mL () 1000 L . 1060 () gmol - Loyola University Chicagoafitch.sites.luc.edu/Gen Chem pdf lectures/13...

1

“A” students work(without solutions manual)~10 problems/night.

Dr. Alanah FitchFlanner Hall [email protected]

Office Hours Th & F 2-3:30 pm

Module #13Solution Properties

Defining types of concentrationsFor mixtures

SOLUTIONS

solute

solvent

Define some measurement scales and units

1. Molarity

2. mole fraction

3. molality

G3: Science Is referential

M molesL

solute

solution=

nn n

A

A BA+

≡.

χ

molality mmoleskg

solute

solvent= =

Method for Conversion from Molarity to molality

molality m moleskg

solute

solvent= =

1. Assume 1 L volume2. Calculate moles of solute

3. Calculate g of solution from d, V

4. Calculate g of solute from M

5. Subtract to get g of solvent

6. Calculate g solute/g solvent

( )M Lg

molgsolution

solute

solutesolute1

⎛⎝⎜

⎞⎠⎟ =

( )dmL

LL gsolution solution solution

101

3⎛⎝⎜

⎞⎠⎟ =

( )molesL

L molessolute

solutionsolute

⎛⎝⎜

⎞⎠⎟ =1

g g gsolution solute solvent− =

Example Density of an aqueous solution of ammonium sulfate is 1.06g/mL and the molarity is 0.886. What is the molality?

( )1061000

1060.g

mLmL

LL gsolution solution

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟ =

( )0 886 1133098

117 9248..

.molL

Lg

molgsolution

solute

solutesolute

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =

1060 117 9248 9421g g gsolution solute soluent− =. .

1. Assume 1 L volume2. Calculate moles of solute

3. Calculate g of solution from d, V

4. Calculate g of solute from M

5. Subtract to get g of solvent

6. Calculate g solute/g solvent

( ) ( )

( ) ( )

2 14 01 28 028 8 1008 8 064

32 07 32 074 4 16 00 64

133098

NHSO

total

. .. .. ..

.

molality mmoleskg

moleskg

solute

solvent

solute

solvent= = = =

08860 9421

0 941..

.

( )molesL

L molessolute

solutionsolute

⎛⎝⎜

⎞⎠⎟ = =1 0886.

Atoms amu total

( )NH SO4 2 4

2

Method for Conversion from molality to Molarity

moleskg

kg molessolute

solventsolvent solute

⎛⎝⎜

⎞⎠⎟ =1

1. Assume 1 kg solvent2. Calculate moles of solute

3. Calculate g of solute from MM

4. Sum masses to get total mass of solution

5. Calculate V of solution from density

6. Calculate moles/V

( ) ( )gmLg

LmL

Lsolutionsolution

solutionsolution

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟ =

103

Mmoles

Lsolute

solution=⎛⎝⎜

⎞⎠⎟

( )molesg

molesgsolute

solute

solutesolute

⎛⎝⎜

⎞⎠⎟ =

1000g g gsolvent solute solution+ =

Example Density of an aqueous solution of KOH is 1.43g/mL and the molality is 14.2. What is the molarity?

1. Assume 1 kg solvent2. Calculate moles of solute

3. Calculate g of solute from MM

4. Sum masses to get total mass of solution

5. Calculate V of solution from density

6. Calculate moles/V

[ ]14 21

1 14 2.

.molesKOHkgwater

kgwater molesKOH⎡

⎣⎢

⎤

⎦⎥ =

[ ]14 23910 16 00 108

797 756.. . .

.molesKOHgKOH

moleKOHgKOH

+ +⎛⎝⎜

⎞⎠⎟ =

g g g g g gsolution H O KOH solution= + = + =2

1000 797 756 1797 756. .

( )1797 7561143

1 25717..

, .gmL

gmLsolution

KOHsolution

KOHsolution

⎛⎝⎜

⎞⎠⎟ =

MmolesL Lsolution

moles KOH

KOHsolution= = =

14 21257

11296.

..,

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14 16Molarity

Mol

aity

or (

dens

ityx1

0)

0

5

10

15

20

25

30

35

40

45

50 w

eigh

t per

cent

DensityMolalityWeight %

Methods of measurement are related but not equivalent “A” students work(without solutions manual)~10 problems/night.




