Brdy 6Ed Ch13 MixturesAtTheMolecularLevel

8/21/2019 Brdy 6Ed Ch13 MixturesAtTheMolecularLevel

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Chapter 13:Mixtures at the Molecular

Level: Properties of Solutions

Chemistry: The Molecular Nature

of Matter, 6E

Jespersen/Brady/Hyslop



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

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A Look Ahead

• We begin by examining diferent types o soluons that can be

ormed rom the three states o maer solid" li#uid" and gas.

We also characteri$e a soluon by the amount o solute present

as unsaturated" saturated" and supersaturated.

%&'.&(

• )ext *e study the ormaon o soluons at the molecularle+el and see ho* intermolecular orces afect the energecs o

the soluon process and solubility. %&'.'(

• We study the our major types o concentraon units,

percent by mass" mole racon" molarity" and molality,and

their intercon+ersions. %&'.-(• emperature in general has a marked efect on the solubility o

gases as *ell as li#uids and solids. %&'./(


http://slidepdf.com/reader/full/brdy-6ed-ch13-mixturesatthemolecularlevel 4/128Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

A Look Ahead

• We see that pressure has no in0uence on the solubility o li#uids

and solids" but greatly afects the solubility o gases. he #uanta+erelaonship bet*een gas solubility and pressure is gi+en by 1enry2s

la*. %&'.3(

• We learn that physical properes such as the +apor pressure"

melng point" boiling point" and osmoc pressure o a soluon

depend only on the concentraon and not the identy o the solutepresent. We 4rst study these colliga+e properes and their

applicaons or nonelectrolyte soluons. %&'.5(

• We then extend our study o colliga+e properes to electrolyte

soluons and learn about the in uence o ion pair ormaon on these

properes. %&'.6(• he chapter ends *ith a brie examinaon o colloids" *hich are

parcles larger than indi+idual molecules that are dispersed in

another medium. %&'.7(



Chapter 13: Solutions

Solution Homogeneous mixture

Composed of solvent and solutes!

Solvent More a"undant component of mixture

Solute(s) Less a"undant or other components! of mixture

E! Lactated #inger$s solution

%aCl& 'Cl& CaCl(& %aC)H*+) in ,ater

*



"hy #o Solutions $orm%

-,o .riving orces "ehind ormation ofSolution

01 2ntropy/.isorder

(1 3ntermolecular orces4hether or not a solution forms depends on

"oth opposing forces

5



Spontaneous Miin& ( gases mix

spontaneously .ue to random motions

Mix ,ithout outside ,or6

%ever separatespontaneously

-endency of system

left to itself& to"ecome increasinglydisordered

Entropy effect7

Gas A Gas B

separate mixed



Spontaneous Miin& Strong driving force in nature System& left to itself& ,ill tend to,ards

most pro"a"le state

'aseous Solutions

2ntropy only driving force 8ttractive 3ntermolecular! forces negligi"le

iui# Solutions

2ntropy is one driving force 8ttractive intermolecular! forces are very

important

9



*ntermolecular $orces (*M$)

8ttractive forces "et,een solute andsolvent hold solution together

Strength of intermolecular attractive forces

depends on +oth solute and solvent

0



*ntermolecular $orces (*M$) 3nitially solute and solvent separate

Solute molecules held together "y 3Ms

Solvent molecules held together "y 3Ms

4hen mixed& for solution to form&

Solvent;to;solute attractions must "e <to attractions "et,een solute alone andsolvent alone

00



"hy Such ifferent -ehavior%

4hen li=uids com"ine

01 Must put in 2nergy to overcome or lessenintermolecular attractive forces "et,eenmolecules

Must push solute molecules apart

Must push solvent molecules apart

0(




"hy Such ifferent -ehavior%

4hen li=uids com"ine

(1 4hen solute and solvent mix or cometogether

Must form ne, intermolecular forces

"et,een solute and solvent #eleases energy

0)




Misci+le iui#s

Misci+le li=uids

( li=uids that are solu"le in each other .issolve in one another in all proportions

orm solution

Strengths of intermolecular attractionsare similar in solute and solvent

Similar polarity

E! 2thanol and ,ater

0>

H+CC

H

H

H

H

H

.

2thanol?polar "ond




*mmisci+le iui#s

-,o insolu"le li=uids

.o not mix @et t,o separate phases

Strengths of 3Ms are different in

solute and solvent .ifferent polarity

E! BenAene and ,ater

0*

C

CC

C

CC

H

H

H

H

H

HBenAene?no polar "ond

f




/ule of Thum+ i0e #issolves i0e

se polar solvent for polar solute se Nonpolar solvent for nonpolar

solute

"hen stren&ths of intermolecularattractions are similar in solute an#solvent, solutions form +ecause net

ener&y echan&e is a+out the same

0D




2rocess of issolution 2olar solutes interact ,ith

and dissolve in polar solvents

4+on#in& solutes interact

,ith and dissolve in 4+on#in& solvents

E! Ethanol in 5ater

Both are polar molecules Both form hydrogen "onds

3Ms of 2t+H and H(+ large

and similar 05

3C C

7

7




E! of Misci+le Solution 4hen ethanol dissolves in ,ater& get H;"onding

3M of solution also large 4hen solutions form& there ,ill "e enough

energy to move 2t+H and H(+ apart so can mix

Solvent and solute are similarE 3M strength! Solution 5ill form

07

3CC

7

7

7

C3C

7

7

2 f i l ti




2rocess of issolution Non4polar solutes interact ,ith and

dissolve in non4polar solvents Both have only London dispersion forces

4hy form solutionF

4hen li=uids com"ine Put ener&y in to overcome 3Ms "et,een

molecules in solute and solvent

4hen solute and solvent mix or come together&form ne, 3Ms "et,een solute and solvent

#eleases G same amount of energy asput in

09

E - i CCl




E! -en8ene in CCl9

CCl> %onpolar 4ea6 London forces

BenAene& CDHD

%onpolar 4ea6 London forces

Similar in strength to CCl>

Small 2 to move apart Small 2 gained for solution 3Ms

.oes dissolve& solution forms

(

E f * i i+l S l ti




E! of *mmisci+le Solution-en8ene in 5ater

BenAene %onpolar

4ea6 London dispersion forces only

4ater

Polar

Strong hydrogen "onding

(0

E of *mmisci+le Sol tion




E! of *mmisci+le Solution-en8ene in 5ater

Solvent and solute are very differentE Costs energy to "rea6 strong H;"onds in H(+

