+ All Categories
Home > Documents > Solutions

Solutions

Date post: 31-Dec-2015
Category:
Upload: alfonso-collins
View: 34 times
Download: 0 times
Share this document with a friend
Description:
Solutions. Solution Basics. Solution = homogeneous mixture in which a solute is dissolved in a solvent Not limited to solids dissolved in water Solutions can be solids, liquids or gases. Key Terms. Solute- the material being dissolved Solvent-the dissolving medium - PowerPoint PPT Presentation
104
Solutions Solutions
Transcript
Page 1: Solutions

SolutionsSolutions

Page 2: Solutions

Solution BasicsSolution Basics

SolutionSolution = = homogeneous mixturehomogeneous mixture in which in which a a solutesolute is dissolved in a is dissolved in a solventsolvent

Not limited to solids dissolved in waterNot limited to solids dissolved in waterSolutions can be solids, liquids or gasesSolutions can be solids, liquids or gases

Page 3: Solutions

SolutionSolution StateState DescriptionDescription

AirAir GasGas OO22 dissolved in N dissolved in N22

Salt waterSalt water liquidliquid Salt dissolved in waterSalt dissolved in water

Seltzer waterSeltzer water liquidliquid COCO22 gas dissolved in water gas dissolved in water

18 k gold18 k gold solidsolid Silver and copper dissolved Silver and copper dissolved in goldin gold

Household Household hydrogen hydrogen peroxideperoxide

liquidliquid Hydrogen peroxide Hydrogen peroxide dissolved in waterdissolved in water

Your catalytic Your catalytic converterconverter

solidsolid Gases dissolved in solidGases dissolved in solid

Humid airHumid air gasgas Water dissolved in nitrogenWater dissolved in nitrogen

Page 4: Solutions

Key TermsKey Terms Solute- the material being dissolvedSolute- the material being dissolved Solvent-the dissolving mediumSolvent-the dissolving medium Solution- homogeneous mixtureSolution- homogeneous mixture Soluble- capable of dissolving (sugar is soluble Soluble- capable of dissolving (sugar is soluble

in water)in water) Insoluble- not capable of dissolving (lead sulfide Insoluble- not capable of dissolving (lead sulfide

is insoluble in wateris insoluble in water Miscible- Miscible- fluidsfluids that mix ( that mix (liquid,gasliquid,gas) ) Immiscible- fluids that don’t mix (oil and water Immiscible- fluids that don’t mix (oil and water

are immiscible in each other)are immiscible in each other)Note: specify solute and solvent: It is not Note: specify solute and solvent: It is not

enough to say salt dissolves. It does enough to say salt dissolves. It does dissolve in water, but not in hexane.dissolve in water, but not in hexane.

Page 5: Solutions

The dissolving processThe dissolving process

The process by which an ionic compound The process by which an ionic compound dissolved in water involves water dissolved in water involves water molecules surrounded (solvating) the ions. molecules surrounded (solvating) the ions. Because the water molecules have Because the water molecules have charges (polar molecule) they can interact charges (polar molecule) they can interact with the ions.with the ions.

Page 6: Solutions

Why is the O end of thewater molecules

“facing” in toward the Na+ ion?

But the H end facesIn toward the Cl- ion?

Cl-

Na+

Page 7: Solutions

What determines how well What determines how well something dissolves?something dissolves?

There must be an interaction between the There must be an interaction between the molecules of solute and solventmolecules of solute and solvent

There is also a tendency for substances to There is also a tendency for substances to mix regardless of interactions (this is mix regardless of interactions (this is related to a concept known as entropy related to a concept known as entropy which we’ll discuss later)which we’ll discuss later)

Page 8: Solutions

Ionic MaterialsIonic Materials

Tend to dissolve in polar solutes (water) Tend to dissolve in polar solutes (water) as the ions can interact with the solute as the ions can interact with the solute moleculesmolecules

The sum of the interactions must The sum of the interactions must overcome the electrical attraction between overcome the electrical attraction between ions (lattice energy).ions (lattice energy).

Lattice energy depends on charge and Lattice energy depends on charge and size.size.

Page 9: Solutions

Lattice EnergyLattice Energy

Electrical attraction depends on charge Electrical attraction depends on charge and distanceand distance

(F = q(F = q11qq22/d/d22)) Ions with +1 or -1 are easier to separate Ions with +1 or -1 are easier to separate

than those with charges of 2 or 3 (+ or -). than those with charges of 2 or 3 (+ or -). That is why the soluble ions are all +1 or -That is why the soluble ions are all +1 or -1 (1 (solubility rules: Nasolubility rules: Na+1+1, K, K+1+1, NO, NO33

-1-1, NH, NH44+1+1

and Cand C22HH33OO22-1-1 are almost always soluble are almost always soluble))

Page 10: Solutions

SizeSize

Here it gets complicatedHere it gets complicated Larger ions are separated by a greater distance Larger ions are separated by a greater distance

so their attractions are less, making them tend to so their attractions are less, making them tend to be more solublebe more soluble

But, smaller ions have more charge But, smaller ions have more charge concentration (nature does not like a lot of concentration (nature does not like a lot of charge or energy concentrated in one place) charge or energy concentrated in one place) which would make small ions more solublewhich would make small ions more soluble

Which factor matters more (it varies a lot and it Which factor matters more (it varies a lot and it depends) That’s a big help, I know.depends) That’s a big help, I know.

Page 11: Solutions

Molecular Compound SolubilityMolecular Compound Solubility

In order for two liquids to mix with each In order for two liquids to mix with each other there be some interactions between other there be some interactions between the molecules.the molecules.

Two polar liquids can mix with each other Two polar liquids can mix with each other (be miscible).(be miscible).

Two nonpolar liquids can be miscible in Two nonpolar liquids can be miscible in each other.each other.

