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

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

Dr. Alanah FitchFlanner Hall [email protected]

Office Hours Th&F 2-3:30 pm

Module #17C:Buffering and Titrations

Holding Proteins intactBy pH constant

mailto:[email protected]

FITCH RulesG1: Suzuki is SuccessG2. Slow me down G3. Scientific Knowledge is ReferentialG4. Watch out for Red HerringsG5. Chemists are LazyC1. It’s all about chargeC2. Everybody wants to “be like Mike”C3. Size MattersC4. Still Waters Run DeepC5. Alpha Dogs eat first

Che

mis

tryG

ener

al

E kq qr rel

1 2

1 2

E kq qr rel

1 2

1 2

Piranhas lurk

What pH is best suitedFor the hydrogen bond?

Hemeglobin

K xa Val COOH, , . 5 0 1 1 0 3

K xa tyr ROH, , . 1 0 1 0 1 0

Review Module 17B

K xa Val COOH, , . 5 0 1 1 0 3

K xa tyr ROH, , . 1 0 1 0 10

Which pH (2, 7, 11) is most favorable for the formation of a hydrogen bond between Val and tyr in hemoglobin

%,

, ,

,

dissoc ia tedA

A H O

KA

aq eq

aq eq aq eq

aaq eq

1 0 0

3

%,

d issoc ia tedH O

Kaq eq

a

10 0

13

pH 2 1 0 2

%

.

d issocia tedCOOH

x

10 010

5 01 1 01

2

3 %

.

.dissocia tedCOOH

x

1 0 01 0

5 01 1 01

33 3 32

3

KaCOOH5.00E-03

pH [H+] %diss2 1.00E-02 33.33333

KaCOOH Ka ROH5.00E-03 1.00E-10

pH [H+] %diss %diss2 1.00E-02 33.33333 1E-067 1.00E-07 99.998 0.099911 1.00E-11 100 90.90909

How does the body maintainThis pH?

Review Module 17B

How does the body maintain pH ~7?Remember it will have to constantly adjust to changes in pHFrom an enormous number of chemical reactions!!!

Control is ultimately based on CO2,(g)

CO H O H COg aq2 2 2 3,

http://www.meddean.luc.edu/lumen/MedEd/MEDICINE/pulmonar/physio/pf4.htm

C P kg CO H2

Henry’s law

Respired (breathed) air PCO2 =0.23 mm HgPressure increases through mechanics of lungAorta PCO2=41.8mmHg

k xm olesL a tmH

3 4 1 0 2.

4 21

7 6 03 4 1 0 1 8 7 1 0

2

2 3mmatmmm

xM

L atmx MCO

. .

All dissolved CO2 goes to carbonic acid

http://www.meddean.luc.edu/lumen/MedEd/MEDICINE/pulmonar/physio/pf4.htm

H CO H O HCO H Oaq aq aq2 3 2 3 3, ,

K xa 1

74 4 10 .

K

HCO H O

H CO

x x

H CO x

x

H COa

eq aq eq

aq eq aq in it aq in it

1

3 3

2 3

7

2 3

2

2 3

1 0

, ,

, , , , , ,

x K H COa aq in it 1 2 3 , ,

H O xaq eq3 ,

p H O xaq eq352 8 6 1 0 4 5 4, lo g . .

This is too low!!! – we would not have theRight amount of ionization on the protein

Carbonic acid is a weak acid (anion is high charge density, Oh Card me PleaSe!)

x K H CO x xa aq in it 1 2 3

7 34 4 1 0 1 87 10, , . . x K H CO x x xa aq in it 1 2 3

7 3 104 4 10 1 8 7 10 8 2 2 8 10, , . . . x K H CO x x x xa aq in it 1 2 3

7 3 10 54 4 10 1 87 10 8 22 8 1 0 2 86 1 0, , . . . .

HCO H O CO H Oaq aq aq3 2 32

3, ,

K xa 2

114 7 1 0 .


K xa 174 4 1 0 .

Will the pH get to the “right” value by the second acidDissociation reaction?

K

CO H O

HCOx x x

x xa

eq aq eq

aq eq2

32

3

3

5

5

2 86 102 86 10

, ,

, ,

..

p H O xaq eq352 2 9 10 4 54, lo g . .

2 8 6 1 0 2 8 6 1 052 2

5 2. .x K xK x x xa a

x x x x K xKa a2 5 5

2 22 8 6 1 0 2 8 6 1 0 0 . .

2 8 6 1 0 2 8 6 1 052

5. .x x K x x xa

x x K x x Ka a2 5

25

22 86 1 0 2 8 6 1 0 0 . .

x x x x x x2 5 11 5 112 8 6 1 0 4 7 10 2 8 6 1 0 4 7 10 0 . . . .

x x x x2 5 1 52 8 6 1 0 1 3 4 1 0 0 . .

x

x x x

2 8 6 10 2 86 1 0 4 1 34 1 02

5 5 2 1 5. . .

xx x x

2 8 6 10 8 1 7 96 10 5 3 6 10

2

5 1 0 1 5. . .

xx x

2 86 1 0 2 8 60 0 0 93 10

2

5 5. .

xx

x

9 37 1 02

4 6 8 1011

11..

