitate may be taken as proof for the exist-ence of a particular ion. For example, the owner of a swimming pool usingsalt chlorination may simply wish to know if there is any salt left in the pool

water. If a small amount of silver nitrate solution is added to a sample of thepool water, a white precipitate will indicate that chloride ions are present. The

white precipitate is silver chloride, formed according to the equation:

Ag +(aq) + Cl−(aq) AgCl(s)

In quantitative analyses we must remember that there is no such thing asa totally insoluble substance. So-called ‘insoluble’ precipitates will vary in

their degree of solubility and this must be taken into account in designingthe experimental procedure. What may appear as a precipitate may in fact besoluble enough to lead to significant losses. The role of temperature and itssubsequent effect on solubility is another crucial factor to consider.

The mole conceptChemists never deal with particles one at a time. In even the smallest-scalechemical reaction, we would be dealing with huge numbers of atoms andmolecules. To cope with this and be able to make the necessary predictions,chemists use a special quantity for counting these particles — the mole.

A chemist weighing out a

substance to determine the number

of moles.


18/28


The mole is defined as ‘the amount of substance that contains the samenumber of particles as there are atoms in exactly 12 g of pure 12C isotope’.This number has been estimated to be 6.02 × 1023 and is known as Avogadro’snumber, a constant with the symbol N A . The abbreviation for the mole is mol.

Chemists therefore count particles such as atoms, molecules and ions usingthe mole, in much the same way as an egg farmer might count his productionin dozens, or a manufacturer of copy paper would sell his product in packetsof 500 (called a ‘ream’). However, while it may be practical to count eggs andpaper into the above groups, the very size of the Avogadro constant means

that this is not practical when dealing with moles. Instead, chemists count by weighing. If the mass of a 1 mole amount (called the molar mass — symbol M )is known, then a sample can be weighed and the mass compared to the molarmass in order to determine the number of moles present.

The molar mass of a substance is obtained by simply adding together therelative atomic masses of each atom that appears in the formula, and showing‘g mol−1’ as the unit.

For example, the molar mass of glucose is:

M (C6H12O6) = (6 × 12.0) + (12 × 1.0) + (6 × 16.0)= 180.0 g mol−1

A useful skill is calculating the percentage composition of a particularelement in a compound. If the formula is known accurately, this can be done

if the molar mass has been calculated. Thus, for the example above, thepercentage of carbon in glucose, by mass, would be:

72.0

180.0 × 100

= 40.0%

The role of stoichiometry inquantitative analysis

As mentioned earlier, quantitative analysis involves the determination of

‘how much’ is present.In Unit 2 you saw how it was possible to use balanced chemical equations

to make predictions concerning the components of a chemical reaction.Such calculations (stoichiometry ) are central to the performance of anyquantitative form of analysis.

Although each situation is different, all stoichiometric calculations willinvolve, at some stage or another, conversion to and from moles. In most casesit will also be noted that you will be linking together two of the substancesfrom the equation, one about which you have some information and oneabout which you will be making predictions. Sample problem 1.7 illustratesboth of these points.

The molar mass of a pure

substance is the mass of

6.02 × 1023 particles of that

substance. These ‘particles’

may be atoms, molecules or,

in the case of ionic substances,

‘formula units’.

To calculate the molar mass,

simply add together the relative

atomic masses of all the atoms

in its formula and show g mol−1

as the unit.

Single moles of mercury, zinc,

silicon, aluminium, sulfur and

bromine


19/28


Revision — calculating the number ofmoles of a substance — a review of formulae• When a mass of a known substance is given:

n =

where m is the mass of the sample (g) and M is the molar mass of thesubstance (g mol−1).