%o strong 3Ms in solution ,ith "enAene tooffset

%o solution forms

( layers& .on$t Mix

((

earnin& Chec0




earnin& Chec0

4hich of the follo,ing are misci"le in ,aterF

()

our Turn;




our Turn;4hich of the follo,ing molecules is solu"le in

CDHDF 81 %H)

B1 CH)%H(

C1 CH)+H.1 CH)CH)

21 CH)Cl

(>




Solutions of Soli#s in iui#s Basic principles remain the same ,hen

solutes are solids

Sodium chloride %aCl! 3onic "onding

Strong intermolecular forces

3ons dissolve in ,ater "ecause ion;dipole forcesof ,ater ,ith ions strong enough to overcome

ion;ion attractions

(*

y#ration of Soli# Solute




y#ration of Soli# Solute 8t edges& fe,er

oppositely charged ions

around H(+ can come in

3on;dipole forces

#emove ion %e, ion at surface

Process continues untilall ions in solution

y#ration of ions Completely surrounded

"y solvent

(D

y#ration vs Solvation




y#ration vs! Solvation y#ration

3ons surrounded "y ,ater molecules Solvation

@eneral term for surrounding solute particle "ysolvent molecules

o polar molecules #issolve in 7F es

8ttractions "et,een solvent and solute dipoles

dipole;dipole interactions! dislodge moleculesfrom solid

Bring into solution

(5

2olar Molecule in "ater




2olar Molecule in "ater

(7

H2O reorients so

Positive Hs are near negative ends of solute Negative Os are near positive ends of solute

Solvation in Nonpolar Solvents%




Solvation in Nonpolar Solvents%

4ax and "enAene 4ea6 London dispersion forces in "oth

4ax molecules 2asily slip from solid

Slide "et,een "enAene molecules

orm ne, London forces "et,een solvent andsolute

(9

eat of Solution




eat of Solution 2nergy change associated ,ith formation of

solution .ifference in intermolecular forces "et,een

isolated solute and solvent and mixture

Cost of mixing

2nthalpy exchanged "et,een system andsurroundings

Molar Enthalpy of Solution (soln) 8mount of enthalpy exchanged ,hen one

mole of solute dissolves in a solvent atconstant pressure to ma6e a solution

)

eat of Solution




eat of Solution

soln < = (positive) Costs energy to ma6e solution

2ndothermic

P2 of system

soln > = (ne&ative) 2nergy given off ,hen solution is made

2xothermic

P2 of system

4hich occurs depends on your system

)0

Mo#elin& $ormation of Solution




Mo#elin& $ormation of Solution

ormation of solution from solid and li=uid

can "e modeled as (;step process Step 1: Separate Solute and Solvent

molecules

Brea6 intermolecular forces 2ndothermic& P2 of system& costE

Step : Mix Solute and Solvent Come together

orm ne, intermolecular forces

2xothermic& P2 of system& profitE

)(

4Step 2rocess




Step 2rocess 8pplication of Hess$ La,

4ay to ta6e ( things ,e can measure and useto calculate something ,e can$t directly measure

soln is path independent

Method ,or6s "ecause enthalpy is state function

soln ? soln @ solute @ solvent

+verall& steps ta6e us fromsolid solute I li=uid solvent final solution

-hese steps are not the ,ay solution,ould actually "e made in la"

))

Enthalpy ia&ram




Enthalpy ia&ram01 Brea6 up solid lattice

Hlattice Lattice enthalpy

P2

01 .issolve gas in solvent

HSolvation Solvation

enthalpy P2

solution ? lattice . Solvation

4hether Hsolution

I or K

depends on values

3n la"& solution formeddirectly

)>

1

#irect

$i& 13 A issolvin& B* in 7




$i& 13!A issolvin& B* in 7Hlattice '3! D)( 6J mol K0

Hhydration '3! K D09 6J mol K0

solution ? latt . hy#r

soln ? D)( 6J mol K0

K D09 6J mol K0 soln ? .13 0 mol @1

ormation of '3a=! is

en#othermic

)*

issolvin& Na-r in 7




issolvin& Na-r in 7Hlattice %aBr! 5(7 6J mol K0

Hhydration %aBr! K5>0 6J mol K0

solution ? latt . hy#r

)D

Hsoln = 728 kJ mol –1

– 741 kJ mol –1

Hsoln = – 13 kJ mol –1

Formation of NaBr a!" is

exothermic

our Turn;




our Turn; 4hen (1* g of solid sodium hydroxide is added to

01 g of ,ater in a calorimeter& the temperatureof the mixture increases "y D1* oC1 .etermine themolar enthalpy of solution for sodium hydroxide1

8ssume the specific heat of the mixture is >107> J g;

0

' ;0

1 81 I>>1D 6J/mol

B1 I>)1* 6J/mol

C1 ;>>1D 6J/mol

.1 ;>)1* 6J/mol

Hsoln ;0(1* g!>107> J g;0 ' ;0!D1* '!

0 6J/0J!/(1* g/>1 g mol;0! ;>>1D 6J

)5

Solutions Containin& iui# Solute




Solutions Containin& iui# SoluteSimilar treatment "ut,ith ) step path

01 Solute expanded to gas

H I

P2

01 Solvent expanded togas

H I

P2

01 Solvation occurs H K

P2

)7

Hsoln = Hsolute + Hsolvent + Hsolvation

*#eal Solution




*#eal Solution +ne in ,hich inter;

molecular attractive

forces are identical soln ? =

Ex. Benzene in CCl

#ll $ondon for%es Hsoln ! "

Step 0 I Step ( KStep )

or

solute . solvent ? @solvation

)9

'aseous Solutes in iui# Solution





+nly very ,ea6 attractions exist "et,een

gas molecules -here are no intermolecular attractions in ideal

gases

4hen ma6ing solution ,ith &as solute 2nergy re=uired to expand soluteE is

negligi"le

Heat a"sor"ed or released ,hen gasdissolves in li=uid has t,o contri"utions:

012xpansion of Solvent solvent

(1Solvation of @as solvation >





@as dissolves in

organic solvent 2ndothermic

Hsolvation

@as dissolves in H(+

2xothermic K!

