Chap 17 No Titrate Calc No Clicker(2)

8/12/2019 Chap 17 No Titrate Calc No Clicker(2)

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Chapter 17:

Additional AqueousEquilibria

Chemistry the Central Science,12thed

Dr. Stacey Gulde

1


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Common-Ion Effect Common ion effectshiftin equilibrium caused

by adding anion involvedin the equilibrium

AKA Le ChteliersPrinciple

If add NaC2H3O2, what happens?

C2H3O2-causes equilibriumto shift Thus, [H+]decrease& pH becomes morebasic

Dissociationof a weakelectrolyte (ie. HC2H3O2) decreaseswhen a strongelectrolyte (ie. NaC

2

H3

O2

) containing acommon ion (ie. C2H3O2

-) is added2

)()((aq) 232232 aqOHCaqHOHHC

C2H3O2-Na+

breaks up into ions


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Sample Problems1. Calculate the pH of a 0.50M HF solution.

3

)(aqHFI

C

E

0.50M 0 0

xx-x

0.50x xx

4108.6 x]50.0[

]][[

x

xxKa

]50.0[

][

108.6

24 x

x

Mx 018.0 ][

H

100]50.0[

]018.0[% xI %6.3% I

(Assumption OK)

]018.0log[pH

74.1pH

Check 5% Rule:

Appendix D

Ignore x

(W. Acid)

)()( aqFaqH


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Sample Problems2. Calculate the pH of a 0.50M HF solution when 0.10M

NaF is added.

4

4108.6 x]50.0[]10.0][[

xxxKa

]50.0[

]10.0][[108.6 4

xx

Mx 0034.0 ][

H

100]50.0[

]0034.0[% xI

%68.0% I

]0034.0log[pH

47.2pH

More basic

)()()( aqFaqHaqHF

I

C

E

0.50M 0 0.10M

xx-x

0.50x 0.10+xx

Ignore x

(W. Acid)

(Common

Ion)

[HF] only:pH=1.74


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Buffer

Solutions Buffersweakacid/base & conjugateacid/base (salt)

Should have similarconcentrations

Sample Prob. #2: 0.50M HF with 0.10M NaF

Naturalbuffers:

Human blood maintains pH of 7.4

Seawatermaintains pH of 8.1-8.3

CalculatingpH of a buffer:

Previous sample problem #2 (pH=2.47)

OR use Henderson-Hasselbalch eq. (H-H eq)

5

)()()( aqFaqHaqHF (W. Acid) (CB)


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Henderson-Hasselbalch Eq

Sample Problem #2 Revisited: Calculate pH of a0.50M HF solution when 0.10M NaF is added.

6

acidbase

pKpH a log

)()()( aqFaqHaqHF Ka= 6.8x10-4

4108.6log xpH

47.2

pH

50.0

10.0log

70.017.3

0.50M 0.10M


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Adding Strong Acids/Bases

to a Buffer Solutions Buffer Purpose: resistpHchange when limited

amountsof strongacid or base is added

Lab Exp. 18

How happen:

Addedacid or base reactsw/something in the buffer

Addedacid or base is limiting

Ex. Buffer:

7

Buffer Demo

HA A-

Buffered soln

)()()( aqAaqHaqHA (W. Acid) (CB)
http://www.youtube.com/watch?v=g_ZK2ABUjvAhttp://www.youtube.com/watch?v=g_ZK2ABUjvA


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Adding a substance:

1. H+added: REACTS w/ ____ in buffer

Producing more ____

2. OH-added: REACTS w/ ____ in buffer

Producing more ____

8

HA A-

Buffered soln

)()()( aqAaqHaqHA (W. Acid) (CB)

H+(aq) +A-(aq) HA(aq)

HA A-

Buffered soln

OH-(aq) + HA(aq) A-(aq)

HA A-

HA A-

A-

HA

HA

A-

+H2O(l)


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Adding Strong Acid/Base

to a Buffer Solution If the addedacid/base is strong, it isALL used up

To calculate pH you use:a) ICE (must use MOLES!)

b) H-Heq.(must use M)

9


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Sample Problems1. Calculate the pH of a 1.0L solution containing 0.10M NaF &

0.50M HF, when 9.5mL of 2.1M HCl is added.(assume minimal/no volume change)

HF(0.50M)

NaF(0.10M)

&

1.0L Solnalready made

BUFFER!

