Mr. Bartelt Presents Solutions and their properties Solutions part 2.

Mr. Bartelt PresentsMr. Bartelt PresentsSolutions and their propertiesSolutions and their properties

Solutions part 2Solutions part 2

Expressions of concentrationExpressions of concentration

Molarity—we already know this one. Molarity—we already know this one. Sadly, there are many other ways Sadly, there are many other ways to express concentration of to express concentration of solutions.solutions.

1.1. Mass PercentMass Percent

2.2. Mole FractionMole Fraction

3.3. MolalityMolality

Mass percentMass percent

As the name suggests this is mass of As the name suggests this is mass of solute (what’s being dissolved) over solute (what’s being dissolved) over mass solution (whole thing)mass solution (whole thing)

Commonly used in the place of Commonly used in the place of molarity for stuff you can buy at Wal-molarity for stuff you can buy at Wal-MartMart– Vinegar (acetic acid and water)Vinegar (acetic acid and water)

– Hydrogen peroxide (HHydrogen peroxide (H22OO22 and water) and water)

ExamplesExamples

Find the molarity of a 5% solution of Find the molarity of a 5% solution of acetic acid (CHacetic acid (CH33COOH) in water. COOH) in water. This is the vinegar you buy at the This is the vinegar you buy at the store. (assume D= 1.00 g/ml)store. (assume D= 1.00 g/ml)

1.1. Assume a solution mass of 100. gAssume a solution mass of 100. g

2.2. Now you know that you’ve got 5.0 g Now you know that you’ve got 5.0 g of acetic acid and 100. ml of water.of acetic acid and 100. ml of water.

ExampleExample

M=mol/VolumeM=mol/Volume– Volume = 100. ml or 0.100 LVolume = 100. ml or 0.100 L– mol= g/MMmol= g/MM– We have 5 grams of acetic acidWe have 5 grams of acetic acid– Acetic acid’s MM is 60.05 g/molAcetic acid’s MM is 60.05 g/mol– 5.0 g / 60.05 g/mol= 0.0833 mol5.0 g / 60.05 g/mol= 0.0833 mol

acid acetic M .83300.100

0.0833M

Next exampleNext example

You can buy 3.00 % Hydrogen You can buy 3.00 % Hydrogen peroxide at the store.peroxide at the store.

I have 1.00 M hydrogen peroxide in I have 1.00 M hydrogen peroxide in the lab.the lab.

Which is more concentrated?Which is more concentrated?

Mole fractionMole fraction

This is similar to what we looked at in our This is similar to what we looked at in our unit on the gas laws.unit on the gas laws.

The above equation assumes a solution The above equation assumes a solution of only two components. If there were of only two components. If there were more then the denominator would have more then the denominator would have additional termsadditional terms

ba

a

NN

Nfraction Mole

ExampleExample

What is the mole fraction if you mixed What is the mole fraction if you mixed equal masses of water and ethylene glycol equal masses of water and ethylene glycol [C[C22HH44(OH)(OH)22]?]?

Assume 100. grams of eachAssume 100. grams of each

watermol 5.5618.0g

1mol * 100.g

0.1627.17

1.61

5.561.61

1.61fraction Mole

glycol ethylene 1.61molg 62.07

mol 1*g 100.

Mole fraction exampleMole fraction example

Concentrated sulfuric acid is 18.0 M Concentrated sulfuric acid is 18.0 M HH22SOSO44. What is the mole fraction water . What is the mole fraction water in 18.0 M sulfuric acid. Assume density in 18.0 M sulfuric acid. Assume density is the same as sulfuric acid 1.84 g/ml.is the same as sulfuric acid 1.84 g/ml.

Assume 1.00 literAssume 1.00 liter Now we have 18.0 mol HNow we have 18.0 mol H22SOSO44

1770gg/mol 98.08*mol 18.0SOH mass 42

watermol 3.89g 18.0

mol 1* waterg 70.g 1770g 1840

178.0 watermol 3.89 SOH mol 18.0

watermol 3.89

42

You tryYou try

Based on the calculations above, Based on the calculations above, determine the mass percent of determine the mass percent of concentrated sulfuric acid.concentrated sulfuric acid.

MolalityMolality At this point it may seem like I’m just At this point it may seem like I’m just

making these names up.making these names up. This is actually a very important expression This is actually a very important expression

of concentration that will prove extremely of concentration that will prove extremely useful when we determine freezing point useful when we determine freezing point depression and boiling point elevation in depression and boiling point elevation in lab on Wednesday.lab on Wednesday.