• When a gas volume at STP or SLC is given:

V m

where V m is the molar volume, measured in litres. The volume of agas can be measured at any temperature and pressure. However, tostandardise such measurements, scientists around the world oftenquote such measurements at either standard temperature and pres-sure (STP — 101.3 kPa, 0°C) or at standard laboratory conditions(SLC — 101.3 kPa, 25°C). Under these conditions, the molar volume of

a gas is 22.4 L mol

−1

or 24.5 L mol

−1

respectively.• When a gas volume, at conditions other than STP or SLC, is given:

n p=

where p is the pressure measured in kilopascals, V is the volumemeasured in litres, T is the temperature in kelvin and R is the generalgas constant (R = 8.31 J K −1 mol−1). Remember that gas pressure canalso be measured in atmospheres (atm) and in millimetres of mercury(mm Hg), where: 760 mm Hg = 1 atm, and 1 atm = 101.3 kPa To obtain the kelvin temperature, you must add 273 to the Celsius

temperature. Mass/volume units (e.g. g L−1, mg L−1) are sometimes used as con-centration units for solutions.

• For solutions, the relevant formula is:

n = cV

where c is the concentration (mol L−1) and V is the volume of solutionused in litres.

substance no. 1

(initial known

information)

substance no. 2

(information being

predicted)

balanced chemical

equation

change

into moles

change

from moles

Although each problem is different, this scheme forms part of nearly all stoichiometric

calculations.

Formulae for calculating the

number of moles of substance

include:

n m

M

n V

V

n pV

T

n cV

R

m

=

=

=

=

Do moreThe mole

Unit: 3

AOS: 1 Topic: 1

Concept: 1

Do moreStoichiometry

Unit: 3AOS: 1

Topic: 1

Concept: 4


20/28


Sample problem: 1.7 A calculation involving the general gas equation

A sample of potassium chlorate is suspected of being contaminated withpotassium chloride. To test the purity of this sample, 7.21 g of it was heatedand the evolved oxygen collected. The volume of oxygen produced was 1.16 L,measured at 101.3 kPa and 18°C.

Assuming that the potassium chlorate was the only source of oxygenpresent, calculate its percentage in the original sample.

Solution:

The equation for the reaction involved is:

2KClO3(s) 2KCl(s) + 3O2(g)

From the information in the question, it is obvious that the oxygen is the‘known’ substance and the potassium chlorate is the ‘unknown’ substance.

pV n T As R=

n pV

T R=

n O 101.3 1.168.31 291

2 )(∴ =×

×

0.0486mol=

From the equation:

∴ n(KClO3) = 2

3 × n(O2)

∴ n(KClO3) = 2

3 × 0.0486

= 0.0324 mol

m(KClO3)=

0.0324×

M (KClO3) = 0.0324 × 122.6

= 3.97 g

Percentage purity = 3.97

7.21 × 100

= 55.1%

Calculations involving excess reactantsIn some cases, the amounts of more than one reactant may be known. Beforethe amount of product can be predicted, it will therefore be necessary to workout which reactant is in excess and which reactant is used up. Once the numberof moles of each reactant has been calculated, the mole ratio (determined from

the reaction equation) is used to calculate the number of moles of all reactantsused in the reaction. The excess reactant no longer takes part in the reaction.

The amount of product may then be forecast, based upon the substancethat is completely consumed (this is called the limiting reagent). The molesof this substance must be used in all future calculations.

Excess reactants are often used to ensure complete reaction of the otherreactants. For example, providing excess air during the combustion of hydro-carbons ensures that all of the hydrocarbon molecules react to producecarbon dioxide and water. The reaction is therefore cleaner and more efficientthan one that produces uncombusted and partially combusted hydrocarbonsas well as carbon dioxide and water.

n KClO

2

3)( n O

3

2 )(=

Do more

Limitingreactant

Unit: 3

AOS: 1

Topic: 1Concept: 5


21/28


Sample problem: 1.8 Determining the limiting and the excess reagents

Calculate the volume of carbon dioxide gas evolved (at SLC) when 20.0 mLof 0.052 mol L−1 sodium carbonate solution is mixed with 22.0 mL of 0.0980 mol L−1 hydrochloric acid.

Solution:

The equation for this reaction is:

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

n(Na2CO3) = cV

= 0.0520 × .

1000

= 0.001 04 mol

n(HCl) = cV

= 0.0980 × 22.0

1000

= 0.002 16 mol

As every 1 mole of Na2CO3 requires 2 moles of HCl, it follows that 0.001 04 molof Na2CO3 will require only 0.002 08 mol of HCl. The HCl is therefore in excessand Na2CO3 is the limiting reagent.