A polar liquid and a nonpolar liquid will not A polar liquid and a nonpolar liquid will not be misciblebe miscible

Page 12: Solutions

WhyWhy Essentially, when molecules mix with each other Essentially, when molecules mix with each other

they are “trading off” interactionsthey are “trading off” interactions Water can form H-bonds with waterWater can form H-bonds with water Methanol can form H-bonds with methanolMethanol can form H-bonds with methanol Water can form H-bonds with methanolWater can form H-bonds with methanol It is okay for water to interact with methanol since It is okay for water to interact with methanol since

the H-bond interaction is equally “favorable” the H-bond interaction is equally “favorable” whether it is between two waters or a water and a whether it is between two waters or a water and a methanolmethanol

Plus, there is an added “bonus” from mixing and Plus, there is an added “bonus” from mixing and increasing entropyincreasing entropy

Page 13: Solutions
Page 14: Solutions

Like Dissolves LikeLike Dissolves Like

In general, polar substances dissolve well In general, polar substances dissolve well in other polar substances (think of ionic as in other polar substances (think of ionic as extremely polar)extremely polar)

Nonpolar substances dissolve well in Nonpolar substances dissolve well in nonpolar substances nonpolar substances

Page 15: Solutions

Kitchen ExamplesKitchen ExamplesWaterWater salt sugar vegetable oil salt sugar vegetable oil

(polar) (ionic) (polar) (nonpolar)(polar) (ionic) (polar) (nonpolar)

What dissolves in water (salt and sugar)What dissolves in water (salt and sugar)

What does not (oil)What does not (oil)

Will salt or sugar dissolve in oil? Why?Will salt or sugar dissolve in oil? Why?

Ammonia dissolves in water so it must be…Ammonia dissolves in water so it must be…

Iodine dissolves in oil, so it must be…Iodine dissolves in oil, so it must be…

Page 16: Solutions

Oil and WaterOil and WaterThey don’t mix because the water They don’t mix because the water

molecules are all so “busy” attracting each molecules are all so “busy” attracting each other with hydrogen bondsother with hydrogen bonds

An interaction between polar water and An interaction between polar water and nonpolar oil could involve London forces nonpolar oil could involve London forces which are less energetically favorable.which are less energetically favorable.

In a sense, the water won’t trade off In a sense, the water won’t trade off hydrogen bonds for London Forceshydrogen bonds for London Forces

They don’t “repel” each otherThey don’t “repel” each other

Page 17: Solutions
Page 18: Solutions

Examples of SolubilityExamples of Solubility

Page 19: Solutions

A Pattern For IllustrationA Pattern For Illustration

The greater the hydrogen bonding The greater the hydrogen bonding capability of a molecule, the more water capability of a molecule, the more water soluble (or miscible) it will be.soluble (or miscible) it will be.

Alcohols (with an –OH) are less soluble as Alcohols (with an –OH) are less soluble as the molecule gets larger (the OH is a the molecule gets larger (the OH is a smaller percentage of the molecule)smaller percentage of the molecule)

Molecules with two OH groups (glycols) Molecules with two OH groups (glycols) are more solubleare more soluble

Page 20: Solutions

NameName FormulaFormula Solubility in water Solubility in water (g/100g H2O) 20(g/100g H2O) 20°°

MethanolMethanol CHCH33OHOH Completely miscibleCompletely miscible

EthanolEthanol CHCH33CHCH22OHOH Completely miscibleCompletely miscible

1-Propanol1-Propanol CHCH33CHCH22CHCH22OHOH MiscibleMiscible

1-Butanol1-Butanol CHCH33CHCH22CHCH22OHOH 8.08.0

1-Pentanol1-Pentanol CHCH33CHCH22CHCH22CHCH22CHCH22OHOH 2.72.7

1-Hexanol1-Hexanol CHCH33CHCH22CHCH22CHCH22CHCH22CHCH22OHOH 0.60.6

Ethylene Ethylene GlycolGlycol

HOHOCHCH22CHCH22OHOH Completely miscibleCompletely miscible

Propylene GlycolPropylene Glycol HOHOCHCH22CHCH22OHOHCHCH33 Completely miscibleCompletely miscible

GlycerolGlycerol HOHOCHCH22CHCHOHOHCHCH22OHOH Completely miscibleCompletely miscible

Page 21: Solutions

Solubility Curves- SolidsSolubility Curves- Solids

Graphs show the limits of how much solute Graphs show the limits of how much solute can be dissolved per 100g solvent at a can be dissolved per 100g solvent at a given temperaturegiven temperature

The lineThe line shows what would make a shows what would make a saturated solutionsaturated solution

Below the lineBelow the line (more solute can be (more solute can be dissolved) = dissolved) = unsaturated solutionunsaturated solution

Above the lineAbove the line (more solute dissolved than (more solute dissolved than is supposed to happen) = is supposed to happen) = supersaturated supersaturated

solutionsolution

Page 22: Solutions

Most solids (not all) are more Most solids (not all) are more soluble as T increases)soluble as T increases)

Page 23: Solutions

How many grams of KBr can dissolve in 100 g water at 60°?

85 grams

How many grams of KBr can dissolve in 200 g water at 60°?Twice as much solvent candissolve twice as much solute = 170 grams

If 60 grams of NaClO3 are placed in 100 grams of water at 40°C,what type of solution is made?

Unsaturated(below the line which sowsa saturated solution)

Page 24: Solutions

Solubility Curves- GasesSolubility Curves- Gases

GasesGases are are more solublemore soluble as as temperature temperature decreasesdecreases and as and as pressure increasespressure increases

Solubility of solids is not really affected by Solubility of solids is not really affected by P (only T)P (only T)

Easy to remember- soda stays “fizzy” in Easy to remember- soda stays “fizzy” in the fridge (low T) and with the cap on the fridge (low T) and with the cap on (higher P)(higher P)

Page 25: Solutions

Temperature EffectsTemperature EffectsTwo points:a) As T increases, solubilityDecreasesb) Solubility is much lowerfor gases than solids (lookat y-axis)

How does T affect aquatic life?(thermal pollution)

Why do active fish like colder water?

