Polyprotic Acids and Bases - Politechnika Gdańska · Polyprotic Acids and Bases . Tadeusz Górecki...

Tadeusz Górecki Ionic Equilibria

Page 63

Polyprotic Acids and Bases


Page 64

Phosphoric acid

HPOHPOH 4243 ][

]][[

43

421

POH

POHHKa

23.21 apK

HHPOPOH 2442

][

]][[

42

2

42

POH

HPOHKa 21.72 apK

HPOHPO 34

24

][

]][[2

4

3

43

HPO

POHKa 32.123 apK

Structure of phosphoric acid molecule:

P

OH

OH

OH

O

Pyrophosphoric acid:

P

OH

O

OH

O P

OH

OH

O

Dependence of equilibrium constant on ionic strength:

For H3PO4:

0

1

043

4201

][

][][

aa KPOH

POHHK

)log()log( 0

0

11 aa pKpK


Page 65

From Davies equation:

III

IpKpK aa 1.02.0

151.020

11

22

42

2240

2][

][][aa K

POH

HPOHK

)log( 20

22

aa pKpK

I

I

IpKpK aa 2.0

151.040

22

The constant factor b can be adjusted to get better agreement.


Page 66

2

33

224

3340

3][

][][

aa K

HPO

POHK

)log()log( 23

0

33 aa pKpK

I

I

IpKpK aa 2.0

151.060

33


Page 67

Temperature dependence of K


8

Fraction (degree) of dissociation (ionization)

Weak monoprotic acid:

][

][

HK

K

C

A

a

a

HA

][

][][1

HK

H

C

HA

aHA

Polyprotic acids: n , where n is the number of protons that the acid has lost.

Monoprotic acid:

][

][0

HK

H

a

][1

HK

K

a

a

Diprotic acid H2S:

HSHSH2 ][

]][[

2

1SH

HSHKa

2SHHS ][

]][[ 2

2

HS

SHKa

CCCSHSSHC 2102

2 ][][][

C

SH ][ 20

C

HS ][1

C

S ][ 2

2

0

1

0

1

2

1

][][

][

]][[

H

C

CH

SH

HSHKa thus

][

101

H

Ka


Page 69

1

22

2

][

][

]][[

H

HS

SHKa thus

2

210

212

][][

H

KK

H

K aaa

Obviously, n

n 1 , hence

1][][ 2

210

100

H

KK

H

K aaa

From this

2112

2

0][][

][

aaa KKHKH

H

2112

11

][][

][

aaa

a

KKHKH

HK

2112

212

][][ aaa

aa

KKHKH

KK

Rules for writing expressions:

1. The denominator for each in a component system HnB is identical.

2. The denominator is a decreasing power series of [H+], starting with [H

+]

n,

and ending with [H+]

0(=1). In each successive term, and additional stepwise

dissociation constant becomes a factor. There is a total of n+1 terms, in

which the last term is anaa KKK ...21 .

3. Since each term in the denominator is proportional to the concentration of a

particular species, the first term [H+]

n forms the numerator for 0 ,

11 ][ n

a HK for 1 , and so on. Thus, n has anaa KKK ...1 21 as the

numerator.


Page 70

Equilibrium diagram plotted using expressions:

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H+]

[H3PO4] [H2PO4-] [HPO4

2-] [PO4

3-]

Species distribution diagram:

0

0.25

0.5

0.75

1

0 2 4 6 8 10 12 14

pH

Fra

cti

on

al co

mp

osit

ion

H3PO4 H2PO4-

HPO42-

PO43-

0 1 2 3


Page 71

Compact distribution diagram:

0

0.25

0.5

0.75

1

0 2 4 6 8 10 12 14

pH

Alp

ha

H3PO4 H2PO4-

HPO42-

PO43-

First line represents 0 , second 10 , third 210 .

Finding pH from equilibrium diagrams:

Proton condition must be fulfilled.