A table can be used to determine which reactant is in excess and how manymoles of the products are produced.

Substance Na2CO3 2HCl 2NaCl CO2 H2O

Mole ratio 1 2 2 1 1

No. of mol 0.001 04 0.002 16

No. of mol in reaction 0.001 04 0.002 08 0.002 08 0.001 04 0.001 04

Excess no. of mol 0.000 08

From the equation:

n(CO2) = n(Na2CO3)

∴ n(CO2) = 0.001 04 mol

∴ V (CO2) at SLC = 0.001 04 × 24.5

= 0.0255 L (or 25.5 mL)

Dilution of solutionsLaboratory work often involves the preparation of solutions by the dilution ofmore concentrated solutions.

For example, sulfuric acid is often purchased at 18 mol L−1 strength. Mostlaboratory uses, however, require much lower concentrations than this.

Since only water (or solvent) is added to the original solution, the numberof moles of substance remains the same. Of course, adding water will increasethe volume and decrease the concentration, but the product n = cV will remainconstant. Therefore the equation c 1V 1 = c 2V 2 is used to calculate the increasedvolume or decreased concentration of the diluted solution.


22/28


Sample problem: 1.9 Determining a volume required for dilution (1)

What volume of 18 mol L−1 sulfuric acid must be used to prepare 2.0 L of2.0 mol L−1 sulfuric acid solution?

Solution:

We need to work out the number of moles present in the diluted solution,then calculate the volume of the undiluted solution that is required to supplythis number of moles. c 1V 1 = c 2V 2

18.0 × V 1 = 2.0 × 2.0

∴ V (undiluted H2SO4) = 4.0

18.0

= 0.22 L

Therefore 220 mL of 18 mol L−1 sulfuric acid will need to be added to(2000 − 220) = 1780 mL of water to make 2.0 L of diluted solution.

Sample problem: 1.10 Determining a volume required for dilution (2)

What volume of water, in mL, must be added to 30 mL of 10 mol L−1 ammoniasolution to make a 0.25 mol L−1 solution of ammonia?

Solution:

n(undiluted ammonia) = cV

=10 30

1000

×

= 0.30 mol

V (diluted ammonia) = nc

=0.30

0.25

= 1.2 L

= 1200 mL

Therefore 1170 mL of water must be added (to make a total of 1200 mL ofdiluted solution).

Revision questions

9. If 1.26 g of KClO3 is heated according to the equation2KClO3(s) 2KCl(s) + 3O2(g), calculate the volume of oxygen that would be evolved at a temperature of 14°C and 100.4 kPa pressure.

10. Calculate the volume of CO2 produced at STP if 40.00 mL of 0.052 mol L−1

sodium carbonate solution is mixed with 11.0 mL of 0.0980 mol L−1 hydro-chloric acid solution.

11. What volume of 2 mol L−1 sodium hydroxide solution is required to produce100 mL of 0.1 mol L−1 solution, and what volume of water will be required?


23/28


Chapter review

Summary• Chemical analysis is the process of determining the

substances present in a test sample. A range of tech-

niques exists for doing this.• The choice of a particular technique depends on

many factors, the most important of which are theproperties of the substance that is being tested for.

• Qualitative analysis is the process of determining what substances are present in a sample.

• Quantitative analysis is the process of determininghow much of a substance is present.

• Chemical equations are a concise way of summar-ising a chemical reaction.

• Ionic equations are an even more concise way ofsummarising a reaction. Ions that remain unchangedcan be cancelled out to leave behind the ionicequation.

• Dissociation and ionisation are two processes thatmust be considered when writing an ionic equation.

• In order to make sense of the huge range of chemicalreactions that exist, chemists classify them intogroups. Three common types of chemical reactionare acid–base reactions, redox reactions and precipi-tation reactions.