Page 26: Solutions

Pressure EffectsPressure Effects

Gas solubility is directly related Gas solubility is directly related (proportional) to Pressure(proportional) to Pressure

This is known as Henry’s lawThis is known as Henry’s law

Page 27: Solutions
Page 28: Solutions

Solution Concentration Solution Concentration DescriptionsDescriptions

Here we (we means you like when your Here we (we means you like when your parents used to say “we’re going to the parents used to say “we’re going to the dentist”) learn ways to describe solution dentist”) learn ways to describe solution concentration and how to do conversions concentration and how to do conversions between these unitsbetween these units

It helps to keep in mind exact definitions of It helps to keep in mind exact definitions of solute, solvent and solutionsolute, solvent and solution

Remember that density = mass/volume Remember that density = mass/volume and that mass = volume X densityand that mass = volume X density

Page 29: Solutions

Concentration UnitsConcentration Units

Mole FractionMole FractionMass PercentMass PercentMolarityMolaritymolalitymolality

Page 30: Solutions

Mole Fraction (Mole Fraction (XX))

Ratio of moles of substance x over total Ratio of moles of substance x over total moles in mixturemoles in mixture

Has Has no unitsno units (moles/moles = (moles/moles = dimensionlessdimensionless))Example: What is the mole fraction of each Example: What is the mole fraction of each

component in a solution of .45 moles of component in a solution of .45 moles of benzene and .65 moles of toluenebenzene and .65 moles of toluene

(it does not matter if you know what benzene (it does not matter if you know what benzene and toluene are)and toluene are)

Page 31: Solutions

Mole Fraction #1Mole Fraction #1

XXbenben = .45/1.1 = .41 = .45/1.1 = .41

XXtoltol = .65/1.1 = .59 = .65/1.1 = .59

Sum of mole fractions = 1Sum of mole fractions = 1

Page 32: Solutions

Example #2Example #2

Example #2: What is the mole fraction of Example #2: What is the mole fraction of sucrose (Csucrose (C1212HH2222OO1111) in a solution made by ) in a solution made by

dissolving 171 g sucrose in 180 grams of dissolving 171 g sucrose in 180 grams of water.water.

XXsucrosesucrose = moles sucrose/total moles= moles sucrose/total moles0.5 moles sucrose/(10.0 moles water +0.5 moles sucrose)=0.5 moles sucrose/(10.0 moles water +0.5 moles sucrose)=

.5/10.5 = .048.5/10.5 = .048

Page 33: Solutions

Mass %Mass %

Mass % solute = grams solute/mass of Mass % solute = grams solute/mass of solutionsolution

Keep in mind that solution = solute + Keep in mind that solution = solute + solventsolvent

An isotonic solution of NaCl is 0.9%, which An isotonic solution of NaCl is 0.9%, which means that for each 100g solution, there means that for each 100g solution, there are 0.9 g NaClare 0.9 g NaCl

Page 34: Solutions

Mass % Example 1Mass % Example 1

How many grams of KCl are needed to How many grams of KCl are needed to make 525.0 g of 3.10% KClmake 525.0 g of 3.10% KCl

Out of the 525g solution, there must be: Out of the 525g solution, there must be: (0.0310)(525.0) g KCl = 16.3 g KC(0.0310)(525.0) g KCl = 16.3 g KC

The rest (525.0-16.3 = 508.7g) is the mass The rest (525.0-16.3 = 508.7g) is the mass of waterof water

You add 16.3 g KCl to 508.7 g water.You add 16.3 g KCl to 508.7 g water.

Page 35: Solutions

Mass % Example 2Mass % Example 2

In an experiment you need 42.0 g of an In an experiment you need 42.0 g of an HCl solution that is 18.0% HCl.HCl solution that is 18.0% HCl.

How many grams of HCl are used?How many grams of HCl are used?Mass of HCl = (42g)(.18) = 7.6 g HClMass of HCl = (42g)(.18) = 7.6 g HCl

Page 36: Solutions

Molarity (M)Molarity (M)

Moles of solute/L solutionMoles of solute/L solutionMost commonly used unit of solution Most commonly used unit of solution

concentration in chemistry (and the one concentration in chemistry (and the one we have already done).we have already done).

Example 1: What is the molarity of a Example 1: What is the molarity of a solution made by dissolving 35.6 g KBr in solution made by dissolving 35.6 g KBr in a solution volume of 400 ml?a solution volume of 400 ml?

Page 37: Solutions

Example 1Example 1

Moles KBr = 35.6 g KBr 1 mole KBr =Moles KBr = 35.6 g KBr 1 mole KBr =

119 g KBr119 g KBr

.299 moles/.400L = .75M (molar).299 moles/.400L = .75M (molar)

Page 38: Solutions

Example 2Example 2

What mass of sucrose (CWhat mass of sucrose (C1212HH2222OO1111) is ) is needed to make 350 ml of .25M solution.needed to make 350 ml of .25M solution.

M= moles soluteM= moles solute L solutionL solution.25M = x moles.25M = x moles .350L.350LX = .0875 mole x 342 g/mole = 29.9 gX = .0875 mole x 342 g/mole = 29.9 gsucrose sucrose

Page 39: Solutions

Volumetric FlaskVolumetric Flask

Glassware used for Glassware used for accurately making accurately making solutionssolutions

(see molarity notes (see molarity notes from ch 4 for from ch 4 for more detail)more detail)

1 mole in 1 L =1 Molarsolution

Page 40: Solutions

Molality (m)Molality (m)

Moles of solute per kilogram of solventMoles of solute per kilogram of solventNote two differences with molarity- mass Note two differences with molarity- mass

instead of volume and solvent instead of instead of volume and solvent instead of solution.solution.

Used because mass is not affected by Used because mass is not affected by Temperature, while volume isTemperature, while volume is

Page 41: Solutions

Molality example 1Molality example 1

What is the molality of a solution made by What is the molality of a solution made by dissolving 67.8 g HF in 780 g water?dissolving 67.8 g HF in 780 g water?

m = moles HFm = moles HF

Kg water (the solvent) =Kg water (the solvent) =

67.8 g HF 1 mole HF =67.8 g HF 1 mole HF =

20.0 g HF .780 Kg20.0 g HF .780 Kg

4.35m (molal)4.35m (molal)

Page 42: Solutions

Molality example 2Molality example 2

What mass of CaClWhat mass of CaCl22 must be dissolved in must be dissolved in

500 g water top make a .5m solution 500 g water top make a .5m solution m = moles calcium chloride/Kg waterm = moles calcium chloride/Kg waterMoles calcium chloride = .5m x .5 Kg =Moles calcium chloride = .5m x .5 Kg = .25 moles CaCl.25 moles CaCl22 110 g = 27.5 g CaCl 110 g = 27.5 g CaCl22

1 mole CaCl1 mole CaCl22

Page 43: Solutions

Conversions Between Conversions Between Concentration UnitsConcentration Units

Keys- the one thing that ties volume and Keys- the one thing that ties volume and mass together is densitymass together is density

Solution = solute + solventSolution = solute + solventLabel everything (unit and id as solute, Label everything (unit and id as solute,

solvent or solution as appropriatesolvent or solution as appropriateWe’ll look at several examples and you’ll We’ll look at several examples and you’ll

get lots of practiceget lots of practice

Page 44: Solutions

molality molality mole fraction mole fraction

An aqueous solution of sucrose (CAn aqueous solution of sucrose (C1212HH2222OO1111) )

has a molality of 0.150m. What are the mole has a molality of 0.150m. What are the mole fractions of sucrose and water in the solution?fractions of sucrose and water in the solution?

