11-Lecture 11 (2015-07-21) (student version) (1).ppt

8/17/2019 11-Lecture 11 (2015-07-21) (student version) (1).ppt

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1

Housekeeping for Tue. 7/21/15- Exam 3 graded; scores posted to Back!oard; "i return at end of cass- Exam 2 corrections due tomorro" #$ed. 7/22%

- H$ 13 #and soutions% posted to Back!oard and &ooge 'ri(e- ) a of acetic acid "ork-up due toda*+

- Exam 3 corrections in pa*; due in cass next Monday #7/27%+


2/58

Titration

, in an acid-!ase titration a soution of kno"n concentration

#titrant% is so"* added to a soution of unkno"n concentration#analyte% from a !urette unti te reaction is compete

"en te reaction is compete "e a(e reaced te endpoint ofte titration

,an indicator ma* !e added to determine te endpoint an indicator is a cemica tat canges coor "en te pH

canges

, "en te moes of H301 moes of 0H te titration as reaced

its equivalence point

, E4ui(aence point not te same as end point+

E4ui(aence point is te theoretical point "en moes acid moes !ase.

End point is experimental point tat approximates e4ui(aence point.


3/58

Titration


4/58

6ndicators

, man* d*es cange coor depending on te pH of te soution

, tese d*es are "eak acids esta!ising an e4uii!rium "it te H20

and H301 in te soution

H-6nd#aq% H20#l %⇔ 6nd−#aq% H301#aq%

) a 6nd-8H3018 / H-6nd8

• p) a of indicator is pH "ere indicator canges coor

• $en using an indicator to monitor a titration "e "ant to matc te

p) a of te indicator "it te pH of te e4ui(aence point as cose*

as possi!e.


5/58

5

9enoptaein

5


6/58

:

et* > > >

H

>a00=

=

= =H

=H

=H

=H

=

=H

=H

=

=H

=H

#=H3%

2 > > > >

>a00=

H301 0H-


7/587

atcing indicators "it titrations

Te e4ui(aence point for te titration of H= "it >a0H occurs at pH 7.

$ic indicators "oud !e suita!e for tis titration?


8/58

T*pes of titrations

@our genera t*pes of titrationsA

- strong acid / strong !ase #!ase added to acid; !ase titrant%

- strong !ase / strong acid #acid added to !ase; acid titrant%

- "eak acid / strong !ase

- "eak !ase / strong acid

titrations sare common traitsC

- pH !efore titrant is added

- pH !efore e4ui(aence point is reaced

- pH at e4ui(aence point #acid or !ase is entire* neutraiDed+%

- pH after e4ui(aence point #excess titrant%


9/58

trong acid / strong !ase titration #>a0H added to H=%

0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

initia pH

pH !efore e4ui(aence point

e4ui(aence

point

pH after e4ui(aence point


10/58


=ase stud*A F.5 >a0H is added to 2FF.F mG of F.5 H=

9ase 1A $at is te initia pH?


11/58



9ase 2A $at is te pH "en 1FF mG of >a0H a(e !een added?

0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H


12/58




uick detourC Gooking at te titration cur(e "e can see tat "e

a(enIt reaced te e4ui(aence point. But "e usua* donIt a(e a

titration cur(e to refer to. 6n fact "e donIt rea* need te titrationcur(e to gauge "ere "e are in te titration cur(e.

$at "e need to kno" isC

- te tota moes of acid !efore "e start to titrate

- te (oume of !ase "e need to add to reac te e4ui(aence point

#in oter "ords te (oume of !ase to react compete* "it a te

acid%


13/58



9ase 3A $at is te pH at te e4ui(aence point?

0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H


14/58



9ase A $at is te pH "en 3FF.F mG #tota% of >a0H a(e !eenadded?

0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H


15/58


Jo. >a0H #mG% pH

F F.3

1FF F.7K

2FF 7

3FF 13

35F 13.