1. H3PO4

][][3][2][][ 34

2442

OHPOHPOPOHH

From the diagram:

][][ 42 POHH


Page 72

H3PO4

2.3

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-]

[H+]


2-] [PO43-]

Numerical solution:

321212

13

21

142][][][

][][

aaaaaa

a

KKKHKKHKH

HCKCPOH

Proton condition

][][ 42 POHH

Thus

321212

13

21

][][][

][][

aaaaaa

a

KKKHKKHKH

HCKH

Neglecting the last two terms in the denominator:

112 ][][ aa CKHKH

Solution: pH = 2.278


Page 73

Solution can also be found through iteration of:

])[(][ 12 HCKH a

2. NaH2PO4

Proton condition:

][][2][][][ 34

2443

OHPOHPOPOHH

From the diagram

][][][ 2443 HPOPOHH


Page 74

NaH2PO4

4.8

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H+]


2-] [PO43-]

Numerical solution:

][][][ 2443 HPOPOHH

Thus

][

][]][[][ 422

1

42

H

POHK

K

POHHH a

a

][][

1][ 422

1

422

POHK

K

POHH a

a

][

][][

421

42212

POHK

POHKKH

a

aa

From the diagram, in the region where the proton condition is fulfilled,

CPOH ][ 42 , thus:

CK

CKKH

a

aa

1

212][

which yields pH = 4.82


Page 75

3. Na2HPO4

Proton condition:

][][][][2][ 344243

OHPOPOHPOHH

In the region where ][ 24HPO is dominant (around pH = 10):

][][][ 3442

OHPOPOH


Page 76

Na2HPO4

9.5

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H+]


2-] [PO43-]

Numerical solution:

][][][ 3442

OHPOPOH

Thus

][][

][]][[243

2

24

H

K

H

HPOK

K

HPOH wa

a

][][

24

232

2

HPO

KKKKH wa

aa

From the diagram, in the region where the proton condition is fulfilled,

CHPO ][ 24 , thus:

C

KKKKH wa

aa2

322][

which yields pH = 9.52.


Page 77

4. Na3PO4

Proton condition:

][][][2][3][ 244243

OHHPOPOHPOHH

In the region where ][ 34PO is dominant (around pH = 13):

][][ 24

OHHPO

Na3PO4

11.9

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H+]


2-] [PO43-]

Numerical solution:

][][ 24

OHHPO

From the diagram:

CPOHPO ][][ 34

24

][

][][

2433

4

H

HPOKPO


Page 78

Thus

CH

KHPO a

][1][ 32

4

3

24

][

][][

aKH

HCHPO

From the proton condition:

][][

][

3

H

K

KH

HC w

a

][][ 32 HK

C

KH a

w

which yields pH = 11.88.

Sketching equilibrium diagrams for polyprotic acids by hand

4-hydroxybenzoic acid

MC BH310

2

3.41 apK

4.92 apK

2112

2

02][][

][][

aaa KKHKH

HCCBH

2112

11

][][

][][

aaa

a

KKHKH

HKCCHB

2112

212

2

][][][

aaa

aa

KKHKH

KKCCB


Page 79

-14

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

pH

log

C

Line for [H2B]:

1. pH < pKa1 < pKa2

CBH ][ 2

CBH log]log[ 2

0]log[ 2

pHd

BHd


Page 80

2. pKa1 < pH < pKa2

2112

2

2][][

][][

aaa KKHKH

HCBH

][1 HKa , hence 2

1 ][][ HHKa

2][ aKH , hence 211 ][ aaa KKHK . Thus:

11

2

2

][

][

][][

aa K

HC

HK

HCBH

1

12

log

log]log[log]log[

a

a

pKpHC

KHCBH

1]log[ 2

pHd

BHd

3. pKa1 < pKa2 < pH

2112

2

2][][

][][

aaa KKHKH

HCBH

][21 HKK aa , hence 2

121 ][][ HHKKK aaa

Thus:

21

2

2

][][

aa KK

HCBH

21

212

2log

loglog]log[2log]log[

aa

aa

pKpKpHC

KKHCBH

2]log[ 2

pHd

BHd


Page 81

Line for [HB-]:

2112

1

][][

][][

aaa

a

KKHKH

HKCHB

1. pH < pKa1 < pKa2

][][

][][ 1

2

1

H

CK

H

HCKHB aa

pHpKC

HKCHB

a

a

1

1

log

]log[loglog]log[

1]log[

pHd

HBd

2. pKa1 < pH < pKa2

0]log[

pHd

HBd

3. pKa1 < pKa2 < pH

1]log[

pHd

HBd

Line for [B2-

]:

1. pH < pKa1 < pKa2

2]log[ 2

pHd

Bd

2. pKa1 < pH < pKa2

1]log[ 2

pHd

Bd


Page 82

3. pKa1 < pKa2 < pH

0]log[ 2

pHd

Bd

Slopes of the lines for a diprotic acid

Species pH < pKa1 < pKa2 pKa1 < pH < pKa2 pKa1 < pKa2 < pH

[H2B] 0 -1 -2

[HB-] 1 0 -1

[B2-

] 2 1 0

Note: sections having slopes of 2 and -2 are usually unimportant, because they

occur at extremely low concentrations!

When Ka values are spread widely:

1

2][

][][

aKH

HCAH

21

1

][

][

][][

aa

a

KH

HC

KH

KCHA

2

22

][][

a

a

KH

KCA

Does not apply when the pKa values are close.

Example: Pyrophosphoric acid H4P2O7

pKa1 = 0.91 pKa2 = 2.10 pKa3 = 6.70 pKa4 = 9.32


Page 83

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[H4P2O7]

[H3P2O7-]

[H2P2O72-]

[HP2O73-]

[P2O74-]

Proton condition:

][][4][3][2][][ 4

72

3

72

2

722723

OHOPOHPOPHOPHH

1.1

-5

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0 1 2 3 4 5

pH

log

C

[H+]

[H4P2O7]

[H3P2O7-]

[H2P2O72-]

[HP2O73-]


Page 84

What is the pH of 0.1 M NaH3P2O7 (NaH3A)?

Proton condition:

][][3][2][][][ 4

72

3

72

2

722724

OHOPOHPOPHOPHH

From the equilibrium diagram:

][][][ 2

24

AHAHH

4321321

2

21

3

1

4

4

4][][][][

][][

aaaaaaaaaa KKKKHKKKHKKHKH

HCAH

Neglecting the last two terms in the denominator (corresponding to [HA3-

] and

[A4-

]:

211

2

2

4][][

][][

aaa KKHKH

HCAH

Similarly,

211

2

1

3][][

][][

aaa

a

KKHKH

HCKAH

211

2

212

2][][

][aaa

aa

KKHKH

KCKAH

Combined with the approximate proton condition:

0][)(][][ 21211

23

aaaaa KCKHKKCKHH

2

21

211

23

][

][][][

HKK

HKKKHHC

aa

aaa


Page 85

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.50 1.60 1.70 1.80 1.90 2.00 2.10pH

C

pH = 1.74

Asymptote

pH = 1.51

Mixtures of polyprotic acids and bases

Same principle, same technique. Use overlaid equilibrium diagrams and proton

conditions for the mixtures.

Example: 0.1 M solution of (NH4)2HPO4

Two systems: H3PO4 (0.1 M) and NH3 (0.2 M!)