0.150 m means 0.150 moles of sucrose for 0.150 m means 0.150 moles of sucrose for each 1 kg of water. That ratio holds no matter each 1 kg of water. That ratio holds no matter what the exact amounts are. 1 kg of water = what the exact amounts are. 1 kg of water = 1000 g = 1000 g water 1 mole water =1000 g = 1000 g water 1 mole water =

18.0 g water18.0 g water

55.6 moles water.55.6 moles water.

Page 45: Solutions

molality molality mole fraction cont’d mole fraction cont’d

Mole fraction of sucrose = moles sucrose =Mole fraction of sucrose = moles sucrose =

total molestotal moles

= 0.150 moles/(0.150 moles + 55.6 moles ) == 0.150 moles/(0.150 moles + 55.6 moles ) =

.00269.00269Mole fraction of water = moles water =Mole fraction of water = moles water =

total moles total moles

= 55.6 moles/(0.150 moles + 55.6 moles) == 55.6 moles/(0.150 moles + 55.6 moles) =

0.9970.997

Page 46: Solutions

Mole fraction Mole fraction molality molality

A solution of sodium bromide has a NaBr A solution of sodium bromide has a NaBr mole fraction of 0.250 and a water mole mole fraction of 0.250 and a water mole fraction of 0.750. What is the molality of the fraction of 0.750. What is the molality of the NaBr in the solution?NaBr in the solution?

The molality is = moles NaBr divided by Kg The molality is = moles NaBr divided by Kg water. We already know there are .250 water. We already know there are .250 moles of NaBr and there would bemoles of NaBr and there would be

0.750 moles H0.750 moles H22O 18g HO 18g H22O 1 Kg =O 1 Kg =

1mole H1mole H22O 1000 gO 1000 g

Page 47: Solutions

Mole fraction Mole fraction molality cont’d molality cont’d

Molality = .250 moles NaBr =Molality = .250 moles NaBr =

0.0135 kg H0.0135 kg H22O O

18.5 m18.5 m

Page 48: Solutions

Mass % Mass % mole fraction mole fraction

What are the mole fractions in an aqueous What are the mole fractions in an aqueous 5.4% NaF solution?5.4% NaF solution?

Assume 1000 gAssume 1000 g5.4% of the mass (54 grams is NaF) and 5.4% of the mass (54 grams is NaF) and 94.6% of the mass (946 g is H94.6% of the mass (946 g is H22O)O)Mole NaF = 54 g NaF 1 mole NaF = 1.29 Mole NaF = 54 g NaF 1 mole NaF = 1.29 42 g NaF42 g NaFMole HMole H22O = 946 g HO = 946 g H22O 1mole HO 1mole H22O = 52.6O = 52.6

18 g H18 g H22OO

Page 49: Solutions

Mass % Mass % mole fraction cont’d mole fraction cont’d

Mole fraction NaF = 1.29/(1.29 +52.6) Mole fraction NaF = 1.29/(1.29 +52.6) = .024= .024

Mole fraction HMole fraction H22O = 52.5/(1.29 + 52.6) =O = 52.5/(1.29 + 52.6) = .976 .976

Page 50: Solutions

Molality Molality mass % mass %

What is the mass % of a .650 m solution of What is the mass % of a .650 m solution of benzene (Cbenzene (C66HH66) dissolved in toluene (C) dissolved in toluene (C77HH88)?)?

This means that we can think of the solution This means that we can think of the solution as being made of .650 moles of benzene as being made of .650 moles of benzene dissolved in 1000 g toluenedissolved in 1000 g toluene

We could use any similar ratioWe could use any similar ratio .650 moles C.650 moles C66HH66 78 g C 78 g C66HH66 = 50.7 g C = 50.7 g C66HH66

1 mole C1 mole C66HH66

Page 51: Solutions

Molality Molality mass % cont’d mass % cont’d

Total mass = 1050.7 gTotal mass = 1050.7 g% C% C66HH66 = 50.7g C = 50.7g C66HH66 /1050.7 g total = /1050.7 g total =

4.8%4.8%% C% C77HH88 = 1000g C = 1000g C77HH88 /1050.7 g total = /1050.7 g total =

95.2 %95.2 %

Page 52: Solutions

molality (m) molality (m) Molarity (M) Molarity (M)

An aqueous solution of KBr has a molality An aqueous solution of KBr has a molality of .298m. What is the molarity of the solution? of .298m. What is the molarity of the solution? The solution density = 1.014 g/ml.The solution density = 1.014 g/ml.

We have .298 moles of KBr in 1000 grams of We have .298 moles of KBr in 1000 grams of water (molality definition)water (molality definition)

The solution mass is equal to the mass of the The solution mass is equal to the mass of the water (1000 grams) plus the mass of the .298 water (1000 grams) plus the mass of the .298 mole KBr. mole KBr.