Water/Salt Solutions 1

3

SOLUTIONSAny two components A and BMixed in any mole fraction

B A

Solvent is usuallyConsidered the “dissolver”

Solute is usuallyConsidered the “dissolved”

nn n

A

A BA+

≡.

χ

nn n

B

A BB+

≡.

χ

χ χA BA

A B

B

A B

nn n

nn n

+ =+

++. .

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1moles of A

mol

e fr

actio

n

A mixture is usually plotted as variation in mole fractions

χ χA BA

A B

B

A B

B B

A B

nn n

nn n

n nn n

+ =+

++

=++. . .

χ χA BA

A B

B

A B

B B

A B

nn n

nn n

n nn n

+ =+

++

=++

=. . .

1

Pure APure B

mixture

SOLUTIONS

solute

solvent

water

MgSO Mg SO HkJ

molsH O

aq aq42

422 912, , .⎯ →⎯⎯⎯ + = −+ − ∆

Any two components A and BMixed in any mole fraction

Solvation Diagrammed as an Example of Hess’s Law

-1376.1kJmole

MgSO Mg SO HkJ

molsH O

aq aq42

422 912, , .⎯ →⎯⎯⎯ + = −+ − ∆

-1284.9kJmole

4

NH NO NH NO H kJs aq aq4 3 4 3 281( ) ( ) ( ) .→ + = ++ − ∆

AquatedAmmonium

http://www.lsbu.ac.uk/water/magic.html

Aquatednitrate

http://www.qtp.ufl.edu/~roitberg/pdf/2002_04.pdf

∆H kJ= +281.

For most salts, heat will increase solubility

NH NO NH NO H kJswater

aq aq4 3 4 3 281( ) ( ) ( ) .⎯ →⎯⎯⎯ + = ++ − ∆

NH NO heat NH NOs aq aq4 3 4 3( ) ( ) ( )+ → ++ −





Organic/SoluteSolutions

SOLUTIONS

solute

solvent

water organic

MgSO Mg SO HkJ

molsH O

aq aq42

422 912, , .⎯ →⎯⎯⎯ + = −+ − ∆

5

What trends do you see here?

sgg

solute

solvent=

What trends do you see here?

Like Dissolves Likesimilar intermolecular forces between

solute-solutesolvent-solvent

imply solute-solvent interaction will be decent

Soluble in water? Soluble in water?

Solute-Solvent interactions: Vitamin C and Lead

Vitamin C is more soluble in the aqueousPhase of the body (urine) than otherVitamins which are stored in fat. Vitamin C is water soluble

so it is excreted from the body

the longest you can last without serious disease without consuming fresh vitamin C is 6 months

Context Slide

Sir Franklin

Context Slide

6

Context Slide

John Hartnell

Context Slide

Context Slide“A” students work(without solutions manual)~10 problems/night.




Gas/Solvent Solutions Raoult’s Law, FreezingPoint, Boiling Pt changesOsmotic Pressure

7

SOLUTIONS

solute

solvent

Liquid Gas Solution

Component AComponent B

Liquid Consider first case where AIs a pure substance

nn n

A

A BA+

≡ =.

χ 1

Liquid PhaseIn a container

AlA Arate escapes

gl ⎯ →⎯⎯⎯⎯⎯Number of escapes will depend on1. Number of molecules with energy>>

intermolecular forces2. Fraction of molecules with that

energy at some the given temperature

3. The Surface area available for escape

4. Escapes independent of what gasis doing

AgA Ag

rate sticking⎯ →⎯⎯⎯⎯⎯ l

Number of returns will depend on1. Intermolecular forces in gas

phase greater than collisionalenergy

2. Fraction of molecules at with less energy than intermolecular forces at that temperature

3. Sticking followed by dropping does not depend upon the surface area

Prediction 11. T8, more escapes, more gas phase A2. T8, less sticking (fewer returns)

nn n

A

A BA+

≡ =.

