1 Ch 12. Properties of Solutions; Mixtures of Substances at the Molecular Level Brady & Senese, 5th...

1

Ch 12. Properties of Solutions; Mixtures of Substances at the

Molecular Level

Brady & Senese, 5th Ed.

2

Index

12.1. Substances mix spontaneously when there is no energy barrier to mixing12.2. Heats of solution come from unbalanced intermolecular

attractions12.3. A substance's solubility changes with temperature12.4. Gases become more soluble at higher pressures12.5. Molarity changes with temperature; molality, weight percentages, and mole fractions do not12.6. Solutes lower the vapor pressure of a solvent12.7. Solutions have lower freezing points and higher boiling points than pure solvents12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal concentrations12.9. Ionic solutes affect colligative properties differently than nonionic solutes

12.1. Substances mix spontaneously when there is no energy barrier to mixing 3

Mixing Processes

• Mixing occurs due to interaction between molecules “like dissolves like”

• As partition is removed, molecules are able to move freely and interact

• Mixed state is statistically more probable


The Process Of Dissolution

• Polar solutes dissolve in polar solvents • Non-polar solutes dissolve in non-polar solvents• Dipoles of solvent may induce dipoles in solute,

effecting dissolution


Miscibility of Liquids

• Liquids that can dissolve in one another are miscible, while insoluble liquids are immiscible

• Ethanol and water are miscible, while benzene and water are not


Learning Check

Which of the following are miscible in water?

water

acetic acidcarbon disulfide

ammonia


Your Turn!

Which of the following are likely to be miscible with water?

A. CH3CH2CH2CH3

B. C6H6

C. CH3CO2H

D. All are expected to be miscible


Dissolution Of An Ionic Compound In Water• Positive end of the dipole

of the water surrounds the anions of the ionic solid, extracting them from the lattice

• Negative end of the dipole orients toward the cations, surrounding and extracting them from the lattice


Dissolution Of A Polar Compound In Water

Dipole of the water interacts with the oppositely charged dipoles of the solid, extracting them from the crystal

12.2. Enthalpy of solution comes from unbalanced intermolecular attractions 10

Enthalpy (Heat) Of Solution

• Heat of solution (Ηsoln ) is the energy exchanged when a solute dissolves in a solvent at constant pressure

• Enthalpy is a state function: the pathway can be written in any way and the result will be the same

• When Ηsoln=0, solution is called an ideal solution

12.2. Heats of solution come from unbalanced intermolecular attractions 11

Dissolution Of An Ionic Solid

• Visualized in steps:

• step1: ionic solid breaks apart into vapor phase lattice energy (U)

• step 2: vapor phase interacts with solvent solvation energy (ΔHsolv); if solvent is water, (Ηhydration)

Ηsoln (ion in water)= U + Ηsolvation


Dissolution: Liquid In Liquid

• Step1: solute expands

• Step2: solvent expands

• Step 3 solute & solvent mix

• If the Ηsoln=0, we have an ideal solution Ηsoln = Η1 + Η2 + Η3


Dissolution: Liquid in Liquid (Ideal)


Dissolution: Gas In Liquid

• step1: expansion of solvent• step2: mixing Ηsoln = Η1 + Η2


Your Turn!

What factor does not affect the value of ΔHsoln ?

A. The polarities of solute and solvent

B. The size of the solute

C. The charge on the solute

D. The temperature of the solution

E. All affect the value

12.3. A substance's solubility changes with temperature 16

Saturated Solutions

• Solute is at equilibrium with the dissolved solute

• Addition of more dissolved solute results in supersaturation and precipitation of excess solid

• The presence of less solute than the solubility results in an unsaturated solution


Solubility Varies With Temperature

• Solubility may increase or decrease with increasing temperature

• The extent to which temperature has an effect is specific to the solute and solvent

• Most gases are less soluble in water at high temperature, while most solids are more soluble


Case Study: Dead Zones

During the industrial revolution, factories were built on rivers so that the river water could be used as a coolant for the machinery. The hot water was dumped back into the river and cool water recirculated. After some time, the rivers began to darken and many fish died. The water was not found to be contaminated by the machinery. What was the cause of the mysterious fish kills?

increased temperature lowered amounts of dissolved oxygen


Effects Of Temperature On Solubility

• Solubility varies with temperature according to the enthalpy of solvation

• The efficiency of a solvation process (K) depends on the enthalpy (ΔH) in Joules, the ideal gas constant (R), and the temperature (T) in Kelvin

• If the dissolution process is endothermic (ΔΗ is +), increasing temperature results in greater efficiency

)11

()ln(212

1

TTR

H

K

K


Your Turn!