From solution mass and density, we can get From solution mass and density, we can get solution volume, which lets us get Molaritysolution volume, which lets us get Molarity

Page 53: Solutions

m m M M

.298 mole KBr 119 g KBr = 35.5 g KBr.298 mole KBr 119 g KBr = 35.5 g KBr

1 mole KBr1 mole KBrMass of solution = mass of solute (KBr) Mass of solution = mass of solute (KBr)

plus mass of solvent (1000 g Hplus mass of solvent (1000 g H22O) = 1036 gO) = 1036 gVolume of solution V = m/D = Volume of solution V = m/D =

1036 g/ 1.014 g/ml = 1022 ml = 1.022 L1036 g/ 1.014 g/ml = 1022 ml = 1.022 L

Molarity = .298 mole KBr/1.022 L = .292 MMolarity = .298 mole KBr/1.022 L = .292 M

Page 54: Solutions

M and mM and m

For dilute solutions molality and Molarity For dilute solutions molality and Molarity are quite close (the solution is almost all are quite close (the solution is almost all solvent so for an aqueous solution volume solvent so for an aqueous solution volume in L and mass in Kg are similar (D = 1.00 in L and mass in Kg are similar (D = 1.00 g/ml or Kg/L)g/ml or Kg/L)

For more concentrated solutions this is not For more concentrated solutions this is not truetrue

Page 55: Solutions

Molarity (M) Molarity (M) molarity (m) molarity (m)

A solution of glucose (CA solution of glucose (C66HH1212OO66) has a ) has a

molarity of 3.67M and a density of 1.27 molarity of 3.67M and a density of 1.27 g/ml. Find the molality of the solution.g/ml. Find the molality of the solution.

3.67M means there are 3.67 moles 3.67M means there are 3.67 moles glucose dissolved in 1000 ml (1 L) of glucose dissolved in 1000 ml (1 L) of solutionsolution. .

The 1000 ml solution has a density of 1.27 The 1000 ml solution has a density of 1.27 g/ml, so it has a mass of 1270 gramsg/ml, so it has a mass of 1270 grams

Page 56: Solutions

M M m m

That 1270 grams of solution is made of That 1270 grams of solution is made of glucose and water. The mass of glucose in glucose and water. The mass of glucose in the solution = 1270 = 661 g glucose + the solution = 1270 = 661 g glucose + mass of water .mass of water .

Mass of water = 1270 g solution – 661 g Mass of water = 1270 g solution – 661 g sucrose (solute) = 609 g water (solvent)sucrose (solute) = 609 g water (solvent)

molality = moles glucose/kg solvent = molality = moles glucose/kg solvent = 3.67 moles/.609Kg = 6.03 m3.67 moles/.609Kg = 6.03 m

Page 57: Solutions

Mass % to MolarityMass % to Molarity

What is the molarity of a 3.7% solution of What is the molarity of a 3.7% solution of glucose with a density of 1.02 g/ml?glucose with a density of 1.02 g/ml?

In 1000 g solution we will have 37 g In 1000 g solution we will have 37 g glucose ( = 0.206 moles) and 963 g water.glucose ( = 0.206 moles) and 963 g water.

1000 g solution have a volume of 1000 g solution have a volume of 1000g/1.02g/ml = 980.4 ml solution1000g/1.02g/ml = 980.4 ml solution

M = mole glucose/L solution =M = mole glucose/L solution = .0206moles/.980L = 0.210 M.0206moles/.980L = 0.210 M

Page 58: Solutions

Other combinationsOther combinations

Given mass %, molality, mole fraction and Given mass %, molality, mole fraction and molarity there are lots of combinations that are molarity there are lots of combinations that are possible.possible.

Keep units and labels clear (not just 2.4 moles, Keep units and labels clear (not just 2.4 moles, but 2.4 moles sucrose for examplebut 2.4 moles sucrose for example

The fact that “ties” mass and volume together is The fact that “ties” mass and volume together is densitydensity

Keep solute, solvent and solution clear (by Keep solute, solvent and solution clear (by labelling)labelling)

Ample practice will be provided!Ample practice will be provided!

Page 59: Solutions
Page 60: Solutions

Don’t Give Up if the first few are difficult!

Page 61: Solutions

Colligative PropertiesColligative Properties

Definition- properties of a solution that Definition- properties of a solution that depend only on the number of dissolved depend only on the number of dissolved particles and not on the identity of the particles and not on the identity of the dissolved particlesdissolved particles

Examples we’ll go overExamples we’ll go overVapor pressure loweringVapor pressure loweringBoiling point elevationBoiling point elevationFreezing point depressionFreezing point depressionOsmotic pressureOsmotic pressure

Page 62: Solutions

Vapor Pressure Vapor Pressure

The vapor pressure of a solution = P from The vapor pressure of a solution = P from the evaporated molecules above a liquid in the evaporated molecules above a liquid in a closed containera closed container

In a mixture the vapor pressure of each In a mixture the vapor pressure of each component is proportional to the mole component is proportional to the mole fraction of that componentfraction of that component

Nonvolatile substances have a vapor Nonvolatile substances have a vapor pressure of zeropressure of zero

Page 63: Solutions

Raoult’s LawRaoult’s Law

The partial pressure of a solvent (PThe partial pressure of a solvent (PAA) = the ) = the mole fraction of the solvent (Xmole fraction of the solvent (XAA) times the ) times the vapor pressure of the pure solvent (Pvapor pressure of the pure solvent (P°°AA))

PPAA = (X = (XAA )(P )(P°°AA))

If there is a volatile (capable of evaporating) If there is a volatile (capable of evaporating) solute (B) then its vapor pressure is : Psolute (B) then its vapor pressure is : PBB = = (X(XBB )(P )(P°°BB))

Page 64: Solutions

Raoult’s LawRaoult’s Law

Or, the amount of vapor pressure lowering Or, the amount of vapor pressure lowering ΔΔPP = (X= (XBB )(P )(P°°AA))

Time for a few examples!Time for a few examples!