χ 1


Predict: 1. T8 6more gas vapor2. T8 6less sticking of

gases more gas vapor

P epure liquidT

HR

vaporization

=−⎡

⎣⎢

⎤

⎦⎥

1 ∆

T ↑

Intermolecularforces

T e T↑ ↑−

;1

T e PTAo↑ ↑ ↑

−

; ;1

P epure liquid

HRT

vaporization

=−⎡

⎣⎢

⎤

⎦⎥

∆

Exponential represents fraction of molecules at temperature T with sufficient energy to break intermolecular forces ()Hvaporization)

nn n

A

A BA+

≡ =.

χ 1

Math for the fraction of molecules with sufficient energyIs on next slide, if desired

From earlier chapter we learned of kinetic energy:

E mvk =12

2

ukT

grmsmolecule

=3

E mkT

mass of moleculek ass of molecule=⎡

⎣⎢

⎤

⎦⎥

12

32

E mkT

mass of moleculek ass of molecule=⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

12

3

EkT

k =⎛⎝⎜

⎞⎠⎟

32

However, only a fraction of molecules at a given emperature that average speed, and therefore,Average energy

If you assume a normal distribution for the “bell curve” you can calculate the fraction of molecules with an energy above the average value

( )f x ex

=−

−⎛⎝⎜

⎞⎠⎟

12

2µσ

µ

f x e eE

kTERT

molecule mole

( ) = =− −

Maxwell’s Distribution

σRT

αEK

Review

8


AlA Arate escapes

gl ⎯ →⎯⎯⎯⎯⎯Number of escapes will depend on1. Number of molecules with energy>>

intermolecular forces2. Fraction of molecules with that

energy at some the given temperature

3. The Surface area available for escape

4. Escapes independent of what gasis doing

AgA Ag

rate sticking⎯ →⎯⎯⎯⎯⎯ l

Number of returns will depend on1. Intermolecular forces in gas

phase greater than collisionalenergy

2. Fraction of molecules at with less energy than intermolecular forces at that temperature

3. Sticking followed by dropping does not depend upon the surface area

Prediction 21. Surface Area 9, less escapes, less gas A2. Surface Area 9, no impact on rate sticking

SOLUTIONS

solute

solvent

Liquid Gas Solution


Liquid One way to diminishSurface area is to add solute toPure substance A

nn n

A

A BA+

≡ <.

χ 1

nn n

A

A BA+

≡.

χ

nn n

B

A BB+

≡.

χ

χ χA BA

A B

B

A B

nn n

nn n

+ =+

++. .

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1moles of A

mol

e fr

actio

n

A mixture is usually plotted as variation in mole fractions

χ χA BA

A B

B

A B

B B

A B

nn n

nn n

n nn n

+ =+

++

=++. . .

χ χA BA

A B

B

A B

B B

A B

nn n

nn n

n nn n

+ =+

++

=++

=. . .

1

Pure APure B

mixture

nn n

A

A BA

A

B A+≡ =

+.χ

991 99

Within every layer of liquid (10x10) there are 99A and 1BHow has the surface Area available for escape changed?

( )NewSurface Area Original Surface AreaA= χ

How will vapor pressure change?

A Arate escapesgl ⎯ →⎯⎯⎯⎯⎯

Predict: 1. T8 6more gas vapor2. Surface area 9 6less gas vapor3. Escapes independent of gas!!

We just made a prediction!

P PA A Ao= χ

Raoult’s Law

10

10

9

P PA A Ao= χ

Raoult’s Law

Pure A

PA

PA0

0

Assumes effect of “B” is to simply block the surfaceNo interaction of “B” (solute) with “A” (solvent)