• Acid–base reactions are characterised by hydrogenions (H+ ions) being transferred from one substanceto another. The substance that donates the hydrogenion is the acid, and the substance that receives it is

the base.• Important acid–base terms are ‘conjugate pairs’,

‘amphiprotic substances’ and ‘polyprotic acids’.• Applied to acids and bases, strength and concen-

tration mean completely different things.• The pH scale is commonly used to measure the level

of acidity or basicity in a particular solution.• Redox reactions involve the transfer of electrons

from one substance to another.• Redox reactions may be described by special types

of ionic equations called half-equations.• Oxidation is the process of losing electrons, while

reduction is the process of gaining electrons.

• An oxidant allows another substance to be oxidisedand is itself reduced. A reductant allows anothersubstance to be reduced, and is itself oxidised.

• Two very useful tools in dealing with redox reactionsare the writing of half-equations, and the calculationof oxidation numbers. Rules exist for both thesetasks.

• Precipitation reactions result from the associationof soluble ions to form an insoluble precipitate.

• The mole concept is a very important and usefulidea in chemistry. This concept allows chemists to

work with convenient amounts of chemicals, whilenever losing sight of the underlying atomic nature ofsubstances.

• The mole can be thought of as a counting unit in

a similar way to the dozen. The number of parti-cles that constitute a mole is called the Avogadroconstant and has a value of 6.02 × 1023.

• A number of formulae exist for estimating thenumber of moles present in a given sample. Whichformula is used depends on the form of the sampleand the information that we are presented aboutit.

• The mass of 1 mole of an element is called themolar mass ( M ). The molar mass of a compoundis obtained by adding together the relative atomicmasses of each atom that appears in the formula.

• The calculation of percentage composition is auseful skill for comparing the amount of an elementin a compound.

• Stoichiometry is the use of chemical equa-tions involving moles to perform quantitativecalculations.

• In some instances, it is necessary to identify theexcess and the limiting reagents before proceeding

with a stoichiometric calculation. Once identi-fied, the calculation is always based on the limitingreagent.

• Calculations involving the dilution of solutions arebased on the formula for calculating the number of

moles in a solution n = cV or c 1V 1 = c 2V 2.

Multiple choice questions 1. Consider the acid–base reaction that is represented

by the following equation:

HCO3−(aq) + H2O(l) H3O

+(aq) + CO32−(aq)

The conjugate pairs present are:A HCO3

−/H3O+, H2O/CO3

2−

B HCO3−/H2O, H3O

+/CO32−

C HCO3−/CO3

2−, H3O+/OH−

D HCO3−/CO3

2−, H3O+/H2O.

2. A solution of pH 5 is diluted by adding constantportions of water to it. After each addition, the newpH is measured. Which of the following statements correctlydescribes the trend in pH values that is obtained?A The pH will decrease.B The pH will increase, but will not go above the

value 7.C The pH will increase by a constant value after

the addition of each portion of water.D The pH will remain at a value of 5.


24/28


3. The oxidation number of chromium in potassiumdichromate, K 2Cr2O7, is:A −2B +1C +6D +12.

4. VO2+ ions can act as an oxidising agent. When this

happens, VO2+

ions may be produced. The half-equation for this reaction may be rep-resented as follows:

v VO2+(aq) + w H+(aq) + x e− y VO2+(aq) + z H2O(l)

The values of v , w , x , y and z are respectively:A 1, 2, 1, 1, 1B 1, 2, 1, 1, 2C 2, 1, 2, 2, 1D 1, 2, 1, 2, 1.

5. Referring to the equation in question 4, which ofthe following statements is correct?A VO2

+ ions have been oxidised.B The oxidation number of V in VO

2

+ is +1.C The equation represents a reduction process.D The H+ ions have been oxidised to H2O.

6. Caustic soda, NaOH, can be made according to theequation:

Na2CO3(aq) + Ca(OH)2(s)2NaOH(aq) + CaCO3(aq)

What mass of caustic soda would be obtained from1 kilogram of sodium carbonate?A 377 g B 755 g C 1000 g D 2000 g

7. Uranium may be prepared by reacting uraniumtetrafluoride with magnesium, according to theequation:

UF4(g) + 2Mg(s) U(s) + 2MgF2(s)

If 100 g of uranium tetrafluoride is reacted with10 g of magnesium, what mass of uranium will beproduced?A 49.0 g B 75.8 g C 98.0 g D 152 g

8. Some of a 12 mol L−1 HCl solution is to be used to

prepare 2.0 L of 0.5 mol L−1 HCl solution. The volume of 12 mol L−1 HCl that will berequired is closest to:A 10 mLB 20 mLC 70 mLD 80 mL.