Let’s make an “intelligent” guess

NO, wouldOnly slightlyIncrease acidity

Kx x

xx xa 2

5

5112 8 6 1 0

2 8 6 1 04 7 1 0

.

..

HCO H O CO H Oaq aq aq3 2 32

3, ,

K xa 2114 7 10 .


K xa 1

74 4 10 .

To tweak blood pH to 7.4 red blood cells (RBC) control [HCO3

-]

HCO x Maq eq blood3

24 0 1 0, .

Will the pH get to the “right” value by the second acidDissociation reaction? p H O xaq eq3

52 2 9 10 4 54, lo g . .

RBC

Value can be controlled by red blood cell


H CO x Maq CO g2 3

3

2

1 8 7 1 0,,

.

K

HCO H O

H COa

eq aq eq

aq eq

1

3 3

2 3

, ,

, ,

HCO x Maq eq RBC3

22 4 1 0, .

KH CO

HCOH Oa

aq eq

eq

aq eq1

2 3

3

3

, ,

,

,

pH H O KH CO

HCOaq eq a

aq eq

eq

lo g lo g,

, ,

,

3 1

2 3

3

pH KH CO

HCOa

aq eq

eq

lo g lo g, ,

,

1

2 3

3

pH pKH CO

HCOa

aq eq

eq

1

2 3

3

lo g, ,

,

pH xxx

lo g . lo g..

4 4 1 01 87 1 02 4 1 0

73

2

pH 6 35 1 10 8 7 4 5. . .

K xa 174 4 1 0 .

pH x lo g . lo g .4 4 10 0 0 77 97

H O KH CO

HCOaq eq a

aq eq

eq

3 1

2 3

3

,

, ,

,

Red Blood Cell

PlasmaHCO3

-=24mmol/LpH=7.4

P mm x MCO blood24 0 1 2 1 0 3

, .

+H CO aq2 3 ,

HCO H Oaq3 3,

5 0 % 2CO bound,

hemoglobin

Carbonic anhydrase

That is the Context for Buffering

1. Must control concentration of both1. Conjugate acid2. Conjugate base

2. So what are the equations, etc. necessaryto determining the buffering of a solution?





Henderson/HassalbachBuffer Equation


The body controls the pH ([H+]) concentration in blood by1. controlling WA H2CO3 via CO2,g

2. controlling WB HCO3- through activity of the red blood cells

3. The WA and WB are linked (conjugates) so that an equilibrium can be written between the two

4. All parameters are constant, leading to constant pH

5. This process is called buffering

pH pKH CO

HCOa

aq eq

eq

1

2 3

3

lo g, ,

,

A general equation can be written

pH pKHA

Aa

aq eq

eq

lo g, ,

,


mL x M = mmoles

mL x mmoles = mmoles mL

moles x 103 mmoles x L = mmoles L moles 103 mL mL

pH pK

mmoleAto ta l mLmmoleHAtota lm L

a

lo g

pH pKmmoleA

to ta l mL

mmoleHAto ta lm L

a

lo g

1

1 pH pKmmoleAmmoleHAa

lo g

pH pKHA

Aa

aq eq

eq

lo g, ,

,

Henderson Hasselbach Equation

OJO this is NOT the “equivalence point”

pH pK a

if mm oleA mmoleHA

Linked (conjugated)

mmole titran t mmole titra ted

pH pKHA

Aa

aq eq

eq

lo g, ,

,

pH pK

A

HAa

eq

aq eq

lo g,

, ,

OR

pH pK pKa a lo g 1 0

pH pKmmoleAmmoleHAa

lo g

Marie the Jewess, 300 Jabir ibn Hawan, 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 Augustin Coulomb 1735-1806

John Dalton1766-1844

B. P. Emile Clapeyron1799-1864

Jacques Charles 1778-1850

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 Ohm 1789-1854

Henri Louis LeChatlier1850-1936

Svante Arrehenius1859-1927

Francois-Marie Raoult1830-1901

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

pH pKAHAa

lo g

Buffer Example 1 what is equilibrium pH of a solution of 1.0 M HF, and 1.0 M NaF? Ka HF = 7.2x10-4

pH pKFHFa

lo g

pH pK a

lo g..

1 01 0

pH pK a lo g 1

pH pK a 0

pH pK a

Do we have linked WA and WB both present?

HF H O H O Faq l aq aq

2 3

_

Yes = buffer

Ka=7.2x10-4

pH x lo g .7 2 1 0 4

pH x lo g . .7 2 1 0 3 1 44

In preceding module 17B we calculated the pH ofA solution of 1.0 M HF and found it to be

pH 1 5 7.What causes the difference?

