kiss chemistry - water

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keep it simple science®

1

but first, an introduction...

Preliminary Chemistry Topic 3

WATERWhat is this topic about?To keep it as simple as possible, (K.I.S.S.) this topic involves the study of:1. THE IMPORTANCE OF WATER

2. STRUCTURE, BONDING & PROPERTIES OF WATER3. THE CHEMISTRY OF AQUEOUS SOLUTIONS

4. HEAT CAPACITY & CALORIMETRY...all in the context of water’s vital role in the natural world

Water... Amazing Stuff!It’s seems very strange to devote an entire topicto just one simple compound.

But water is so important in so many ways...

WATER...• covers most of the Earth• controls weather and climate• carves landscapes• makes up 75% of all living things• dissolves things• absorbs heat

...and much more

all life onEarth

depends on

WATER

WATER

is involved in our life & leisure

Chemically...You already know several different ways todescribe and visualise the water molecule...

H2O

H

H O

HH

O

In this topic you will learn more about the

Chemical Bondingwithin and between water molecules, and how

this is responsible for water’s many unique properties.

You will learn about the Chemistry of Solubilityand how we measure concentrations, and

calculate with

Molarity of solutions.

Finally, as a lead-in to Topic 4, you will learnabout water’s quite remarkable Heat Capacity,

and how we use it in the technique of

Calorimetryfor measuring energy changes

during chemical processes.




2

Temperature,Heat Energy

&Specific

Heat Capacity

Rolesof Wateron Earth

Basic Properties:Density,

m.p. & b.p. PolarCovalentBonding

Calorimetry& Calorimeters

Water’sHeat Capacity

& Life on Earth.

Endothermic&

ExothermicChanges

Heat of

Solution

MeasuringConcentration;

“Molarity”

Ionic Solutions&

Equations

How Ionic & PolarSubstances

Dissolve

Bonding in Water and SimilarMolecules

Dipoles&

Hydrogen Bonds

Water as aSolvent

Precipitation Reactions

WATER

Importance ofWater

Structure, Bonding&

Propertiesof Water

Chemistry ofAqueousSolutions

Heat Capacity&

Calorimetry

CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome students find that memorising the OUTLINE of a topic helps them learn and remember the concepts and important facts. As you proceed through the topic,

come back to this page regularly to see how each bit fits the whole. At the end of the notes you will find a blank version of this “Mind Map” to practise on.

Dynamic Equilibriumin a

Saturated Solution

Water in the “Spheres” of the EarthIn the Atmosphere, water is present as watervapour, and as tiny liquid droplets in the clouds.

In percentage terms, water makes up between1% and 5% of the air, varying with time, placeand weather.

In the Lithosphere, water makes up about 10%of the solid Earth. Although solid rock andminerals seem perfectly dry, there is often waterincorporated into the crystal lattice structure ofmany minerals. When volcanoes erupt, a hugeamount of steam is released as the rocks aremelted.

The Hydrosphere is, of course, nearly all water.Oceans contain about 3% dissolved salt, but theice caps, rivers and lakes are virtually 100%water.

In Living Things, water makes up about 75% ofevery life-form, but jellyfish or watermelons aremore like 95%

The Many Roles of Water on EarthWater is essential in all living things because it is

• a solvent for all the chemicals in a living cell, andthe medium in which all the chemical reactions occur.• a reactant or product in many biological reactions,such as photosynthesis and cellular respiration.• a transport medium for carrying substances, suchas when food, oxygen, etc. are carried in the blood.• a shock-absorber and support structure. Manyplants and simple animals (e.g. worms) rely on waterpressure in their tissues to hold their body in shape.Our brain and other body parts are cushioned bywater-based body fluids.• a habitat (place to live) for many species. Waterhabitats have very stable temperatures because ofwater’s ability to absorb heat with little temperaturechange.

Water is a majorfactor in globalclimate, weatherand the shaping oflandforms.

• The “water-cycle” produces all rain, hail andsnowfall.

• Water is the main agent of erosion, carving outthe valleys and wearing down the mountains, creating the landscapes.

• Water can absorb, transport and release vastamounts of heat energy. The ocean currentslargely control global climates by re-distributing heat world-wide.

For humans & civilization, water is a majorresource:

• for drinking, cooking, washing and recreation.• for crop irrigation and farming.• in industry as a solvent, cleaning and

cooling agent.• for transport by boat and ship.• for generating hydro-electricity.

3

1. THE IMPORTANCE OF WATERkeep it simple science

®



Glacier

Water in the landscape

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RevisionSolutions, Solute and Solvent

A SOLUTION is a mixture, usually of a solid (the“SOLUTE”) and a liquid (the “SOLVENT”).

The solute and solvent particles areintimately associated so that the mixturecannot be separated by filtration, and the

solute will never settle to the bottom. We saythe solute is “dissolved” in the solvent.

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4

Basic Properties of WaterYou may have done some practical work to investigate some of the basic properties of water.

Water’s Density “Anomaly”For almost every pure substances the solid ismore dense than the liquid. Water is theopposite... liquid water has a higher densitythan ice. How do we explain this?

In solid water (ice) the molecules form a “molecularlattice”. Each molecule is held rigidly in place.

When ice melts to form liquid water, the moleculeshave enough energy to move around freely.However, they are still very close together, and infact they “wriggle” in even closer to each otherthan when rigidly arranged in the solid lattice. Nowthere is the same mass of particles crammed intoless space... higher density.

Since solid ice has a lower density, it floats in liquid water.



Density of Liquid & Solid WaterDensity is the ratio between the mass of a

substance and the space (volume) itoccupies. All pure substances have a fixed

and characteristic density.

Density changes with temperature. Water achieves its highest density at 4oC.

This value is 1.00 g/mL

Density = Mass Volume

D = m V

A simple methodis to weigh an

empty, drymeasuring

cylinder, then fillwith water. Readthe volume of

water accuratelythen re-weigh toget the mass of

water.

Ideally, you wouldrepeat these

measurementswith different

volumes of water.

For ice, you need to weigh it quickly before itmelts. If the ice cubes really are cubes orrectangular prisms, you might measurelength, width and height, then calculate thevolume.

Typical Results

Liquid water: Mass = 245 g Volume = 250mL

D = m = 245 = 0.98 g/mLV 250

Ice: Mass = 33 g Volume = 36mL

D = m = 33 = 0.92 g/cm3

V 36

Note: When measuring volume, we normallymeasure liquids in millilitres (mL) and solidsin cubic centimetres (cm3). For practicalpurposes these are equal volumes.

Melting & Boiling PointsPure water melts at 0oC, and boils at 100oC, undernormal “1 atmosphere” of pressure. (As you may know, the celsius temperature scale isbased on the m.p. & b.p. of water.)

Under different pressures, or if impure, the m.p.and b.p. will change. For example, on a highmountain water boils at a much lower temperaturebecause of the lower air pressure.

You may have done experiments to find out theeffect of impurities on the boiling point. A commonexperiment is to boil water with, and without, anadditive such as salt and measure the boilingtemperature. It will usually be found that the boilingtemperature rises by several degrees with solutedissolved in it.

Longer Response QuestionsMark values shown are suggestions only, and are togive you an idea of how detailed an answer isappropriate. Answer on reverse if insufficient space.

5. (6 marks)List 2 important roles of water:

a) in living things.

b) as a factor in weather , climate & geography.

c) as a resource for human civilization.

6. (4 marks)a) What is meant by saying that water has a“density anomaly”?

b) Explain, in terms of bonding and particlearrangements, why water has a densityanomaly.

Water is important in controllingp)............................ and weather, and is a majoragent of q)................................

For human society, water is a major resource fordrinking, r)........................, washing ands).......................... and is used for cropt).................................. in farming. In industry it isa u).......................... and cleaning and coolingagent.Water is important for generatingv)...................................................

At normal atmospheric pressure, pure watermelts at w).........oC and boils at x)..........oC. Itsdensity is y)...............g/mL, but (very unusually)the density of ice is z)..........................(more/less) than liquid water. This is because inthe liquid, the molecules actually getaa)............................ to each other than whenrigidly lined up in the molecularab)............................. of ice.

5




In the atmosphere, water is present asa)............................. and as tiny liquid droplets inthe b)........................... In the lithosphere, watermakes up about c)..........% of solid rock,embedded as part of the d)................... structureof many minerals. The e)........................................is almost 100% water, most of it liquid in thef)...................., but some as solid ice in the polarg).................. and glaciers. In living things, watermakes up about h)......% of every living cell.

Water is essential to all living things because itis the i)............................ for all the chemicals oflife and it is involved in many importantj)......................................... Water k).......................substances around the body and it l)....................and ............................. body organs.

Water is a m)............................. for many plants &animals to live in. Water environments havevery n).......................... temperatures, because ofwater’s ability to o)......................... ......................without much temperature change.

Practice Problems on DensityUse D = m/V

1. Complete the table of valuesSample Mass Volume Density

(g) (mL or cm3)A 150 17 (a).............B 22 18 (b).............C 65 (c)............. 13.5 D 210 290 (d).............E (e)............. 85 3.1F 6.8 f)............... 1.2

2. a) Which 2 samples in the table might be thesame substance? Explain your answer.

b) Which substance in the table would float inwater? Explain.

Multiple Choice3. The water content of the lithosphere is mainlyin the form ofA. water vapourB. liquid waterC. solid iceD. water of crystallisation in minerals

4. The “density anomaly” of water is that:A. the density of ice is higher than liquid water.B. the density of water is exactly 1.00 g/mL.C. ice floats in water.D. water’s density is extremely high.

Worksheet 1 Water: Importance & Basic Properties Fill in the blank spaces Student Name..........................................

Worksheet 2 Practice Questions & Test Questions section 1Student Name..........................................

Water the Weirdo!We are so familiar with the everyday propertiesof water that we do not realize how unusual andstrange water is, until we make a carefulcomparison with other, similar compounds.

Some of these properties will be studied in thistopic, but here’s a preview:

The Strange Properties of Water • Abnormally high m.p. and b.p.• Abnormally high viscosity and surface tension• Abnormally high Heat Capacity• Unusual Density anomaly (already described)

... when compared to similar sized molecules.

Why? It’s all a matter of bonding...