9. The percentage, by mass, of sulfur in sulfuric acidis:A 33%B 65%

C 66%D 75%.

10. A sodium hydroxide solution has a concentrationof 0.10 mol L−1. Expressed in g L−1, this concen-tration is the same as:A 0.040 g L−1

B 0.40 g L−1

C 4.0 g L−1

D 40 g L−1.

Review questions

Chemical equations

1. Classify the following analyses as either qualitativeor quantitative.(a) Analysis of the propellent from a spray can by

an instrument, to check the claim that it doesnot contain chlorofluorocarbons, CFCs.

(b) Adding a piece of ‘testape’ to a diabetic’s urineto estimate the glucose level present.

(c) Instrumental checking of the level of mercuryin a sample of fish.

(d) Analysis of a hair-colouring preparation ‘con-taining less than 1.5% aromatic nitro amines’,to check the claim.

2. A number of qualitative tests are used to identifyvarious substances. Find out the tests that are usedto identify the following.(a) Carbon dioxide gas(b) Starch(c) Oxygen gas

(d) Protein in a food sample(e) Hydrogen gas(f) Water

3. Balance the equations below by inserting thenecessary coefficients.(a) Mg + HCl MgCl2 + H2(b) CH4 + O2 CO2 + H2O(c) H2S + O2 SO2 + H2O(d) Fe2O3 + CO Fe + CO2(e) C x H y + O2 CO2 + H2O(f) After you have balanced them, what important

piece of information is still missing from theabove equations?

4. In each of the following reactions, a precipitate isformed. Write the full balanced equation for eachreaction and then write the corresponding ionicequation. (The precipitate formed in each case isshown in brackets.)(a) Silver nitrate solution is added to sodium chlo-

ride solution. (Silver chloride)(b) Barium chloride solution is added to sodium

sulfate solution. (Barium sulfate)(c) Lead nitrate solution is added to potassium

chromate solution. (Lead chromate)


25/28


(d) Sodium phosphate solution is added to silvernitrate solution. (Silver phosphate)

(e) Dilute hydrochloric acid is added to leadnitrate solution. (Lead chloride)

5. A VCE student is planning a formal to celebratethe end of examinations. After some initialdiscussions with the management about the

seating arrangements, she decides to seat herguests (11 boys and 9 girls), as shown in figure(i) below.

A few days beforehand, however, the manage-ment contact her and advise her that the proposedseating arrangement is no longer possible. Shetherefore works out a second arrangement, whichis shown in figure (ii) below.(a) In what ways is the rearrangement of the guests

similar to a chemical reaction?(b) Using appropriate symbols, summarise this

rearrangement as a ‘chemical equation’.

(i)G B

BB

G

B G

BG

B

G B

BB

G

G B

GB

G

(ii)

B G B G B

G B G B G

B B

B

B

G

G

G G B B

Mole concept

6. Calculate the amount of substance (in mol), in thefollowing samples.

(a) 24.7 g of ethanol(b) 123.6 g of aluminium(c) 45.5 g of sulfur dioxide(d) 157.3 g of sodium hydroxide(e) 2.35 L of methane gas at STP(f) 42.9 L of CHCl3 at 100 000 Pa and 23°C(g) 24.2 mL of 1.3 mol L−1 sodium chloride

solution(h) 4.0 L of 2.5 mol L−1 sulfuric acid solution

7. Why is a ‘mole’ interpretation of a chemical equa-tion the same as a ‘molecule’ interpretation?

Acid–base reactions

8. Identify the conjugate acid/base pairs from thefollowing equations.(a) CH3COOH(l) + H2O(aq)

CH3COO−(aq) + H3O

+(l)(b) HCO3

−(aq) + OH−(aq) H2O(l) + CO32−(aq)

(c) NH3(g) + H

2O(l) NH

4

+(aq) + OH−(aq)(d) HSO3

−(aq) + H2O(l) H2SO3(aq) + OH−(aq)

9. Write conjugate acids for the following bases.(a) NH2

−

(b) Cl−

(c) HSO4−

(d) SO42−

10. Write conjugate bases for the following acids.(a) HF(b) HNO3(c) H2O(d) HSO3

−

11. What is the difference between:

(a) a strong acid and a concentrated acid?(b) a weak acid and a dilute acid?