Hsolvation N

>0

our Turn;




our Turn;

-he solu"ility of a su"stance increases ,ithincreased temperature if:

81 OHsolution

B1 OHsolution NC1 OHsolution

>(

Solu+ility




y Mass of solute that forms saturated solution ,ith

given mass of solvent at specified temperature

3f extra solute added to saturated solution& extra

solute ,ill remain as separate phase

solubility= g solute

100 g solvent

>)

Effect of T on Solu+ility of Soli#s an#




yiui#s in iui# Solvent

3f heat is a"sor"ed ,hen solute dissolves&solu"ility ,hen -

3f energy is released ,hen solute dissolves&solu"ility ,hen -

4hen apply stressE to e=uili"rium "y -!&

e=uili"rium ,ill shift so as to relieve a"sor"s orminimiAes! stress

eChDteliers 2rinciple

soluteundissolved

+ energy ←→ solute

dissolved

soluteundissolved

←→ solute

dissolved + energy

>>

Solu+ility of Most Su+stances




y*ncreases 5ith Temperature

Most su"stances"ecome moresolu"le as -

8mount solu"ility aries considera"ly

.epends on

su"stance

>*

Effect of T on 'as Solu+ility in iui#s




Solu+ility of &ases usually

as T

Ta+le 13! Solu"ilities of Common @ases in 4ater

>D

Case Stu#y: ea# Fones




.uring the industrial revolution& factories ,ere

"uilt on rivers so that the river ,ater could "eused as a coolant for the machinery1 -he hot,ater ,as dumped "ac6 into the river and cool,ater recirculated1 8fter some time& the rivers"egan to dar6en and many fish died1 -he,ater ,as not found to "e contaminated "ythe machinery1 4hat ,as the cause of the

mysterious fish 6illsF

>5

3ncreased temperature& lo,eredamounts of dissolved oxygen

Effect of 2ressure on 'as Solu+ility




Solu+ility as 2

"hy% P means a"ove

solution for gas

@as goes intosolution

#elieves stress onsystem

Conversely& solu"ility as P

Soda in can

>7

Effect of 2ressure on 'as Solu+ilityl " " h d




G! 8t some P& e=uili"rium exists "et,een vapor phase andsolution

ratein ? rateout -! in P puts stress on e=uili"rium

fre=uency of collisions so ratein < rateout

More gas molecules dissolve than are leaving solution

C! More gas dissolved

#ateout ,ill until #ateout #atein and e=uili"rium

restored

>9

enrys a5




y Pressure;Solu"ility La,

Concentration of gas in li=uid at any giventemperature is directly proportional to partialpressure of gas over solutionE

C &as ? 0 3 2 &as - is constant!

C &as concentration of gas

2 &as partial pressure of gas

0 3 HenryQs La, constant

ni=ue to each gas

-a"ulated*

enrys a5




y -rue only at lo, concentrations and

pressures ,here gases do %+- react ,ithsolvent

8lternate form

C 1 and 2 1 refer to an initial set of conditions

C 6 and 2 6 refer to a final set of conditions

C 1 P 1

=C 2 P 2

*0

E! 1 Hsin& enrys a5




Calculate the concentration of C+( in a soft drin6 that

is "ottled ,ith a partial pressure of C+( of * atm overthe li=uid at (* RC1 -he Henry$s La, constant for C+(

in ,ater at this temperature is )10( 0( mol/Latm1

C CO2=k H (CO 2) P CO 2

*(

)10( 0( mol/Latm T *1 atm

10*D mol/L

=!16 molI

"hen un#er A!= atm pressure

E! 1 Hsin& enrys a5




Calculate the concentration of C+( in a soft drin6

after the "ottle is opened and e=uili"rates at (* RCunder a partial pressure of C+( of >1 0> atm1

*)

C# & 1.#

1" ' mol$%

&'en open to air

C 2=(0 . 156 mol/L ) (4 . 0×10−4 atm )

5.0atm

C 1

P 1=C

2

P 2C 2=

P 2C 1

P 1

earnin& Chec0




4hat is the concentration of dissolved nitrogenin a solution that is saturated in %( at (1 atmF

6 H 71>(U05 M / atm!

*>

V Cg6 HPg

V Cg 71>(U05 M / atm! U (1 atm

V Cg015 U0

D

M

our Turn;f 0




Ho, many grams of oxygen gas at 01 atm,ill dissolve in 01 L of ,ater at (* oC ifHenry$s constant is 01) x 0;) M atm;0 at thistemperature F

81 1>( g

B1 10) g

C1 1>( g

.1 1(0 g

21 (1> g

**

Solu+ility of 2olar vs! Nonpolar 'ases@ l l ith l " d h




@as molecules ,ith polar "onds are much moresolu"le in ,ater than nonpolar molecules li6e oxygen

and nitrogen C+(& S+(& %H) +(& %(& 8r

orm H;"onds ,ith H(+

Some gases have increased solu"ility "ecause theyreact ,ith H(+ to some extent

E! C+(a= ! I H(+ H(C+)a= ! HIa= ! I HC+) K

a= !

S+(a= ! I H(+ H(S+)a= ! HIa= ! I HS+) Ka= !

%H)a= ! I H(+ %H>Ia= ! I H+ Ka= !