Molality is defined as moles solute over Molality is defined as moles solute over kilograms of solvent.kilograms of solvent.

solvent

solute

kg

nMolality

ExampleExample

What is the molality of battery acid?What is the molality of battery acid?– Battery acid is sulfuric acid and water with a Battery acid is sulfuric acid and water with a

mass % of 33.5% acid.mass % of 33.5% acid. Assume 100.g of solution HenceAssume 100.g of solution Hence

– 33.5 g H33.5 g H22SOSO44 and 66.5 g water and 66.5 g water

I now needI now need– moles Hmoles H22SOSO44 (easy) (easy)

– Kilograms water (easy)Kilograms water (easy)

acid sulfuric mol .3420g/mol 98.08

g 33.5n

42SOH

ExampleExample

I now needI now need– moles Hmoles H22SOSO44 (easy) (easy)

– Kilograms water (easy)Kilograms water (easy)

waterkg 0.06651000g/kg

66.5gkg OH2

acid sulfuric mol .3420g/mol 98.08

g 33.5n

42SOH

m 14.5kg 0.0665

mol 0.342m

NoteNote

These are terms that need to be These are terms that need to be memorizedmemorized

Molarity (M) = moles/literMolarity (M) = moles/liter Mass % = massMass % = masssamplesample/mass/masstotaltotal

Mole fraction = nMole fraction = nsamplesample/n/ntotaltotal

Molality (m) = nMolality (m) = nsamplesample/kg/kgsoventsovent

– You need to work many problems to get You need to work many problems to get this downthis down

Like dissolves likeLike dissolves like

We have not discussed Lewis dot structures We have not discussed Lewis dot structures yet but we need to discuss the fundamental yet but we need to discuss the fundamental principal of solubility.principal of solubility.

That priciple is “like dissolves like”That priciple is “like dissolves like”– This means that polar solvents are good at This means that polar solvents are good at

dissolving polar molecules and ionic compoundsdissolving polar molecules and ionic compounds– Additionally, non-polar solvents are good at Additionally, non-polar solvents are good at

dissolving non-polar molecules.dissolving non-polar molecules. We will investigate this in a quick activityWe will investigate this in a quick activity But first the structuresBut first the structures

EntropyEntropy

Let’s look at water and hexane (CLet’s look at water and hexane (C66HH1414))

Water has O-H bonds Water has O-H bonds which are highly polarwhich are highly polar

Hexane has C-H bonds Hexane has C-H bonds which are non-polarwhich are non-polar

Determining “likes”Determining “likes”

You don’t really need to know how to You don’t really need to know how to draw the Lewis structure to draw the Lewis structure to determine if a substance is polar or determine if a substance is polar or non-polar.non-polar.

A good rule of thumb isA good rule of thumb is– if your molecule is exclusively carbon if your molecule is exclusively carbon

and hydrogen it’s non-polarand hydrogen it’s non-polar– if it contains carbon, hydrogen, and if it contains carbon, hydrogen, and

nitrogen or oxygen as well, it’s polar.nitrogen or oxygen as well, it’s polar.

Determining “likes”Determining “likes”

There are countless examples There are countless examples contrary to this rule, but you can’t be contrary to this rule, but you can’t be expected to know them so you expected to know them so you should be fine.should be fine.

We will look at something you should We will look at something you should be able to figure out now however.be able to figure out now however.

If the molecule is extremely large If the molecule is extremely large and only contains one oxygen or and only contains one oxygen or nitrogen the region of polarity will nitrogen the region of polarity will not be large enough to make it polar.not be large enough to make it polar.

Polar “tails”Polar “tails” Fatty acids are long chains of hydrocarbons with Fatty acids are long chains of hydrocarbons with

polar “tails” on the ends.polar “tails” on the ends.

The gray and black spheres are hydrogen and carbon respectively. The red spheres are oxygen.

NOTE: The size of the non-polar “tail” dominates the polarity of the molecule and hence the molecule is considered non-polar.

Hence oil and water don’t mix.

Another notable exceptionAnother notable exception

COCO22 has 2 carbon oxygen bonds has 2 carbon oxygen bonds which is highly polar, but they which is highly polar, but they happen to be exactly opposite each happen to be exactly opposite each other in the molecule.other in the molecule.

Hence they “cancel” each other’s Hence they “cancel” each other’s polarity making the molecule non-polarity making the molecule non-polar.polar.

Polar or non-polarPolar or non-polar

CHCH33OH (methanol)OH (methanol)


CHCH44 (methane) (methane)


Carbon tetrachlorideCarbon tetrachloride


CC1212HH2222OO1111 (sugar) (sugar)


Any ionic compoundAny ionic compound


Graphite (carbon)Graphite (carbon)

Now we’re readyNow we’re ready

Today’s lab will focus on the principle Today’s lab will focus on the principle of “like dissolves like”.of “like dissolves like”.