Tings to keep in mind "en "orking a titration pro!emC

- tota L of moes of acid #or !ase%

- L of moes of acid #or !ase% eft after addIn of titrant; eps *ou

decide if *ouI(e reaced te e4ui(aence point *et

- at e4ui(aence point no moes of acid #or !ase% eft+

- after e4ui(aence point pH is due to excess titrant


16/58

0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

$eak acid / strong !ase titration #>a0H added to H=2H302%

initia pH

pH !efore e4ui(aence point; *buffering region!*

e4ui(aence

point

pH after e4ui(aence point

1:


17/58


=ase stud*A F.5 >a0H is added to 2FF.F mG of F.5 H=2H302


17


18/58



9ase 2A $at is te pH "en 1FF mG of >a0H are added?

0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

1K


19/58



1M


uick detourC Gooking at te titration cur(e "e can see tat "e

a(enIt reaced te e4ui(aence point. gain "e usua* donIt a(e

a titration cur(e to refer to. nd again "e donIt rea* need te

titration cur(e to gauge "ere "e are in te titration cur(e.

$at "e need to kno" isC

- te tota moes of acid !efore "e start to titrate

- te (oume of !ase "e need to add to reac te e4ui(aence point#in oter "ords te (oume of !ase to react compete* "it a te

acid%

- this is exactly the same as what we did with the strong acid /

strong base titration


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0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

Haf-e4ui(aence

point+

t af-e4ui(aance point acid8 conNugate !ase8.

pH p) a+

Buffering region is p) a

/- 1.

Buffering region

2F

k id / ! i i # dd d %


21/58




0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

E4ui(aence point

21

k id / ! i i # 0 dd d = 0 %


22/58




0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

E4ui(aence point

>otice tat te pH is not 7+ Tis

is !ecause of te e4uii!rium of

acetate as a "eak !ase at te

e4ui(aence point.

22

$ k id / ! i i #> 0H dd d H= H 0 %


23/58

9ase A $at is te pH "en 3FF.F mG #tota% of >a0H a(e !een

added?



0

2

4

6

8

10

12

14

0 100 200 300 400

Vol. NaOH (mL)

p H

23

$ k id / ! i i #> 0H dd d H= H 0 %


24/58

Jo. >a0H #mG% pH

F 2.521FF .7

2FF M.1

3FF 13

35F 13.

Tings to keep in mind for a "eak acid/strong !ase titrationC

- tota L of moes of acid #or !ase%

- "at e4uii!rium is taking pace in te !uffer region? $eak acid or "eak !ase?

- "at e4uii!rium is taking pace at te e4ui(aence point? $eak

!ase or "eak acid?

- !e*ond e4ui(aence point cacuations are same as strong acid/

strong !ase titration


2


25/58

25

atcing indicators "it titrations

Te e4ui(aence point for te titration of H=2H302 "it >a0H occurs at O

pH M. $ic indicators "oud !e suita!e for tis titration?


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outions to exercises


27/58




What’s going on?

Te initia pH is te pH of te soution !efore "e add an* titrant.

ince H= is a strong acid te pH is Nust te *dronium

concentration.

H308 H=8 F.5

pH -og #F.5 %

0.


28/58




What’s going on?

Before "e start our cacuations etIs Nust do a 4uick detour to

determine te total num!er of moes of acid tat *ouIre starting"it. Tis "i te us t"o tingsA

- "at (oume of !ase "e need to reac te e4ui(aence point;

- o" muc acid is eft o(er after "eI(e added our 1FF mG of !ase6nitia moes of H= H=8 x Jo. H=

#F.5 moes / G% x #F.2FF G%

F.1 moes 1FF mmoes


29/58




What’s going on?

rigt so o" muc !ase do "e need to add to react

compete* "it 1FF mmoes of our acid? Te neutraiDationreaction so"s tat >a0H reacts 1A1 "it H=C

H= >a0H→ >a= H20

o "e need 1FF mmoes of !ase. $eI use te moarit* of >a0H todetermine te (oumeCF.1FF moes >a0H x #1 G / F.5 moes% F.2 G 00 m"


30/58




What’s going on?

rigt+ >o" "eIre read* to cacuate te pHC

@rom te pre(ious cacuation "e determined tat "e need to add

2FF mG of our !ase to reac te e4ui(aence point. 6n pase 2

"eI(e on* added 1FF mG of !ase. #o "e a(enIt reaced te

e4ui(aence point. $eI sti a(e H= eft o(er.

