chapter 10 solution lecture solution ecture notes

8/18/2019 Chapter 10 Solution Lecture solution ecture Notes

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Chapter 10

Properties of Solutions

17.1 Solution Composition

17.2 The Thermodynamics of Solution Formation (skip)

17. Factors !ffectin" Solu#ility

17.$ The %apor Pressures of Solutions

17.& 'oilin"Point le*ation and Free+in"Point ,epression

17.- smotic Pressure

17.7 Colli"ati*e Properties of lectrolyte Solutions

17./ Colloids GAS

SOLID LIQUID

Freezing

Melting

S u b l

i m a t i o

n

D e p

o s i t i o

n

C o n d e n s a t

i o n

E v a p o r a t i o n


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Colli"ati*e Properties of Solutions

For Colligative properties the difference

#eteen a pure sol*ent and dilute solutiondepends only on the number of solute

particles present and not on their

chemical identity.

3amples

4 Vapor Pressure Depression

4 Boiling Point Elevation

4 Melting Point Depression 4 Osmotic Pressure


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Lowering of Vapor Pressure

4 %apor Pressure of a sol*ent a#o*e a dilute solutionis alays less than the *apor pressure a#o*e the

pure sol*ent.

Elevation of Boiling Point

4 The #oilin" point of a solution of a non*olatilesolute in a *olatile sol*ent alays e3ceeds the

#oilin" point of a pure sol*entBoiling

• liuid in euilibrium "it# its vapor at t#e e$ternal pressure%

Boiling Point

• &apor press ' e$ternal pressure

Normal boiling point

• Vap press. = 1 atm


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Phase diagrams for pure water red

lines!

and for an a"ueous solution

containing a nonvolatile

solution #lue lines!.

Elevation of BoilingPoint $ VaporPressure Depression

∆T = K bm

solute

∆T is the boiling point elevation

K b is molal boiling - point elevation constant

msolute

is the molality of the solute in solution


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Solution Composition

Solution( A #omogeneous mi$ture )mi$ed at level o* atoms

mole+ules or ionsSolvent(

Solute(

,#e ma-or +omponent

,#e minor +omponent

,#e solute and solvent +an be an.

+ombination o* solid ) s/0 liuid )l /0 and

gaseous ) g / p#ases%

Dissolution( ,"o )or more/ substan+es mi$ at t#e level o*

individual atoms0 mole+ules0 or ions%


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Solution Composition

Mass percentage (weight percentage):

mass percentage of the component =

X 100mass of component

total mass of mi!t"re

Mole fraction: ,#e amount o* a given +omponent )in moles/

divided b. t#e total amount )in moles/

12 ' n2#)n2 3 n4/ *or a t"o +omponent s.stem

14 ' n4#)n2 3 n4/ ' 2 $ 12 or 12314'2

Mass %raction& Mole %raction& Molalit' an

Molarit'


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Molalit'msolute '

moles solute per 5ilogram solvent

' moles per 5g or )mol 5g62/

Molarit' (biochemists pa' attention)

csolute '

moles solute per volume solution

' moles per liter of solution )mol L62/


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Factors !ffectin" Solu#ility

1. 5olecular 6nteractions

4 Review chapter 4 4 Polar molecules ater solu#le hydrophilic (ater

lo*in")

.". %itamins ' and C atersolu#le

4 8onpolar molecules solu#le in nonpolarmolecules hydropho#ic (ater fearin")

.". %itamins ! , 9 and fatsolu#le


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Factors !ffectin" Solu#ility of :ases

1. Structure ffects

2. Pressure ffects


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;enry


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:#en t#e partial pressure o* nitrogen over a sample o* "ater at

2;% atm0 t#e +on+entration o* nitrogen in t#e "ater is

?%@ $ 2>6B mol L62% Compute 8enr.9s la" +onstant *or nitrogen in

"ater at t#is temperature%Given

PN2

= 9.20 atm

cN2 = [N2] = 5.76x10−3

mol/LHenry's Law

PN2

= k N2 X N

2

X N2

= nN2

nN2

+ nH2O

≈nN2

nH2O


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)iter1ass"meNe!t

%in(

*i+en

X

P

,

rearrange

X, P

-

N

-N

-N

-N

-N

-N

=

=

:#en t#e partial pressure o* nitrogen over a sample o* "ater at

2;% atm0 t#en t#e +on+entration o* nitrogen in t#e

"ater is ?%@ $ 2>6B mol L62% Compute 8enr.9s la" +onstant *or

nitrogen in "ater at t#is temperature%

OO n

n

nn

n

X

X

-

-N

-

-N

-N

-N

-N

-N

-N

/--

N

-N

, P

)awsenr'