Bonding in Molecular Compounds of Hydrogen

To understand water, we need to compare it toother, similar sized, covalent moleculescontaining hydrogen:

Methane CH4Structural formula

Lewis Formula

The Lewis Formula, and the structural formula,would suggest that the molecule is a flat, box-shape. However, the pairs of electrons in eachcovalent bond always try to get as far away fromeach other as possible, and in 3-dimensions thisresults in a tetrahedron shape (a regular, triangularpyramid with 4 points as far apart as possible).

Each point of a tetrahedron is as far away from the other 3 as it can get.

In the methane molecule,each covalent bond (andtherefore each hydrogenatom) is as far away fromthe other 3 as itcan get.

Ammonia NH3Structural

formula

Lewis Formula

In this molecule, the 4 pairs of electronssurrounding the nitrogen atom are also at the

points of a tetrahedron.

However, one pair is notinvolved in a covalentbond... it is an “unbondedpair”, but still occupies apoint of the tetrahedron.

The result is that the ammonia molecule is a triangular

pyramid shape.

Water H2OStructural

formula

Lewis Formula

In the water molecule there are twounbonded pairs occupying 2 of the

points of the tetrahedron.

Therefore, the water moleculeis bent. The diagramssuggest a 90o right anglebetween the hydrogen atoms, but in 3-D it is more like 105o.

Hydrogen sulfide H2Shas exactly the same bonding geometry aswater. The central sulfur atom is larger thanoxygen, but otherwise the molecules are verysimilar.

Lewis formula

Structural formula

6

2. STRUCTURE, BONDING & PROPERTIES OF WATERkeep it simple science

®



H

H

H

H C

H

H

HH C Covalent bonds

(shared pairs ofelectrons)

H

HH NH

HH N

H

H O

H

H S

H

H O

H

H S

7

Comparing the Properties of These Compounds

The 4 compounds CH4, NH3, H2O and H2S are ofcomparable size and bonding. Now we comparetheir melting and boiling points, and how theseare related to their relative “molecular weights”.

Compound Molecular m.p. b.p.Weight (oC) (oC)

CH4 16 -183 -162

NH3 17 -78 -33

H2O 18 0 100

H2S 34 -86 -60

Usually, the m.p.’s & b.p.’s of comparablesubstances show a steady increase as theatomic or molecular weight increases.

This graph shows that both water and ammoniahave unusually high melting and boiling points.Water especially has values way above those ofcomparable molecules.

Why? What’s going on?

In Topic 1 you learned how the properties ofm.p. & b.p. are controlled by the bonding withinsubstances.

Covalent molecules are held together internallyby strong covalent bonds (“intra-molecularbonds”). These however, are not the bonds thatmust be overcome to melt or boil the substance.

It’s the forces between the molecules (“inter-molecular bonds”) that must be overcome tomelt or boil a molecular substance.

In water, it these forces are unusually strong!

Polar Covalent BondingTo understand water better, you must learn moreabout covalent and ionic chemical bonding.

Up to this point, you have seen these types ofbonding as quite different things. Now you mustrealize that they are really different degrees of thesame thing.

An analogy might help... Imagine 2 people sharingsome lollies. If both people are very fair about it, andneither dominates or intimidates the other, thesharing will be equal:

An ionic bond can be thought of as the lolly-sharing between a hungry bully and a wimp whohates lollies:

Now you must learn that there is also a situation(or a whole heap of situations) in between theseextremes, where the lollies will be shared, butperhaps not evenly.

Sharing,but notevenly.

In chemical bonding, this kind of sharing iscalled a “Polar Covalent Bond” and occurswhen electrons are shared between 2 atomswith quite different values for Electronegativity.(This was introduced in Topic 2... revise)




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This is like a “purecovalent bond”where electrons

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A “Pure” Covalent Bondoccurs when electrons are

shared evenly.

In a “Polar Covalent Bond”the sharing is not even.

The electrons areattracted more to one

atom than the other.

This causes the bond (and perhaps the entire molecule) tobecome electrically “polarised”. The electric charge is notevenly distributed. One end has a greater concentration ofelectrons and has a slight negative charge (δδ−−), while theother end becomes slightly positive (δδ++). The Greek letterdelta (δδ) is used to denote a “small amount” of something,in this case electric charge. The molecule is called a“dipole”, meaning it has 2 poles.

δδ ++δδ −−

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8

Polar Bonds Create Inter-Molecular Forces

The charges on each end of a molecular dipoleare only a fraction of the size of the charges onan ion, but they do cause electrical forces tooccur between nearby molecules.

These forces are calledDipole-DipoleForces

It is these forces which are the “inter-molecularforces” that hold the molecules together in the solidstate. These are the forces which must be overcomewith thermal energy in order for the solid to melt.These are the forces which determine the m.p. andb.p. of a molecular substance.

The strength of the dipole-dipole force variesaccording to the degree of polarity of thecovalent bond (how evenly or unevenly theelectrons are being shared) and also variesaccording to the shape of the molecule. In somesubstances the forces are very weak, in othersquite strong.

The strongest dipole-dipole forces are about 1/3as strong as a full-scale ionic bond. These occurwhenever hydrogen atoms are bonded toOxygen, Nitrogen or Fluorine, and are called...

Hydrogen BondsOxygen, Nitrogen and Fluorine are all small,strongly electronegative atoms. Hydrogen iseven smaller, and once the electrons are“sucked away” from it in the polar bond, thehydrogen atom is really a “naked” proton.

The result is an especially strong set of partialcharges, a powerful dipole, and strong inter-molecular force, which attracts nearbymolecules to each other. These especiallystrong dipole-dipole attractions are called

“Hydrogen Bonds”.

Hydrogen Bonding in WaterIn the water molecule the covalent bonds arevery polar, so the atoms develop especiallylarge partial charges. Each molecule is a dipole,and strong inter-molecular “Hydrogen Bonds”attracts each molecule to its neighbours.

It is this network of hydrogen bonds that holdsthe molecules in a rigid lattice in the solid state.

The Hydrogen Bonding is the reason that icehas such a high melting point, compared toother molecules. (Ammonia also has relativelyhigh m.p. & b.p... same reason!)

Once melted to a liquid, the molecules can movearound, but “cling” to each other because of thehydrogen bonds. The molecules even “wriggle”closer to each other and the density increases.

To boil water to a gas, the molecules must beable to totally break free from the hydrogenbonds. This requires considerable energy, sowater has an unusually high boiling point,compared to other molecules.




δδ ++

δδ ++

δδ −−

δδ ++ δδ ++

δδ −−

δδ ++

δδ ++δδ −−

δδ ++

δδ −−

δδ ++

δδ −−

δδ ++δδ −−

δδ ++δδ −−

δδ ++δδ −−

δδ ++

δδ −−

O, N or F atomH atom

Polar Covalent Bond Hydrogen Bond

Intra-mmolecularCovalent Bondswithin molecules

Inter-mmolecularHydrogen Bonds

betweenmolecules

It is the HYDROGEN BONDING

between water moleculeswhich explains all of water’sweird and unusual properties

9

As well as the “Density Anomaly” and the veryhigh m.p. and b.p., water has other propertieswhich, compared to other similar sizemolecules, are quite extraordinary.

You may have done simple practical work todemonstrate these:

Surface Tension is a phenomenon wherea liquid acts as if it has a “skin” at the surface.In most liquids the effect is small, but water hasa relatively strong surface tension.

Technically, the metal is NOT floating.

The explanation is, as usual, hydrogen bonding.Water molecules have a network of forcesattracting them to each other. At the surface,this network of force resists penetration andcan support objects. They will sink if pushedthrough the “skin”.

Surface tension is also the reason that waterforms droplets.

The surface tension network of forces tries topull the droplet into a spherical shape. Gravityoften squashes them to form the typical“teardrop” shape.

Viscosity is another phenomenon you mayhave experimented with.

Viscosity is a measure of how “sticky” or“thick” a liquid is. Technically it is measured asthe resistance of a liquid to flowing through athin tube, but it can be thought of as how easyor difficult it is for things to move through theliquid.

You may have droppedmarbles into various liquidsand compared the rates atwhich they fell, as a way toobserve viscosity differences.

Liquids like oil are veryviscous, so you may get theidea that water has a lowviscosity. Yes it has,compared to oil, but that’s notreally a fair comparison.

In fact, when the viscosity ofwater is compared to liquidswith similar size molecules,water’s viscosity is very high.

Why? It’s that hydrogen bondingagain...

The hydrogen bonds betweenwater molecules cause themto “cling” to each other, andmake it much more difficultfor a moving object to movethrough the liquid.

The high viscosity of water has had a majorimpact on the evolution of any aquatic animalswho need to move quickly to catch food orescape predators.

Fast moving aquatic animals are alwaysstreamlined in shape and equipped withpowerful tails or flippers for propulsion.




A piece of metalbeing supportedon the surface

tension of water

More of Water’s Unusual Properties

Beads of dew in a spider’s web




10

Water has a number of unusuala)............................... including abnormally highm.p & b.p. and the “density anomaly”. These areall due to the b)................................. between themolecules.

When an atom has 4 pairs of electrons around it(as is the case in most covalent molecules) eachpair tries to stay c).............................................................. as possible. The result isthat each pair lies at one of the points of ad).................................. This is why methane is atetrahedral shaped molecule. In ammonia, thecentral e)........................... atom is bonded to 3f)............................. atoms but also has ang)............................ electron pair occupying onepoint of the tetrahedron. Therefore, themolecule is a h)................................. shape. Inwater, the oxygen atom has 2 pairs ofi)........................................ occupying 2 points ofthe tetrahedron. This results in the moleculebeing j)........................ (shape)

In a “pure” covalent bond, the electrons areshared k)............................... An ionic bondoccurs when the “sharing” is totally uneven sothat ions form. In between these extremes thereare “l).............................. covalent bonds” inwhich the sharing is m).................................. Theresult is that n)............................ electric charges(denoted by the greek letter “δδ”) are producedon the molecule because of the unevendistribution of electrons.

The molecule is said to be a o).............................because it has 2 electric poles.The small charges on the o)..............................are not as large as the charge on an ion, but docreate forces of p)............................. betweeneach molecule and its neighbours. Theseq)................. - .................... forces tend to holdmolecules together. These are the r)..............-molecular forces which must be overcome for asubstance to change s)...........................

When hydrogen is bonded to atoms oft)......................., ....................... or ..........................the forces are especially strong. These arecalled “u)................................ Bonds”.

Water is such a molecule. The molecules arestrongly attracted to each other by theu)............................. bonds. This means that them.p. & b.p. are abnormally v)..................compared to similar sized molecules.