HF H O H O Faq lHF push to righ t

aq aq 2 3

_

LeChatlier’s Principle

Effect of added NaF is to reduce the ionization of HF

pH 1 5 7.

pH 3 14.HF H O H O Faq lHF push to righ t

aq aq 2 3

_

NaF H O F Naaq lcom plete

aq aq 2

Buffer Example 2:What is the pH of 1.0 M H2CO3 in the presence of 1.0 M NaHCO3

pH pKAHAa

lo g


What is in solution? Are the chemical species linked by an equilibrium reaction?

Solution contains a weak acid in the presence of it’s linked (conjugate) weak base

pH x

M NaHCO

M H CO

lo g( . ) lo g4 4 1 01

17 3

2 3

K xa 174 4 10 .

pH 6 3 6 1. log

pH 6 3 6 0 6 3 6. .

Buffer Example 3:What is the pH of a solution containing 0.25 M NH3 (Kb = 1.8x10-5) and 0.40 M NH4Cl?

pH pKAHAa

log

NH H O NH OHaq aq aq3 2 4,

What is in solution? Are the chemical species linked by an equilibrium reaction?

Solution contains a weak acid in the presence of it’s linked (conjugate) weak base

pH x

M NH

M NH

aq

aq

log(5. ) lo g

.

.

,

,

55 100 25

0 401 0 3

4

K xb 1 8 1 0 5.

pH

9 25

0 250 4

. lo g..

pH 9 2 5 0 2 0 4 9 0 5. . .

or NH H O NH H Oaq aq aq4 2 3 3

,

KKKa

w

b

KKK x

xaw

b

1 0

1 8 1 05 5 56 1 0

1 4

51 0

..K

KK xa

w

b

1 01 8 1 0

14

5.





4 Kinds of ProblemsReviewed


Limiting Reagent must involve SA or SB

WA

pH pK

A

HApK

A

HAa

eq

eq

a

in it

in it

lo g lo g

HA H O A H Oaq aq aq

2 3

K

H O A

HA

x x

HA x

x

HAa

aq aq


3

7 21 0

, , ,

WB

A H O OH HAaq aq aq

2

K

OH HA

A

x x

A x

x

Ab

aq aq eq


,

, , ,

1 0 7 2

A H O OH HAaq aq aq

2


2 3

HA NaOH A H O Naaqcom ple te ly L im iting reagen t

aq aq aq ,2

Na A HC l HA C l N aaq aqcom ple tely L im iting agen t

aq aq aq , R e

Which reagent is completely consumed?Which remains?

HC l NaOH C l H O Naaqcompletely L im iting reagen t

aq aq aq ,2

Equi

libriu

mWe now know how to do 4 types of problems

WA +conjugate(linked) WB

Calculate the concentration of Na+, Cl-, CH3COO-, and CH3COOH in solution when 10.0 mL of 1.0 M HCl is added to 6.0 mL of 2.0 M NaAcetate acid (NaCH3COO). Each solution is obtained from made from a stock solutions at 1 atm pressure

To refresh your memory on Limiting Reagent Problems

What do we know What do we want What is a red herring?10.0 mL 1.0 M HCl5.0 mL 2.0 M NaCH3COO1 atm

Conc.

Classify the chemical substances added to solution:Strong Acid (SA)Strong Base (SB)Strong electrolyte (SE)Weak Acid (WA)Weak Base (WB)Other or unknown (OU)

Protonated low q/r anion (No Clean Socks) HNO3, HCl, H2SO4

OH +low q/r Cation (Group 1 and 2) NaOH, KOH, Ca(OH)2…Anion and Cation, low q/r NaCl, KNO3….

Protonated hi q/r anion (Oh Card me PleaSe)Contains R group, anion hi q/r with cation low q/r

H2CO3…..Na2CO3

Calculate the concentration of Na+, Cl-, CH3COO-, and CH3COOH in solution when 10.0 mL of 1.0 M HCl is added to 6.0 mL of 2.0 M NaAcetate acid (NaCH3COO). Each solution is obtained from made from a stock solutions at 1 atm pressure

To refresh your memory on Limiting Reagent Problems

mL M mmoles

1 0 01 0

111

11

1 03

3.

.mL

moleL

molem ole

m olem ole

mmolesHC lHC l

HC l

H

HC l

CH COOH

HCH COOH

H CH COO CH COOHaq aqcom plete ly L im iting agen t

aq 3 3

, R e

6 02 0

1

1

111

1 23

3

3

3

3

3

3

3.

.mL

moleL

mole

m olem olem ole

mmolesNaCH COO lN aCH COO

NaCH COO

CH COO

NaCH COO

CH COOH

CH COOCH COOH

HCl can produce

SA (HCl) is limiting reagent, completely consumed, 0mmoles

SA/WB

WB can produce


aq aq aq , R eDetermine the Limiting Reagent

WA formed is 10 mmoles How much WB remains?