Page 86

Proton condition:

][][][][][2][ 3

434243

OHPONHPOHPOHH

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H+]

[H3PO4]

[H2PO4-]

[HPO42-]

[PO43-]

[NH4+] [NH3]

8.1

-5.5

-4.5

-3.5

-2.5

-1.5

-0.5

6 7 8 9 10 11

pH

log

C

[OH-]

[H3PO4]

[H2PO4-] [HPO4

2-]

[PO43-]

[NH4+] [NH3]


Page 87

Degree of formation n

Average number of protons bound to the acid anion. For H3PO4:

C

POHPOHHPOn

][3][2][ 4342

2

4

Using fractions of ionization:

012 32 n

Degree of dissociation: nN , where N is the maximum number of protons

that can be bound to the acid. For H3PO4, N = 3, hence the degree of

dissociation is n3 :

)32()(3133 0123210 nn

123 233 n

0

0.5

1

1.5

2

2.5

3

0 2 4 6 8 10 12 14

pH

n 3-n

H3PO4

H2PO4-

HPO42-

PO43-

When the pKa values are less separated (e.g. for pyrophosphoric acid):


Page 88

1.5

2

2.5

3

3.5

4

0 1 2 3 4 5 6

pH

n

H4A

H3A-

H2A2-

Titration of a polyprotic acid with a strong base

Charge balance: ][][3][2][][][ 32

2

OHAHAAHNaH

Mass balance:

ba

bb

VV

VCNa

][

Mass balance:

k

k

ba

aa AHVV

VCAHAAHAH ][][][][][ 32

23

k

k AH

AH

][

][ 21 , hence

ba

aa

k

kVV

VCAHAH 112 ][][ , etc.

Thus:


Page 89

][

32][ 321

H

K

VV

VC

VV

VCH w

ba

aa

ba

bb

Monoprotic acid:

][][ 1

H

K

VV

VC

VV

VCH w

ba

aa

ba

bb

For H3PO4, we can also write:

][

3][

H

K

VV

VCn

VV

VCH w

ba

aa

ba

bb

Fraction titrated

aa

bb

VC

VC

Iterative form: ][][3

HOHVC

VVn

aa

ba

Non-iterative form: ][][

][][3

HOHC

HOHn

a

In both cases, when C is large compared to [H+] and [OH

-], n 3


Page 90

0

0.5

1

1.5

2

2.5

3

0 2 4 6 8 10 12 14

pH

3-n

0

2

4

6

8

10

12

14

0 1 2 3 4

pH


Page 91

Titration error

1.. epeT (first equivalence point)

Near the equivalence point,

ba

ba

aa

ba

CC

CC

VC

VV

Also, 0232)()3(13 nnn

Thus

ep

ep

w

ba

ba

epep HH

K

CC

CC][

][21 023

At the second equivalence point, 2 , and the titration error is given by

2ep:

ep

ep

w

ba

ba

epep HH

K

CC

CC][

][

222 013

Systems involving gas phase

CO2:

.)()(22

aqCOgasCO 2

.)]([2 COH

PKaqCO 464.1H

pK

32

.)( HCOHaqCO .)]([

]][[

2

3

1

aqCO

HCOHK

a

363.61

apK

2

33COHHCO

][

]][[

3

2

3

2

HCO

COHK

a 329.10

2

apK

The second reaction is often broken into two steps:

3222.)( COHOHaqCO .)]([][

2

†

32aqCOKCOH 97.2† Kp

332

HCOHCOH ][

]][[

32

3

1

†

COH

HCOHK

38.31

† Kp


Page 92

At equilibrium, [H2CO3] is only 0.1% of [CO2(aq.)], and can be neglected in

equilibrium calculations.

5.7

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H+]

[CO2(aq)]

[HCO3-]

[CO32-

]

-8.5

-7.5

-6.5

-5.5

-4.5

-3.5

-2.5

-1.5

-0.5

3 5 7 9 11

pH

log

C

[OH-] [H+]

[CO2(aq)]

[HCO3-]

[CO32-

]

CT

"true"[H2CO3]

Polyprotic Acids and Bases - Politechnika Gdańska · Polyprotic Acids and Bases . Tadeusz Górecki...

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Transcript of Polyprotic Acids and Bases - Politechnika Gdańska · Polyprotic Acids and Bases . Tadeusz Górecki...