*D

Case Stu#y




4hen you open a "ottleof seltAer& it fiAAes1 Ho,should you store it toincrease the time "efore

it goes flatF

*5

@ases are more solu"le at

lo, temperature and highpressure1 Cap it and cool it1

Concentration




4hat units ,e use depends on situation

Stoichiometric calculations Molarity

Hnits: mol/L

2ro+lem: M varies ,ith temperature Jolume varies ,ith temperature

Solutions expand and contract ,hen heated

and cooled

3f temperature in#epen#ent concentrationis needed& must use other units

*7

M =mol of solute

L of solution

Temperature *nsensitiveC t ti




Concentration

1! 2ercent Concentrations 8lso called percent "y mass or percent "y

,eight

-his is sometimes indicated W,/,! ,here ,Estands for ,eight

-he ,/,!E is often omitted

percent by mass =mass of solutemass of solution ×100

*9

E! 2ercent +y Mass




4hat is the percent "y mass of %aCl in a

solution consisting of 0(1* g of %aCl and 5*1g ,aterF

wt NaCl

=14. 3 aCl

D

wt NaCl

=12. 5 g

(12.5 + !5. 0 ) g

×100

percent by mass=

mass of solute

mass of solution×

100

earnin& Chec0 S t i t i ll ) *W lt d h d it




Sea,ater is typically )1*W sea salt and has a densityof 01) g/mL1 Ho, many grams of sea salt ,ould "e

needed to prepare enough sea,ater solution to fill aD(1* L a=uariumF

4hat do ,e need to findF

D(1* L F g sea salt4hat do ,e 6no,F

)1* g sea salt 0 g solution

01) g soln 01 mL solution

0 mL 01 L

D0

62 .5 L ×1000 mL

1 L ×

1. 03 g soln

1. 00 mL soln×

3 . 5 g sea salt

100 g soln

? K1=

3

& sea salt

More Temperature *nsensitiveConcentration Hnits




Concentration Hnits

Molality (m ) %um"er of moles of solute per 6ilogram solvent

8lso Molal concentration

3ndependent of temperature

m vs1 M

Similar ,hen # 01 g/mL

.ifferent ,hen # 01 g/mL or# NN 01 g/mL

molality =m= mol of solute

kg of solvent

D(

E! Concentration Calculation3f you prepare a solution "y dissolving (* )7 g of 3 in




3f you prepare a solution "y dissolving (*1)7 g of 3( in

*1 g of ,ater& ,hat is the molality m ! of the

solutionF


(*1)7 g F m

4hat do ,e 6no,F (*)17 g 3( 0 mol 3(

m mol solute/6g solvent

*1 g 1* 6g

Solve it

D)

25 .3" g #2×

1 mol #2

253 ." g #2

×1

0 .5000 $g %ater

1( mol 3( /6g ,ater =!=== m

E! Concentration Calcn! (cont)4hat is the molarity M! of this solutionF -he density




4hat is the molarity M ! of this solutionF -he densityof this solution is 01 g/mL1


(*1)7 g F M

4hat do ,e 6no,F

(*)17 g 3( 0 mol 3(

M mol solute/L soln

01 g soln 0 mL soln

Solve it

D>

25 . 3" g #2×1 mol #2

253 . " g #2

× 1

525 . 3" g solnr × 1 . 00 g

1.00 mL×1000 mL

1 L

109) mol 3( / L soln =!1L=3 M

E! M an# m in CCl94hat is the molality m! and molarity M! of a solution




4hat is the molality m! and molarity M ! of a solutionprepared "y dissolving (*1)7 g of 3( in *1 g of CCl>F

-he density of this solution is 01*9 g/mL14hat do ,e need to findF

(*1)7 g F m

4hat do ,e 6no,F (*)17 g 3( 0 mol 3(

m mol solute/6g solvent

*1 g soln 1* 6g soln

Solve it

D*

25 .3" g #2×

1 mol #2

253 . " g #2

×1

500 .0 g %ater ×

1.00 g

1.00mL×

1000 mL

1 L

1( mol 3(

/6g CCl>

=!=== m

E! M an# m in CCl9 (cont)4h t i th l it M! f thi l ti F -h d it




4hat is the molarity M ! of this solutionF -he densityof this solution is 01*9 g/mL1


(*1)7 g F M

4hat do ,e 6no,F

(*)17 g 3( 0 mol 3(

M mol solute/L soln

01*9 g soln 0 mL soln

g of soln g 3( I g CCl> *1 g I (*1)7 gSolve it

DD

25 . 3" g #2×

1 mol #2

253 ." g #2

×1

525 .3" g soln×

1.5& g

1.00mL×

1000 mL

1 L

1)) mol 3( /L soln =!3=3= M

Convertin& +et5een ConcentrationsCalculate the molarity and the molality of a >1W




y yHBr solution1 -he density of this solution is 01)7

g/mL1

3f ,e assume 01 g of solution& then >1 g of HBr1

3f 0 g solution& then

mass 7 01 g soln K >1 g HBr D1 g H(+

mol '(r =40 .0g '(r

"0.&1 g '(r/ mol =0.4&4 mol '(r

D5

40 . 0 '(r =wt%=40. 0g '(r

100 g solution ∗100

m= mol '(r

kg of '2O

m= mol '(r

kg '2O =

0.4&4 mol '(r

0.0600 kg '2O =".24 m

Convertin& +et5een Concentrations(cont!)




( )

%o, Calculate Molarity of >W HBr

Vol )oln=mass

*ensity =100 g

1.3" g /mL

M = 0.4&4mol '(r

!2.46 mL solution ×1000mL

1L

D7

M = mol '(r

L solution

mol HBr & ()4*4 mol

? !96 m

? 6! M

our Turn;4hat is the molality of *1W ,/,! sodium




4hat is the molality of *1W ,/,! sodiumhydroxide solutionF

81 1* m

B1 01(* m

C1 1(* m

.1 (* m 21 * m

D9

MM(&Imol) 7: 1!=O Na7: 9=!==

01 g soln *1 g %a+H I *1 g ,ater

50 .0 g aO'×1 mol aO'

40. 00 g aO'

×1 m

mol/$g

×1000 g

1 $g

×1

50. 0 g %ater ? A m

our Turn; 4hat is the molarity of the *W,/,! solution if its




y / !density is 01*(9 g/mLF

81 09 M

B1 01(* M

C1 019 M .1 15D M

5

our Turn; 4 Solution



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 50

(

0 mol %a+H g soln*: g %a+H x x 01*(9>:1: g mL

0 0::: mL

x x J 090:: g soln L

or

*: g H +mmol 01*(9 g(* x x 09g soln 0:: g soln mL

M

M

7ther Temperature *nsensitiveConcentration Hnits




Concentration Hnits

Mole $raction

Mole P

5(

χ A=

+ mol ,

-otal moles of all components

mol %A= χ A∗100

Colli&ative 2roperties Physical properties of solutions




Physical properties of solutions

.epend mostly on relative populations ofparticles in mixtures

.on$t depend on their chemical identities

Effects of Solute on Japor 2ressure of Solvents

Solutes that can$t evaporate from solution arecalled nonvolatile solutes

$act: 8ll solutions of nonvolatile solutes have

lo,er vapor pressures than their pure solvents

5)