To refresh your memory on Limiting Reagent ProblemsCalculate the concentration of Na+, Cl-, CH3COO-, and CH3COOH in solution when 10.0 mL of 1.0 M HCl is added to 6.0 mL of 2.0 M NaAcetate acid (NaCH3COO). Each solution is obtained from made from a stock solutions at 1 atm pressure

H CH COO CH COOHaq aqcom pletely L im iting agen t

aq 3 3

, R eSA/WB

WB initial 12 mmoles; WB consumed:

Initial-consumed=12-10=2mmoles WB left

Concentrations after the limiting reagent reaction

CH COOHmmoles

mL mLMCH COOH

3

1 010 0 6 0

0 6 2 53

. .

.

CH COOmmoles

mL mLM

CH COOH3

2 010 0 6 0

0 1 253

.

. ..

10 01 0

111

11

11

103 3

3

3.

.mL

moleL

m olem ole

m olem ole

mo lemole

mmolesHC lHC l

HC l

H

HC l

CH COOH

H

CH COO

CH COOHCH COO

Na

mL MmL mL

M

6 2

1 0 0 6 00 7 5

. ..

C l

mL MmL mL

M

1 0 1 0

1 0 0 6 00 6 25

.. .

.

SA (HCl) is limiting reagent, completely consumed, 0mmoles

WA formed is 10 mmoles

To refresh your memory on Limiting Reagent ProblemsConcentrations

In the preceding problem what kind of equilibrium problem remains?

You have 3 choices:WAWBWB and its conjugate WA (or WA and its conjugate WB)

We have present a weak acid (WA) and it’s linked (conjugate) weak base (WB)

CH COO M3 0 1 2 5 .

CH COOH M3 0 6 25 .Buffer equilibrium problem

CH COOHmmoles

mL mLM

CH COOH3

1 01 0 0 6 0

0 6 2 53

. .

.

CH COOmmoles

m L mLM

CH COOH3

2 01 0 0 6 0

0 12 53

.

. ..

Na

mL MmL mL

M

6 2

1 0 0 6 00 7 5

. ..

C l

mL MmL mL

M

10 1 0

1 0 0 6 00 6 25

.. .

.




A feast of titrations



We can combine Limiting reagent problems followed by an equilibrium problem to do a titrationTitrations allow us to determine the conc of an unknown

Titrations involve 1 of 3 LR followed by 1 of 3 Eq Limiting Reagent must involve SA or SB

WA HA H O A H Oaq aq aq

2 3

K

H O A

HA

x x

HA x

x

HA in ita

aq aq

aq eq aq in it aq

3

7 21 0

, , ,

WB

A H O OH HAaq aq aq

2

K

OH HA

A

x x

A x

x

Ab

aq aq eq


,

, , ,

1 0 7 2

WA +conjugate(linked) WB

A H O OH HAaq aq aq

2


2 3

HA NaOH A H O Naaqcom ple te ly L im iting reagen t

aq aq aq ,2


aq aq aq , R e

Which reagent is completely consumed?Which remains?

HC l NaOH C l H O Naaqcompletely L im iting reagen t

aq aq aq ,2

Equi

libriu

m

pH pK

A

HApK

A

HAa

eq

eq

a

in it

in it

lo g lo g

010 72 InitKKxx

0 20. M acetic ac id

Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL; Ka acetic acid = 1.8x10-5

Main Dish: Limiting Reagent Problem

CH COOH OH CH COO H O3 3 2

Solution contains no CH3COO-

Solution contains CH3COOHSolution contains no NaOH Side dish: WA

K

H O A

HA

x x

HA x

x

HA in ita

aq aq

aq eq aq in it aq

3

7 210

, , ,

1 8 1 00 2 0

52

..

xx 1 8 1 0 0 2 0 1 8 97 1 05 3. . .x x x

x 1 0 7 ?

x 0 2. ?

0 2 00 0 0 18 9 7

0 19 8 1 0 2 6 3

..

.

Sig fig, round to 0.2

H x xaq 1 8 9 7 1 0 3.

pH x lo g .1 8 9 7 1 0 3

pH 2 7 2.

If you prefer – do an ICE chart

limiting

K x try assum ptionsa 1 8 1 0 5. ;

0 20 40. M mLacetic acid ace tic ac id 0 2 0 4 01

13

3

. M mLmmolemmoleace tic acid ace tic ac id

CH COO

CH COOH

0 2 0 4 0

11

83

3

3. M mLmmolemmole

mmoleCH COOacetic ac id acetic acidCH COO

CH COOH

0 4 0 011

03

3. M mLmmolemmole

mmoleCH COONaOH NaOHCH COO

NaOH

Equilibrium Calculation.H2O H+ OH-

55.5 10-710-7

CH3CO2H H+ CH3CO2-

init 0.200 10-70change -x +x +xassume x<<0.2 x>>10-7

equil 0.2 x x

check? yes.Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL;Ka = 1.8x10-5

Rx 1

Rx 2

pH x lo g ( . ) .1 8 9 1 0 2 7 23

1 8 1 00 2 0

52

..