Another result of the hydrogen bonding is thatwater has a very strongw)........................................ which acts like a“skin” and can support small objects which willx)............................. if placed under the surface.

Water also has a relatively highy)........................... due to the way the moleculescling to each other. Because of this, manyaquatic animals are z)........................................ toallow easier movement through the water.

Worksheet 3 Structure & Bonding in WaterFill in the blank spaces Student Name..........................................

Worksheet 4 Test Questions section 2Student Name..........................................

Multiple Choice1. The compound H2Se can be represented by the Lewis Formula shown.The covalent molecule contains 2 unshared pairs ofelectrons. You would expect its shape to be:A. linearB. tetrahedralC. triangular pyramidD. bent

2.“Hydrogen Bonds” are likely to occur withinsubstances in which hydrogen atoms arebonded to atoms of:A. oxygen, chlorine and carbon.B. nitrogen, oxygen and fluorine.C. sulfur, oxygen and chlorine.D. fluorine, chlorine and bromine.

3. Water tends to form droplets because of its:A. high viscosity.B. high surface tension.C. density anomalyD. high boiling point.

HH Se


4. (6 marks)The diagram represents 2 watermolecules.

a) Add labels to identify:

i) a covalent bond.ii) a hydrogen bond.

iii) the partial charges (δδ+ , δδ-) on one molecule.

b) Explain what is meant by a “polar covalentbond”.

c) Explain how the presence of hydrogen bondsis responsible for water’s relatively high m.p. &b.p.




Water as a SolventPerhaps the main reason that water is soimportant to living things, and in the study ofChemistry is that it is a great solvent.

This doesn’t mean that everything will dissolvein water... far from it. You may have doneexperimental work to try to find any generalrules about which substances will, or will not,dissolve in water. Generally, it all depends onthe type of bonding within the substance.

Ionic Compounds are (generally) solublein water, and all because water molecules are polar.

Ionic compounds are composed of a strongionic crystal lattice. It requires a hightemperature to melt this lattice, but watermolecules can dissolve the crystal bysurrounding each ion and detaching it from thelattice.

Notice how the the (+ve) ions are surrounded bywater molecules which are presenting the (δδ-)end of their dipole to the ion. The (-ve) ions aresurrounded by molecules presenting the (δδ+)end of the dipole.

With each ion surrounded by dozens of watermolecules, the attraction between the ions is“blanketed” and the individual ions can nolonger get close enough to each other for theircharges to bond them together.

An ionic compound in solution is made up offree moving, separate, hydrated ions.

Covalent Molecular Substancesmay, or may not dissolve, in water depending ontheir own polar nature, and on how large themolecules are.

If the solute molecules are themselves polar,they will generally dissolve, because the watermolecules will surround each molecule,attracted by dipole-dipole forces.

In the case of ethanol (CH3CH2OH) (alcohol) thewater molecules form hydrogen bonds with theethanol molecules which contain the highlypolar -OH chemical group.

There are many covalent molecules like this,with -NH or -OH groups on the molecule,including all the alcohols and the “sugars”such as sucrose (table sugar).

Small, non-polar covalent molecules such asiodine (I2), oxygen (O2) and nitrogen (N2) willdissolve in water, but only in small amounts...we say they are “sparingly soluble”. Thesemolecules do not have any dipole charges toattract a water molecule and become“hydrated”, but they are so small and have suchsmall “dispersion forces” holding them to eachother, that they can simply spread out, in smallnumbers, among the water molecules.

Larger non-polar molecules will NOT dissolve inwater. They are too large to simply disperseamong the water molecules, and there are nodipoles for the water molecules to associatewith or form hydrogen bonds.

These substances include petrol, oils andwaxes, and are often described as“hydrophobic” (= water hating/fearing) becausethey will not mix with water.

11

3. THE CHEMISTRY OF AQUEOUS SOLUTIONS

Na+

Na+

Cl-

Cl-

Na+

Na+

Cl-

Na+Na+

Cl-

Na+

Cl-

Cl-

Cl-

Na+

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δδ++

δδ++δδ++

δδ++

δδ++

δδ−−

δδ++

δδ++

δδ++

δδ−−

δδ−−

δδ−−

δδ−−

δδ−−

δδ−−

δδ−−

δδ−−

Ethanol CH33CH22OH

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“Dispersion Forces”are extremely weak attractive forces that exist

within all substances. Among non-polar moleculesthey are the only inter-molecular forces acting to

hold the molecules together. This is why suchsubstances have very low m.p. & b.p. Knowledge

of how and why these forces arise is not required for this course.

12

A Special Case to Know About... Hydrogen Chloride

In the pure state, the compound hydrogen chloride(HCl) is composed of small polar molecules:

Despite the dipole-dipole attractions, the m.p. &b.p. are quite low, so pure HCl is a gas at roomtemperature.

You would expect that these molecules woulddissolve in water, but they do much more thanjust dissolve... they interact so strongly withwater that the molecules ionise and becomeseparate H+ and Cl- ions.

Hydrogen chloride dissolved in water is, ofcourse, hydrochloric acid. This is more than justdissolving in water because the molecule hasionised... what was a polar covalent bond hasbecome ionic, due the the influence of the polarwater molecule.

This equation describes the dissolving of HClgas to form hydrochloric acid.

Water as a Solvent (continued)

Covalent Network Substanceslike the elements Silicon and Carbon, andcompounds like silicon dioxide SiO2 (themineral silica), are crystal lattices of atomsbonded together covalently.

Since the bonds are non-polar, or only slightlypolar, water molecules are not attracted, and thesubstance will NOT dissolve.

Compounds with Very Large MoleculesLiving cells produce many very large molecules,each containing perhaps tens of thousands ofatoms. Some, like cellulose (in plant cell walls)contain many polar groups, and watermolecules will be attracted and form hydrogenbonds. However, the cellulose molecules areoften linked together by their own hydrogenbonding, and covalent “cross-linking”, and it isimpossible for the huge molecules to be takeninto solution.

Cellulose is therefore insoluble, but is describedas being “hydrophilic” (= water loving) becausewater will cling to it, wet it and soak into it verywell.

Some protein molecules will dissolve if theyhave a folded, “globular” shape that allowswater molecules to surround them. This is thecase with enzyme proteins, which are dissolvedin the water inside a cell, or in the blood.

Other proteins, like keratin (in hair and skin) arein long chains that cross-link to others. They willnot dissolve, but are hydrophilic.

Plastics, such as polyethylene, are composed ofhuge molecules too. Most are non-polar, andmay be cross-linked with each other. They tendto be insoluble in water and are generallyhydrophobic.




δδ−−δδ++

δδ−−δδ++

δδ−−δδ++

δδ−−δδ++HCl molecules

Cl- ++

δδ++

δδ++δδ−−δδ−−

δδ−−

δδ−−

δδ++

δδ++

HCl((gg))

molecule

Separate, hydrated ionsCl-((aaqq)) and H++

((aaqq))

HCl(g) H+(aq) + Cl-(aq)

(aq) means “aqueous”.This is Latin for “in water”.

In an equation it meansdissolved and hydrated by

water molecules.

Molecules in thegas state.

“Like dissolves like”... water is polar, so it dissolves:• ionic compounds• polar molecules (unless too large)• very small non-polar molecules (sparingly)

13

Ionic SolutionsWhen an ionic compound dissolves in water, the crystallattice disintegrates and the (+ve) and (-ve) ions becomeseparately hydrated to form the solution.

The positive (+ve) ions are collectively called“cations”. Negative (-ve) ions are “anions”.

You need to be able to write an equation todescribe the dissolving of any ionic compound. More examples:

Notice that the equation must balance in termsof the ratio of the ions. In this case there are 2nitrate ions for each magnesium ion. Noticealso that the total of (+ve) charges is the sameas the total of (-ve) charges.

Dilute, Concentrated, SaturatedIf you dissolved a pinch of salt in a bucket ofwater this is a “dilute” solution, meaning that itcontains very little solute compared to theamount of solvent.

If you dissolved a heaped spoonful of salt in a glassof water the solution is “concentrated”... it has quitea lot of solute compared to the amount of solvent.

There is a limit to how much solute can be dissolvedin a given amount of solvent. When this limit isreached, and the solution contains as much solute asit can hold, it is said to be “saturated”.

Different compounds have different solubilities,and this can change with temperature, but as anexample, at 25oC a salt-water solution issaturated when about 36g of salt have dissolvedin each 100mL of water. We say the solubility ofsalt is 36 g/100mL, or simply 36 % m/v.

(“% m/v” means “percentage mass to volume”and refers to the measurement of grams (mass)in 100mL (volume).

This is not the only way we can measure theconcentration of a solution...

the Mole is Back!! (soon)

Dynamic Equilibrium in a Saturated Solution

If you keep adding and stirring salt into wateruntil the solution is saturated, you reach a“dynamic equilibrium” between the ions still inan undissolved, solid, crystal lattice, and thosein the solution as separate, hydrated ions.For simplicity in this diagram, the watermolecules have been left out.

Since dissolving and precipitating occur at thesame rate, the concentration of the solutiondoes not change, and the amount ofundissolved solid remains the same. At themacroscopic level, it seems that nothing ishappening, but down at the atomic level thingsare moving... ions constantly dissolving intosolution and precipitating back out of it again.This is known as a “Dynamic Equilibrium”

This double-arrow symbol indicates that thereaction is occurring in both directions, at thesame rate, in dynamic equilibrium.

Many chemical reactions reach this state.




Dissolving of sodium chloride:

NaCl(s) Na+(aq) + Cl-(aq)

Dissolving of magnesium nitrate:

Mg(NO3)2(s) Mg2+(aq) + 2NO3

-(aq)

Dissolving of aluminium chloride:

Al2Cl3(s) 2Al3+(aq) + 3Cl-(aq)

UUnnddiissssoollvveedd ssoolliidd

IIoonnss iinn aassaattuurraatteeddssoolluuttiioonn

Ions dissolve from the lattice intosolution, while dissolved ions leave

the solution and join the lattice AT THE SAME RATE

Dynamic Equilibrium in a saturated salt solution:


TThhiiss iioonnddiissssoollvveess

TThhiiss iioonnpprreecciippiittaatteess

WORKSHEET at end of section

14

Precipitation ReactionsSilver nitrate (AgNO3) is soluble:

Sodium chloride is soluble:

If you mix these 2 solutions together, you arereally mixing water containing 4 separate ions...Na+, Cl-, Ag+ & NO3

-.