/aoultQs a5

apor pressure of solution 2 e=uals




apor pressure of solution& 2 soln & e=uals

product of mole fraction of solvent& R solvent &and its vapor pressure ,hen pure& 2 solvent

8pplies for dilute solutions

P solution= vapor pressure of te solution X

solvent = mole fraction of te solvent

P solvent ∘ =vapor pressure of pure solvent

P solution= X

solvent P

solvent ∘

5>

Glternate form of /aoults a5 Plot of 2 soln vs1 R solvent




soln solvent

should "e linear

Slope

*ntercept

Change in vaporpressure can "eexpressed as

sually more interested in ho, solute$s mole fractionchanges the vapor pressure of solvent

ΔP =cange in P =( P solvent ∘ − P

solution)

5*

ΔP = X solute P solvent∘

P solvent ∘

E! 'lycerin (usin& /aoults a5)

@lycerin C H + is a nonvolatile nonelectrolyte ,ith a




@lycerin& C)H7+)& is a nonvolatile nonelectrolyte ,ith a

density of 01(D g/mL at (* RC1 Calculate the changein vapor pressure as (* RC of a solution made "yadding *1 mL of glycerin to *1 mL of ,ater1-he vapor pressure of pure ,ater at (* RC is ()17

torr1-o solve use:

irst ,e need Xsolute& so ,e need mole glycerin

and mole H(+1

ΔP = X solute

P solvent ∘

5D

E! 'lycerin (cont!)

1 l C H OMole glycerin



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 55

50.0mL C3

H "

O3

×1.26 g

mL×

1mol C3 H

"O

3

&2.1g C3 H "O3

¿0.6"4 mol C3 H

"O

3

500.0mL '2O×1.00 g

mL×1mol '

2O

1".02 g '2O =2!.!5mol '2O

X C

3

H "

O3

=0.6"4 mol

(2!.!5+0.6"4)mol =2 . 4 0 6×10

−2

ΔP = X solute

P solvent ∘ =(2 . 4 0 6×10−2)×23." torr

g y

Mole ,ater

Mole fraction glycerin

? =!A3 torr

E! 'lycerin (cont!)4hat is the final pressureF




4hat is the final pressureF

Can solve t,o ,ays:

+r

P solution

= P solvent ∘ − ΔP

57

X H 2O=

2!.!5mol

(2!.!5+0.6"4 )mol =0.&!5&

P solution

= X solvent

P solvent∘

=(0.&!5&)×23." torr =23.2 torr

P soln

=23." torr −0.5!3 torr =23.2torr

ΔP = P solvent ∘ − P

solution

earnin& Chec0 -he vapor pressure of (;methylhexane is )5197D torr at



Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

0*RC1 4hat ,ould "e the pressure of the mixture of 571

g (;methylhexane and 0* g naphthalene& ,hich is nearlynon;volatile at this temperatureF

P =(0 ."6&×3!. &"6 torr )

mole 2metyleane=!". 0 g

100 . 2 g/mol =0 .!! " 4 mol

59

(solution = )solvent(o

solvent

& ++)(2 torr = 33 torr

mole naptalene=15 g

12".1! g/mol=0.1 1 ! mol

X 2−metyleane

=0 . !! " 4 mol

0 . !! " 4 mol + 0 . 11 ! mol=0 . "6&

naphthaleneC0H7

MM 0(7105

(;methylhexaneC5H0D

MM 0::1(

"hy Nonvolatile Solute o5ers Japor 2ressure




-o evaporate& molecule must have enough

'inetic 2nergy to escape surfaceYsay 0W +nly those molecules escape

Set up e=uili"rium "et,een li=uid and vapor

8dd solute to solvent to get (W ,/,! solution %o, only 0W of 7W solvent can escape or 17W

of all molecules

So vapor pressure "ecause fraction ofsolvent molecules capa"le of leavingsolution

7

"hy Nonvolatile Solute o5ers Japor2ressure

G L f l l l i li id h




G! Lots of solvent molecules in li=uid phase

#ate of evaporation and condensation high-! e,er solvent molecules in li=uid

#ate of evaporation lo,er

8t e=uili"rium& fe,er molecules in gas phase

apor pressure lo,er

70

Solutions That Contain T5o orMore Jolatile Components



Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter 6E

%o, vapor contains molecules of "othcomponents

Partial pressure of each component 8 and B isgiven "y #aoult$s La,

-otal pressure of solution of components 8 andB given "y .alton$s La, of Partial Pressures

P A= X

A P

A

∘

P = X

P

∘

P total

= P A+ P

= X

A P

A∘ + X

P

∘

7(

$or *#eal, T5o Component Solutionof Jolatile Components



Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter 6E 7)

P A= X A P A∘

P = X

P

∘

P -otal

= P A+ P

= X

A P A

∘+ X

P

∘

E! -en8ene an# Toluene

Consider a mixture of "enAene& CDHD& and




Consider a mixture of "enAene& CDHD& and

toluene& C5H7& containing 01 mol "enAeneand (1 mol toluene1 8t ( RC& the vaporpressures of the pure su"stances are:PR

"enAene 5* torr

PRtoluene (( torr

8ssuming the mixture o"eys #aoult$s la,&,hat is the total pressure a"ove thissolutionF

7>

E! -en8ene an# Toluene (cont!)1! Calculate mole fractions of G an# -




! Calculate partial pressures of G an# -

3! Calculate total pressure

P !en"ene

= X !en"ene

∗ P !en"ene∘ =0.33×!5 torr =25 torr

P toluene

= X toluene

∗ P toluene∘ =0.6!×22torr =15torr

7*

X !en"ene=

1 .0 mol

(1 . 0+2 . 0 )mol =0 . 33 benene

X toluene=

2 . 0mol

1 . 0+2 . 0 )mol =0 . 6! toluene

P total

= P !en"ene

+ P toluene

¿(25+15 )torr =40 torr

earnin& Chec0 -he vapor pressure of (;methylheptane is




())19* torr at **RC1 );ethylpentane has avapor pressure of (51D7 at the sametemperature1 4hat ,ould "e the pressure ofthe mixture of 571g (;methylheptane and 0*

g );ethylpentaneF

(;methylheptaneC7H07

MM 00>1() g/mol

);ethylpentane

C5H0D

MM 01( g/mol

7D

(solution = )A(oA + )B(o

B

earnin& Chec0 mole 2metyleptane=

!".0 g

114 23 g/mol =0 . 6" 2" mol




P =(0."2!×233 . &5 torr )+(0.1!3×20! .6" torr )