Buffer:

0.50M 0.10MH+(aq) + F-(aq)HF(aq)ADD HCl

H+(aq)

RXN:

+ F-(aq)HF

(aq)

Added: H+(aq) reacts w/

W. Acid!


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0.50M HF, when 9.5mL of 2.1M HCl is added.(assume minimal/no volume change)

+0.020-0.020

0.08mol 0.52mol

-0.020

0

I

C

E

a) ICE Stoichiometry: (MOLES)Change by lower reactant value

(0.10M)(1.0L)(2.1M)(0.0095L)

0.10mol0.020 mol

(0.50M)(1.0L)

0.50mol

1.0L 1.0L

b) H-H: (M)

AcidBase

pKpH a log

4108.6log xpH

81.017.3 pH

36.2pH

52.008.0log

H+(aq) + F-(aq) HF(aq)

RXN:


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0.50M HF, when 9.5mL of 2.1M NaOH is added.(assume minimal/no volume change)

12

Buffer=

a) ICE Stoich.: (MOLES) b) H-H eq (M)

0

+0.0200-0.020

0.48mol 0.12mol0

-0.020

0

I

C

E

(0.50M)(1.0L)(2.1M)(0.0095L)

0.50mol0.020 mol

(0.10M)(1.0L)

0.10mol

1.0L 1.0L

0.50M 0.10M

AcidBase

pKpH a log

4108.6log xpH48.012.0log

60.017.3 pH

57.2pH

OH-(aq)+ HF(aq) +F-(aq)H2O()

H+(aq) + F-(aq)HF(aq)

Added: OH-(aq) reacts w/


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pH Comparison 0.50M HF: pH = 1.74

Buffer0.50M HF & 0.10M NaF: pH = 2.47 pH inc. (more basic) due presence of a common ion

Buffer0.50M HF & 0.10M NaF w/added9.5mL of

2.1M HCl: pH = 2.35 pH dec. (more acidic) due additionof acid

Buffer0.50M HF & 0.10M NaF w/added9.5mL of2.1M NaOH: pH = 2.59

pH inc. (more basic) due additionof base 13


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Sample Problem(What ifVolume changes?)

1. Calculate the pH of a solution formed by mixing 65mL of0.20M NH4Cl with 75mL of 0.15M NH3. W. Base!

NH3

NH4Cl

NH3

NH4Cland

75mL0.15M65mL

0.20M

65mL+75mL

??M

Buffer:

To Find Initial:Convert original M to moles,

then back toMusing Total Vol

=140.mL


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1. Calculate the pH of a solution formed by mixing 65mL of0.20M NH4Cl with 75mL of 0.15M NH3.

BUFFER, MUSTuse ICE table!!

15

)(NH4 aq

0-x

0.093+x0

+x

0.080-x

I

C

E

(0.20M)(065L)(0.15M)(0.075L)

0.013mol0.01125 mol0.140L

)(NH3 aq O(l)H2

0 0

0.140L

0.080M 0.093M

+x

x

5

108.1

x]080.0[

]][093.0[

x

xx

Kb

]080.0[

]][093.0[108.1

5 xx

Mxx5

105.1

][

OH

W. Base!

)(OH aq


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1. Calculate the pH of a solution formed by mixing 65mL of0.20M NH4Cl with 75mL of 0.15M NH3.

BUFFER, MUSTuse ICE table!!

16

)(NH4 aq

0-x

0.093+x0

+x

0.080-x

I

C

E

(0.20M)(065L)(0.15M)(0.075L)

0.013mol0.01125 mol0.140L

)(NH3 aq O(l)H2

0 0

0.140L

0.080M 0.093M

+x

x

W. Base!

)(OH aq 100080.0

105.1%

5

xx

I

%019.0% I

]105.1log[5

xpOH

82.4pOH

82.400.14 pH

18.9pH


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Sample Problem (Volume changes?)2. Calculate the pH of a solution formed by mixing 65mL of

0.20M NH4Cl with 75mL of 0.15M NH3& adding 15.3mL

of 0.17M NaOH.