However, silver chloride (AgCl) has an extremelylow solubility, so the mixture of ions may containAg+ ions and Cl- ions at concentrations way abovethe saturation concentration of AgCl. The ions willimmediately form an ionic crystal lattice and solidAgCl will precipitate from the solution, until thecorrect dynamic equilibrium of solid and solutionis re-established.

This is an ionic equation describing exactlywhat happened. On the left is the mixture of ionsthat were brought together in the 2 solutions.The Ag+ and Cl- ions have combined to formsolid AgCl, while the other 2 ions have stayed insolution, unchanged... they are “spectators”.

We can leave out the spectators to see theessential change that occurred:

This is a net ionic equation.

Notice that it is simply the reverse of theequation for the dissolving of silver chloride.

Ionic equations can be tricky to balance. Ifinsoluble PbCl2 is formed by precipitation of ions,the net ionic equation is:

Notice that 2 Cl- ions are needed. If these weredelivered in a sodium chloride solution, then tobalance everything, 2 Na+ ions must be present inthe full ionic equation.

You may have done experimental work assuggested by this photo, to discover anypatterns regarding which ions are often involvedin precipitation reactions, and which mostly stayin solution.

The results of such experiments are oftensummarised by a list of “Solubility Rules”. In keepingwith the K.I.S.S. Principle, here is a simplified version:

If you learn these “rules” you can predict whatwill happen when 2 ionic solutions are mixed:

Example 1Mix solutions of barium hydroxide & potassiumiodide.Prediction: No reaction. There is no combinationof any of these ions which will form an insolubleprecipitate.

Example 2Mix solutions of potassium carbonate withcopper(II) sulfate.Prediction: A precipitate of copper(II) carbonatewill form.

Net ionic equation: Cu+2

(aq) + CO32-

(aq) CuCO3(s




AgNO3(s) Ag+(aq) + NO3

-(aq)


Precipitation of solid silver chloride:

Na+(aq)+ Cl-(aq) + Ag+

(aq) + NO3-(aq) AgCl(s) + Na+

(aq) + NO3-(aq)

Cl-(aq) + Ag+(aq) AgCl(s)

2 Cl-(aq) + Pb2+(aq) PbCl2(s)

2Na+(aq)+ 2Cl-(aq)+ Pb2+

(aq)+ 2NO3-(aq) PbCl2(s) + 2Na+

(aq) + 2NO3-(aq)

IonicSolutions

in dropperbottles

Spot-TTestPlate

Solubility RulesMostly Soluble (and stay in solution)

Na+ & K+ always NO3- always

Cl- Br- & I- (except with Ag+ & Pb2+)

SO42- (except with Ag+ Pb2+ & Ba2+)

Mostly Insoluble ( and form precipitates)

CO32- (except with Na+ & K+)

OH- (except with Na+ K+ Ba2+ Ca2+)


15

Measuring Concentrations With the Mole

Earlier, the idea of measuring the concentrationof a solution was introduced. One way to do thisis to measure the mass of solute in each 100mLof solution (%m/v). However, although this isfairly common, it is not the standard way toexpress or measure concentrations.

The Mole is Back!For reasons that will become obvious later, thestandard method for measuring concentrationsof solutions is in moles per litre (molL-1)

Why Have Different Concentration Measurements?

Simple: it’s a matter of convenience, for the particulartask being done.

In an industrial situation it might be required to mixup a salt solution for pickling olives (for example). Tomake it easy and efficient, the instructions might be

“1 kg of salt to every 10 litres of water” or some such.

In this case the units of concentration would bekilograms per litre (kgL-1).

In another situation, it might be convenient to use%m/v.

In Chemistry, it is usually best to measure in molL-1

(“molarity”) because this allows easy conversions ofmass, volumes of gases and volumes of solutions,when chemical reactions are involved.

Technique For Making SolutionsOne important laboratory technique is that ofmaking up a solution to a requiredconcentration.

The first step is to calculate the mass of soluterequired to make the desired solution, as inExample Problem 2, on the left of this page.

Once this exact mass is weighed out, thetechnique is:

Note that to make 500mL of solution you do NOTadd 500mL of water. You make the volume of thesolution up to 500mL... yes, there IS adifference!

Once a solution is prepared this way, othersolutions can be made from it by takingmeasured quantities, and diluting themappropriately.




Concentration = number of moles (of solute)(of solution) Volume (of solution)

c = n V

Units of measurementc in moles per litre (molL-1)n in moles (mol), and remember that n = m V in litres (L) MM

Example Problem 1If 12.00g of pure solid NaCl was dissolved in water,and made up to 250.0mL (0.2500 L) of solution, whatis the molar concentration (“molarity”) of thesolution?

Solution:Step 1. Find the number of moles.

MM(NaCl) = 58.44gn = m/MM = 12.00/58.44

= 0.2053 molStep 2. Calculate concentration.

c = n/V = 0.2053/0.2500concentration = 0.8214 molL-1

Example Problem 2What mass of potassium iodide is required toprepare 150.0mL (0.1500 L) of solution with aconcentration of 0.2000 molL-1?

Step 1. How many moles are required to get thisconcentration?

c = n/V so n = cV = 0.2000 x 0.1500= 0.03000 mol

Step 2. What mass is this? MM(KI) = 166.0g

n = m/MM, so m = n x MM= 0.03000 x 166.0

mass = 4.980g

Add water to flask to fill it tothe mark.

(Use a dropper to avoid over-shooting)

Insert stopper & mix well.

VVoolluummeettrriicc FFllaasskk


Dissolve Solute in a smallamount of (pure) water in a

clean beaker

Carefully transfer solutioninto a Volumetric Flask.Rinse beaker with smallamounts of water & add

washings to flask

16

Diluting to a Desired Concentration

A common procedure in the Chemistrylaboratory is to have chemical solutionsalready prepared to a knownconcentration, and dilute them to newconcentrations as needed.

To calculate the new concentration, orto calculate the volume needed to get adesired concentration, use thefollowing relationship:

Equipment for Diluting SolutionsYou may have done practical workin the laboratory to learn how tocarry out a dilution.

The required volume (calculatedas in Example Problem 2 on theleft) is measured by pipette andtransferred to a volumetric flask.

Pure water is added to the mark.

These are bulbpipettes whichmeasure accuratelya single volume e.g. 25.00mL

For odd amounts(like 5.6mL) use agraduated pipette.

Concentration of Ions in SolutionWhen an ionic compound dissolves in water theionic lattice disintegrates as the individual ionsare hydrated and taken into the solution. What isthe concentration of the individual ions?

If the compound contains ions in a 1:1 ratio thisis a very simple situation. For example, considerthe dissolving of salt, sodium chloride:


If the solution has a concentration of (say) 0.5 molL-1,then the concentration of the Na+ ions is 0.5 molL-1

and the concentration of the Cl- ions is 0.5 molL-1 aswell.

However, if magnesium chloride (MgCl2)dissolvesthere are 2 chloride ions for every 1 magnesium ion.If the concentration of the solution was 0.5 molL-1,then the individual ion concentrations are:

MgCl2(s) Mg2+(aq) + 2Cl-(aq)

0.5 molL-1 0.5 molL-1 1.0 molL-1

In a 0.5 molL-1 solution of aluminium sulfate theconcentrations would be:

Al2(SO4)3(s) 2Al3+(aq) + 3SO4

2-(aq)

0.5 molL-1 1.0 molL-1 1.5 molL-1




c1V1 = c2V2

(or cV = constant)

c1 = concentration of original solution, in molL-1

V1 = volume of original solution used, in L **c2 = concentration of diluted solution, in molL-1

V2 = volume of diluted solution made, in L **

** It actually doesn’t matter what units you use, so longas you are consistent throughout the calculation. Inthe examples below, volumes are in mL.

Example Problem 1If 25.00mL of a solution of concentration0.3750molL-1 was diluted to a new volume of500.0mL, what is the concentration of thediluted solution?

Solutionc1V1 = c2V2, so c2 = c1V1/V2

= 0.3750 x 25.00/500.0∴∴ c2 = 0.01875 molL-1

(1.875 x 10-2)Example Problem 2It is required to make 250.0mL of a solutionwith concentration 5.000x10-3 molL-1, from a“stock solution” with concentration0.2250molL-1. What volume of the stocksolution should be measured for dilution?

Solutionc1V1 = c2V2, so V1 = c2V2/c1

= 5.000x10-3 x 250.0/0.2250∴∴ V1 = 5.555 mL

(In fact, you would not be able to measuresuch a precise volume by pipette. Appropriateanswer is really 5.6 mL)

Bulb Pipettes

Something worth knowing:In Chemistry, square brackets around a formula is

shorthand for “molar concentration of...”e.g. [NaCl] means “molar concentration of NaCl”WORKSHEET at end of section




17

Mass, Volume & Concentration in Precipitation ReactionsArmed with a knowledge of molarity you can now link calculations involving concentration of

solutions to masses and even gas volume quantities.

Example Problem 115.00mL of 0.3055 molL-1 solution of lead(II) nitratewas treated as follows:An excess of potassium iodide solution was added,causing a precipitate. The solid precipitate wascollected by filtration, dried and then weighed.

What substance, and what mass, was collected?

(Note: an “excess” of something means that thequantity added was more than enough to ensure acomplete reaction)

SolutionStep 1: use the “Solubility Rules” to figure out whatsubstance precipitated, then write a balancedequation for the reaction.

Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

Step 2: find how many moles of Pb(NO3)2 werepresent in the 15mL (0.015 L) of solution.

c = n/V, so n = cV = 0.3055 x 0.01500n(Pb(NO3)2) = 4.5825 x 10-3 mol

Step 3: find how many moles of PbI2 wereprecipitated.

The balanced equation shows the mole ratio is 1:1, so n(PbI2) = 4.5825 x 10-3 mol

Step 4: convert moles to mass. MM(PbI2) = 461.0gn = m/MM, so m = n x MM

= 4.5825x10-3 x 461.0∴∴ m(PbI2) = 2.113g

Note:The working above assumes 100% precipitation ofthe lead ions. Technically, a small fraction of thelead ions would stay in the solution, so not quite allof it would precipitate. However, the solubility ofPbI2 is very low, so for simplicity (K.I.S.S.) we’reassuming complete precipitation.