#o pH "i !e due to te *dronium from te remaining H=.

Ho" muc acid "i !e eft o(er?


31/58




$alculations%

oes of >a0H added >a0H8 x Jo. >a0H


F.F5 moes 5F mmoes

ince H= reacts "it >a0H in a 1A1 manner 5F mmoes of H=

are neutraiDed.

oes H= remaining 1FF mmoes - 5F mmoes 5F mmoes


32/58




$alculations%

0) amost tere. >o" "e Nust need to cacuate te pH !ased on

te amount of remaining *dronium.

H308 moes remaining acid / new (oume+

$e need to keep track of our (oumeA

initia (oume 2FF mG

1FF mG of >a0H added

>e" (oume #2FF 1FF% mG 00 m"


33/58




$alculations%

0) amost tere. >o" "e Nust need to cacuate te pH !ased on

te amount of remaining *dronium.

H308 moes remaining acid / new (oume+

5F mmoes / 3FF mG

F.17

pH -og #F.17 %

0.&'


34/58




What’s going on?

t te e4ui(aence point te amount of !ase is e4ua to te amount

of acid. Tat is moes !ase moes acid.

(ll the acid has been neutrali)ed!

Te on* source of acidit* is from te autoioniDation of "ater.

) " H3080H-8 1.F x 1F-1

ince H308 0H-8 for pure "ater H30

8 1.F x 1F-7

pH -og #1.F x 1F-7% &


35/58




@or te titration of a strong acid "it a strong !ase #or (ice (ersa%

te pH at te e4ui(aence point "i always !e 7.

Tis is not te case for "eak acids and "eak !ases as "eI seesort*.


36/58




What’s going on?

t tis point "eIre adding excess !ase. Te pH "i !e due to te

excess *droxide.

Ho" muc excess *droxide do "e a(e?

Tere are t"o "a*s "e can figure tis outA

- cacuate te tota moes of !ase tat a(e !een added; su!tract

te amount of !ase needed to neutraiDe te acid

- or cacuate te (oume of !ase re4uired to neutraiDe te acid;

te excess !ase "i come from te (oume of !ase tatIs eft o(er


37/58




$alculationsA

-'etermining amount of excess !aseC

$e started "it 1FF mmoes of acid. o "e need 1FF mmoes of

!ase to neutraiDe te acid.

moes of !ase in 3FF mG F.5 >a0H ?

moes >a0H >a0H8 x Jo. >a0H


F.15F moes or 15F mmoes


38/58




$alculationsA

Excess moes !ase 15F mmoes 1FF mmoes 5F mmoes

0H-8 moes excess !ase / new (oume

$e started "it 2FF mG of acid; "eI(e added 3FF mG of !aseC

new (oume #2FF 3FF% mG 00 m"0H-8 5F mmoes / 5FF mG F.1

p0H -og #F.1 % 1

pH 1 - p0H 1 - 1 +

k id / ! i i # 0 dd d = 0 %


39/58




What’s going on?

Te initia pH is te pH of te soution !efore "e add an* titrant.

ince acetic acid is a "eak acid it exists in e,uilibrium !et"een its

acid form and its conNugate !ase formA

H=2H302 H20 PQ =2H302- H30

Te pH is determined !* te e,uilibrium concentration of*dronium.

3M

$ k id / ! i i #> 0H dd d H= H 0 %


40/58




$alculations%

et up an 6=E ta!e "it F.5 as te initia concentration of acetic

acid and F as te initia concentration of acetate and *dronium

H=2H3028 =2H302-8 H30

8

6nitia F.5 F F

=ange -x x x

E4uii!rium F.5 - x x x

F

$ k id / ! i i #> 0H dd d H= H 0 %


41/58




$alculations%

et up e4uii!rium expressionA

) a =2H302-8H30

8 / H=2H3028

1.K x 1F-5 x#x% / #F.5 x%; in(oke Rx is smaS approximation

x2 F.5 x #1.K x 1F-5%

x 3 x 1F-3

H308 3 x 1F-3

pH -og H308 -og #3 x 1F-3 % .