17

O(l)H)(NH 23 aq

-0.0026-0.0026

0.01385mol0mol

+0.0026

0.0104mol

I

C

E

(0.15M)(075L)(0.20M)(0.065L)

0.01125mol0.013 mol

0.155L

)(NH4 aq )(OH- aq

0

VT=65mL+75mL+15.3mL=155.mL

0.155L

0

0

a) ICE Stoich: (MOLES)

(0.17M)(0.0153L)

0.0026 mol

0.00671M 0.0894M

b) H-H: (M)

AcidBase

pKpH a log

14

101

xKK ba

145

101108.1

xxKa

10106.5 xKa

Try on ownBuffer=


0.15M 0.20MNH4

+(aq) + OH-(aq)NH3(aq)+ H2O(aq)

(NH3is a base, K

b= 1.8x10-5)

Must solve for Ka


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Sample Problem (Volume changes?)2. Calculate the pH of a solution formed by mixing 65mL of

0.20M NH4Cl with 75mL of 0.15M NH3& adding 15.3mL

of 0.17M NaOH.

18

O(l)H)(NH 23 aq

-0.0026-0.0026

0.01385mol0mol

+0.0026

0.0104mol

I

C

E

(0.15M)(075L)(0.20M)(0.065L)

0.01125mol0.013 mol

0.155L

)(NH4 aq )(OH- aq

0

VT=65mL+75mL+15.3mL=155.mL

0.155L

0

0

a) ICE Stoich: (MOLES)

(0.17M)(0.0153L)

0.0026 mol

0.00671M 0.0894M

b) H-H: (M)

AcidBase

pKpH a log

Try on ownBuffer=


0.15M 0.20MNH4

+(aq) + OH-(aq)NH3(aq)+ H2O(aq)

125.025.9 pH

0671.00894.0

log

38.9pH

10

106.5log

xpH


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Buffer Capacity Buffer Capacityamount of acid/baseadded

before a significantpH change occurs

In demo video: needed to count drops of NaOHadded Took ~30 drops to turn green

Took ~35 drops to turn blue/violet

Depends on conc.of acid/base used to prepare buffer Which has a greater buffering capacity?

19

1 MHF & 1 M NaF0.1M HF & 0.1 M NaF

More initialacid/baseto react with

addedacid/base


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Buffer Range Buffer range - pHrange where buffer is effective

Want conc.of acid/base & conjugate to be similar

Ex: 0.50M HF & 0.10M NaF

If conc of 1 component in buffer is > 10x the conc of theother, buffering action is poor

Choose an acidwhose pKais close to the desired pH

20

pKaofunitpH1Range


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Questions:1. Which of the following will have the worstbuffering

range?

A. 0.1M HA and 0.1M A-

B. 0.2M HA and 0.1M A-

C. 0.1M HA and 1.5M A-

D. 1.0M HA and 1.0M A-

21

Difference ismore than 10x


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Questions:2. You wish to make a buffered solution with a pH=7.0,

which acid will be the best to use?

A. HNO2: Ka = 4.5x10-4 pKa= 3.35

B. HClO: Ka= 3.0x10-8 pKa= 7.52

C. H3PO4: Ka= 7.5x10-3 pKa= 2.12

D. H2CO3: Ka= 4.3x10-7 pKa= 6.37

E. HBrO: Ka= 2.5x10-9 pKa= 8.60

22


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Acid/Base Titration Introduced in Chapter 4

Indicatorschemicals that changecolorwhen a solution turns fromacidic to basic

Phenolphthalein

Acid = Clear

Base = Pink

23


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Acid/Base Titration Titration CurvepHvs. volumeof titrantadded

Titrant- liquid in buret

Equivalence Pointstoichiometric# of moles of acid =

# mols base reacted

Endpointwhere indicator changescolor permanently

Just past the equivalence point

24

TitrationCurve

Equiv. pt

Endpt
http://www.youtube.com/watch?v=yirkozUyG74http://www.youtube.com/watch?v=yirkozUyG74http://www.youtube.com/watch?v=yirkozUyG74http://www.youtube.com/watch?v=yirkozUyG74http://www.youtube.com/watch?v=yirkozUyG74


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pH at Various

Titration Points 2 types of titrations

1. Strong/Strong

2. Weak/Strong

What is happening to pH @ 4 different points

A. InitialpH

B. Beforeequivalence pt

C. @equivalence pt

D. Afterequivalence pt

25


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pH at VariousTitration Points

1. Strong Acid

&Strong Base

26

1 Strong Ac Strong Base:


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1. StrongAc - StrongBase:(NaOH added to HCl)