xx

1 8 1 0 0 2 0 1 8 97 1 05 3. . .x x x

Q

CH CO H

CH CO HK

in it in it

in it

r

3 2

3 2

70 1 00 2

,

.Because rx goes to rightSign of x is neg for reactants

K x try assum ption sa 1 8 1 0 5. ;




0 40 511

2 03

3. .M mLmmolemmole


NaOH

0 2 0 401

18 0

3 3

3

3

3. .M mLmolem ole

mmoleCH COOCH COOH CH COOHCH COO

CH COOH

Solution contains 2.0 mmole CH3COO-

2.0 mmole CH3COOH used to make 2 mmole CH3COO-

CH COOmmole

m L mLM

CH COO

CH COOH NaOH3

2 0

4 0 50 0 44 43

3

.

.

CH COOHmmole mmole

mL mLM

CH COOH NaOH3

8 0 2 040 5

0 1 3 3 33

. ..

Limiting Reagent

Side Dish: Buffer

mL pH0 2.72



CH COOmmole

mL mLM

CH COO

CH COOH NaOH3

2 0

40 50 04 443

3

.

.

CH COOHmmole mmole

m L mLM

CH COOH NaOH3

8 0 2 04 0 5

0 1 33 33

. ..

pH pKAHAa

lo g

pH x

lo g( . ) lo g..

1 8 100 04 4 40 13 3 3

5

pH 4 27.

pH 4 7 4 0 5 2 2. .

Invoke Rule: Chemists Are Lazy

pH pK

mmolesAto ta l m L

mmolesHAto ta l m L

a

lo g

pH pKmmolesAmmolesHAa

log

pH

4 7426

4 27. lo g .

mL pH0 2.725 4.27




0 4 0 1 011

4 03

3. .M mLmolem ole


NaOH

0 2 0 401

18 0

3 3

3

3

3. .M mLmolem ole


CH COOH


Limiting Reagent

Side Dish: buffer


lo g

pH

4 7 44

8 44 7 4 1 4 7 4. lo g . lo g .

mL pH0 2.725 4.2710 4.74

Solution contains 8.0-4.0 mmole CH3COOH




0 4 0 1 911

7 63

3. .M mLmolem ole


NaOH

0 2 0 401

18 0

3 3

3

3

3. .M mLmolem ole


CH COOH


8 CH3COOH – 7.6 mmole OH = 0.4 mmole

Limiting Reagent

Side Dish: buffer


lo g

pH

4 7 4

7 68 7 6

4 747 60 4

4 7 4 1 27 8 7 6 02. lo g.

.. lo g

.. . .

mL pH0 2.725 4.2710 4.7419 6.02

titration of acetic acid

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30mL 0.400 NaOH

Cal

cula

ted

pH

Difference between “exact” and non-exactsmall

K

A x xHA x

x xx

pH

pHmmolesAmmolesHA

a

1 0 0 1 2 8 100 06 7 7

6 08

4 7 7 6 0 4

7 7..

; .

. lo g .




0 4 0 2011

83

3. M mLmolem o le


NaOH

0 2 0 401

18 0

3 3

3

3

3. .M mLmolem ole


CH COOH

Solution contains 8 mmole CH3COO-

8 mmole CH3COOH-8 mmole OH = 0 mmole CH3COOH

Both Are Limiting ReagentsUSED UP

Side Dish: acetate = WB

CH COO H O CH COOH OHaq aq aq3 2 3

mL pH0 2.725 4.2710 4.7419 6.02


Solution contains 8 mmole CH3COO-

CH COO H O CH COOH OHaq aq aq3 2 3

CH COOmmole

mL mLaq3

84 0 2 0

0 1 3 3 3

.

K

HA OH

A

x x

A x

x

Ab

aq eq aq


,

, , ,

1 0 7 2

KKK x

xbw

a

1 0

1 8 1 05 5 5 1 0

1 4

1 51 0

.. 5 5 5 1 0 0 1 33 3 8 5 9 1 01 0 6. . ~ .x x x

OH x x~ . 8 5 9 1 0 6

pH x 1 4 8 5 9 1 0 8 9 36lo g . .

x

x

0 1 3 3 3

1 0 7

. ?

? ? ?

mL pH0 2.725 4.2710 4.7419 6.0220 8.93

If you prefer – do an ICE chart5 5 5 1 00 1 3 3 3

1 02

..

xx

Kb “small”, assumps. Ok.

Calculate equilibriumH2O H+ OH-

55.5 10-7 10-7

CH3CO2- CH3CO2H OH

init conc: 0.133 0 10-7

change -x +x +xassume (not much reaction to right, Kb small)

x << 0.133 x>> 10-7equil 0.133 x x

Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL pH pOH x 14 8 59 10 8 936log ( . ) .

Q

CH CO H OH

CH COKin it in it

in it

r

3 2

3 2

70 1 00 2

,.

Sign of x for reactants is -

KKK x

xbw

a

1 0

1 8 1 05 5 5 1 0

1 4

151 0

..