mol#l.23!1045Nc

atm6.-0P

*i+en

≈

=

=

=

=


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:#en t#e partial pressure o* nitrogen over a sample

o* "ater at 2;% atm0 t#en t#e

+on+entration o* nitrogen in t#e "ater is ?%@ $ 2>6B

mol L62% Compute 8enr.9s la" +onstant *or nitrogen

in "ater at t#is temperature%%in(

*i+en

X

P

,

-N

-

N

-N

atm2>$C%CA

2%>B@C$2>

atm;%4>

Find

Given

1

75

2>$2%>B@C

2CgDmol

2>>>gDl

molDl?%@A$2>

n

n1

<

<

4 E

4 E

4 E

<

B

O4

8

4 E

4 E

=

===

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=≈

−

−

−


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1. Structure ffects

2. Pressure ffects

. Temperature ffects for !>ueous Solutions

The solu#ility of some

solids as a function oftemperature.

,#e aueous solubilities o* most solids

in+rease "it# in+reasing temperature0

some de+rease "it# temp%Endothermic % heat is a#sor#ed #y the

system thin& evaporation of water' or

melting of ice!

E(othermic % heat is evolved #y the systemthin& fire' or free)ing of water!.


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1. Structure ffects

2. Pressure ffects

. Temperature ffects for !>ueous Solutions

The solu#ility of some gases in

ater as a function of

temperature at a constant

pressure of 1 atm.


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,#e greatest gas solubilit. *or a gas insolution is predi+ted under "#at

+onditions

2/ lo" ,0 lo" 7

4/ lo" ,0 #ig# 7

B/ #ig# ,0 lo" 7


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,#e greatest gas solubilit. *or a gas insolution is predi+ted under "#at

+onditions

2/ lo" ,0 lo" 7

-) low 7& high P

B/ #ig# ,0 lo" 7


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8ccoring to enr's aw& the sol"bilit'of a gas in a li9"i

2/ depends on t#e polarit. o* t#e liuid

4/ depends on t#e liuids densit.

B/ remains t#e same at all temperatures


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A++ording to 8enr.s La"0 t#e solubilit. o* agas in a liuid

2/ depends on t#e polarit. o* t#e liuid

4/ depends on t#e liuids densit.

B/ remains t#e same at all temperatures

) increases as the gas press"re abo+e

the sol"tion increases

?/ de+reases as t#e gas pressure above t#e

solution in+reases


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*he Person Behind the +cience

,rancois-Marie aoult /012-/32/!

4ighlights 4 1//- ?aoult@s la the partial pressure

of a sol*ent *apor in e>uili#rium ith asolution is proportional to the ratio of thenum#er of sol*ent molecules to non

*olatile solute molecules. 4 allos molecular ei"hts to #edetermined and pro*ides thee3planation for free+in" point depression

and #oilin" point ele*ation.

Moments in a Life 4 ?aoult as a prominent mem#er of the

"roup hich created physical chemistryincludin" !rrhenius 8ernst *an t@;off

Planck.

Psoln 5 6solventP7solvent

,or ideal

solutions

http://scienceworld.wolfram.com/chemistry/RaoultsLaw.htmlhttp://scienceworld.wolfram.com/chemistry/FreezingPointDepression.htmlhttp://scienceworld.wolfram.com/chemistry/BoilingPointElevation.htmlhttp://scienceworld.wolfram.com/chemistry/BoilingPointElevation.htmlhttp://scienceworld.wolfram.com/chemistry/FreezingPointDepression.htmlhttp://scienceworld.wolfram.com/chemistry/RaoultsLaw.html


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?aoult


22/37

?aoult


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%apor pressure for a solution of to *olatile li>uids.