1

$ k id / ! i i #> 0H dd d H= H 0 %


42/58




What’s going on?

Before "e start our cacuations etIs Nust do a 4uick detour to

determine te total num!er of moes of acid tat *ouIre starting"it. Tis "i te us t"o tingsA

- "at (oume of !ase "e need to reac te e4ui(aence point;

- o" muc acid is eft o(er after "eI(e added our 1FF mG of !ase

6nitia moes of H=2H302 H=2H3028 x Jo. H=2H302


F.1 moes 1FF mmoes

$ k id / t ! tit ti #> 0H dd d t H= H 0 %


43/58



What’s going on?

rigt so o" muc !ase do "e need to add to react

compete* "it 1FF mmoes of our acid? Te neutraiDationreaction so"s tat >a0H reacts 1A1 "it H=2H302C

H=2H302 >a0H→ >a=2H302 H20

o "e need 1FF mmoes of !ase. $eI use te moarit* of >a0H todetermine te (oumeCF.1FF moes >a0H x #1 G / F.5 moes% F.2 G 00 m"


t id / t ! tit ti #> 0H dd d t H=%


44/58



What’s going on?

rigt+ >o" "eIre read* to cacuate te pHC

@rom te pre(ious cacuation "e determined tat "e need to add

2FF mG of our !ase to reac te e4ui(aence point. 6n pase 2

"eI(e on* added 1FF mG of !ase. #o "e a(enIt reaced te

e4ui(aence point. $eI sti a(e H=2H302 eft o(er.

#o pH "i !e due to te *dronium from te remaining

H=2H302


$eak acid / strong !ase titration #>a0H added to H= H 0 %


45/58


What’s going on?

H=2H302 is !eing neutraiDed !* >a0HA

H=2H302 >a0H→ >a=2H302 H20

$e sti a(e H=2H302 eft o(er - !ecause "e a(enIt reaced te

e4ui(aence point.

>eutraiDation of acetic acid creates its conNugate !ase acetate.We’-e created a buffer!

*#o we can use the endersonasselbach e,uation to

determine the p!*



5



46/58


$alculations%

Tota moes acetic acid acetic acid8 x Jo. cetic acid

#F.5 % x #F.2 G%

1FF mmoes

oes >a0H added >a0H8 x Jo. >a0H

#F.5 % x #F.1 G%

5F mmoes



:



47/58


$alculations%

H=2H302 >a0H→ >a=2H302 H20

oes >a0H moes acetate formed moes acetic acid consumed

o moes acetate 5F mmoes

oes acetic acid remaining 1FF mmoes 5F mmoes

5F mmoes



7



48/58


$alculations%

pH p) a og #-8 / H8%

-og #1.K x 1F-5% og #5F mmoes acetate / 5F mmoes

acetic acid%

.7 og #1%

1.&1



$e can repace te concentration terms "it

L of moes !ecause te (oumes in eac of teconcentrations "i cance eac oter out+

K



49/58


What’s going on?

H=2H302 >a0H→ >a=2H302 H20

t te e4ui(aence point all te acid as !een neutraiDed. Tat is

a te acetic acid as !een con(erted to acetate.

2ather than pure water we ha-e a solution of a wea3 base 4the

acetate anion56

=2H302- H20 PQ H=2H302 0H

-

$eI set up an 6=E ta!e to determine te e4uii!rium

concentration of *droxide.



M



50/58


$alculations

=2H302- H20 PQ H=2H302 0H

-

$atIs te concentration of acetate?

oes acetate 1FF mmoes F.1 moes

Jo. oution ?

6nitia (o. 2FF mG

$at (o. of >a0H a(e "e added to get to e4ui(aence point?