114 . 23 g/mol

75

( = #3" torr

mole 3etylpentane= 1 5 g1 00 . 2 g/mol

=0 .1 4 &! mol

X 3etylpentane

=0 .14 &! mol

(0 . 6" 2"3 mol +0 . 1 4 &! mol)=0.1!3

X 2metylpentane

=0 . 6" 2 "3 mol

(0 . 6" 2 "3 mol +0 .14 &! mol )

=0 . " 2 !

our Turn;n;hexane and n;heptane are misci"le in a large degreeand "oth volatile 3f the vapor pressure of pure




and "oth volatile1 3f the vapor pressure of pure

hexane is 0*01(7 mm Hg& and heptane is >*1D5 at(*Z& ,hich e=uation can "e used to determine themole fraction of hexane in the mixture if the mixture$svapor pressure is 0>*1* mm HgF

81 X0*01(7 mmHg! 0>*1* mmHg

B1 X0*01(7 mmHg! I X!>*1D5 mm Hg! 0>*1*mmHg

C1 X0*01(7 mmHg! I 0 K X!>*1D5 mm Hg! 0>*1* mm Hg

.1 %one of these

77

Solutes also Gffect $ree8in& an#-oilin& 2oints of Solutions

$acts:




$acts:

$ree8in& 2oint of solution al,ays o5er than pure solvent

-oilin& 2oint of solution al,ays i&her

than pure solvent"hy%

Consider the phase diagram of H(+

Solid& li=uid& gas phases in e=uili"rium

Blue lines

P vs1 -

79

2ure "ater -riple Point -P!




8ll ) phases exist ine=uili"riumsimultaneously

Pure H(+

.ashed lines at 5D torr0 atm! that intersectsolid/li=uid and

li=uid/gas curves @ive - for reeAing

Point P! and BoilingPoint BP!

9

*( *B(T2

6=

SolutionEffect of Solute Solute molecules stay

in solution only




in solution only %one in vapor %one in solid

Crystal structure preventsfrom entering

Li=uid/vapor num"er solvent

molecules entering vapor

%eed higher - to get allli=uid to gas

Line at higher - alongphase "order red!

90

SolutionEffect of Solute -riple point lo,er




and to left

! Solid/li=uid

Solid/li=uid line to leftred!

Lo,er - all alongphase "oundary

Solute 6eeps solvent

in solution longer Must go to lo,er - to

form crystal

9(

$ree8in& 2oint epression an#-oilin& 2oint Elevation

Solution




Solution

+"serve BP and P over pure solvent Presence of solute& depresses P and elevates BP

-oth

T f an#

T + #epen# on relative amounts ofsolvent an# solute

Colli&ative properties

-oilin& 2oint Elevation (T + ) in "oiling point of solution vs1 pure solvent

$ree8in& 2oint epression (T f )

in freeAing point of solution vs1 pure solvent

9)

$ree8in& 2oint epression (

Tf )

Tf ? iBf m




Tf ? iB f m

,here

Tf ? (Tfp Tsoln)

m concentration in Molality

B f molal freeAing point depression constant

nits of RC/molal

.epend on solvent& see -a"le 0)1)

i num"er of particles per formula unit 0 for molecular compounds

9>

-oilin& 2oint Elevation (

T+)

T+ ? iB+ m




T+ ? iB + m

,here

T+ ? (Tsoln T+p)

m concentration in Molality

B + molal "oiling point elevation constant

nits of CIm

.epend on solvent& see -a"le 0)1)

i num"er of particles per formula unit 0 for molecular compounds

9*

Ta+le 13!3 B f an# B +



Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter 6E 9D

E! $ree8in& 2oint epression2stimate the freeAing point of a permanent type ofantifreeAe solution made up of 0 g ethylene




antifreeAe solution made up of 01 g ethylene

glycol& C(HD+(& MM D(15! and 01 g H(+ MM 071(!1

100 .0g C2 H

6O

2×

1mol C2 H

6O

2

62.0! g C2 H

6O

2

m= mol solute

kg solvent =1.611mol C 2 H 6O2

0.100 kg %ater

95

-f ' f m 017D RC/m ! U 0D100m

-f -fp -soln!)1 RC 1 RC K -soln

-soln 1 RC K )1 RC

01D00mol C(HD+(

0D100m C(HD+(

)1 RC

@3=!= C

our Turn; 4hen 1(* g of an un6no,n organic compound is

added to (*1 g of cyclohexane& the freeAing point



Jespersen/Brad /H slop Chemistr The Molec lar Nat re of Matter 6E

of cyclohexane is lo,ered "y 01D oC1 ' f for thesolvent is (1( oC m;01 .etermine the molar mass ofthe un6no,n1

81 ** g/mol

B1 )( g/mol

C1 )0* g/mol

.1 0(D g/mol

97

oo

- ' :1(*: g

C

01D CJ(:1( x :1:(* 6g

0(D g/mol

f f m

M4

m

M4

E! -oilin& 2oint Elevation 8 (1 g sample of a large "iomolecule ,as dissolvedin 0* g of CCl> -he "oiling point of this solution




in 0*1 g of CCl>1 -he "oiling point of this solution

,as determined to "e 5517* RC 1 Calculate the molarmass of the "iomolecule1 or CCl>& the B + *15

RC/m and BPCCl> 5D1* RC 1

Δ# != $ !m m=

Δ# !