Boiling Point elevation

ln ,,

PT

HRA b

o

b Ao

vaporization=

−⎡

⎣⎢

⎤

⎦⎥

1 ∆

ln PP

HR T T

vaporization2

1 2 1

1 1⎡

⎣⎢

⎤

⎦⎥ =

−⎡

⎣⎢

⎤

⎦⎥ −⎡

⎣⎢

⎤

⎦⎥

∆

Clausius-Clapyeron Equation

ln ,

, , ,

PP

HR T T

A bp

A bpo

vaporization

A bp A bpo

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

−⎡

⎣⎢

⎤

⎦⎥ −⎡

⎣⎢⎢

⎤

⎦⎥⎥

∆ 1 1

PP

nn n

A

AA

A

A B0 = =

+χ

ln lnP PT

HRpure liquid A

o vaporization= =

−⎡

⎣⎢

⎤

⎦⎥

1 ∆

( )ln, ,

11 1

− =−⎡

⎣⎢

⎤

⎦⎥ −⎡

⎣⎢⎢

⎤

⎦⎥⎥

χBvaporization

A bp A bpo

HR T T

∆

χ χA BB

A B

nn n

= − =+

1

( ) ( )( ) ( )( )ln ,

, ,

,

, ,

1− =−⎡

⎣⎢

⎤

⎦⎥ −⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

χBvaporization A bp

o

A bp A bpo

A bp

A bpo

A bp

HR

T

T T

T

T T

∆

If P small

( ) ( )( )ln , ,

, ,

1− =−⎡

⎣⎢

⎤

⎦⎥

−−

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

χBvaporization A bp

oA bp

A bp A bpo

HR

T T

T T

∆

If numbersSimilar althoNot the same

( )χB

vaporization A bpo

A bp

A bpo

HR

T T

T=

−⎡

⎣⎢

⎤

⎦⎥

−−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∆ , ,

,

2

( )n

n nH

R TTB

B A

vaporization

A bpo

A bpo

+=

−⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∆∆

,

,2

n nB A<<

nn

k TB

AA bpo= ∆ ,

( )n

nkg

mole

kkg

mole

TB

AA

A

A

A

A bpo

⎛⎝⎜

⎞⎠⎟=

⎛⎝⎜

⎞⎠⎟∆ ,

molal k TB A bpo= ' ,∆

1k

molal TB A bpo

' ,= ∆

K molal Tb B A bpo= ∆ ,

I hate drivations

∆P PA B Ao= χ


K molal Tf B A mpo= ∆ ,

KC

molalb

o

= 052.

KC

molalf

o

= 186.

Boiling Point elevationFreezing Point depression

Constants for water

SOLUTIONS

solute

solvent

Liquid Gas Solution


Liquid

nn n

A

A BA+

≡ <<<<<.

χ 1

The other extreme is to

10

Liquid Phase of B In a container

When “A” is completely surrounded by “B”

A Ain B liquidrate escapes

g⎯ →⎯⎯⎯⎯⎯

1. Number of possible A/B collisions (Partial Pressure of A) (PA)

2. Intermolecular forces of A/B in gas phase greater than collisionalenergy

3. Fraction of molecules with intermolecular energy greater than collisional energy increases as Tgoes down.

4. No dependence on surface area

A Agrate sticking

in B liquid⎯ →⎯⎯⎯⎯⎯

1 Surface area for A 2 Escapes of A from B when

energy is larger than A/B intermolecular forces

Predictions1. Greater PAmore collisions with B,

greater condensation, greater solution concentration

2. Greater T less collisions in whchintermolecular forces allow sticking, less solution concentration

Liquid Phase of B In a container

When “A” is completely surrounded by “B”

C k Pg H A=

A/B intermolecular forces

χ AAP

h=

Math (slide afternext)

Henry’s Law

Solubility in water

C k Pg H A=Prediction 1

Prediction 2:As temp goes up, fewerGas phase collisions betweenA/B allow them to stick,Less condensation

if n nA B<<

P kn

n nAA

A B=

+⎛⎝⎜

⎞⎠⎟

kH = Henry’s constant

( )P break A BenergyA A= − χ

P kA A= χ

P knnA

A

B=

⎛⎝⎜

⎞⎠⎟

( )P k

mn

n

nmn

AB

B

A

BB

B

=⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

'

P knmA

A

B=

⎛⎝⎜

⎞⎠⎟' ' Molality

When mb in kg

1k

Pnm

k PAA

BH A' '