Negati+e e+iation

' solute6solvent attra+tions H solvent6solvent attra+tions

Positi+e e+iation

' solute6solvent attra+tions solvent6solvent attra+tions

,or non-ideal +olutions


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boiling point : ∆T = K bm

solute

freezing point : ∆T = K

f

m

solute


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Osmotic Pressure

Fourth Colli"ati*e Property

6mportant for transport ofmolecules across cellmem#ranes calledsemipermea#le mem#ranes

Osmotic Pressure = Π

Π B M ?T

ΠV B n ?T

Molarit. )M/ ' molesL or n&

PV = n;7


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Osmotic Pressure

The normal flo of sol*ent into the solution

(osmosis) can #e pre*ented #y applyin" ane3ternal pressure to the solution.

smotic Pressure useful for

,eterminin" the Molar Mass ofprotein and other macromolecules

small concentrations cause

lar"e osmotic pressures

Can pre*ent transfer of all soluteparticles

,ialysis at the all of most

plant and animal cells


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Dialysis8 ?epresentation of the functionin"

of an artificial kidney

! cellophane (polymeric)

tu#e acts as the semi

permea#le mem#rane

Purifies #lood #yashin" impurities

(solutes) into the

dialy+in" solution.


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A dilute aueous solution o* a non6disso+iating +ompound

+ontains 2%2; g o* t#e +ompound per liter o* solution and

#as an osmoti+ pressure o* >%>4 atm at a temperature o*

B@=C% Compute t#e molar mass o* t#e +ompound%

Strategy

1) useΠ= MRT to find M (mol/L)

2) Recall that # of moles =mass

mwt

3) Rearrange mwt =g

mole=

gLmole

L

=1.19

M


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A dilute aueous solution o* a non6disso+iating

+ompound +ontains 2%2; g o* t#e +ompound per liter

o* solution and #as an osmoti+ pressure o* >%>4 atm

at a temperature o* B@=C% Compute t#e molar mass o*t#e +ompound

S olution

1) use Π= MRT or M = Π

RT

c = Π

RT=

0.0288 atm

(0.0820 L atm mol−1K−1)(37 + 273.15K)

M = 1.132x10−3 mol/L

2) Rearrange M =g

mole=

gL

moleL

M =1.19

gL

1.132x10−3 mol/L= 1.05x103g/mol


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*he Person Behind the +cience

9.4. van:t 4off /0;


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Colligative Properties of Electrolyte +olutions

le*ation of 'oilin" Point

ET# B m 9#Ghere m B molality

(5olality is moles of solute per kilo"ram of sol*ent)

The ffect of ,issociation

ET# B i m 9#i B the num#er of particles released into the

solution per formula unit of solute

e.g.' =aCl dissociates into i 5 <

e.g.' =a dissociates into i 5 1

< =a? ? / +O>-

e.g.' acetic acid a wea& acid and wea&

electrolyte! does not dissociate i 5 /

alsoDepression of

,ree)ing Point

ETf B m 9f

ETf B i m 9f


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:#i+# aueous solution "ould be e$pe+ted to #avet#e #ig#est boiling point

2/ >%2>> m EaCl

4/ >%2>> m CaCl4

B/ >%>> m Fe)EO

B/

B

%>> m Fe)EOB/4

?/ >%>> m Co)SO


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:#i+# aueous solution "ould be e$pe+ted to #ave t#e#ig#est boiling point

2/ >%2>> m EaCl ET# B )4/)>%2>>/ 9# ' >%4>> 9#

4/ >%2>> m CaCl4 ET# B )B/)>%2>>/ 9# ' >%B>> 9#

/) 0.0 m Fe)EOB/4 ET# B )B/)>%>>/ 9# ' >%4 9#

?/ >%>> m Co)SO%>>/ 9# ' >%2> 9#

le*ation of 'oilin" Point

The ffect of ,issociation

ET# B i m 9#


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ColloidsH Colloidal ,ispersions

Colloids are lar"e particles dispersed insolution 4 1nm to 1000 nm in si+e 4 .". :lo#ular proteins &00 nm

3amples 4 pal (ater in solid Si2)

4 !erosols (li>uids in :as) 4 Smoke (solids in !ir) 4 5ilk (fat droplets I solids in ater) 4 5ayonnaise (ater droplets in oil)

4 Paint (solid pi"ments in li>uid) 4 'iolo"ical fluids (proteins I fats in ater)

Characteristics 4 =ar"e particle si+e colloidsH translucent cloudy

milky) 4 Small particle si+e colloidsH can #e clear


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Colloidal ,ispersions

4 Tyndall ffect =i"ht Scatterin"


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chapter 10 solution lecture solution ecture notes

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