A. Initial pH:beforeadding base

NoNaOH added, onlyHCl present

pH =Very acidicB. Beforeequivalencept:

HCl & NaOH react

Baselimiting

Acidexcess

pH=slowlymore basic,b/c less acid in excess

Raises rapidlynear

equivalence pt.27

)(l(aq)a)(a)( 2 lOHCNaqOHNaqHCl

A

100mL0.50M

?mL0.10M

pH=0.30

BpH=0.44

pH=2.17

1 Strong Acid Strong Base:


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C. At equivalencept: HCl & NaOH react

Moles acid=Moles of base

CompletelyNeutralized pH = 7.00

How get pH when only

NaCl & H2O present?- Waterdissociates

28

C

H2O (l) H+(aq) + OH-(aq) 14101 xKw

M101]OH[]H[ 7 x

1. Strong Acid-Strong Base:(NaOH added to HCl)

pH=7.00

)(l(aq)a)(a)( 2 lOHCNaqOHNaqHCl 100mL0.50M

?mL0.10M

1 Strong Acid Strong Base:


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D. After equivalencept: HCl & NaOH react

Acid limiting

Completely consumed

Base excess

pH = slowly more basic

29

D

1. Strong Acid-Strong Base:(NaOH added to HCl)

)(l(aq)a)(a)( 2 lOHCNaqOHNaqHCl 100mL0.50M

?mL0.10M

pH=11.82pH=13.20

1 St A id St B


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mL NaOH pH

0 0.30

10.0 0.44

49.0 2.17

50.0 7.00

51.0 11.82

80.0 13.22

1. StrongAcid - StrongBaseSummary: (NaOH is added to HCl)

Why does pH change drasticallynear equivalence point ()?

pH dependent on MOLESpresentAFTERreaction!

1 St A id St B


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At equivalencepoint: OHH

nn1

2

3

4

MOLESpresentAFTERreacting0.050

0.040

0.001

00

0

0

0

0

0

0.001

0.030

OHNaClNaOHHCl 2

1. StrongAcid - StrongBaseSummary: (NaOH is added to HCl)

StepmL

NaOH

n H+

(after rx n)

n OH-

(after rxn )

1 0

2 10.0

2 49.0

3 50.0

4 51.0

4 80.0

What does a


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What does aStrong Base-Strong Acid

Titration Look Like?


33/63

Question: Where is the end point for the following titration

curve?

33


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pH at VariousTitration Points

2. Weak Acid

&

Strong Base

34

2 Weak Acid - Strong Base:


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A. Initial pH:beforeadding base

NoNaOH added, onlyHC2H3O2present

pH =acidic

35

A

)()((aq)(aq) 2232232 lOHaqOHCOHOHHC

2. WeakAcid - StrongBase:(NaOH added to HC2H3O2)

100mL0.50M

?mL0.10M

pH=2.52



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B. Beforeequivalence pt: HC2H3O2& NaOH react

Baselimiting

Acidexcess pH =slowlymore basic,

b/c less acid in excess

Raises rapidlynear

equivalence pt.

A BUFFER soln!

HC2H3O2& conjugate

base (C2H3O2-

) present 36

B

OHOHCOHOHHC 2232232


100mL0.50M

?mL0.10M

pH=5.69



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C. At equivalencept: HC2H3O2 & NaOH react

Moles acid=Moles of base

NOT completely neutralized!!

Have conjugatebase (C2H3O2-)

pH > 7

37

C

)(OH)(OHHCO(l)H)(OHC 2322232 aqaqaq

OHOHCOHOHHC 2232232


100mL0.50M

?mL0.10M

pH=9.25



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D. After equivalence pt: HC2H3O2 & NaOH react

Acidlimiting

Completely consumed Baseexcess

pH = gets more basic

D

OHOHCOHOHHC 2232232


100mL0.50M

?mL0.10M

pH=11.82

pH=13.20


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0 20 40 60 80 1000

2

4

6

8

10

12

14

pH

Volume of NaOH added (mL)

Titration Curve

Comparisons1. Initial pH:

S/Sis lower, more acidic

DissociatescompletelymoreH+in soln

2. After Equivalence:

Curves are the same Amount of excess

OH-is the same

W. acid S. base

S. acid S. base


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0 20 40 60 80 1000

2

4

6

8

10

12

14

pH


Titration Curve

Comparisons3. @ Equivalence point:

S/Smore rapidrise

Volumeneeded is same

pHdifferent S/S = 7.00

W/S > 7

- b/c conjugate base

present

VNaOH(HAc) =VNaOH(HCl)

W. acid S. base

S. acid S. base


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0 20 40 60 80 1000

2

4

6

8

10

12

14

pH


Titration Curve

Comparisons4. Before Equivalence point:

W/Sis a bufferedsoln!