Page 65: Solutions

Example 1Example 1What is the vapor pressure of a solution of 90 g What is the vapor pressure of a solution of 90 g

glucose (molar mass 180 g/mole) dissolved in glucose (molar mass 180 g/mole) dissolved in 100 g water? The vapor pressure of water at 100 g water? The vapor pressure of water at room T = 24 mm Hgroom T = 24 mm Hg

There are 0.50 mole glucose and 5.56 moles There are 0.50 mole glucose and 5.56 moles waterwater

Get mole fraction of each componentGet mole fraction of each componentXXGlcGlc = 0.5 moles glc/6.06 moles total = 0.083 = 0.5 moles glc/6.06 moles total = 0.083

XXH2OH2O = 5.56 moles water/6.06 moles glc = 0.917 = 5.56 moles water/6.06 moles glc = 0.917

Page 66: Solutions

ContinuingContinuingThe vapor pressure of the solution = The vapor pressure of the solution = P = (XP = (Xwaterwater)(P)(P°°waterwater) = (.917)(24 mm Hg) = ) = (.917)(24 mm Hg) =

21.8 mm Hg21.8 mm HgThe vapor pressure was lowered by 2.2 The vapor pressure was lowered by 2.2

mm Hgmm HgThe glucose is nonvolatile, so it contributes The glucose is nonvolatile, so it contributes

0 to the total pressure0 to the total pressureWhat is both components are volatile? (I What is both components are volatile? (I

hear you ask)hear you ask)

Page 67: Solutions

Water and Alcohol MixtureWater and Alcohol Mixture

““Consider a solution made from 100 grams of Consider a solution made from 100 grams of water and 100 grams of ethanol (Cwater and 100 grams of ethanol (C22HH66O)”. What O)”. What is the vapor pressure above the solution at room is the vapor pressure above the solution at room T. The vapor pressure of pure water is 24 mm T. The vapor pressure of pure water is 24 mm Hg and of pure ethanol is 80 mm HgHg and of pure ethanol is 80 mm Hg

Math and science books always say things like Math and science books always say things like “consider a spherical cow..” I don’t know why.“consider a spherical cow..” I don’t know why.

In this case we have two liquids, each of which In this case we have two liquids, each of which ha its own vapor pressureha its own vapor pressure

We need the mole fraction of each componentWe need the mole fraction of each component

Page 68: Solutions

Water/Alcohol Cont’dWater/Alcohol Cont’d

We have 5.56 moles water and 2.17 moles ethanol.We have 5.56 moles water and 2.17 moles ethanol. XXwaterwater = 0.719 and X = 0.719 and Xetheth = 0.281 = 0.281 The pressure above the liquid due to water =The pressure above the liquid due to water = PPwaterwater = X = XwaterwaterPP°°water water = (0.719)(24 mmHg) == (0.719)(24 mmHg) = 17.3 mm Hg17.3 mm Hg PPetheth = X = XethethPP°°eth eth = (0.281)(80 mmHg) = = (0.281)(80 mmHg) = 22.5 mm Hg22.5 mm Hg And the total P of the vapor = 39.8 mmHgAnd the total P of the vapor = 39.8 mmHg

Page 69: Solutions

Water Alcohol (Again)Water Alcohol (Again)

The vapor has a mole fraction of The vapor has a mole fraction of XXwaterwater = 17.3/39.8 = 0.43 = 17.3/39.8 = 0.43XXethanolethanol = 22.5/39.8 = 0.57 = 22.5/39.8 = 0.57 In the original liquid the mole fraction of In the original liquid the mole fraction of

water was triple that of ethanol but in the water was triple that of ethanol but in the vapor the ethanol (the more volatile vapor the ethanol (the more volatile component) has been enriched.component) has been enriched.

This leads to a way to separate This leads to a way to separate components of a mixture called ….components of a mixture called ….

Page 70: Solutions

Fractional DistillationFractional Distillation

Page 71: Solutions

Solvent plus non-volatile solute(salt dissolved in water)

Pvap is loweredPure solvent

Page 72: Solutions
Page 73: Solutions

Freezing Point DepressionFreezing Point Depression

Adding a solute to a liquid lowers the Adding a solute to a liquid lowers the freezing point of the solution relative to the freezing point of the solution relative to the pure solventpure solvent

The change in the freezing point,The change in the freezing point,ΔΔTTff is is

equal to a constant, Kequal to a constant, Kff (unique to each (unique to each

solvent) times the molality of the dissolved solvent) times the molality of the dissolved solute, Csolute, Cmm. (molal concentration). (molal concentration)

ΔΔTTf f = K= KffCCmm

Page 74: Solutions

Sample CalculationSample CalculationPure benzene freezes at 5.46Pure benzene freezes at 5.46°C. What is the °C. What is the

freezing point of solution made by freezing point of solution made by dissolving .5 moles of toluene in 1000 g dissolving .5 moles of toluene in 1000 g benzene (giving a molal concentration, Cm = benzene (giving a molal concentration, Cm = 0.5m) (K0.5m) (Kff for benzene = 5.07 °C/m). K for benzene = 5.07 °C/m). Kff values values are always given.are always given.

ΔΔTTff = K = KffCCmm = (5.07 °C/m)(0.5m) = 2.54°C = (5.07 °C/m)(0.5m) = 2.54°CThe The freezing point is loweredfreezing point is lowered by 2.54°C to by 2.54°C to

5.46°C – 2.54°C = 2.98 °C5.46°C – 2.54°C = 2.98 °CThe calculation gives the The calculation gives the changechange in freezing in freezing

point.point.

Page 75: Solutions

DrivewaysDriveways

Why do we throw salt (or some other ionic Why do we throw salt (or some other ionic compound) on driveways in the winter?compound) on driveways in the winter?

The solution of salt water has a lower The solution of salt water has a lower freezing point that pure water, making it freezing point that pure water, making it harder to refreeze as it cools off.harder to refreeze as it cools off.

Why is calcium chloride better than Why is calcium chloride better than sodium chloride?sodium chloride?

Page 76: Solutions

DrivewaysDriveways

Each mole of NaCl contributes two moles of Each mole of NaCl contributes two moles of dissolved particles (ions) Nadissolved particles (ions) Na++ and Cl and Cl--

Each mole of dissolved CaClEach mole of dissolved CaCl22 give three give three

moles of dissolved ions one Camoles of dissolved ions one Ca+2+2 and 2 Cl and 2 Cl-1-1..MgClMgCl22 is used b/c you get more moles/ton is used b/c you get more moles/ton

(smaller molar mass)(smaller molar mass) It ends up being a balance between It ends up being a balance between

effectiveness and cost.effectiveness and cost.