=⎛⎝⎜

⎞⎠⎟ =

C kPg A=

( )kn

break A B energy molalHB

B

=−

Henry’s Law

For the math hungry

11

Henry’s LawP

Raoult’s LawSlope is a functionOf interactions

Slope is A function of

interactions

Pure solvent

χ A

C k Pg H A= P PA A Ao= χ

Raoult’s LawVapor Pressure

Partial Pressure

Everything in betweenIs hard to predict

0

10

20

30

40

50

60

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00Mole Fraction Dioxin in Water

P (m

m H

g) Raoult’s LawReflects dioxin-dioxin interactionsSurface area for escape

Henry’s Law

Reflects dioxin-water interactionsPartial Pressure of dioxin

Cl

Cl

O

O

Cl

Cl

Viktor Yushchenko2004, UkrainianOpposition candidateFor president, Poisoned by dioxin

Context Slide

Cl

Cl

O

O

Cl

Cl

Cl

Cl

O

O

Cl

Cl

Pdioxin

PA=PAo

VP

nn n

Dioxin

water DioxinDioxin+

≡.

χ

Marie the Jewess, 300 Jabir ibnHawan, 721-815

Galen, 170 Jean Picard1620-1682

Galileo Galili1564-1642

Daniel Fahrenheit1686-1737

Evangelista Torricelli1608-1647

Isaac Newton1643-1727

Robert Boyle, 1627-1691

Blaise Pascal1623-1662

Anders Celsius1701-1744

Charles AugustinCoulomb 1735-1806

John Dalton1766-1844

B. P. Emile Clapeyron1799-1864

Jacques Charles1778-1850

Germain Henri Hess1802-1850

Fitch Rule G3: Science is Referential

William ThompsonLord Kelvin, 1824-1907

James Maxwell1831-1879

Johannes D.Van der Waals1837-1923

Justus von Liebig (1803-1873

Johann Balmer1825-1898

James Joule(1818-1889)

Johannes Rydberg1854-1919

Rudolph Clausius1822-1888

Thomas Graham1805-1869

Heinrich R. Hertz,1857-1894

Max Planck1858-1947

J. J. Thomson1856-1940

Linus Pauling1901-1994

Werner Karl Heisenberg1901-1976

Wolfgang Pauli1900-1958

Count Alessandro GA A Volta, 1747-1827

Georg Simon Ohm1789-1854

Henri Louis LeChatlier1850-1936

Svante Arrehenius1859-1927

Francois-MarieRaoult

1830-1901

William Henry1775-1836

Gilbert N Lewis1875-1946

Fritz Haber1868-1934

Michael Faraday1791-1867

Luigi Galvani1737-1798

Walther Nernst1864-1941

Lawrence Henderson1878-1942

Amedeo Avogadro1756-1856

J. Willard Gibbs1839-1903

Niels Bohr1885-1962

Erwin Schodinger1887-1961

Louis de Broglie(1892-1987)

Friedrich H. Hund1896-1997

Fritz London1900-1954

An alchemist

Ludwig Boltzman1844-1906

Richard AC E Erlenmeyer1825-1909

Johannes Bronsted1879-1947

Thomas M Lowry1874-1936

James Watt1736-1819

Dmitri Mendeleev1834-1907

Marie Curie1867-1934

Henri Bequerel1852-1908

Rolf Sievert,1896-1966

Louis Harold Gray1905-1965

Jacobus van’t Hoff1852-1911

Same concept for

A Asrate escapes

g⎯ →⎯⎯⎯⎯⎯

A Agrate sticking

s⎯ →⎯⎯⎯⎯⎯10

10

Solid phaseIn a box

Sublimation

12

Barrier (memrane)Only water (not solute)Is allowed to pass

Imagine: Two volumes of water1. is “pure”

2. the second has large salt mole fraction

3. There is a barrier between themwhichis impermeableto theelectrolytes

4. Surfacearea for waterescape isless in cube2

nn n

electrolyte

water electrolytes+

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = =

.0 1χ

nn n

electrolyte

water electrolytes+

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = =

..0 2 2χ

5. Escapes 1> Escapes 26. Water moves 1 to 27. As water in 2 increases

Pressure builds upagainst membrane

[ ]π = =n RT

VB RTB

Osmotic pressure (pressure driving water from one compartment to another) is directly related to the concentration of the solute, B.