>a0H8 moes >a0H / Jo. >a0H

Jo. >a0H moes >a0H / >a0H8 F.1 moes / #F.5 moes/G%

F.2 G 2FF mG

g # 2 3 2%


5F



51/58


$alculations

=2H302- H20 PQ H=2H302 0H

-

$atIs te concentration of acetate?

oes acetate 1FF mmoes F.1 moes

Jo. oution ?

6nitia (o. 2FF mG

2FF mG of F.5 >a0H added to reac e4ui(aence point

>e" (oume 2FF mG 2FF mG 100 m"

=2H302-8 F.1 moes / F. G 0.25 M

g # 2 3 2%


51



52/58


$alculations

=2H302- H20 PQ H=2H302 0H

-

=2H302-8 F.1 moes / F. G F.25

>o" "e can set up our 6=E ta!eC

g # 2 3 2%


=2H302-8 H=2H3028 0H

-8

6nitia F.25 F F

=ange -x x x

E4uii!rium F.25 - x x x

52



53/58

$alculations%

Before "e set up our e4uii!rium expression "e need to remem!er

tat "eIre deaing "it te e4uii!rium of a "eak !ase in "ater.

What’s the e,uilibrium constant?

$e need te ) ! for acetate as a "eak !ase. But o" do "e get ) !?

$eI use te reationsip !et"een ) " ) a acid ) ! conNugate !ase A

) " ) aacid x ) ! conNugate !ase

) ! conNugate !ase ) " / ) a acid

) ! acetate ) " / ) a acetic acid; ) a acetic acid 1.K x 1F-5

-1 -5 -10


g # 2 3 2%


53



54/58

$alculations%

et up e4uii!rium expressionA

) ! H=2H30280H-8 / =2H302

-8

5.: x 1F-1F x#x% / #F.25 x%; in(oke Rx is smaS approximation

x2 F.25 x #5.: x 1F-1F%

x 1.2 x 1F-5

0H-8 1.2 x 1F-5

p0H -og 0H-8 -og #1.7 x 1F-5 % .M

pH 1 - p0H 1 - .K 7.+


g # 2 3 2%


5



55/58


added?What’s going on?

t tis point "eIre adding excess !ase. Te pH "i !e due to te

excess hydroxide.

$at a!out te acetate? $* isnIt it contri!uting?

ddition of excess *droxide "i pre-ent acetate from con(erting

to its conNugate acid form. $e can expain tis using Ge =teierIs

9rincipe. =2H302

- H20 PQ H=2H302 0H

-

g # 2 3 2%


55



56/58


added?What’s going on?

ince te pH is due to te excess *droxide "e a(e te exact

same scenario as "e did "it te titration of H= "it >a0H. Te

cacuations are exact* te same. @or competeness 6I repeattem.

Ho" muc excess *droxide do "e a(e?

Tere are t"o "a*s "e can figure tis outA- cacuate te tota moes of !ase tat a(e !een added; su!tract

te amount of !ase needed to neutraiDe te acid

- or cacuate te (oume of !ase re4uired to neutraiDe te acid;

te excess !ase "i come from te (oume of !ase tatIs eft o(er

g # 2 3 2%


5:



57/58


added?$alculationsA

-'etermining amount of excess !aseC

$e started "it 1FF mmoes of acid. o "e need 1FF mmoes of !ase to neutraiDe te acid.

moes of !ase in 3FF mG F.5 >a0H ?

moes >a0H >a0H8 x Jo. >a0H


F.15F moes or 15F mmoes

g # 2 3 2%


57



58/58


added?$alculationsA

Excess moes !ase 15F mmoes 1FF mmoes 5F mmoes

0H-8 moes excess !ase / new (oume$e started "it 2FF mG of acid; "eI(e added 3FF mG of !aseC

new (oume #2FF 3FF% mG 00 m"

0H-8 5F mmoes / 5FF mG F.1

p0H -og #F.1 % 1

pH 1 - p0H 1 - 1 +

g 2 3 2=ase stud*A F.5 >a0H is added to 2FF.F mG of F.5 H=2H302

11-Lecture 11 (2015-07-21) (student version) (1).ppt

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Transcript of 11-Lecture 11 (2015-07-21) (student version) (1).ppt