Kb “small”, assumps. Ok.

5 5 5 1 00 1 33

102

..

xx

5 5 5 10 0 1 33 8 5 9 1 010 6. . .x x x




0 4 0 2 111

8 43

3. .M mLmolem ole


NaOH

OH

mL Mexcess

excess

1 0 4 0

40 20 10 0 06 5 57

..

Limiting Reagent

mL pH0 2.725 4.2710 4.7419 6.0220 8.9321 11.82

OHexcess

8 4 840 20 1

0 0 0 65 5 7.

.

Or

p OHexcess

lo g .0 0 06 5 57

p OHexcess

2 1 8.

pH pOH 1 4 14 2 1 8 11 82. .

Side dish: seconds of the mainExcess OH

0 2 0 401

18 0

3 3

3

3

3. .M mLmolem ole


CH COOH




0 4 0 2 511

103

3. M mLmolem ole


NaOH

OH

mL Mexcess

excess

5 0 40

40 20 50 0 3 0 7 69

..

Limiting Reagent

mL pH0 2.725 4.2710 4.7419 6.0220 8.9321 11.8225 12.48

OHexcess

10 840 2 0 5

0 03 0 76 9.

Or

p OHexcess

lo g .0 0 30 7 69

p OHexcess

1 511.

pH pOH 14 14 1 5 11 1 2 48. .

Side dish: seconds of the mainExcess OH

0 2 0 401

18 0

3 3

3

3

3. .M mLmolem ole


CH COOH


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30mL 0.400 NaOH

Cal

cula

ted

pH

½ way buffer

Buffer

WeakAcid

Weak Base

Strong BaseRegions of A WEAK ACID titration curve

Defining some points on the Titration Curve

Tells us identity of acid

½ way to the equivalence point

pH pK a


lo g

HA OH Aaq aq aq

½ of HA is converted to A-

mmoleHAmmoleHA mmoleAin itia l

rem a inng fo rm ed2

pH pK a lo g 1

pKa is a clue to the identity of theacid

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100mL base added

pH

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100mL base added

pH

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100mL base added

pH

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100mL base added

pH

CH COOH pK a3 4 7 4; .Valen ine RCOOH pK a; . 2 3 0

HF pK a; . 3 14 NH pK a4 9 2 55 ; .

12 eq p t pH pK a;

Titrate 50 mL 1.0 M WA with1.0 M NaOH

ClueI.D.Of acid


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30mL 0.400 NaOH

Cal

cula

ted

pH

½ way buffer Eq. pt.

Buffer

WeakAcid

Weak Base

Strong BaseRegions of A WEAK ACID titration curve

Defining some points on the Titration Curve

This is whatTells us the conc.

0

2

4

6

8

10

12

14

0 20 40 60 80 100 120 140 160 180 200mL strong base added

pH

0

2

4

6

8

10

12

14


pH

0

2

4

6

8

10

12

14


pH

0

2

4

6

8

10

12

14


pH

0

2

4

6

8

10

12

14


pH

0

2

4

6

8

10

12

14


pHTitrate __mL of 1.0 M Acetic Acid with 1.0 M NaOH

13 25 50 75 100 125

You should have at least 2 observationsWhat are they?

Observation 1?Observation 2?

We found the equivalence point (large pH change) occurred when we added 20 mL of 0.40 M NaOH to 40 mL of an unknown conc of acetic acid. What was the original concentration of the acetic acid?

mmole titran t mmole titra tedeq p t eqp t.

2 0 0 4 0 4 0mL M mL x.

x

mL MmL

20 0 40

4 0.

x M 0 2 0.

or

V M V M

Be sure to account forMultiple protons or hydroxides

2 0 0 4 0 4 0. M

M 2 0 0 4 0

4 00 2

..

OJO, notpH pK a

mmoleA mmoleHA




Indicators




0

2

4

6

8

10

12

14

0 5 10 15 20 25 30mL 0.400 NaOH

Cal

cula

ted

pH


Buffer

WeakAcid

Weak Base

Strong Base

Choose an Indicator so there will be a color change at the equivalencepoint

Regions of A WEAK ACID titration curve

8.93

Indicators:

plum juice (tannic acid); red wine

pink

O

OH OH

O

O

OH

OH

OH

O

O O

O

O

O

OH

OH

O

O

OHOH

OH

O

OH OH

O

O

OH

OH

OH

O

O

OHOH

O

O

OH

OH

OH

O

OH

OH

O

O

OH

OH

OH

O

OH OH

O

O

OH

OH

OH

O

O O

O

O

O

OH

OH

O

O

OHOH

OH

O

OH OH

O

O

OH

OH

OH

O

O

OHOH

O

O

OH

OH

OH

O

OH

OH

O

O

OH

OH

OH Removeprotons

How can we “see” the pH jump so we know where the end point is?