You will note this in your lab and You will note this in your lab and answer questions that relate to the answer questions that relate to the topic of like dissolves like.topic of like dissolves like.

Henry’s lawHenry’s law We will not do any calculations with We will not do any calculations with

Henry’s law in this class.Henry’s law in this class. Henry’s law states that the concentration Henry’s law states that the concentration

of dissolved gas in a solution is directly of dissolved gas in a solution is directly proportional to the partial pressure of the proportional to the partial pressure of the gas above the solution.gas above the solution.

This explains how carbonation works.This explains how carbonation works. When brewing beer yeast breaks sugar When brewing beer yeast breaks sugar

down into alcohol and carbon dioxide. If down into alcohol and carbon dioxide. If the bottle is capped and the COthe bottle is capped and the CO22 is trapped is trapped and builds up pressure.and builds up pressure.

This pressure drives COThis pressure drives CO2 2 into the solution into the solution and makes the beer carbonated.and makes the beer carbonated.

Temperature effects on solutionsTemperature effects on solutions

Typically an increase in temperature Typically an increase in temperature will result in greater solubility of will result in greater solubility of solids, but not always.solids, but not always.

The opposite is true of gasses. The The opposite is true of gasses. The higher the temperature, the less gas higher the temperature, the less gas can be maintained in solution.can be maintained in solution.

Click here if you must know more.Click here if you must know more.

Why do things dissolve?Why do things dissolve? Disorder!!! If something is dissolved Disorder!!! If something is dissolved

in something else in something else The dissolution process makes a The dissolution process makes a

more disordered system more disordered system Now we need to look at Le Now we need to look at Le

Chatelier’s PrincipleChatelier’s Principle

Le Chatelier’s PrincipleLe Chatelier’s Principle

Le Chatelier’s Principle applies to systems Le Chatelier’s Principle applies to systems that are at equilibrium (that are at equilibrium (ΔΔG=0)G=0)

We’ll look at boiler scale in great detail to We’ll look at boiler scale in great detail to learn this new principle.learn this new principle.

Equilibrium systemsEquilibrium systems

CaCOCaCO33 is “insoluble” is “insoluble”

– This doesn’t mean that NO CaCOThis doesn’t mean that NO CaCO33 will be in will be in solution, it just means that very little will be in solution, it just means that very little will be in solution.solution.

– But still, a little is dissolvedBut still, a little is dissolved

CaCa2+2+ + CO + CO332-2- CaCO CaCO33

If water contains CaIf water contains Ca2+2+ ions (and nearly all ions (and nearly all tap water does) it will bond with any COtap water does) it will bond with any CO33

2-2- present and form limestone in a pipe. Not present and form limestone in a pipe. Not good.good.

More systemsMore systems If water contains carbonate ions this reaction If water contains carbonate ions this reaction

will also take placewill also take place COCO33

2-2-(aq)(aq) + CO + CO2(aq)2(aq) + H + H22OO(l)(l) 2HCO 2HCO33

--

What happens to the solubility of COWhat happens to the solubility of CO22 at at higher temperatures?higher temperatures?

If COIf CO22 is removed from our system, the is removed from our system, the equation will shift to the left to compensate equation will shift to the left to compensate for the loss of COfor the loss of CO22..

Now we have extra CONow we have extra CO332-2- and this reaction and this reaction

proceedsproceeds


Le Chatelier's PrincipleLe Chatelier's Principle If a chemical system at equilibrium experiences a If a chemical system at equilibrium experiences a

change in concentration, temperature, volume, or change in concentration, temperature, volume, or total pressure, then the equilibrium shifts to total pressure, then the equilibrium shifts to partially counter-act the imposed change.partially counter-act the imposed change.

COCO332-2-

(aq)(aq) + CO + CO2(aq)2(aq) + H + H22OO(l)(l) 2HCO 2HCO33--


If T goes up If T goes up CO CO22 goes down goes down CO CO332-2- and and

HH22O increase to compensate. The addition O increase to compensate. The addition of COof CO33

2-2- then drives the production of more then drives the production of more CaCOCaCO3 3 and your pipes get clogged up.and your pipes get clogged up.

Colligative propertiesColligative properties Colligative propertiesColligative properties are properties of are properties of

solutions that depend on the number of solutions that depend on the number of particles in a given volume of solvent and particles in a given volume of solvent and not on the mass of the particles.not on the mass of the particles.

We can use colligative properties to We can use colligative properties to determine the molar mass of substances.determine the molar mass of substances.

This is very usefulThis is very useful.. We’re going to look specifically at:We’re going to look specifically at:

– Boiling point elevationBoiling point elevation– Freezing point depressionFreezing point depression

You can change the freezing point?You can change the freezing point?