Page 77: Solutions

ΔΔTb = imKb Tb = imKb

i = “van’t Hoff factor” which is one for a i = “van’t Hoff factor” which is one for a molecular compound (does not separate molecular compound (does not separate into ions when dissolving) but depends on into ions when dissolving) but depends on the number of ions for an ionic substancethe number of ions for an ionic substance

See boiling water example coming upSee boiling water example coming up

Page 78: Solutions

Boiling Point ElevationBoiling Point Elevation

Adding a solute to a liquid raises the Adding a solute to a liquid raises the boiling pointboiling point

The equation = The equation = ΔΔTTbb = K = KbbCCmm where K where Kbb is a is a

constant for each solventconstant for each solventSince PSince Pvapvap is lower boiling point is raised. is lower boiling point is raised.

Page 79: Solutions
Page 80: Solutions

Salting Water for CookingSalting Water for Cooking

So how much do you raise the boiling So how much do you raise the boiling point of water when you put salt in it for point of water when you put salt in it for cooking?cooking?

Example- let’s say you have about about a Example- let’s say you have about about a gallon of water (call it 4.0L or 4000 g) and gallon of water (call it 4.0L or 4000 g) and you toss in a huge handful 100g of NaCl. you toss in a huge handful 100g of NaCl. What is the boiling point going to be?What is the boiling point going to be?

We need the molality of the solutionWe need the molality of the solution

Page 81: Solutions

Cooking ExampleCooking Example

We have 100 g NaCl = 1.72 moles of NaCl We have 100 g NaCl = 1.72 moles of NaCl but it dissolves and ionizes to give 3.44 but it dissolves and ionizes to give 3.44 moles of dissolved ionsmoles of dissolved ions

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

3.44 moles dissolved particle/4.00 Kg 3.44 moles dissolved particle/4.00 Kg solute =solute =

0.86 molal dissolved ions0.86 molal dissolved ionsΔΔTb = KTb = KbbCCmm = (0.512°C/m)(0.86m) = = (0.512°C/m)(0.86m) =0.44°C so…0.44°C so…

Page 82: Solutions

FinallyFinally

The boiling point is raised by 0.44The boiling point is raised by 0.44°C and °C and the solution boils at 100.44°C. the solution boils at 100.44°C.

Not enough to matterNot enough to matterSalt is added to water for taste not to Salt is added to water for taste not to

make the water boil “hotter”make the water boil “hotter”Notice that a dissolved solute raises the Notice that a dissolved solute raises the

boiling point and lowers the freezing pointboiling point and lowers the freezing point

Page 83: Solutions

Anti-freezeAnti-freeze

Adding ethylene glycol (or a similar Adding ethylene glycol (or a similar compound) to your radiator lowers the compound) to your radiator lowers the freezing point and it also raises the boiling freezing point and it also raises the boiling point.point.

You are adding anti-freeze but it could just You are adding anti-freeze but it could just as well be called “anti-boil”as well be called “anti-boil”

Schrodinger’s cat, Perhaps ?(safety story aboutAntifreeze)

Page 84: Solutions

Application (and popular test item)Application (and popular test item)

Finding the molar mass of a solute by Finding the molar mass of a solute by freezing point depression or boiling point freezing point depression or boiling point elevation.elevation.

A solution is made by dissolving .154 A solution is made by dissolving .154 grams of a molecular (i = 1) substance in grams of a molecular (i = 1) substance in 26.87 g water. The resulting solution 26.87 g water. The resulting solution freezes at -0.57freezes at -0.57°C. What is the molar °C. What is the molar mass of the substance? (Kmass of the substance? (Kff = 1.86°C/m) = 1.86°C/m)

Page 85: Solutions

Plan of AttackPlan of Attack

From change in freezing point and KFrom change in freezing point and Kff get get

molality.molality.From molality and g solvent, get moles From molality and g solvent, get moles

solutesoluteFrom moles solute and mass of solute, get From moles solute and mass of solute, get

molar massmolar mass

Page 86: Solutions

SolutionSolution

ΔΔTTff=K=KffCCmm 0.57°C = (1.86°C/m)(C 0.57°C = (1.86°C/m)(Cmm))The molality, Cm = 0.306 mThe molality, Cm = 0.306 m0.306 m = x moles solute0.306 m = x moles solute 0.02687 kg water 0.02687 kg water X = 0.00823 moles soluteX = 0.00823 moles soluteMolar mass = 0.154 g =Molar mass = 0.154 g = 0.00823 moles 0.00823 moles = 18.7 g/mole= 18.7 g/mole

Page 87: Solutions

Osmotic PressureOsmotic Pressure

Osmosis = transport of ions or molecules Osmosis = transport of ions or molecules through a membrane to equalize solute through a membrane to equalize solute concentration of the two sides of the concentration of the two sides of the barrier (membrane) (highbarrier (membrane) (high low) low)

Osmotic pressure is the pressure that Osmotic pressure is the pressure that when applied to the solution can stop when applied to the solution can stop osmosisosmosis

ππ = MRT (M = Molarity), R = gas constant, = MRT (M = Molarity), R = gas constant, and T in Kelvin) and T in Kelvin)

Page 88: Solutions
Page 89: Solutions
Page 90: Solutions

Sample CalculationSample Calculation

What is molar mass of a solution made by What is molar mass of a solution made by dissolving 0.803 grams of a protein molecule dissolving 0.803 grams of a protein molecule in 120 ml solution if the osmotic pressure = in 120 ml solution if the osmotic pressure = 5.12 mm Hg at 205.12 mm Hg at 20°C°C

ππ = MRT = MRT5.21 mm Hg = (M )(62.4 L mm Hg )(293 K) =5.21 mm Hg = (M )(62.4 L mm Hg )(293 K) =

mole Kmole K

= 2.85 x 10= 2.85 x 10-4-4MM

Page 91: Solutions

ContinuingContinuing

2.85 x 102.85 x 10-4 -4 M = moles solute =M = moles solute =

L solutionL solution

2.85 x 102.85 x 10-4-4M = x moles solute =M = x moles solute =

.120 L.120 L

X = 3.42 x 10X = 3.42 x 10-5-5 moles moles

0.803 g/3.42 x 100.803 g/3.42 x 10-5-5 moles = 2.35 x 10 moles = 2.35 x 1044 g/mole g/mole