Example Problem 2To measure the concentration of salt in a40.0mL seawater sample, an excess of silvernitrate solution was added to precipitate allthe chloride ions. The precipitate wascollected by filtration, dried and weighed. Itsmass was 2.76g

a) What substance was precipitated?b) Write a net ionic equation for the reaction.c) Write a full ionic equation for the reaction.d) Calculate the number of moles of

precipitate collected.e) How many moles of chloride ions must

have been in the seawater sample?f) Calculate the molar concentration of salt in

the seawater.

Solution

a) From “Solubility Rules”: Silver chloride, AgCl

b) Ag+(aq) + Cl-(aq) AgCl(s)

c) Ag+(aq)+ NO3

-(aq)+ Na+

(aq)+ Cl-(aq)

AgCl(s)+ NO3-(aq)+ Na+

(aq)

d) n = m/MM MM(AgCl) = 143.35g= 2.76 / 143.35

n(AgCl) = 0.0193 mol

e) Mole ratio in equation is 1:1∴∴ n(Cl-) = 0.0193 mol

f) c = n/V= 0.0193 / 0.040 (40mL = 0.040 L)

c(NaCl) = 0.481 molL-1

Example Problem 3 A little revision of Topic 2What volume of hydrogen gas (measured at SLC)could be produced from the complete reaction of50.0mL of 1.50 molL-1 hydrochloric acid withmagnesium?

SolutionAs usual, start with a balanced equation:

Mg(s) + 2HCl(aq) H2(g) + MgCl2(aq)

Moles of HCl present in the solution:c = n/V, so n(HCl) = cV = 1.50 x 0.050 = 0.075 mol

Moles of H2: equation shows mole ratio = 2:1∴∴ n(H2) = 0.075/2 = 0.0375 mol

Volume of H2: (remember 1 mole = 24.8 L at SLC)vol(H2) = 0.0375 x 24.8 = 0.930 L (930 mL)



Write ionic equations (showing states) for thedissolving of each compound in water.

a) potassium bromide

b) calcium sulfate

c) lithium nitrate

d) magnesium iodide

18



Worksheet 5 Practice ProblemsIonic Equations for Dissolving Student Name..........................................

Worksheet 6 Practice ProblemsPrecipitation Reactions Student Name..........................................

e) aluminium nitrate

f) ammonium chloride

g) iron(II) nitrate

h) copper (II) sulfate

i) calcium hydroxide

1. Predicting PrecipitatesUse the “Solubility Rules” to predict the result ofmixing each pair of ionic solutions.

To answer, write “No Reaction”, or name thecompound which would form a solid precipitate.

a) sodium sulfate & barium nitrate

b) potassium hydroxide & iron(II) chloride

c) calcium sulfate & sodium hydroxide

d) lead(II) nitrate & potassium chloride

e) magnesium bromide & silver nitrate

f) potassium chloride & sodium carbonate

g) sodium carbonate & magnesium chloride

h) copper(II) sulfate & sodium carbonate

i) barium nitrate & copper(II) sulfate

2. Ionic EquationsFor each of the combinations in Q1 which wouldreact to form precipitates:

i) write the full ionic equation, and balance.

ii) write the net ionic equation

1.What is the molar concentration of eachsolution? C = n/V

Mass of solute dissolved in Volume of Solutioni) 15.80g of potassium nitrate 0.200 L

ii) 3.66g of copper(II) sulfate 500mL

iii) 127g of sodium chloride 1.50 L

iv) 85.6g of lead(II) nitrate 3,000mL

v) 2.35g of lithium bromide 250mL

2. How many moles of solute are in

i) 2.00 L of a 0.400 molL-1 solution?

ii) 0.450 L of a 1.25 molL-1 solution?

iii) 50mL of a 0.025 molL-1 solution?

iv) 2.00mL of a 0.0035 molL-1 solution?

v) 0.050 L of a 2.25 molL-1 solution?

19




Worksheet 7 Practice ProblemsMolarity Calculations Student Name..........................................

Worksheet 8 Practice ProblemsMolarity & Mass Calculations Student Name..........................................

1. What is the “molarity” of a solution if:

i) 2.50 mol is dissolved in 0.750 L of solution?

ii) 0.025 mol is dissolved in 0.050 L of solution?

iii) 0.35 mol is dissolved in 100 mL of solution?

iv) 1.2x10-3mol is dissolved in 4.0 L of solution?

v) 0.95 mol is dissolved in 200mL of solution?

2.What mass of solute is required to make eachsolution?

Solute Concentration (molL-1) Volumei) aluminium chloride 0.028 0.050 L

ii) sodium sulfate 0.400 250 mL

iii) calcium hydroxide 3.75x10-5 2.50 L

iv) potassium bromide 1.50 25.0 mL

v) copper(II) nitrate 0.800 100 mL

1.What is the concentration of the diluted solutionif:

i) 25.0 mL of 0.100 MolL-1 solution was diluted to1.00L?

ii) 5.00 mL of 1.25 MolL-1 soln. diluted to 100mL?

iii) 2.5 mL of 0.025 MolL-1 soln diluted to 0.50L?

iv) 8.6 mL of 0.500 MolL-1 soln diluted to 50mL?

v) 10.0mL of 5.35x10-3 MolL-1 sol. diluted to250mL?

2.In each case, what volume of the “stocksolution” is needed to make up the givenvolume at the required concentration?

Stock Solution Volume Concentration(molL-1) Required Required (molL-1)

i) 1.50 1.00L 0.250

ii) 6.00 100mL 0.500

iii) 0.250 250mL 1.00x10-4

iv) 0.500 50mL 0.010

v) 0.875 10.0mL 0.500

20




Worksheet 9 Practice ProblemsDiluting Solutions Student Name..........................................

Worksheet 10 Practice ProblemsMass & Conc. in Reactions Student Name..........................................1. An excess of sodium sulfate solution was addedto 25.0mL of a 0.500 molL-1 solution of bariumnitrate. Assuming complete precipitation,calculate the mass of dried precipitate whichcould be collected.

2. The same reaction as in (a) was used to analysethe concentration of a solution of sodiumsulfate. A 10.0mL sample of the solution wastaken, and an excess of barium nitrate solutionwas added. The mass of dried precipitatecollected was 1.27g.

What was the concentration of the sodiumsulfate solution?

3.A precipitate of silver carbonate was collectedfrom 50.0mL of 0.500molL-1 solution of silvernitrate, by adding an excess of potassiumcarbonate solution.

i) Find the mass of dried silver carbonatecollected.

ii) (Some Revision!) If this silver carbonate washeated and decomposed, what mass of silvermetal, and what volume of carbon dioxide gas(at SLC) would be formed?

Multiple Choice1. If small, non-polar molecules were mixed withwater, you would expect them to:A. dissolve, as separate hydrated ions.B. dissolve, hydrated by hydrogen bonding.C. dissolve, dispersed at low concentration.D. not dissolve at all.

2. The correct ionic equation for the dissolvingof solid aluminium chloride is:A. AlCl3(s) AlCl3(aq)

B. AlCl3(s) Al3+(aq) + Cl-(aq)

C. AlCl3(s) Al3+(aq) + 3Cl-(aq)

D. AlCl3(s) 3Al+(aq) + Cl-3(aq)

3. If solutions of potassium carbonate and calciumnitrate were mixed together, you would observe:A. no reaction.B. a precipitate of potassium nitrate.C. a precipitate of calcium carbonate.D. a precipitate of potassium calcide.

4. A saturated solution of lithium bromide is inthe same beaker with solid crystals of lithiumbromide. The concentration of ions in solutiondoes not change over time because:A. the rates of dissolving and precipitating

are the same.B. there can be no further dissolving because

the solution is saturated.C. all chemical processes have ceased

at equilibrium.D. the lithium & bromide ions are at

equal concentrations.

5. The number of moles of sodium ions in500mL of a 2.0 molL-1 salt solution, is the sameas the number of moles of chloride ions in:A. 500mL of a 2.0molL-1 solution of MgCl2.B. 1 L of a 1.0molL-1 solution of MgCl2.C. 2 L of a 1.0molL-1 solution of MgCl2.D. 250mL of a 2.0molL-1 solution of MgCl2.

6. To make 2.0 L of potassium bromide solutionto a concentration of 0.1 molL-1 would requireda mass of KBr closest to:A. 2 g B. 10g C. 20 g D. 120g

7.In the process of accurately preparing 250mL ofa solution to a specified concentration, youwould need to accurately:A. measure 250.0mL by pipette.B. add water to the mark in a 250mL

volumetric flask.C. fill a 250mL measuring cylinder.D. use a 250mL graduated beaker.


8. (6 marks)Write ionic equations to describe:

a) the dissolving of calcium chloride in water.

b) the precipitation of calcium chloride fromwater. (For example, if a solution wasevaporated to dryness)

c) the situation of a saturated solution ofcalcium chloride, in contact with solid calciumchloride.

9. (10 marks)a) Predict what would happen if solutions ofpotassium carbonate and lead(II) nitrate weremixed together.

b) Write a full ionic equation to describe thereaction.

c) The lead(II) nitrate solution had aconcentration of 0.500 molL-1 and 20.0mL wasused. Calculate the mass of precipitate formed.

(Assume complete precipitation, after adding an excess ofpotassium carbonate solution.)

d) Write a symbol equation to describe thedecomposition of the dried precipitate, giventhat a pure metal is formed, and a mixture of 2gases.

e) Calculate the total gas volume (measured atSLC) formed by the decomposition of thequantity of precipitate formed in part (c).

10. (6 marks)You have been given the task of preparing500mL of a solution of potassium iodide (KI)with a concentration of 0.250 molL-1, from thesolid pure chemical.

Describe the steps of the procedure, includingthe exact mass you would use, and any points oftechnique to ensure accuracy.

21





Temperature, Heat Energy & Heat Capacity

When heat is added to any substance, whatreally happens is that the particles(atoms/ions/molecules) move faster. In solidsthe particles just vibrate more quickly, in liquidsor gases they actually move around faster.

What we measure and understand as “temperature”is really a measurement of the average kinetic energy(movement) of the particles.

Not all particles speed up equally when heat is added:

If you do the same thing to water:

The temperature of the water does not changemuch when heat is added.

(Explanation: it’s those “sticky” polar moleculesagain! Water molecules cling to each other byhydrogen bonding. This means they are hard toaccelerate, and it takes more energy to make themspeed up.)