$ !=

(!!."5−!6.50) ∘C 5.0! ∘C /m =0.26"4m

99

m=mol solute

kg solventmol solute=0.26"4m×0.0150kg CCl

4

¿4 . 0 2 6×10

−3

mol MM

!iomoleule=

2.00 g biomoleucle

4 . 0 2 6×10−3mole=4&! g /mol

earnin& Chec0 8ccording to the Sierra[ 8ntifreeAe literature&the freeAing point of a >/D solution of Sierra




the freeAing point of a >/D solution of SierraantifreeAe and ,ater is K > R1 4hat is themolality of the solutionF

2- solution=m× $ fp(0−(−20.)∘C = X ×

1."6∘C

m

0

R ? 11m

- 017 -

C I )(

K >R 017 X I )(

X K (1 RC

- -fp K -soln

earnin& Chec0 3n the previous sample of a Sierra[ antifreeAemixture& 0 mL is 6no,n to contain >( g of the




& g

antifreeAe and D1 g of ,ater1 4hat is the molarmass of the compound found in this antifreeAe if ithas a freeAing point of K >RF

mol solute=11 m×0 .060 $g solvent

00

rom "efore:

R ? 11 m

MM solute

? 69 &Imol

1DD mol solute

=42 g solute

0. 66 mol solute

m=mol solute$g solvent

earnin& Chec0 3n the previous sample of a Sierra[ antifreeAemixture the freeAing point is K >R 4hat ,ill




mixture& the freeAing point is K > 1 4hat ,ill"e its "oiling pointF

2- bp=m× $ bp

(0∘C −(−20∘C ))# boil

−100∘C =1."6∘C/m0.51 ∘C/m

0(

rom "efore:

K >R K (1 RC-reeing oint

-(oiling oint

= $ fp $ bp

=m

T+oilin& ?1=A C

-fp=m× $

fp

our Turn;Beer is 6no,n to "e around a *W ethanolC(H*+H! solution ,ith a density of 0 * g/mL




C(H*+H! solution ,ith a density of 01* g/mL1

4hat is its expected "oiling pointF ' "1*0R/m!

81 0ZCB1 00ZC

C1 0(ZC

.1 0)ZC21 %ot enough information given

0)

MM: H(+0710*)? C(H*+H>D1D9

Mem+ranes an# 2ermea+ilityMem+ranes

Separators




2x1 Cell ,alls 'eep mixtures organiAed and

separated

2ermea+ility 8"ility to pass su"stances

through mem"rane

Semipermea+le

Some su"stances pass& othersdon$t

Mem"ranes are semipermea"le

Selective0>

Mem+ranes an# 2ermea+ility .egree of permea"ility depends on type of

mem"rane




Some pass ,ater only

Some pass ,ater and small ions only

Mem"ranes separating t,o solutions of #ifferent concentration -,o similar phenomena occur

.epends on mem"rane

ialysis

4hen semipermea"le mem"rane lets "oth H(+ andsmall solute particles through Mem"rane called dialyAing mem"rane

'eeps out large molecules such as proteins0*

7smosis7smotic Mem+rane

Semipermea"le mem"rane that lets only solvent




molecules through7smosis

%et shift of solvent molecules usually ,ater!

through an osmotic mem"rane .irection of flo, in osmosis&

Solvent flo,s from dilute to more concentrated side

lo, of solvent molecules across osmotic mem"rane

concentration of solute on dilute side concentration of solute on more concentrated side

0D

7smosis an# 7smotic 2ressure




G! 3nitially& Soln B separated from pure ,ater& 8& "yosmotic mem"rane1 %o osmosis occurred yet

-! 8fter a ,hile& volume of fluid in tu"e higher1 +smosishas occurred1

C! %eed "ac6 pressure to prevent osmosis osmoticpressure1

05

7smotic 2ressure 2xact "ac6 pressure needed to prevent osmotic

flo, ,hen one li=uid is pure solvent1




= p

"hy #oes osmosis eventually stop% 2xtra ,eight of solvent as rises in column

generates this opposing pressure

4hen enough solvent transfers to solution so that,hen osmotic pressure is reached& flo, stops

3f osmotic pressure is exceeded& then reverseprocess occursYsolvent leaves solution

#everse osmosisYused to purify sea ,ater

07

Euation for 7smotic 2ressure 8ssumes dilute solutions




? iM/T osmotic pressure

i num"er of ions per formula unit

0 for molecules

M

molarity of solution

Molality& m & ,ould "e "etter& "ut M simplifies

2specially for dilute solutions& ,here m M

T 'elvin -emperature / 3deal @as constant

17(*5 Latmmol0' 0

09

Jant off Euation Since

M =mol

=n




Su"stitute into osmotic pressure e=uation

@et van$t Hoff 2=uation for osmotic

pressure

J ? in/T J volume in (

n moles 3dentical to 3deal @as La,

But ,ith 2 osmotic pressure!

L V

00

7smometer 3nstrument to measure osmotic pressure

ery important in solutions used for"iological samples




"iological samples

*sotonic solution Same salt concentration as cells

Same osmotic pressure as cells

ypertonic solution Higher salt concentration than cells

Higher osmotic pressure than cells

4ill cause cells to shrin6 and dehydrate

ypotonic solution Lo,er salt concentration than cells

Lo,er osmotic pressure than cells

4ill cause cells to s,ell and "urst000

E! 7smotic 2ressure2ye drops must "e at the same osmotic pressureas the human eye to prevent ,ater from moving




into or out of the eye1 8 commercial eye dropsolution is 1)(5 M in electrolyte particles1 4hat isthe osmotic pressure in the human eye at (* RC F

& =0.32! M ∗0.0"206 L⋅atm

$ ⋅mol ∗2&" $ =".00atm

00(

? M/T T '! (*RC I (5)10*

E! Hsin&

to #etermine MM-he osmotic pressure of an a=ueous solution of certainprotein ,as measured to determine its molar mass1 -he




solution contained )1* mg of protein in sufficient H(+ toform *1 mL of solution1 -he measured osmotic pressureof this solution ,as 01*> torr at (* RC 1 Calculate the molarmass of the protein1

M = &

'# =

1.54 torr 1 atm

!60 torr

(0.0"206 L⋅atm

$ ⋅mol )2&" $

=".2"×10−5 mol

L

mol = M ∗ L= ".2"×10−5 M )∗5.00×10−3 L=4.14×10−!mol

00)

MM = g

mol =

3.50×10−3 g

4.14×10−!mol =".45×103 g /mol

earnin& Chec0: 7smosis 8 solution of .*4& *W dextrose CDH0(D! in

,ater is placed into the osmometer sho,n at




right1 3t has a density of 01 g/mL1 -hesurroundings are filled ,ith distilled ,ater14hat is the expected osmotic pressure at(*RCF