Page 92: Solutions

Main UseMain Use

Osmotic pressure problems are mainly Osmotic pressure problems are mainly used for large molar mass substances used for large molar mass substances (proteins and other macromolecules) when (proteins and other macromolecules) when the Molarity or molality (they are almost the Molarity or molality (they are almost the same for such dilute solutions) would the same for such dilute solutions) would be so small that there would be no way to be so small that there would be no way to measure a change in freezing or boiling measure a change in freezing or boiling point changepoint change

ΔΔTTff = K = KffCCmm = (1.86°C/m)(2.85 x 10 = (1.86°C/m)(2.85 x 10-4-4m)=m)=5.3 x 105.3 x 10-4-4°C (freezes at -0.00053°C)°C (freezes at -0.00053°C)

Page 93: Solutions

Another ApplicationAnother Application

You could combine the molar mass You could combine the molar mass determination with empirical formula info determination with empirical formula info to get a fun-filled problem (as in a hw to get a fun-filled problem (as in a hw example or two and who knows where example or two and who knows where else…)else…)

Page 94: Solutions

ColloidsColloids

A dispersion of particles of one substance A dispersion of particles of one substance into another substanceinto another substance

Particles are larger than in a true solution Particles are larger than in a true solution (large enough to reflect light) but smaller (large enough to reflect light) but smaller than can be seen with a light microscope.than can be seen with a light microscope.

Colloids do not “settle” out over time as a Colloids do not “settle” out over time as a suspension does.suspension does.

Page 95: Solutions

Tyndall EffectTyndall EffectColloid particles are large enough to scatter or Colloid particles are large enough to scatter or

reflect light (also the case for particles in a reflect light (also the case for particles in a suspension, such as muddy water). Particles suspension, such as muddy water). Particles in a solution are too small to reflect a beam of in a solution are too small to reflect a beam of light.light.

COLLOID SOLUTION

TyndallEffect

Page 96: Solutions

Colloid examplesColloid examples Fog, whipped cream, shaving cream, Fog, whipped cream, shaving cream,

mayonnaisemayonnaise Liquid droplets dispersed in a gas = aerosol Liquid droplets dispersed in a gas = aerosol

(fog)(fog) Liquid droplets dispersed in a liquid = emulsion Liquid droplets dispersed in a liquid = emulsion

(mayo)(mayo) not not

Solid dispersed in liquid = sol (jelly)Solid dispersed in liquid = sol (jelly)

Page 97: Solutions

More ColloidsMore Colloids

If the If the continuous phasecontinuous phase of a colloid is water then of a colloid is water then we can look at the strength of attraction between we can look at the strength of attraction between the the dispersed phasedispersed phase and the water: and the water: Strong = Strong = hydrophillic colloidhydrophillic colloid (often hydrogen bonding (often hydrogen bonding

occurs)occurs) No attraction = No attraction = hydrophobic colloidhydrophobic colloid (these are (these are

unstable and will separate eventually- but a very long unstable and will separate eventually- but a very long time)time)

Coagulation happens when the dispersed phase Coagulation happens when the dispersed phase aggregates (often by interaction with ions)aggregates (often by interaction with ions)

Page 98: Solutions

Association ColloidsAssociation Colloids

Micelle = colloidal particle formed in water Micelle = colloidal particle formed in water by the association of molecules or ions by the association of molecules or ions with a hydrophobic (nonnpolar) end and a with a hydrophobic (nonnpolar) end and a hydrophilic end (polar end).hydrophilic end (polar end).

Soap in water is an exampleSoap in water is an example

Page 99: Solutions

SoapSoap

Page 100: Solutions

How Soap Works (simplified)How Soap Works (simplified)

Page 101: Solutions

DetergentsDetergents

Work on the same principle except they Work on the same principle except they don’t interact with dissolved ions in water don’t interact with dissolved ions in water (hard water) so they don’t form a “soap (hard water) so they don’t form a “soap scum”- that’s the technical termscum”- that’s the technical term

Both soaps and detergents lower surface Both soaps and detergents lower surface tension, by disrupting hydrogen bonds) tension, by disrupting hydrogen bonds) allowing water to interact with the micelles allowing water to interact with the micelles and “wash away” dirtand “wash away” dirt

Page 102: Solutions

Wrap-UpWrap-Up

Start with the basicsStart with the basicsMake sure you can do all of the Make sure you can do all of the

calculations on basic solution calculations on basic solution concentration unitsconcentration units

When doing conversions between ways of When doing conversions between ways of describing concentration, label everything describing concentration, label everything with a unit and an identity (ex: 0.35 moles with a unit and an identity (ex: 0.35 moles KI) KI)

Page 103: Solutions

Need Help?Need Help?

http://wine1.sb.fsu.edu/chm1046/notes/Solnhttp://wine1.sb.fsu.edu/chm1046/notes/SolnProp/SolnProc/SolnProc.htmProp/SolnProc/SolnProc.htm

http://www.chem.purdue.edu/gchelp/solutiohttp://www.chem.purdue.edu/gchelp/solutions/character.htmlns/character.html

http://www.towson.edu/~ladon/concas.htmlhttp://www.towson.edu/~ladon/concas.htmlhttp://cwx.prenhall.com/bookbind/pubbooks/http://cwx.prenhall.com/bookbind/pubbooks/

hillchem3/chapter12/deluxe.htmlhillchem3/chapter12/deluxe.htmlhttp://www.sciencegeek.net/Chemistry/http://www.sciencegeek.net/Chemistry/

taters/chap13SolutionConcentration.htmtaters/chap13SolutionConcentration.htm

Page 104: Solutions

More Help More Help

http://www.chem.lsu.edu/lucid/tutorials/soluhttp://www.chem.lsu.edu/lucid/tutorials/solubility/Solubility.htmlbility/Solubility.html

http://chemed.chem.purdue.edu/genchem/thttp://chemed.chem.purdue.edu/genchem/topicreview/bp/ch15/colligative.phpopicreview/bp/ch15/colligative.php

http://library.thinkquest.org/C006669/data/Chttp://library.thinkquest.org/C006669/data/Chem/colligative/colligative.html?tqskip1=1hem/colligative/colligative.html?tqskip1=1

httphttp://dbhs.wvusd.k12.ca.us/webdocs/Solutions://dbhs.wvusd.k12.ca.us/webdocs/Solutions/Intro-to-ColligativeProp.html/Intro-to-ColligativeProp.html


Recommended