Acid& conjugate basepresent

TitrationComparison

(extra)

W. acid S. base

S. acid S. base
http://www.youtube.com/watch?v=xRVTdDrrtLchttp://www.youtube.com/watch?v=xRVTdDrrtLchttp://www.youtube.com/watch?v=xRVTdDrrtLchttp://www.youtube.com/watch?v=xRVTdDrrtLc


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Titration Curve

Comparisons Differences morepronounced, the weakerthe acid

Weaker acids have more basicequivalence points

Can determine Kaof aweakacid from thetitration curve

In a bufferedsystem: AcidBase

pKpH a log

0apKpH


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Titration of

Polyprotic Acids Polyprotic acidcontains 1 or more ionizable H

atoms

Ex: H3PO3=

H2C4H4O6=

Ionization occurs 1 H atom at a time,thus multipleequivalence pts exist

Ex: H3PO3

43

3H+

2H+

H2PO3-(aq)

H2PO3-(aq) HPO32-(aq)

(aq)H(aq)POH 33

(aq)H


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Titrating w/Indicators How do you know what indicatorto use?

Ideallyit should change at/nearthe equivalence pt.

Since pH changes so fast here, as long indicator changesanywhere on rapid rise of titration curve, all is ok

44


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Titrating w/Indicators Are the following goodindicators for the titration

shown?

45

1. Phenolphthalein: 2. MethylRed:

Yes, change colorin rapid rise

No, doesnt change colorin rapid rise


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Titrating w/Indicators What if not given a picture diagram for color change?

Indicators are acids thus have a Ka

Indicators change color when pHof the system equalstheir pKa!

Want pKa of Indicator to be as close to pH of

endpoint/equivalence as possible 46

IndHHInd

Color 1 Color 2


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Questions:1. Which indicators would work for the titration shown?

47

Both,b/c on rapid decline


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Questions:2. A laboratory worker is titrating an unknown solution of a

weak acid with a strong base. Which indicator shouldthey choose given the following titration curve?

A. Pentamethoxy red: pKa = 1.8

B. Congo red: pKa = 4.0

C. Azolitmin: pKa = 6.5

D. Phenolphthalein: pKa = 9.0

E. Trinitrobenzoic acid: pKa = 12.7

48


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Solubility Equilibria Some ions when combined form insolublesalts

PrecipitationRxns (Chap 4, solubility Rules)

Saturated solutioncontains undissolvedsolute

However, somedoes dissolve (SLIGHTLY!)

49

PbI2


50/63

Solubility Equilibria Solubility-Product Constant, Kspequilibrium

constant for a slightlysoluble(or nearlyinsoluble)

ioniccompound Degreeto which a solid is soluble in water

NOsolidsor liquidsin equilibrium expression!

Appendix D

50

)(2I)(Pb(s)PbI 22 aqaq

22

sp ]][I[PbK

5107.1 x


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Sample Problem If 0.0012 mols PbI2(s) dissolves in 1.0 L water, what is

its Kspvalue?

51

22 ]][[ IPbKsp0.0012M 0 0

2(0.0012)0.0012-0.0012

0M 0.0012M 0.0024M

2]0024.0][0012.0[spK

9

109.6

xKsp

I

C

E

)(2I)(Pb(s)PbI 22 aqaq


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52

Solubility vs.

Solubility Product, Ksp Solubilitygrams dissolved to form a saturated

solution Units = grams/L

Molar Solubilitymolesdissolved to form asaturatedsolution Units = mols/L

Kspequilibrium constant for ionic solid & itssaturatedsolution


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53

Sample Problem If the Kspof BaF2is 1.7x10

-6, calculate the molarsolubility.