H(tannic)H+ + In-

greenHIn

http://antoine.frostburg.edu/chem/senese/101/features/water2wine.shtml

Core structure= cyanidin

Red cabbage primarily

Connects electrons through ring,

BlueberriesRaspberriesCranberriesRed cabbageRed wines

CAS 528-58-5

http://antoine.frostburg.edu/chem/senese/101/features/water2wine.shtml

http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/p26_anth-e.htm

http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/p26_anth-e.htm

Hydrangea flower color based on available aluminumCation in the soil

Indicators:

plum juice (tannic acid)

HIn H+ + In-

pink green

Observed color? When is it green vs pink?

K

H InH Ina

KH

InH In

a

KH

moles greento ta l vo l

m oles p inkto ta l vo l

m oles greenm oles p ink

a

Ratio: 10 Green/1 Pink Ratio: 10 Pink/1Green

Observed color? depends on mole ratioRule of thumb (10/1 or 1/10)

KH

moles greenm oles p ink

p inka

1 0

KH

moles greenmoles pink

greena

10 KHa

1 0

1 0 K Ha

pK p Ha ( lo g )1 0 pH pKgreen a 1

lo g 1 0 pK pHa

pH pKp ink a 1

pH pKco lor change a ind ica to r , 1

pH pK a 1

Common Indicators (besides grape juice)

Name pHcolor change = pKa +/- 1 pKa

methyl violet 0-2thymol blue 1-3bromophenol blue 3-5methyl red 4.5-6.5 5phenol red 6.5-8bromothymol blue 6-8 7.1phenolphthalein 8-10 9alizarin yellow R 10-12

Color change over 2 pH unitsChange centered over pKa

Common Indicators (besides grape juice)Name pHcolor change = pKa +/- 1 pKa


Biggest pH change in our calculation was centered at pH 8.93

Name pHcolor change = pKa +/- 1 pKa


Titrate 40 mL 0.20 acetic acid with 0. 40 M NaOH 0, 5, 10, 19, 20, 21,25 mL Can omit for BLB


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30mL 0.400 NaOH

Cal

cula

ted

pH


Choose an Indicator so there will be a color change at the equivalencepoint




Two more example titrationsWB/SASA/SB



A quick and dirty (WA/SB) titration curve.

mL added problem pH01/2 to equivalence ptequivalence ptafter equivalence pt

Can you predict how a WB/SA titration curve will work out?

WA <pKa

buffer pKaWB

SB >7

mL added problem pOH pH01/2 to equivalence ptequivalence ptafter equivalence pt

titrate WB with SA

Example titrate 50 mL 0.1 M NH3 0.2 M HCl

quick and dirty: 0 mL, 1/2 equivalencepoint, 1 mL before eq. pt., eq. pt., 1 ml after

Example Titration 2

WB <pKb

K xb 1 8 1 0 5.

buffer pKbWA

SA <7

>pKa=pKa

0 mL HCl added; WB

NH3 + H2O NH4+ + OH-

init. conc. 0.1 0 10-7

change -x +x +xassume? Kb = 1.8x10-5, not much reaction, yes

0.1> x x>10-7

equil. 0.1 x x

equilibrium chart

K x

x xx

xb

1 8 1010

0 1 0 15

7 2

.. .

1 8 10 0 15 2. .x x

x x 1 8 10 0 15. .

x x 1 34 10 3.

pOH x lo g . .1 34 1 0 2 8 73

pH pOH 1 4 11 1 3.

mL pH0 11.13

Q

NH OH

NHK x

in it in it

in itb

4

3

750 1 0

0 11 8 1 0

,

,.

.

need to know eq. pt.

Example titrate 50 mL 0.1 M NH3 0.2 M HCl; include 1 mL and 1 mL after eq.pt.

mL pH0 11.1312.5 9.25

50 0 1 0 23mL MNH x MHC l. .

x mL eq p t 2 5

12

252

1 2 5eq ptmL

mL. . .

pH pKeq pt a1 2/ .

K K K KKK

ORpK pK

w a b aw

b

a b

;

1 4

K xb NH, .3

1 8 10 5

pK xb lo g . .1 8 10 4 74 45

1 4 4 74 4 pK a .

pK a 9 2 5.

pH pKeq pt a1 2 9 2 5/ . .

In case you are wondering?

pOH pKeq p t b1 2 4 74 4 7/ . .

1 4 1 4 4 74 4 9 2 51 2 pOH pHeq p t/ . . .

Eq pt = 25 mL; pKaa=9.25 (from preceding slide)=9.25 (from preceding slide)

1 mL before eq. pt = 24mL.; LR prob4.8 mmoles NH4

+

0.2 mmoles NH3

Buffer

pK a 9 2 5.


lo g

pH

9 25

0 24 8

. lo g..

pH 9 25 1 38 7 8 6. . .


mL pH0 11.1312.5 9.2524 7.86

0 2 0 2 4 4 8 4 8. . .M mL mmoleH added mmolesHA form edHC l HC l

5 0 0 10 4 8 0 23

mL M mmoles mmolesNH. . .