1. This provides a convenient way to measure molar masses.

2. Has huge physiological implications

Water will move from the pure solvent to the solutecontaining container.

Cucumber Prune

Visualization

Cucumber placed inhigh salt solution

Aka =Pickle

Prune placed in pure water

Aka = Plume

When not well regulated the extracellular fluid increases andedema (bursting of cells) results

Context

13

A Student who did NOT take Gen Chem

Context

Typical ion concentrations in vertebrates and invertebratesIon Cell (M) Blood (M)K+ 0.139 0.004Na+ 0.012 0.145Cl- 0.004 0.116HCO3

- 0.012 0.029X- 0.138 0.009Mg2+ 0.0008 0.0015Ca2+ <0.0000002 0.0018

Water motion across cell membranes is driven byosmotic pressure related to the saltconcentration gradient

Bumpy Horse blood cells are dessicating= “crenation”

Normal horse blood http://images.google.com/imgres?imgurl=http://www.vetmed.auburn.edu/distance/clinpath/morphol/sa10.jpg&imgrefurl=http://www.vetmed.auburn.edu/distance/clinpath/morphol1/&h=80&w=120&sz=3&hl=en&start=10&tbnid=jyQ0sNBf3dS1HM:&tbnh=59&tbnw=88&prev=/images%3Fq%3Dcrenation%26svnum%3D10%26hl%3Den%26client%3Dfirefox-a%26channel%3Ds%26rls%3Dorg.mozilla:en-US:official%26hs%3Dk0K%26sa%3DGcells

Normal RBC

Hemolysis RBCRupture of RBC

www.healthenlightenment.com/hemolysis.jpeg

Normal horseRed blood cells

The equations

P PA A A= 0χ

C kPg A=


K molal Tf B A mpo= ∆ ,

∆P PA B Ao= χ

Raoult’s Laws

Derived fromRaoult’s Law

Henry’s Law

Colligative PropertiesDepend on Solute Conc.

Depends primarily on solvent

[ ]π = =n RT

VB RTB

14





Example CalculationsRaoult’s Law, Henry’s Law,Boiling Pt elevationFreezing Pt depression

EXAMPLE Calculations

Example A solution contains 102 g of sugar, C12H22O11, in 375 g of water. Calculate the vapor pressure lowering at 25 oC (vp of pure water = 23.76 mm Hg

Example Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henry’s law constant for CO2 in water at this temperature is 3.1x10-2 mol/L@atm.

Example: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution? Kb of water is 0.512 (oC/m) and Kf of water is 1.86 (oC/m).

Example The solubility of pure nitrogen in blood at body temperature, 37 oC,And one atmosphere nitrogen is 6.2x10-4 M. If a diver breathes air (χN2 = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen inhis blood.

Osmotic Pressure Example Calculation: What is the osmotic pressure related to a 0.10 M CaCl2 solution?

Example 1 A solution contains 102 g of sugar, C12H22O11, in 375 g of water. Calculate the vapor pressure lowering of water at 25 oC (vapor pressure of pure water = 23.76 mm Hg

What do we know?sugar in water (sugar= solute; water= solvent)

What equations might apply?Vapor pressure lowering

What information is irrelevant? (Red herring)

∆P PA B Ao= χ

none.

Example 1 A solution contains 102 g of sugar, C12H22O11, in 375 g of water. Calculate the vapor pressure lowering of water at 25 oC (vapor pressure of pure water = 23.76 mm Hg

χ2 1P Po = ∆

χsolutesolute

solvent solute

molesmoles moles

=+

Molar mass: 12C = 12x12= 14422H = 2211O = 11x16 = 176

total = 342

[ ]

[ ] [ ]χsugar

sugarsugar

sugar

waterwater

watersugar

sugar

sugar

gmol

g

g molg

gmol

g

=

⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢

⎤

⎦⎥ +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

1021

342 30

375 118 02

1021

342 30

.

. .

χsugarsugar

water sugar

molmol mol

=+

=0 298

208 0 2980 0141

.. .

.

[ ][ ]0 0141 2376 0 335. . .mmHg mmHg=

15

Example 2 Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henry’s law constant for CO2 in water at this temperature is 3.1x10-2 mol/L@atm.