Redox reactions

12. The definitions for oxidants and reductants, onthe one hand, and acids and bases on the other,are sometimes confused by the less careful chem-istry student. Prepare a short comparison of theseterms and definitions to assist such a student.

13. In their initial studies of redox reactions, chemistrystudents often find it hard to distinguish betweenthe terms oxidation and oxidant on the one hand,and reduction and reductant on the other. Prepare a short paragraph to explain the differ-ence between each pair of terms.

14. Calculate the oxidation number of the boldedelement in each of the following.(a) Mn2+

(b) N2H4(c) CrO4

2−

(d) MnO4−

(e) N2(f) SO3

2−

(g) Na

(h) C2H415. Calculate the oxidation number of the bolded

element in each of the following.(a) Na 2O(b) CaCl2(c) NaClO4(d) K 2Cr2O7(e) LiH(f) H2O2(g) Na2O2(h) BrF5


26/28


16. Below are shown the equations for a number ofreactions. Which of these are acid–base reactionsand which are redox?(a) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)(b) H−(aq) + H2O(l) H2(g) + OH

−(aq)(c) HS−(aq) + OH−(aq) S2−(aq) + H2O(l)(d) Pb(s) + PbO2(s) + 4H

+(aq) + 2SO42−(aq)

2PbSO4(s)+

2H2O(l)17. Which of the following equations demonstratesthe redox properties of sulfuric acid?(a) 2NaOH(aq) + H2SO4(aq)

Na2SO4(aq) + 2H2O(l)(b) NaCl(s) + H2SO4(aq)

NaHSO4(aq) + HCl(aq)(c) Zn(s) + 2H2SO4(aq)

ZnSO4(aq) + 2H2O(aq) + SO2(g)

(d) 2NH3(g) + H2SO4(aq) (NH4)2SO4(aq)18. Write half-equations for:

(a) the oxidation of NH3 to NO2(b) the reduction of OCl− to Cl−

(c) the reduction of SO42− to SO2(d) the oxidation of CH4 to CO2.

19. Use the half-equation method to derive the overallequation for the production of nitrogen dioxidefrom concentrated nitric acid and copper.

20. Dichromate ions, Cr2O72−, react with iodide ions, I−,

to produce chromium ions, Cr3+, and iodine, I2.(a) Write the half-equation for the production of

chromium ions from the dichromate ions andidentify this process as oxidation or reduction.

(b) Write the half-equation for the productionof iodine from iodide ions and identify this

process as oxidation or reduction.(c) From your answers to parts (a) and (b), derivethe overall equation for this reaction.

(d) Which substance is the oxidant?(e) Which substance is the reductant?

Stoichiometry

21. Aluminium can be used in flashbulbs to producean intense burst of light. Aluminium oxide is pro-duced in the process. The equation for the reactionis:

4Al(s) + 3O2(g) 2Al2O3(s)

Calculate:(a) the mass of aluminium oxide produced from

5.0 g of aluminium(b) the volume of oxygen consumed at 15°C and

pressure of 101 300 Pa.22. As an emergency procedure, the Apollo 13 astro-

nauts used lithium hydroxide to remove carbondioxide from the interior of their crippled space-craft as it returned from the Moon. Like all hydrox-ides, lithium hydroxide forms the appropriate metalcarbonate when it reacts with carbon dioxide.

(a) Write the equation for the reaction betweenlithium hydroxide and carbon dioxide.

(b) Calculate the mass of carbon dioxide that couldbe removed per kilogram of lithium hydroxide.

A possible alternative chemical for thisprocess might have been the more readilyavailable sodium hydroxide.