“Specific Heat Capacity” is a measure ofhow much heat energy (in joules) is required tochange the temperature of 1 gram of asubstance, by 1oC. The units of Heat Capacity are, therefore, joulesper degree per gram (J/oC/g)

Comparison of Some Specific Heat Capacities

Substance Heat Capacity (J/oC/g)

Water 4.18

Typical Metal 0.3 (approx)

Other Liquid SolventsEthanol (alcohol) 2.44Acetone 2.17Petrol (mixture) 2.2 (approx)

Note that water’s Heat Capacity ismuch higher than most other

substances.This is another of water’s “weird”

and unusual properties.

Measuring Heat Energy ChangesWhen any substance gains or loses heat, theamount of heat energy involved depends upon:

• the amount of substance. i.e the mass.• the Specific Heat Capacity of that substance.• the temperature change.

22

4. HEAT CAPACITY & CALORIMETRYkeep it simple science

®



1000 joulesof

Heat Energy

Temperature riseabout25ooC

110000 ggrraammss ooff CCooppppeerr

1000 joulesof

Heat Energy

Temperature riseabout2ooC

110000 ggrraammss ooff WWaatteerr

ΔΔH = - m C ΔΔT

ΔΔH = change in heat energy, in joules (J)m = mass of substance, in grams (g)C = Specific Heat Capacity, in J/oC/gΔΔT = temperature change, in oC

Notes• The Greek letter delta (ΔΔ) means “change in...”

• Chemical Data Sheets may give Heat Capacities for1 kilogram of substance instead of 1 gram. No problem; just divide by 1,000.

• Why is there a negative sign??For technical reasons (explained later) if thetemperature goes up, the energy change isconsidered negative. If temperature drops (negativetemp. rise), the energy change is consideredpositive. The negative sign in the equation takes care of this.

Example Problem 1How much energy is needed to raise thetemperature of 50.0g of water by 12.0oC?Specific Heat Capacity of water = 4.18 J/oC/g

SolutionΔΔH = - mCΔΔT

= - 50.0 x 4.18 x 12.0= - 2,508 J

(In this non-chemical situation the (-ve) sign canreally be ignored. The energy required is 2.51 x 103 J (2.51 kJ))

Example Problem 2If 10,000 J of heat energy was added to 100g ofethanol (Specific Heat Capacity = 2.44 J/oC/g) whatwould be the temperature rise?

SolutionSince the temperature will rise, technically theenergy is a negative quantity, so ΔΔH = - 10,000J

ΔΔH = - mCΔΔT, so ΔΔT = ΔΔH/(-m x C)

= -10,000/(-100 x 2.44)= 41.0 oC

i.e. Temp. will rise by 41oC





23

CalorimetryCalorimetry is a technique used to measure theenergy change occurring during chemicalprocesses. The word is derived from the“calorie”, a unit for heat energy no longer inuse. The equipment used to make energymeasurements is called a “calorimeter”.

(Since we now use joules for our energy unit,maybe we should call it a “joulemeter”)

Since many chemical processes occur in water,and because water has such a high SpecificHeat Capacity (i.e. it can absorb lots of energywith little temperature change) calorimetry oftenuses water as the “working fluid” or mediumused to absorb the heat energy.

Exothermic Reactions (“Exo”= to go out)are the reactions that produce and releaseenergy.

The amount of energy involved is the “delta-H”for the process, and is measured per mole of thesubstance(s) involved. When the chemicals loseenergy, the temperature in a calorimeter rises,because the energy release heats up the waterin the calorimeter. This is why, when thetemperature rises, the energy quantity isconsidered negative... the chemicals involvedhave LOST this amount of heat energy.

Endothermic Reactions (“Endo” = to go in)are the reactions that absorb energy... thosewhere you must supply energy to make ithappen.

The “delta-H” for this change is consideredpositive because the chemicals have gainedenergy. The temp. change is negative, becausethe calorimeter temp. drops.

Simple Laboratory Calorimeter

Thermometermeasurestemperature change

Copper Beakerreaction container

Polystyrene bodyand lid prevents heatloss/gain with thesurroundings

Prac. Work: Heat of SolutionYou may have carried out experiments tomeasure the energy change that occurs whenionic compounds dissolve in water.

General Method:Use a calorimeter to measure the temperaturechange in a measured mass of water, when ameasured mass of a solid dissolves.

You can then calculate:• the energy change occurring

(for the quantities used) and then

• the energy change per gram of soluteand then

• the energy change per mole of solute.

Typical Results for dissolving Potassium hydroxide.

Mass of water placed in calorimeter = 100gMass of potassium hydroxide dissolved = 4.50gInitial temperature of water = 21oCFinal temperature of solution, after dissolving = 28oC

∴∴ Temperature change, ΔΔT = 7.0oC

Calculations:ΔΔH = - mCΔΔT

= - (100 + 4.5) x 4.18 x 7.0= - 3, 058 J for the dissolving of 4.50g

Energy per gram: ΔΔH = - 3,058/4.50 = -679 J per gram

Energy per mole: MM(KOH) = 56.1gΔΔH = -679 x 56.1 = - 38,100 J per mole

∴∴ Heat of Solution = - 38.1 kJmol-1 (exothermic)

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RReeaaccttaannttss EEnneerrggyy LLeevveell

EEnneerr

ggyy CC

oonnttee

nntt

PPrroodduuccttss hhaavveeMMOORREE eenneerrggyy

EEnneerrggyyaabbssoorrbbeedd bbyycchheemmiiccaallss

dduurriinnggrreeaaccttiioonn

RReeaaccttaannttssEEnneerrggyy LLeevveell

EEnneerr

ggyy CC

oonnttee

nntt

PPrroodduuccttss hhaavveeLLEESSSS eenneerrggyy

EEnneerrggyyrreelleeaasseedd bbyycchheemmiiccaallss

dduurriinnggrreeaaccttiioonn

= ΔΔHnegative

= ΔΔHpositive




24

Thermal PollutionSome industries, especially coal-burning ornuclear power stations produce large amountsof waste heat.

In some places, these plants are situated besidelakes or the sea so that the water can be usedfor cooling the equipment. Typically, lake wateris pumped through the equipment, then hotwater discharged back into the lake.

This isthermal

pollution,and is verydestructive

to aquatichabitats.

Effect of Extra Heat on Aquatic Life

The main problem is a matter of solubility.

Oxygen, and other gases, are only slightlysoluble in water. Aquatic organisms are totallydependent on this low concentration ofdissolved gases for their survival.

The problem is, that the solubility of gasesdecreases as the temperature rises. If the watertemperature rises by as little as 5oC, thedissolved oxygen concentration drops by 20%and fish begin to suffocate.

Not only that, but increased temperatures caninterfer with the normal breeding cycles andalter the delicate balance between populationsof food plants, disease microbes, parasites, etc.Habitat destroyed!

“Heat of Solution” is the common namefor the energy change that occurs when 1 moleof a solute dissolves in water.

ΔΔHsol is negative if energy is released.(exothermic: the calorimeter temperature rises)Examples: soluble hydroxides (e.g. NaOH, KOH)

ΔΔHsol is positive if energy is absorbed.(endothermic: the calorimeter temperature falls)Examples: ammonium nitrate (NH4NO3),

ammonium chloride (NH4Cl)

Limitations of CalorimetryWhen you use a simple calorimeter to measure anenergy change in the laboratory, there are a numberof assumptions and approximations involved.

• It is assumed that the calorimeter itself does notabsorb a significant amount of the heat energy of thereaction. This source of error is minimised by usinga copper reaction vessel, since the very low SpecificHeat Capacity of copper means it absorbs littleenergy.

• It is assumed that there is no heat lost or gainedbetween the calorimeter and the surroundings. Thissource of error can be minimised by good heatinsulation of the calorimeter.

• It is assumed that the Specific Heat Capacity of thesolution reacting in the calorimeter is the same aswater. i.e. C = 4.18 J/oC/g. For many solutions this isnot quite true, but (generally) the error this causes isvery small.

A serious limitation of many calorimetry experimentsin school laboratories is the poor precision of theusual lab. thermometers. Usually these can only beread to the nearest 0.5oC, and if the temperaturechange is only a few degrees, the % error is huge.Serious calorimetry needs thermometers with aprecision of at least 0.1oC.

Water’s Heat Capacity & Life on Earth

The fact that water has a remarkably high SpecificHeat Capacity is of enormous significance to weather,climate and life on Earth.

It means that, on a hot day, the ocean or a lakecan absorb a large amount of energy from theSun without much temperature change. The airand the land may get very hot, but the watertemperature changes very little. In cold weather,the air and land can get really cold, but the waterchanges only a little.

This means that water habitats have very stabletemperatures and do not change much from dayto night, or even summer to winter.

Aquaticorganisms donot needcomplextemperaturecontrolmechanismsbecause theirhabitat remainsquite stable.

More importantly, the oceans absorb andtransport (via ocean currents) huge quantities ofheat from the tropics towards the poles. Thishas the effect of cooling the tropical areas andwarming the temperate regions, and generallyevening-out the Earth’s temperature.

Without water, very little of the Earth wouldhave liveable temperatures. Without the“moderating effect” of water, the tropics wouldbe too hot for life, and the temperate regionswould be too cold.


Temperature is a measure of theaverage a)........................... energy of theparticles within a substance. When heatis added, the particles b)........................................

However, some substances requiremore heat energy than others for thesame temperature change. Thisdifference is measured by the propertyof c)...........................................................which has units of d)...............................Water has a very e).............................(high/low) value.

The amount of heat energy involved inany change is given by the formulaf).......................................

A g).............................. is a device formeasuring energy changes. Water isoften used as the “working fluid”because of its high h)..................................... Capacity.

i)............-thermic changes releaseenergy, so the temperature in thecalorimeter j).......................... Thechemicals in the reaction havek).......................... energy, so the energyvalue is considered l)........................ (+/-)

m)..........-thermic changes absorbenergy, so the temperature in thecalorimeter n).......................... Thechemicals in the reaction haveo)........................ energy, so the energyvalue is considered p)...................... (+/-)

The energy change involved indissolving a solution is called the“q).............................................................”

25



Worksheet 12 Heat Capacity & CalorimetryFill in the blank spaces Student Name..........................................

Calorimetry has a number of limitationsand sources of error; the calorimeteritself may r)....................................... Thiserror is minimized by using a reactioncontainer with a very low s)........................................