The solubility of a substances increases with increased temperature if:

A. ΔHsolution >0

B. ΔHsolution <0

C. ΔHsolution =0

12.4. Gases become more soluble at higher pressures 21

Pressure Effects On Solubility Of Gases

• Cg=kHPg

C = concentration of dissolved gas (M) kH = Henry’s Constant

P = pressure applied to system (mm Hg)kH (M/mm Hg)

N2 8.42×10 -7

O2 1.66×10-4

CO2 4.48×10-5

• Gases are all more soluble at higher pressures (the cause of “the bends”)


Learning Check

What is the concentration of dissolved nitrogen in a solution that is saturated in N2 at 2.0 atm

kH= 8.42×10 -7 (M / mm Hg)

•Cg=kHPg

•Cg= 8.42×10 -7 (M / mm Hg) × 2.0 atm × 760 mmHg/atm

•Cg=1.3 ×10-3 M


Your Turn!

When you open a bottle of seltzer, it fizzes. How should you store it to increase the time before it goes flat?

A. Heat it and pressurize it

B. Cool it and pressurize it

C. Heat it and reduce the pressure

D. Cool it and reduce the pressure

12.5. Molarity changes with temperature; molality, weight percentages, and mole fractions do not

24

Units Of Concentration

• Molarity (M) = moles solute / L solution changes with Temperature

• Molality (m) = moles solute/kg solvent

• mole fraction (X) X = moles component/ total moles

• Percent by mass (%) (mass solute / mass solution)*100


25

Units Of Very Low Concentrations

• Parts per million (ppm) μg solute/mL soln

• Parts per billion (ppb) ng solute/ mL soln

• In extremely dilute solutions mostly solvent is present

• When the solvent is water (d≈1g/mL) thus for ppm ≈ μg solute/g soln

• 1/106 magnitude difference leads to the name 1 part per 1 billion


26

Organize Your Thoughts!

• All concentration units are a ratio of information

• Develop a sense of the data that you have available

Solute Solvent Solution Solution Volume

Mass

Mole

Reference MM g/mol

MM g/mol

d (g/mL)


27

Learning Check: What Does Molarity Tell Us?

M=moles solute/L solution. What are the m, X, % and ppm concentration of a 1.0M solution of KCl with a density of 0.99 g/mL

Solute

KCl

Solvent

(H2O)

Solution SolutionVolume

Mass (g)

Mole

Reference 74.5510

g/mol

18.0153 g/mol

0.99 (g/mL)

1 L1.0

74.55 990

50.815

915.44

51.815

X = 0.019m = 1.1 % =7.5 ppm=7.5(104)


28

Learning Check: What Does Molality Tell Us?

m=moles solute/kg solvent.What are the M, X, % and ppm concentration of 1.0 m KCl with a density of 0.98 g/mL

Solute

KCl

Solvent

(H2O)


Mass (g)

Mole

Reference 74.5510

g/mol

18.0153 g/mol

0.98 (g/mL)

1.0

100074.55 1074.55

55.51 56.51

1096 mL

=1.096 L

% = 6.9 ppm =6.9×104M = 0.91 X = 0.018


29

Learning Check: What Does Mole Fraction Mean?

Xsolute = moles solute/moles total. What are the M, m, % and ppm concentration of a solution that has XKCl = 0.060 with a density of 0.87 g/mL

Solute

KCl

Solvent

(H2O)


Mass (g)

Mole

Reference 74.5510

g/mol

18.0153 g/mol

0.87 (g/mL)

10.060

4.473

0.94

16.93 21.40324.601 mL

=.024601 L

% = 21 ppm =1.8×105M =2.4 m = 3.5


30

Learning Check: What Does % Mass Tell Us

%=(mass solute/mass solution) x 100. What are the M, m, X and ppm concentration of a 1.05 % KCl solution with a density of 1.15 g/mL

Solute

KCl

Solvent

(H2O)


Mass (g)

Mole

Reference 74.5510

g/mol

18.0153 g/mol

1.15 (g/mL)

1001.05

.0140843

98.95

54.9255 55.066486.957 mL

=.086957 L

X = 2.26×10-4 ppm =1.21×104M =0.162 m = 0.142


31

Your turn!