22 ]][[ FBaKsp)(2)((s)2

2 aqFaqBaBaF

26]2][[107.1 xxx

364107.1 xx

Mxx

3

105.7

[Ba2+]

1:1 ratio of Ba2+:BaF2,molar solubility of BaCl2= 7.5x10

-3M

x 0 0

2(x)+x-x

0M x 2x

I

C

E


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Factors Affecting Solubility1. Common Ion EffectAdding a common ion

decreasesthe solubilityof the partially soluble salt

AKA Le ChteliersPrinciple

Adding BaCl2(Ba2+) or NaF (F-) shifts equilibrium

54

)(2)((s) 22 aqFaqBaBaF


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Sample Problem What is the molar solubility of BaF2in 0.15M NaF?

55


0 0.15

x 2x

x 0.15+2x

22 ]][[ FBaKsp

6107.1 x

26]15.0][[107.1 xx

Mxx5

1056.7

I

C

E

2]215.0][[ xx x

-x

0M

small

Just BaCl2= 7.5x10-3M

Decreased


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Factors Affecting Solubility2. pH:Almostany ioniccompound will dissolve if the

pH of the solution is altered

Solubilityincreasesif acidis added to a basicanion

Ex: Add HNO3to BaF2: H+from HNO3reacts w/F

-from BaF2

Adding more acid (H+) shifts equilibrium

56

)(aqH )( aqF )(aqHF22 2

)(2)((aq)2)(2

2 aqHFaqBaHsBaF



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Factors Affecting SolubilityComplexes

3. In the presence of suitable Lewis bases, metalionscan form complexes:

57

(aq)S(aq)FeFeS(s) 22

(aq)CNFe(aq)6CN(aq)Fe -462

Ksp= 6.31x10-18

Kf= 1.00x1035

(aq)S(aq)CNFe(aq)6CNFeS(s)-2-4

6

K = 6.31x10

17

f for formation(of a complex).

More in Chapter 23, but will use this idea in Experiment 19!


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Precipitation &Separation of Ions

How do you know if a precipitatewill form?

Need to compare Kspto your actualconditions Qsp

IF:

Qsp= Ksp: Equilibrium (saturated soln)

Qsp> Ksp: Pptoccurs

Qsp< Ksp: Dissolutionoccurs

58


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Sample Problems

1. [Ca2+] in human blood is 0.0025M & Calcium oxalate (CaC2O4)makes kidney stones. If Ksp= 2.3x10

-9for calcium oxalate,will it ppt in a patient with [C2O4

-2] = 1.0x10-6?

59

]][[ 2422

OCCaQsp

]100.1][0025.0[ 6 xQsp9105.2 xQsp

Qsp> Ksp

Not at equilibrium, use Qsp

Ppt occurs & Kidney stones form (OUCH!!)

Sample exercises (in book)17.15 & 17.16

very helpful!

)()((s)2

42

2

42 aqOCaqCaOCaC


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Sample Problems

2. A solution contains 1.0x10-2M Ag+& 2.0x10-2M Pb2+.When Cl-is added, both AgCl (Ksp=1.8x10

-10) & PbCl2(Ksp=1.7x10

-5) can precipitate. Which precipitates first?

60

]][[ ClAgKsp22 ]][[ ClPbKsp

Solve for [Cl-]

10108.1 x

8108.1][

xCl

225 ]][100.2[107.1 Clxx2

109.2][

xClAnything abovethis will ppt

AgCl ppt first, b/c smaller [Cl-]

]][100.1[ 2 Clx

Anything abovethis will ppt

ua tat ve Ana ys s


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Determines presenceorabsenceof metalions

Noinfo about amount

5 groups

1. Insoluble Cl-

2. InsolubleAcid S-

3. Insoluble Base S-&OH-

4. Insoluble PO4-3

5. Soluble alkali metal& NH4

+ions

61

y(selective precipitation)

(LAB 22)


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Sample Question

1. Which ion will precipitate whenphosphate ion is added to a

solution containing all of theions below

A. K+

B. Ba2+

C. Na+

D. NH4+

62

Summary Chap. 17


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Summary Chap. 171. Common Ion Effect

2. Buffers:

Calculationsa) Ice Table

b) Henderson-Hasselbalch eq.

3. Adding Strong Acid or Base to a buffer

RXN occurs! ICE table (MOLES)

H-H eq. (M)

4. Acid/Base Titration

S/S

S/W

Compare titration curves

5. Solubility equilibria

6. Qualitative analysis

acidbasepKpH a log

Documents

Chap 17 No Titrate Calc No Clicker(2)