Heat may be lost or gained between thecalorimeter and the t)...............................This error is minimized byu)..................................... the calorimeter.

It is assumed that the reacting solutionhas the same v)................................................... as water. This is anapproximation, but only causes aw)........................ error.

Experimental error often comes from thelack of precision of thex)............................................

Water’s very high S.H.C. is of greatsignificance to the Earth’sy).......................... and .....................

Water habitats have veryz)............................. temperatures, andthe ocean currents aa).............................huge amounts of heat,ab).............................. the tropics andwarming the ac)..............................regions.

Thermal pollution is the release ofad)............................... into aquatichabitats. This is destructive, mainlybecause the ae)......................................of gases (such as af)..........................)becomes much ag)............................. athigher temperatures.

WHEN COMPLETED, WORKSHEETSBECOME SECTION SUMMARIES

The +/- signs may be ignored in these questions.

1. Calculate the amount of heat energy involved to:

i) heat 50.0g of water from 20oC to 50oC.C(water) = 4.18 J/oC/g.

ii) cool 400g of water from 95oC to 10oC.

iii) heat a swimming pool containing 560 tonnesof water (1 tonne = 1x106 gram) from 12oC to28oC.

iv) heat 100g of copper (C = 0.39 J/oC/g) from10oC to its melting point, 1,085oC.

v) cool a 100 kg car engine (steel, C = 0.45J/oC/g) from 120oC to 20oC.

2. Calculate the Final Temperature(to the nearest degree) if:

i) 250g of water at 20oC absorbs 72,000 J of heat.

ii) 5.00 kJ of energy was extracted from 80.0g ofwater at 25oC.

iii) 1 L of water (= 1kg mass) at 4oC absorbs10,000J.

iv) a 5.00kg lump of steel at -25oC absorbs20,000J.

v) 20.0g of ethanol (C = 2.44 J/oC/g) at 30oCloses 1.2kJ.

26




Worksheet 13 Practice ProblemsHeat Calculations Student Name..........................................

Worksheet 14 Practice ProblemsHeat of Solution Student Name..........................................

(+/- sign important!)1. Find the Molar Heat of Solution

i) When 5.85g of ammonium nitrate dissolved in100mL of water (100mL water = 100g) in a calorimeter,the temperature went from 24oC to 11oC.

ii) 8.42g of sodium carbonate was dissolved in100mL of water in a calorimeter. Thetemperature increased by 9.5oC.

2. Find the Final Temperaturei) The “heat of solution” for sodium hydroxide islisted as ΔΔHsol (NaOH) = -41.6 kJmol-1.

If 10.0g of NaOH was dissolved in enough waterto make 250g of solution, what would the finaltemperature be? The initial temperature = 18oC.

ii) The “heat of solution” for ammonium chlorideis listed as ΔΔHsol (NH4Cl) = +15.2 kJmol-1.

If 18.5g of NH4Cl was dissolved in 150mL ofwater (initially at 22oC) what would be the finaltemperature?

Multiple Choice1.The Specific Heat Capacity values for 4substances are shown.

Substance S.H.C.(J/oC/g)A 0.95B 2.5C 1.2D 2.1

Which substance (A, B, C or D) would have thelargest temperature rise, if equal amounts ofheat energy were added to equal masses ofeach substance?

2.An exothermic process is considered to have anegative value for the energy change, because:

A. the temperature in the reaction containerdrops.

B. you need to put energy into the system.C. it is the opposite of an endothermic change.D. the chemicals have lost energy in the change.

3.“Thermal Pollution” damages ecosystemsmainly because:

A. living things cannot tolerate the temperature rise.

B. less oxygen can dissolve in warmer water.C. the extra heat speeds up plant growth.D. more dirt dissolves in warm water,

so erosion increases.

Longer Response QuestionMark values shown are suggestions only, and are togive you an idea of how detailed an answer isappropriate. Answer on reverse if insufficient space.

4. ( marks)Using a polystyrene cup as a simple calorimeter,a student added 50.0mL of water and measuredits temperature to be 18oC. She weighed out4.27g of lithium hydroxide and dissolved it in thewater. The water temperature rose to amaximum of 45oC.

a) Showing all working, calculate the molar Heatof Solution (including sign) for lithiumhydroxide.

b) Later, she looked up a Chemical Data Bookand found the “accepted” value for ΔΔHsol(LiOH).It was a significantly larger amount of energythan the experimental results gave.

Suggest 2 reasons why.

27





FOR MAXIMUM MARKS ALWAYS SHOW

FORMULAS & WORKING, APPROPRIATE PRECISION & UNITS

IN ALL CHEMICAL PROBLEMS




28

CONCEPT DIAGRAM (“Mind Map”) OF TOPICSome students find that memorising the OUTLINE of a topic helps them learn and

remember the concepts and important facts. Practise on this blank version.

WATER


29

Answer SectionWorksheet 1a) water vapour b) cloudsc) 10% d) crystale) hydrosphere f) oceansg) ice caps h) 75%i) solvent j) biological chemical reactionsk) transports l) supports and cushionsm) habitat n) stableo) absorb a lot of heat p) climateq) erosion r) cookings) recreation t) irrigationu) solvent v) hydro-electricityw) 0oC x) 100oCy) 1.00 g/mL z) lessaa) closer ab) lattice

Worksheet 21.a) 8.8 g/cm3 b) 1.2 g/cm3

c) 4.8 cm3 d) 0.72 g/cm3

e) 264g f) 5.7 cm3

2. a) Substances B & F, because they have the samedensity.b) Substance D, because density is less than water.3. D 4. C5.a) It is the solvent for life chemicals.

It transports substances in blood. b) Main agent of erosion.

Ocean currents distribute heat... controls climate.c) Crop irrigation

Making hydro-electricity 6.a) For most substances, the solid has a higherdensity than the liquid. Water is the opposite.

b) Water ice is a molecular lattice in which themolecules are held in a regular array by hydrogenbonding. When ice melts, the molecules can movearound, but the strong hydrogen bonds still act.Molecules are attracted so strongly that they“wriggle” in even closer to each other than in thesolid lattice arrangement. This results in a smallervolume, and a higher density.

Worksheet 3a) properties b) forces/bonding/H-bondsc) as far apart d) tetrahedrone) nitrogen f) hydrogeng) unshared h) pyramidi) unshared electrons j) bentk) evenly l) polarm) uneven n) partialo) dipole p) attractionq) dipole-dipole r) inter-s) statet) oxygen, nitrogen or fluorineu) Hydrogen v) highw) surface tension x) sinky) viscosity z) streamlined



Worksheet 41. D 2. B 3. B4. a) diagram labels

b) Polar covalent bond is formed when a pair ofelectrons is sharedunevenly.

c) Hydrogen bonds attract themolecules to each other verystrongly, so it requires a lot of heatenergy to break up the solid lattice (melting) or toallow particles to fly freely in the gas state. Hencethe m.p. & b.p. are relatively high.

Worksheet 5a) KBr(s) K+

(aq) + Br-(aq)

b) CaSO4(s) Ca2+(aq) + SO4

2-(aq)

c) LiNO3(s) Li+(aq) + NO3-(aq)

d) MgI2(s) Mg2+(aq) + 2I-(aq)

e) Al(NO3)3(s) Al3+(aq) + 3NO3

-(aq)

f) NH4Cl(s) NH4+

(aq) + Cl-(aq)

g) Fe(NO3)2(s) Fe2+(aq) + 2NO3

-(aq)

h) CuSO4(s) Cu2+(aq) + SO4

2-(aq)

i) Ca(OH)2(s) Ca2+(aq) + 2OH-

(aq)

Worksheet 61.a) barium sulfate b) iron(II) hydroxidec) no reaction d) lead(II) chloridee) silver bromide f) no reactiong) magnesium carbonateh) copper(II) carbonatei) barium sulfate

2.a) i) 2Na+

(aq) + SO4-2

(aq) + Ba+2(aq) +2NO3

-(aq)

BaSO4(s) + 2Na+(aq) + 2NO3

-(aq)

ii) Ba+2(aq) +SO4

-2(aq) BaSO4(s)

b) i) 2K+(aq) + 2OH-

(aq) + Fe+2(aq) +2Cl-(aq)

Fe(OH)2(s) + 2K+(aq) + 2Cl-(aq)

ii) Fe+2(aq) + 2OH-

(aq) Fe(OH)2(s)

c) no reaction

d) i) 2K+(aq) + 2Cl-(aq) + Pb+2

(aq) +2NO3-(aq)

PbCl2(s) + 2K+(aq) + 2NO3

-(aq)

ii) Pb+2(aq) +2Cl-(aq) PbCl2(s)

Covalentbonds

Hydrogenbond

δδ++δδ++

δδ−−


30

2. i) moles required: n = cV = 0.028 x 0.050 = 0.0014 mol

mass: m = n x MM MM(AlCl3) = 133.33g= 0.0014 x 133.33 = 0.19g

ii) moles required: n = cV = 0.400 x 0.250 = 0.100 mol

mass: m = n x MM MM(Na2SO4) = 142.05g= 0.100 x 142.05 = 14.2g

iii) moles: n = cV = 3.75x10-5 x 2.50 = 9.375x10-5 mol

mass: m = n x MM MM(Ca(OH)2) = 74.096g= 9.375x10-5 x 74.096 = 6.95 x10-3g

iv) moles: n = cV = 1.50 x 0.0250 = 0.0375 mol

mass: m = n x MM MM(KBr) = 119.0g= 0.0375 x 119.0 = 4.46g

v) moles: n = cV = 0.800 x 0.100 = 0.0800 mol

mass: m = n x MM MM(Cu(NO3)2) = 187.57g= 0.0800 x 187.57 = 15.0g

Worksheet 91. use c2 = c1V1/V2 in each case.i) 0.100 x 25.0/1000

c2 = 0.00250 (2.50x10-3) molL-1

ii) 1.25 x 5.00/100c2 = 0.0625 (6.25x10-2) molL-1

iii) 0.025 x 2.5/500c2 = 0.000125 (1.25x10-4) molL-1

iv) 0.500 x 8.6/50c2 = 0.086 (8.6x10-2) molL-1

v) 5.35x10-3 x 10.0/250c2 = 0.000214 (2.14x10-4) molL-1

2. Use V1 = c2V2/c1 in each case. (all vols in mL)i) 0.250 x 1000/1.50 V1 = 167 mLii) 0.500 x 100/6.00 V1 = 8.33 mLiii) 1.00x10-4 x 250/0.250 V1 = 0.100 mLiv) 0.010 x 50/0.500 V1 = 1.00 mLv) 0.500 x 10.0/0.875 V1 = 5.71 mL

Worksheet 10(Equations written in “molecular” form to save space)1.Na2SO4(aq) +Ba(NO3)2(aq) BaSO4(s) + 2NaNO3(aq)

n(Ba(NO3)2) = cV = 0.500 x 0.0250 = 0.0125 mol.