Which of the following corresponds to a 3.5M solution of NaCl with a density of 0.997 g/mL?

m XNaCl %

A. 0.0035 0.074 21

B. 3.5 0.080 0.21

C. 0.0035 0.074 0.21

D. 3.5 0.074 21

MM H2O: 18.0153; NaCl: 58.443

12.6. Solutes lower the vapor pressure of a solvent 32

Raoult’s Law

• Vapor pressure of a liquid varies as a function of purity

• X= mole fraction of solvent P0= vapor pressure of pure solvent

• Psolution=XsolventP0solvent

• Psolution=XAP0A+XBPB

0

Where A and B are both volatile components.


Learning Check

The vapor pressure of 2-methylheptane is 233.95 torr at 55°C. 3-ethylpentane has a vapor pressure of 207.68 at the same temperature. What would be the pressure of the mixture of 78.0g 2-methylheptane and 15 g 3-ethylpentane?

2-methylheptaneChemical Formula: C8H18Molecular Weight: 114.23

•Psolution=XAP0A+XBP0

B

•mole 2-methylheptane : 78.0g/114.23 g/mol = 0.68283 mol

•mole 3-ethylpentane: 15g/100.2 g/mol = 0.1497 mol

•X2-methylheptane=0.8202 ; X3-ethylpentane =1-0.8202 = 0.1798

3-ethylpentaneChemical Formula: C7H16Molecular Weight: 100.2

torr207.680.17 torr233.950.82P 9802 P = 230 torr


Learning Check

The vapor pressure of 2-methyl hexane is 37.986 torr at 15°C. What would be the pressure of the mixture of 78.0g 2-methylhexane and 15 g naphthalene which is nearly non-volatile at this temperature?

2-methylhexaneChemical Formula: C7H16Molecular Weight: 100.2naphthalene

Chemical Formula:

C10H8Molecular Weight: 128.17

•Psolution=XsolventP0solvent

•mol 2-methylhexane: 78.0g/100.2 g/mol = 0.778443 mol

•mol naphthalene: 15 g/128.17 g/mol = 0.11703

•X2-methylhexane = 0.869309

•Psolution = 0.869309 ×37.986 torr

•P=33.02 torr


Your Turn!

n-hexane and n-heptane are miscible in a large degree and both volatile. If the vapor pressure of pure hexane is 151.28 mm Hg and heptane is 45.67 at 25º, which equation can be used to determine the mole fraction of hexane in the mixture if the mixture’s vapor pressure is 145.5 mm Hg?

A. X(151.28 mmHg)=145.5 mmHg

B. X(151.28 mmHg) + (X)(45.67 mm Hg) = 145.5 mmHg

C. X(151.28 mmHg)+(1-X)(45.67 mm Hg)=145.5 mm Hg

D. None of these

12.7. Solutions have lower freezing points and higher boiling points than pure solvents 36

Solute Effects On Phase Changes:

• Regardless of the identity of the dissolved particles, the presence of an impurity will result in a change in the boiling point and freezing point.

• The effect is solely dependent on the nature of the solvent, a factor labeled K, and the concentration of particles present (m)

• ΔT=mK boiling point elevation ΔT=Tmix-Tpure

freezing Point Depression ΔT=Tpure-Tmix


Effects Of Impurities On Phase Changes


Some BP/FP Constants

Solvent Normal BP (°C)

Kbp

(°C/m)Normal FP(°C)

Kfp (°C/m)

Water 100.00 +.51 0.0 1.86

Acetic Acid 1118.3 +3.07 16.6 3.57

Benzene 80.2 +2.53 5.45 5.07

Camphor 207 +5.611 178.4 37.7

Chloroform 61.20 +3.63 - -

Cyclohexane 80.7 2.69 6.5 20.0


Learning Check

According to the Sierra™ Antifreeze literature, the freezing point of a 40/60 solution of sierra antifreeze and water is -4 °F. What is the molality of the solution?

fpsolution KmΔT

mC1.86

m?C.)20(0(

11=m

-4°F = 1.8 (°C) + 32

-20. °C


Learning Check:In the previous sample of a Sierra™ antifreeze mixture,

100 mL is known to contain 42 g of the antifreeze and 60. g of water, what is the molar mass of the compound found in this antifreeze if it has a freezing point of -4°F?

fpsolution KmΔT

mC1.86

m?C.)20(0(

10.75=m

from before:

-4°F = 1.8 (°C) + 32 =-20. °C m 10.solvent kg 0.060

solute mol75

650 g/mol solute

0.6452 mol solute


Learning Check:

In the previous sample of a Sierra™ antifreeze mixture, the freezing point is -4°F? What will be its boiling point?

bp

solution boiling

pf

solution freezing

bpsolution boiling and

pfsolution freezing

K

ΔTm

K

ΔT

ion,concentrat same theisit If

Km ΔT

KmΔT

from before:

-4°F = 1.8 (°C) + 32 =-20. °C

mC/0.51

mC/1.86

C100-T

C))(-20-C(0

mix

mKK

ΔTΔT

pb

fp

PointBoiling

PointFreezing

T=105 °C


Your Turn!