& = 0. 2!! mol

L × 0. 0"205! L⋅atm

mol⋅ $ ×2&"

5g C6 H

12O

6

100g solution ×

1 . 0g soln

mL soln ×

mol C6 H

12O

6

1"0.16g ×

1000mL

L = M

00>

& =iM'# i 0 as dextrose is molecular

? atm

earnin& Chec0 or a typical "lood plasma& the osmotic pressure at"ody temperature )5RC! is *>9 mm Hg1 3f the




y p ! g

dominant solute is serum protein& ,hat is theconcentration of serum proteinF

!.11!atm=

7 mol

L ×

0. 0"205! L⋅atm

mol⋅ $ ×310.15

00*

540& mm 'g×

1atm

!60 mm 'g = &

& = M'#

M ? =!= M

our Turn;Suppose that your tap ,ater has (* pp"pp" 0/0&&& or 0U0 K9! of




pp / & & & !dissolved H(S & and that its density is a"out

01 g/mL1 4hat is its osmotic pressure at(*RCF

81 1*7 atm

B1 1D> atm

C1 1*9 atm

.1 107 atm

00D

MM: H(S )>15D

Colli&ative 2roperties of ElectrolyteSolutions iffer

B f H(+! 017D CIm




2xpect 01 m solution of %aCl to freeAe at017D C

8ctual freeAing point )1)5 C

G t,ice expected -

"hy % Colligative properties depend on

concentration num"er! of particles 0 %aCl dissociates to form ( particles

%aCl %aIa= ! I Cla= !

005

Colli&ative 2roperties of ElectrolyteSolutions epen# on Num+er 7f *ons

8ctual concentration of ions (1 m




Started ,ith 01 m %aCl! %o, use this to calculate T -f 017D RC/m (1 m )15( RC

+r T final ? T initial @ T f 1 K )15( RC

? @ 3! C

%ot exactly to actual Tf 3!3 C

-his method for ions gives rough estimate ifyou assume that all ions dissociate 0W1

007

"hy isnt this Eact for Electrolytes% 8ssumes 0W dissociation of ions

2lectrolytes don$t dissociate 0W& especially in




concentrated solutions Some ions exist in ion pairs

Closely associated pairs of oppositely charged ions that"ehave as a single particle in solution

So& fe,er particles than predicted

/esult: P. and BP2 not as great as expected

8s you go to more dilute solutions& electrolytes

more fully dissociated and o"serve P and BPcloser to calculated value1

Model ,or6s "etter at dilute concentrations

009

vant off $actor ? i Scales solute molality to correct num"er of

particles




Measure of dissociation of electrolytes van$t Hoff factor is e=uivalent to percent

ioniAation

3n general& it varies ,ith concentration see-a"le 0)1>& page D((!

i= ( Δ#

( )

measure*( Δ#

( )

calc* as nonelectrolyte

0(

Ta+le 13!9 vant off $actors vs!Concentration



Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 0(0

Note:

01 as concentration & io"served iexpected

(1 MgS+> much less dissociated than %aCl or 'Cl

"hy oes M&S79 issociate essThan NaCl or BCl%

MgS+> Mg(I a=! I S+>( a=!




(I/( ions rather than 0I/0 ions

Larger charge means greater attractive forces"et,een oppositely charged ions

So for Gl279 5oul# 5e epect more or less#issociation than M&S79%

8lP+> 8l)I a=! I P+>) a=!

Larger charges )I/)! expect greaterattractions and less dissociation

0((

Nonelectrolytes Some molecular solutes produce 5ea0er

colligative effects than predicted "y their molal




concentrations 2vidence of solute molecule clustering or

associatin&

#esult: only \ num"er of particles expected"ased on molality of solution& so Tf only \

,hat expected

,-H. C

,

, H

2 ,-H. C

,

, H

,-H.C

,

,H

0()

Nonelectrolytes

,-H. C

,

2 ,-H. C

,

,-H.C

,H




or

/esult: MM dou"le ,hat is expected

]particles^ \ ,hat is expected an#

Tf only \ ,hat expected

So siAe of solute particles is important Common ,ith organic acids and alcohols

, H , H ,

0(>

earnin& Chec0 3n preparing pasta& ( L of ,ater at (*C are com"ined,ith a"out 0* g salt %aCl& MM *71>>g/mol! and the

l ti " ht t " il 4h t i th t d " ili




solution "rought to a "oil1 4hat is the expected "oilingpoint of the ,aterF

# −100∘C =2 ion

mol ×

0 .123m

1 ×

0.51∘C

m

0(*

ΔT=i mK bpmass of ,ater volume U density ( mL U 01 g/mL

(g ,ater ( 6g

mol %aCl 0* g / *71>> g/mol 1(*DD5 mol

m%aCl 1(*DD5 mol / (6g 10() m

T ? 1==!1 C

Case Stu#ySuppose you run out of salt1 4hat mass ofsugar C0(H((+00& MM)>(1) g/mol! added to

( L f t ld i th t t f




( L of ,ater ,ould raise the temperature of,ater "y 10 RCF

0(D

ΔT=i mK bp

mass of ,ater volume U density ( mL U 01 g/mL

(g ,ater ( 6g

1)9(0* mol mass sucrose / )>(1) g/mol

109D m R / (6g

0.10∘C =1 ion

mol× X ×

0.51∘C

m R 109D m

R 1)9(0* mol

mass of sucrose ?13= &

Colli&ative 2roperties SummaryColli&ative properties #epen# on num+er

of particles i /




i mapp /mmolecular

#aoult$s La,

reeAing Point .epression

Boiling Point 2levation

+smotic Pressure

Must loo0 at solute an# see if molecular

or ionic 3f molecular& i 0

3f ionic& must include i > 1 in e=uations

0(5

Colli&ative 2roperties #aoult$s La,

P l ti X l t P l t

∘




reeAing Point .epression

T f ? iB f m

Boiling Point 2levation

T + ? iB + m

+smotic Pressure

? iM/T

P solution= X solvent P solvent

0(7

Brdy 6Ed Ch13 MixturesAtTheMolecularLevel

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Transcript of Brdy 6Ed Ch13 MixturesAtTheMolecularLevel