Equilibrium prob?

A- remaining

NH4+ NH3 + H+

init conc .066 0 10-7

change -x +x +xKa = 10-14/(1.8x10-5) = 5.5x10-10

Assump? 0.066>>x x~10-7

equil 0.066 x x+10-7

Eq. pt. Problem; WA

Kx

xa

10

1 8 105 5 10

1 4

51 0

..

5 5 1 0

100 6 6

1 00 6 6

107 7

.. .

xx x

xx x

5 5 1 0 0 6 6 1 01 0 7 2. .x x x

3 6 6 1 0 1 011 7 2. x x x

x x x2 7 1110 3 6 6 10 0 .

x

x

1 0 1 0 4 1 3 6 6 1 02 1

7 7 2 11.

xx

10 1 0 1 46 10

2

7 1 4 1 0.

xx

1 0 1 2 1 1 0

2

7 5.

x x 6 1 0 6 pH x lo g .6 1 0 5 2 26

conc. SA = conc. H+ = 2.63x10-3

pH = -log(2.63x10-3)pH = 2.5


1 0 25 0 2 5 1

2 6 3 10 3mL M

x M Hexcess HC l.. [ ]

1 mL past eq. pt.; 26 mL; SA

50 0 1

50 25 15

7 63 4

4

mL M mmolemL

NHNH NH.

[ ]/

mL pH0 11.1312.5 9.2524 7.8625 5.2226 2.5

Eq pt = 25 mL; pKa=9.25 (from preceding slide)a=9.25 (from preceding slide)

OR, limiting reagent

2 6 0 2 52 4mL M mmoleNHs HC l.

50 0 1 5 0 4mL M mmoleNHs HC l. LimitingReagent

2mmole excess H

25 0 2 5 1

2 63 1 0 3mmolex

.

conc. SA = conc. H+ = 1.25x10-2 pH =-log(1.25x10-2)pH=1.9

5 0 25 0 2 5 5

1 25 1 0 2mL M

x M Hexcess .. [ ]

5 mL past eq. pt.mL pH0 11.1312.5 9.2524 7.8625 5.2226 2.530 1.9

50 mL 0.1 NH3 titrated with 0.2 M HCl

0

2

4

6

8

10

12

0 5 10 15 20 25 30mL HCl

pH

bufferWB

WA

SAeq pt

1/2 eqpt

Choose an indicator?



Biggest pH change is near 5.5

Example Titration 3: We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOHMain Course: SA/SB Side courses: more of sameMmoleHNO3

mL OHadded

MmolesOH added

Mmole H+

excessMmoleOH excess

Total Vol [H+]Or[OH-]

5 0 0 5-0=5 0 25+0 5/25=0.2

5 20 2 5-2=3 0 25+20=45 3/45=0.066

5 49 4.9 5-4.9=0.1 0 25+49=74 0.1/74=0.00135

5 50 5 5-5=0 0 25+50=75 H2O

5 51 5.1 5-5=0 5.1-5=0.1 25+51=76 0.1/76=0.00136

5 75 7.5 5-5=0 7.5-5=2.5 25+75=100 2.5/100=0.025

5 100 10 5-5=0 10-5=5 25+100=125 5/125=0.04

0 20 2 5. M mL 0 10. M mL

init

Example Titration 3: We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOHMain Course: SA/SB Side courses: more of same[H+]Or[OH-]

5/25=0.2 pH=-log(0.2)=0.0698

3/45=0.066 pH=-log(0.066)=1.17

0.1/74=0.00135 pH=-log(0.00135)=3.17

H2O pH=7

0.1/76=0.00136 pOH=-log(0.0013)=2.88 pH=14-pOH=11.12

2.5/100=0.025 pOH=-log(0.025)=1.602 pH=14-pOH=12.398

5/125=0.04 pOH=-log(0.04)=1.39 pH=14-pOH=12.61

Titration of 50 mL 0.2 M HNO3

0

2

4

6

8

10

12

14

0 20 40 60 80ml 0.1 M NaOH

pH

Where is largest pH?What indicatorshould we use?

We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOH



Biggest pH change is near 7

Titration of 50 mL 0.2 M HNO3

0

2

4

6

8

10

12

14

0 20 40 60 80ml 0.1 M NaOH

pH

SA

SB

yellow

blue

We have 25.0 mL of 0.200 M HNO3. What will be the pH after we add 0, 20, 49, 50, 51, 75, 100 mL of 0.100 M NaOH

Color change defines large pH change





END


Tannins – represent linked cyanidins

Red wine - antioxidants

http://www.micro-ox.com/chem_tan.htm

Food Source Anthocyanins Glycoside form

apple cyanidin monoglycoside blueberry malvidin monoglycoside

petuniclin monoglycoside

delphinidin monoglycoside

cranberry cyanidin monoglycoside peonidin monoglycoside

red cabbage cyaniclin diglycoside strawberry pelargonidin monoglycoside

cyanidin monoglycoside

Documents

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