C kPg g=

( )C xmol

L atmatm

molLg =

−⎛⎝⎜

⎞⎠⎟ =−31 10 4 0 01242. . .

Sig figs

Solute?Solvent?Equation?Red Herrings?

Example 3The solubility of pure nitrogen in blood at body temperature, 37 oC,and one atmosphere of nitrogen is 6.2x10-4 M. If a diver breathes air (χN2 = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen inhis blood.

C k Pg H g=

Solute?Solvent?Equation?Red Herrings?

N2Blood

Example 3The solubility of pure nitrogen in blood at body temperature, 37 oC,and one atmosphere nitrogen is 6.2x10-4 M. If a diver breathes air (χ N2 = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen inhis blood.

C k Pg H g=

P X PN N total2 2=

Unknown, how can weGet it?

( )6 2 10 14. x M k atm− =

( )( )P atm atmN20 78 2 5 195= =. . .

( )C xM

atmatm x Mg =

⎡⎣⎢

⎤⎦⎥

=− −6 2 10 195 1209 104 3. . .

C x Mg = −12 10 3.

C kPg g=Unknown how can we get it?

Dalton’s Partial Pressure Law!

( )kx Matm

xM

atm= =

−−6 2 10

16 2 10

44.

.

Example 4: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution? Kb of water is 0.512 (oC/m) and Kf of water is 1.86 (oC/m).

[ ]molalitymol solutekg solvent

=

∆TC

molalitymC H O Cb

oo=

⎛⎝⎜

⎞⎠⎟ =0512 362 1852 6 2. . .

The radiator will boil at 101.85 oC

[ ]∆T k molalityb b=

[ ]∆TC

molalitymolalityb

o

=⎛⎝⎜

⎞⎠⎟0512.

[ ]( )

molalitykgC H O

gkg

molC H OgC H O

kg solvent=

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟100

101

162 072 6 2

32 6 2

2 6 2.

.[ ]( )

molalitykgC H O

gkg

molC H OgC H O

gH Okg

g

=

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

100101

162 07

44501

10

2 6 2

32 6 2

2 6 2

2 3

..[ ]

( )molality

kgC H Og

kgmolC H O

gC H O

gH Okg

g

molC H Okg

m=

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= =100

101

162 07

44501

10

1614 450

3622 6 2

32 6 2

2 6 2

2 3

2 6 2

.. .

..

[ ]∆TC

molalitymolalityf

o

=⎛⎝⎜

⎞⎠⎟186.

[ ]∆T k molalityf f=

∆TC

molalitymC H O Cf

oo=

⎛⎝⎜

⎞⎠⎟ =186 362 6 732 6 2. . .

The radiator will freeze at 32-6.73 oC=25.27 oC

16

Osmotic Pressure Example Calculation: What is the osmotic pressure related to a 0.10 M CaCl2 solution?

[ ] [ ]π = =⎡⎣⎢

⎤⎦⎥

⋅⋅

⎡⎣⎢

⎤⎦⎥

=B RTmolL

L atmmol K

K atm3 010 0 0821 298 7 2. . .

[ ]π = =n RT

VB RTB

Where did this come from?

CaCl Ca Clstrong electrolye completeaq aq22 2⎯ →⎯⎯⎯⎯⎯⎯⎯⎯⎯ ++ −

3 moles solute

[ ]π = B RT[ ] [ ]π = =⎡⎣⎢

⎤⎦⎥

⋅⋅

⎡⎣⎢

⎤⎦⎥

B RTmolL

L atmmol K

K3 010 0 0821 298. .

In order to control this pressure the kidneyregulates the amount of Ca2+, Na+ and other ions in the blood

Lead can cause significantBlood pressure problems by interferingKidney regulation of ions

sgg

solute

H O=

2

KggOWsolute Loc ol

solute Lwater= / tan

/

Set a reference state to compare all chemicalssolubility in water (s)solubility between water and octanol (Kow)KH

Why would thisBe important?