Beer is known to be around a 5% ethanol (C2H5OH) solution with a density of 1.05 g/mL. What is its expected boiling point?( Kf=0.51°/m)

A. 100ºC

B. 101ºC

C. 102ºC

D. 103ºC

E. Not enough information given

MM: H2O=18.0153; C2H5OH=46.069

12.8. Osmosis is flow of solvent through a semipermeable membrane due to unequal concentrations

43

Osmosis• When two solutions are separated by a semi-permeable

membrane, solvent molecules flow from areas of low concentration to areas of high concentration

• As this occurs, the height of liquid rises in the higher concentration solution, building up “Osmotic pressure” (π)


44

Relative Concentration Terms In Osmosis

• Hypotonic solutions have lower ion concentrations than the cells.

• Hypertonic solutions have higher ion concentrations than the cells

• Isotonic solutions have the same ion concentration as the cells


45

Osmosis

• π=MRT the concentration, is in molarity, M T=Temperature, in Kelvin R=Ideal Gas Constant, 0.082057 L·atm/mol·K

• The basis for kidney function, rising sap, and dialysis


46

Learning Check: Osmosis

A solution of D5W, 5% dextrose (C6H1206) in water is placed into the osmometer shown at right. It has a density of 1.0 g/mL. The surroundings are filled with distilled water. What is the expected osmotic pressure at 25°C?

298KKmol

atm L0.082057

L

mol 0.277

ML

1000mL180.16g

OHC molsoln mLsoln 1.0g

solution 100gOHC 5g 61266126

atm 7

TMR


47

Learning Check

For a typical blood plasma, the osmotic pressure at body temperature (37°C) is 5409 mm Hg. If the dominant solute is serum protein, what is the concentration of serum protein?

K.103Kmol

atm L0.082057Lmol ?

atm117.7 15

Hgmm 760

1atm Hgmm 5409

Lmol 0.280

M

TMR


48

Dialysis

• Pores on the semi-permeable membrane may be of varied size

• In dialysis, the pores are fairly large, allowing transfer of solvent, ions, and small proteins

• Larger cells, such as red blood cells are prevented from passing through the pores

• The dialysis bath may be enriched in substances lacking in the blood, and is hypotonic in waste products in the blood

• Exchange of vital components may be made


49

Your Turn!

Suppose that your tap water has 250 ppb of dissolved H2S , and that its density is about 1.0 g/mL. What is its osmotic pressure at 25°C?

A. 0.00058 atm

B. 0.064 atm

C. 0.059 atm

D. None of these

MM: H2S =34.076

0.21 atm

12.9. Ionic solutes affect colligative properties differently than nonionic solutes 50

Ionic Solutes Affect Colligative Properties Differently Than Non-ionic Solutes

• substances that ionize make more particles in a solution than their own concentration suggests

• i is a factor that demonstrates how many ions are formed per formula unit or molecule

• the apparent molality of particles is then im.

calculated

measured

T

Ti

i

mK

mK

ΔT

ΔT

ionization no assuming calculated

measured


Learning Check

In preparing pasta, 2 L of water at 25°C are combined with about 15 g salt (NaCl, MM= 58.44g/mol) and the solution brought to a boil. What is the expected boiling point of the water?

ΔT=imKbpmass of water =volume ×density =2000 mL ×1.0 g/mL

=2000g water = 2 kg

mol NaCl = 15g / 58.44 g/mol

mol NaCl = 0.25667

m=0.25667 mol / 2kg

=0.123

m

CCT

51.0

1

m1.0

mol

ion 2100

23

T=100.1 °C


Case Study

Suppose you run out of salt. What mass of sugar (C12H22O11, MM=342.30 g/mol) added to 2 L of water would raise the temperature of water by 0.10 °C?

ΔT=imKbpmass of water =volume ×density =2000 mL ×1.0 g/mL

=2000g water = 2 kg

0.39215 mol = ?g / 342.30 g/mol

mass sucrose =130 g

0.196 m=? mol / 2kg

0.39215mol

m

CmC

51.0?

mol

molec 110.0

m=.196

Date post:	19-Dec-2015
Category:	Documents
Upload:	luke-reeves
View:	212 times
Download:	0 times

1 Ch 12. Properties of Solutions; Mixtures of Substances at the Molecular Level Brady & Senese, 5th...

Documents