∴∴ n(BaSO4) = 0.0125 mol. (mole ratio 1:1)

m(BaSO4) = n x MM MM(BaSO4) = 233.37g= 0.0125 x 233.37

mass BaSO4 = 2.92g

2.Na2SO4(aq) +Ba(NO3)2(aq) BaSO4(s) + 2NaNO3(aq)

mass(BaSO4) collected = 1.27g MM(BaSO4) = 233.37g

n(BaSO4) = m/MM = 1.27/233.37 = 0.00544 mol.

∴∴ n(Na2SO4) = 0.00544 mol (mole ratio = 1:1)

c(Na2SO4) = n/V = 0.00544/0.0100 (10mL = 0.01 L)

Concentration Na2SO4 = 0.544 molL-1.



Worksheet 6 (cont)2. (continued)e) i) Mg+2

(aq) + 2Br-(aq) + 2Ag+

(aq) +2NO3-(aq)

2AgBr(s) + Mg+2(aq) + 2NO3

-(aq)

ii) Ag+(aq) +Br-

(aq) AgBr(s)

f) no reaction

g) i) Mg+2(aq) + 2Cl-(aq) + 2Na+

(aq) + CO3-2

(aq)

MgCO3(s) + 2Na+(aq) + 2Cl-(aq)

ii) Mg+2(aq) + CO3

-2(aq) MgCO3(s)

h) i) Cu+2(aq) + SO4

-2(aq) + 2Na+

(aq) + CO3-2

(aq)

CuCO3(s) + 2Na+(aq) + SO4

-2(aq)

ii) Cu+2(aq) + CO3

-2(aq) CuCO3(s)

i) Cu+2(aq) + SO4

-2(aq) + Ba+2

(aq) + 2NO3-(aq)

BaSO4(s) + Cu+2(aq) + 2NO3

-(aq)

ii) Ba+2(aq) +SO4

-2(aq) BaSO4(s)

Worksheet 71. use c = n/V in each case.i) c = 2.50/0.750 = 3.33 molL-1

ii) c = 0.025/0.050 = 0.50 molL-1

iii) c = 0.35/0.100 = 3.5 molL-1

iv) c = 1.2x10-3/4.0 = 3.0x10-4 molL-1

v) c = 0.95/0.200 = 4.75 molL-1

2. use n = cV in each case.i) n = 0.400 x 2.00 = 0.800 molii) n = 1.25 x 0.450 = 0.563 moliii) n = 0.025 x 0.050 = 1.25 x10-3 moliv) n = 0.0035 x 0.00200 = 7.0 x10-6 molv) n = 2.25 x 0.050 = 0.113 mol

Worksheet 81. i) n = m/MM MM(KNO3) = 101.1g

= 15.80/101.1 = 0.1563 molc = n/V

= 0.1563/0.200 = 0.781 molL-1

ii) n = m/MM MM(CuSO4) = 159.62g= 3.66/159.62 = 0.02293 mol

c = n/V= 0.02293/0.500 = 0.0459 molL-1

iii) n = m/MM MM(NaCl) = 58.44g= 127/58.44 = 2.173 mol

c = n/V= 2.173/1.50 = 1.45 molL-1

iv) n = m/MM MM(Pb(NO3)2) = 331.22g= 85.6/331.22 = 0.2584 mol

c = n/V= 0.2584/3.000 = 0.0861 molL-1

v) n = m/MM MM(LiBr) = 86.841g= 2.35/86.841 = 0.02706 mol

c = n/V= 0.02706/0.250 = 0.108 molL-1

31

Worksheet 12a) kinetic b) move/vibrate fasterc) Specific Heat Capacityd) J/oC/ge) high f) DH = -mCDTg) calorimeter h) Specific Heati) Exo- j) risesk) lost l) negativem) Endo- n) fallso) gained p) positiveq) Heat of Solution r) absorb heats) specific heat capacity t) surroundingsu) insulating v) specific heat capacityw) minor/very small x) thermometery) life forms and climate z) stableaa) transport ab) coolingac) temperate ad) waste heatae) solubility af) oxygenag) lower

Worksheet 131. Simple Heat Calculationsa) Use ΔΔH = mCΔΔT in each case.i) ΔΔH = 50.0 x 4.18 x 30 = 6270 J = 6.27 kJ.ii) ΔΔH = 400 x 4.18 x 85 = 142,120 J = 142 kJiii)ΔΔH = (560 x 106) x 4.18 x 16 = 3.75 x 1010 Jiv)ΔΔH = 100 x 0.39 x 1075 = 41,925 J = 42 kJv) ΔΔH = 100,000 x 0.45 x 100 = 4.5 x106 J

2. Use ΔΔT = ΔΔH/mCi) ΔΔT = 72,000/(250 x 4.18) = 69o final T = 89oCii) ΔΔT = 5,000/(80.0 x 4.18) = 15o final T = 10oCiii)ΔΔT = 10,000/(1000 x 4.18) = 2o final T = 6oCiv)ΔΔT = 20,000/(5000 x 0.45) = 9o final T = -16oCv) ΔΔT = 1,200/(20.0 x 2.44) = 25o final T = 5oC

Worksheet 141.i) ΔΔH = - mCΔΔT

= - (100 + 5.85) x 4.18 x (-13)= +5,752 J for the dissolving of 5.85g

Energy per gram: ΔΔH = +5752/5.85 = +983 J per gramEnergy per mole: MM(NH4NO3) = 80.05g

ΔΔH = +983 x 80.05 = +78,691 J per mole Heat of Solution = +78.7 kJmol-1 (endothermic)

ii) ΔΔH = - mCΔΔT = - (100 + 8.42) x 4.18 x 9.5= - 4,305 J for the dissolving of 8.42g

Energy per gram: ΔΔH = - 4,305/8.42 = -511 J per gram

Energy per mole: MM(Na2CO3) = 106.0gΔΔH = -511 x 106 = - 54,166 J per mole

Heat of Solution = - 54.2 kJmol-1 (exothermic)2.i) ΔΔHsol (NaOH) = -41.6 kJmol-1. MM=40.00g

Energy change for 10.0g = -41.6 x 10.0/40.00 = -10.4 kJ = -10,400 J

ΔΔH = - mCΔΔT, so ΔΔT = ΔΔH/(-mC)= -10,400/(- 250 x 4.18)= + 9.95o

final temp = 28oC (to nearest degree)

ii) ΔΔHsol (NH4Cl) = +15.2 kJmol-1. MM= 53.49gEnergy change for 18.5g = +15.2 x 18.5/53.49

= +5.257 kJ = +5,257 JΔΔH = - mCΔΔT, so ΔΔT = ΔΔH/(-mC)

= +5,257/(- (150+18.5) x 4.18)= - 7.46o

final temp = 15oC (to nearest degree)

Worksheet 10 (cont.)3. 2AgNO3(aq) + K2CO3(aq) Ag2CO3(s) + 2KNO3(aq)

i) n(AgNO3) = cV = 0.500 x 0.0500 = 0.0250 mol.

∴∴ n(Ag2CO3) = 0.0125 mol. (mole ratio 2:1)

m(Ag2CO3) = n x MM MM(Ag2CO3) = 275.81g= 0.0125 x 275.81

mass Ag2CO3 = 3.45g

ii) 2Ag2CO3(s) 4 Ag(s) + 2CO2(g) +O2(g)

from previous part,moles of Ag2CO3 decomposed = 0.0125 mol.

∴∴ moles of silver formed = 0.0250 mol (ratio = 2:4)

m(Ag) = n x MM = 0.0250 x 107.9 = 2.70g

moles of CO2 formed = 0.0125 mol (mole ratio = 2:2)

volume of CO2 = 0.0125 x 24.8 = 0.310 L (310 mL)

Worksheet 111. C 2. C 3. C 4. A 5. D 6. C 7. B8.a) CaCl2(s) Ca2+

(aq) + 2Cl-(aq)

b) Ca2+(aq) + 2Cl-(aq) CaCl2(s)

c) CaCl2(s) Ca2+(aq) + 2Cl-(aq)

9.a) A precipitate of insoluble lead(II) carbonate wouldform.

b) Pb2+(aq) + 2NO3

-(aq) + 2K+

(aq) + CO32-

aq)

PbCO3(s) + 2NO3-(aq) + 2K+

(aq)

c) Moles of PbCO3: n = cV= 0.500 x 0.020= 0.0100 mol.

Mass of PbCO3: m = n x MM (MM=267.21g)= 0.0100 x 267.21= 2.67 g

d) 2PbCO3(s) 2Pb(s) + 2CO2(g) + O2(g)

e) Moles of PbCO3 decomposed = 0.0100 (from part c)moles of gases: CO2 = 0.0100 mol. O2 = 0.0050 mol.Total gas moles = 0.0150 mol.

Volume = 0.0150 x 24.8 = 0.372 L (372 mL)

10.Moles of KI needed:

n = cV = 0.250 x 0.500 = 0.125 mol.

Mass KI needed: m = n x MM MM(KI) = 166.0g= 0.125 x 166.0= 20.75g

Procedure: Weigh out the solid and dissolve in purewater in a beaker. Transfer liquid to a 500mLvolumetric flask. Rinse beaker with several smallamounts of extra water and add washings to flask.Add water to the 500mL mark, drop-wise near the end.Insert stopper and invert flask to mix.





32

Worksheet 151. A 2. D 3. B

4.a) ΔΔH = -mCΔΔT

= -(50 + 4.27) x 4.18 x 27= -6,125 J for 4.27g

Heat per gram = -6,125/4.27 = -1,434 J/gMM(LiOH) = 23.95g

Heat per mole = -1,434 x 23.95 = -34,343 J/mol

ΔΔHsol(LiOH) = -34.3 kJ mol-1.

b) The calorimeter absorbs some of the heatreleased, and this has not been included in thecalculation.

Some of the heat released may have been lost to thesurroundings, hence giving a lower measured value.



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