(c) Write the equation for the reaction betweensodium hydroxide and carbon dioxide.(d) From (c), calculate the mass of carbon dioxide

that can be removed per kilogram of sodiumhydroxide.

(e) Use your answers to (b) and (d) to suggest areason for the choice of lithium hydroxiderather than sodium hydroxide in a spacecraft.

(f) Derive the ionic equations for both thereactions mentioned in this question.

23. Limestone is an important raw material forindustry. When heated in a kiln, it decomposes toform quicklime, CaO, and carbon dioxide. A typical

kiln is shown in the figure below.(a) If limestone containing 83.5% calcium car-

bonate is used, calculate the mass of quicklimethat would be produced from 100 tonnes ofthis limestone.

(b) Calculate the mass of the above limestone that would be required to produce 60.0 tonnes ofquicklime.

(c) This process is sometimes called ‘lime burning’. Why is this not a correct term to use?

limestone

gasburner

gasburner

quicklime

A typical limestone kiln. The production of lime from

such kilns was one of Australia’s earliest industries.

24. Silicon carbide is used to coat objects such as drillbits and saw blades in order to make them more

wear resistant. It is produced by reacting carbonand silicon dioxide in electric furnaces, accordingto:

3C(s) + SiO2(s) SiC(s) + 2CO(g)


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If 1 tonne of carbon and 1 tonne of silicon dioxideare heated, how much silicon carbide will beproduced?

25. Elemental phosphorus is produced by treating rockphosphate, Ca3(PO4)2, with a mixture of carbonand silicon dioxide, according to the followingequation:

2Ca3(PO4)2(l) + 6SiO2(l) + 10C(s)P4(g) + 6CaSiO3(l) + 10CO(g)

In this process, the rock phosphate is rarely pure,and various unavoidable losses occur, reducingthe efficiency of the process.(a) If the rock phosphate used contains 90%

calcium phosphate, how much rock phos-phate is needed to produce 1 tonne (1000 kg)of phosphorus, P4?

(b) If the process is only 95% efficient, what isthe total mass of reagents needed to produce1 tonne of phosphorus?


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Exam practice questions

In a chemistry examination you will be required to write a number

of short and extended response questions. Your answers will be

assessed on how well you:

• use your knowledge and the information provided

• communicate using relevant chemistry terminology and

concepts

• present a logical, well-structured answer to the question.

EXTENDED RESPONSE QUESTIONS

1. Give the oxidation number of bromine in each of the

following species.

(a) Br2

(b) Br−

(c) OBr− 3 marks

2. Magnesite, MgCO3, is the principal magnesium ore in Australia. This ore can beanalysed by reacting it with hydrochloric acid solution.

(a) Calculate the percentage by weight of magnesium in magnesite.

(b) Write an equation for the reaction of magnesite with the hydrochloric acid.

An ore sample is thought to contain about 0.250 g of magnesium carbonate.

(c) Calculate the volume of 0.100 mol L−1 hydrochloric acid that would be expected to

react with the ore.

(d) Calculate the volume of carbon dioxide that would be produced at STP during the above analysis. 8 marks

3. Calcium oxide (quicklime) is a useful chemical in agriculture. It is made in large

amounts by heating calcium carbonate (which is found in limestone).

As part of their process controls, companies producing this chemical would

commonly carry out a number of reactions in their laboratories. These would include

reacting both the initial reactant and the final product with hydrochloric acid, as

well as identifying the carbon dioxide gas also produced by reaction with calcium

hydroxide solution (limewater).

(a) Write an equation for the decomposition of the calcium carbonate to produce

calcium oxide.(b) Write an equation for the reaction of calcium carbonate with hydrochloric acid.

(c) Given that calcium oxide is a basic oxide, write an equation for its reaction with

hydrochloric acid.

(d) Given that carbon dioxide is an acidic oxide, write an equation for its reaction with

calcium hydroxide solution. 4 marks

Unit: 3AOS: 1

Topic: 1

PracticeVCAA examquestions

Unit: 3

AOS: 1

Topic: 2

PracticeVCAA examquestions

unit 3 chemical pathways

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