Your paper for a compound of interest

find, listing source of datasKowKH

Reference Data for Risk Assessment

Hint, think aboutVit. C and Vit A

C kPg g=

octanolCH3 OH

Context Slide: Preparation for your papers





Solid/solid solutionsAs an example of importanceOf freezing point depression

SOLUTIONS

solute

solvent

Atin

Bcopper

17

A Different Kind of Phase Diagram

100%A100%B

Temp

Liquid

Solid

nn n

A

A BA+

≡.

χnn n

B

A BB+

≡.

χ

Same idea of mole fraction

Context Slide Bronze phase diagramas Tin is added to Copper

Liquid Phase

Solid phases

Solid+Liquid

Context Slide

Bronze phase diagramas Tin is added to Copper

Liquid Phase

Solid+Liquid

These representvarious unit cell structures (face centered cubic, etc.)adopted

1. TemperatureRange of early“campfires”

2. Grind up and mix tin to copper and you can lower the melting point

3. Bonus point: final materialharder than copper!!

4. Many early economicsOf societiesCollapsed as fuel (forests) ran out

5. England’s interest in controlling IrelandWas initially over wood supply

Context Slide Egyptian Glass how did they do it?

Campfire Temps

Viscosity = measures flowlow viscosity = easy flow

Additives toThe quartz (SiO2) allowThe glass to Melt (becomesLess viscous)At campfireTemperatures.

1300B.C.

Context Slide

18

Sugar of Lead =Pb(CH3COO)2

Pb2+2

••

••

••••

••

−

= − −

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

O

O

C

H

H

C H

Where does the negative charge reside?Is this acetate anion q/r large or small?Is this acetate anion charge dense or not?Will acetate “hold on” to Pb2+ or will water win?

Context

Pb2+

Renal biopsy26 year oldShip paint stripper

Context

Failure of kidneys leads to deposition of Excess uric acid to joints = Gout

Immersion of the body Up to head creates a Pressure gradient in the blood volume towards the head; Body compensates with increased urine to lower Blood volume to brain; Increased urination leads to increased loss Of electrolytes (lead)

Context

Tetraethyl lead poisoning is highly acute leading very rapidly to edemaWhy and what does this have to do with osmotic pressure?

Initial symptoms are hallucinationscreepy bugs, etc.

Context

19

( )Pb CH CH3 2 4

Tetraethyl lead has what kind of structure?Hint: Valence Shell Electron Pair Repulsion

model

•••

−

•• •

• ••

−•

••C

C

Pb

C

C

Valence electrons of atoms 5(4) 20single bonds 2(4) 8remainder to Cs 12

VSEPR = AX4=tetrahedral

Floppy CH3CH2chain buries Pb

Tetrahedral arrangement isimportant because the leadis buried beneath the organicCH3CH2 chains - making itlipid soluble.

Context

Chloride ion getsburied in thecarbon chain center, making it lipid soluble

( )Pb CH CH3 2 4

•••

−

•• •

• ••

−•

••C

C

Pb

C

C

Floppy CH3CH2chain buries Pb

Bond strength- is the Pb-C bond as strong as a true fully covalent C-C bond?

NO!!!Bond enthalpiesC-C 368Pb-C 206.7kJ/mol

( ) ( )Pb CH CH Pb CH CH CH CH3 2 4 3 2 3 3 2→ +

( ) ( )Pb CH CH Pb CH CH CH CH3 2 4 3 2 3 3 2→ ++•

•Carbon getsthe electron

( ) ( )Pb CH CH Cl Pb CH CH Cl3 2 3 3 2 3

+ −+ →

Context

Triethyllead carries chloride ions across the cell membranedisrupting the energy cycle of the cell and the osmoticpressures - ultimately results in cell death by “popping” (edema)

Cl

PV= energy

Context

Typical ion concentrations in vertebrates and invertebratesIon Cell (M) Blood (M)K+ 0.139 0.004Na+ 0.012 0.145Cl- 0.004 0.116HCO3

- 0.012 0.029X- 0.138 0.009Mg2+ 0.0008 0.0015Ca2+ <0.0000002 0.0018

Context

20


Alanah FitchFlanner Hall [email protected]



END

mL () 1000 L . 1060 () gmol - Loyola University Chicagoafitch.sites.luc.edu/Gen Chem pdf lectures/13...

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