Of course you can. Ever added salt Of course you can. Ever added salt to snow to make it melt?to snow to make it melt?

Probably not because you only see Probably not because you only see snow twice a decade, but back in my snow twice a decade, but back in my homeland of Chicago it’s done every homeland of Chicago it’s done every winter.winter.

The salt is added to increase the The salt is added to increase the freezing point of the water and hence freezing point of the water and hence melts the ice.melts the ice.

Freezing point depressionFreezing point depression ΔΔT = KT = Kffmmsolution solution m=m=molalitymolality ΔΔT stands for the change in freezing T stands for the change in freezing

point. point. KKf f is a substance specific constantis a substance specific constant We’re only doing this for water so We’re only doing this for water so

just memorize that just memorize that – KKf f = 1.86 (ºC•kg/mol) for freezing point= 1.86 (ºC•kg/mol) for freezing point

– KKb b = 0.51 (ºC•kg/mol) for boiling point= 0.51 (ºC•kg/mol) for boiling point

ExamplesExamples

What freezing point would you What freezing point would you expect from a solution of 120.0 ml of expect from a solution of 120.0 ml of water and 25.0 g of sugar (Cwater and 25.0 g of sugar (C1212HH2222OO1111))

ΔΔT = KT = Kffmmsolutionsolution

First find molality of solutionFirst find molality of solution– Need moles sugarNeed moles sugar– Need kg waterNeed kg water

ExamplesExamples


– Need moles sugarNeed moles sugar– Need kg water 120. mlNeed kg water 120. ml 0.120 kg water 0.120 kg water

mol 0731.0g/mol 342

25.0g sugar mol


– 0.0731 mol sugar0.0731 mol sugar– 0.120 kg water0.120 kg water

ΔΔT = (1.86 ºC•kg/mol)(0.609mol/kg)T = (1.86 ºC•kg/mol)(0.609mol/kg) ΔΔT = 1.13 ºCT = 1.13 ºC The freezing point will drop by 1.13 ºCThe freezing point will drop by 1.13 ºC

kg

mol0.609

kg 0.120

mol 0.0731m

Boiling pointBoiling point These problems are the exact same except These problems are the exact same except

the Kthe Kff is now a K is now a Kbb. The b stands for boiling.. The b stands for boiling.

ΔΔT = KT = Kbbmmsolutionsolution K Kb b = 0.51 (ºC•kg/mol)= 0.51 (ºC•kg/mol) What boiling point elevation would you What boiling point elevation would you

expect from the addition of 20.0 grams of expect from the addition of 20.0 grams of NaCl to 200.0 ml of water?NaCl to 200.0 ml of water?

Answer you’ll get:Answer you’ll get:

ΔΔT = 2.62 ºCT = 2.62 ºC

But this answer is wrongBut this answer is wrong

Why’s it wrongWhy’s it wrong

When you dissolve sugar, the molecule When you dissolve sugar, the molecule doesn’t break up. The moles of sugar are doesn’t break up. The moles of sugar are the moles dissolved.the moles dissolved.

When you dissolve salt, you produce both When you dissolve salt, you produce both NaNa+ + and Cland Cl-- ions. ions.

The BPE and FPD are affected by the # of The BPE and FPD are affected by the # of particlesparticles, and thus when one mole of , and thus when one mole of NaCl is dissolved an mole of NaNaCl is dissolved an mole of Na+ + and Cland Cl-- ions are produced for a total of 2 moles of ions are produced for a total of 2 moles of particles in solution.particles in solution.

New equationNew equation

ΔΔT = KT = Kbb•i•m•i•msolutionsolution

i in this case would be 2i in this case would be 2– For NaFor Na22SOSO44 it would be 3 it would be 3

– For Al(NOFor Al(NO33))33 it would be 4 it would be 4

What boiling point elevation would you What boiling point elevation would you expect from the addition of 20.0 grams expect from the addition of 20.0 grams of NaCl to 200.0 ml of water?of NaCl to 200.0 ml of water?

We’ll confirm this in labWe’ll confirm this in lab

We’re going to dissolve 20.0 g of salt We’re going to dissolve 20.0 g of salt in 200.0 ml of water in lab tomorrow in 200.0 ml of water in lab tomorrow and determine the freezing point.and determine the freezing point.

Now you know what the freezing Now you know what the freezing point should be.point should be.

But this is only the “calculated” But this is only the “calculated” freezing point, we’ll see what the freezing point, we’ll see what the experimental freezing point really is.experimental freezing point really is.

But why does the FP depress?But why does the FP depress?

Think back to ShivaThink back to Shiva What has more disorder, salt water What has more disorder, salt water

or regular water?or regular water?

Mr. Bartelt Presents Solutions and their properties Solutions part 2.

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