tadeusz górecki ionic equilibria acid-base equilibria · acid-base equilibria brønsted-lowry: an...

Tadeusz Górecki Ionic Equilibria

Page 20

Acid-Base Equilibria

Brønsted-Lowry: an acid is a proton donor, a base is a proton acceptor.

HBaseAcid

Neutral molecules (H3PO4, H2O), cations ( 4NH ) and anions (H2PO4

-) can all

behave as acids.

Example:

HNHNH 34

Substances which can behave both as acids and as bases: ampholytes, or

amphiprotic substances (e.g. H2O, SH-).

baseacidSHSH

2

acidbaseSHSHH 2

Free protons cannot exist in any solvent, thus the above reactions are

simplifications. In reality:

OHNHOHNH 3324

Energy required to dissociate H3O+ to H2O and H

+: 258 kcal/mol


1

Equilibrium constant for acid dissociation:

AOHOHHA 32

][

]][[

HA

AHKa

Base protonation:

OHBHOHB 2

][

]][[

B

OHBHKb

Relationship between Ka and Kb:

wba KKK

b

wa

K

KK

a

wb

K

KK

Lewis: an acid is an electron pair acceptor; a base is an electron pair donor.

________________________________________________________________

Strength of acids and bases

24324 SOOHOHHSO

33232 HCOOHOHCOH

CNOHOHHCN 32

1221 baseacidbaseacid

2

a 101.0HSO

SOOHK

][

]][[

4

2

43

7

a 104.4COH

HCOOHK

][

]][[

32

33


Page 22

10

a 104.0HCN

CNOHK

][

]][[ 3

Larger value of Ka means that the acid is stronger, thus:

HCNCOHHSO 324

Ion product of water:

OHHOH2

Equilibrium constant using activities:

OH

OHH

a

aaK

2

0

Activity of water is by thermodynamic convention proportional to the mole

fraction of water in the solution. In dilute solutions it is close to 1.

Activity of water can be included in the constant:

0][][ wOHHKOHHaa

"Concentration" constant:

wKOHH ]][[

/00 2

w

OH

w Ka

KK


Page 23

At 50°C, pKw = 13.26, and the neutral point is pH = 6.63. At 25°C in 3 M

NaClO4 pKw = 14.18, and the neutral point is pH = 7.09.


Page 24

Non-aqueous solvents:

2433 NHNHNHNH

At -60°C, the equilibrium constant is:

3224 10]][[ NHNHK

Thus, the pH scale (defined as -log[NH4+]) in liquid ammonia ranges from 0 to

32.

pH of a strong acid

Initially PH, or "potential of hydrogen", defined as

PH = -log CH

Today's definition of pH:

)]log([log HapapH HH

General approach

Example: HCl

Mass balance: HACCl ][

Ion product of water: 1410]][[ wKOHH

Charge balance: ][][][ ClOHH

Solution: HAw C

H

KH

][][

This is a quadratic equation, which applies always.

When CHA >> 10-7

, [H+] = CHA ([OH

-] is negligibly small)

At higher ionic strength, activity coefficient should be used.


Page 25

Strong base:

Example: NaOH

Mass balance: bCNa ][


Charge balance: ][][][ OHNaH

Solution: ][

][

H

KCH w

b

Basic solution, thus [H+]<<[OH

-], and in general

b

w

C

KH ][

pH = pKw + log Cb

____________________________________

Example: pH of 7102 M solution of NaOH

0][][ 2 wb KHCH

2

4][

2wbb KCC

H

LmolH /1014.4][ 8 pH = 7.38

Simplified equation: pH = 7.30

____________________________________


Page 26

pH of strong acid/base as a function of concentration:

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

0 2 4 6 8 10 12 14

pH

log

Clog C(acid) log C(base)

Mixture of a strong acid and a strong base

Example: HCl and NaOH

Mass balance: aCCl ][

Mass balance: bCNa ][


Charge balance: ][][][][ OHClNaH

][][

H

KHCC w

ba

When the acid and the base are neutralized:

LmolKOHH w /10][][ 7


Page 27

Titration of Strong Acids and Bases

Volume of the system changes, thus amounts must be taken into mass balances

rather than concentrations.

Example: titration of HCl with NaOH:

Mass balance: aaba VCVVCl )]([

Mass balance: bbba VCVVNa )]([


Charge balance: ][][][][ OHClNaH

][][

H

K

VV

VC

VV

VCH w

ba

aa

ba

bb

At the equivalence point, CaVa=CbVb (1:1 stoichiometry) and [H+]=[OH

-]

Before the equivalence point, [H+]>>[OH

-]:

ba

bbaa

VV

VCVCH

][

After the equivalence point, [H+]<<[OH

-]:

ba

aabbw

VV

VCVC

H

K

][

aabb

baw

VCVC

VVKH

)(

][


Page 28

In the vicinity of the equivalence point (Ca = 0.1 M, Va = 50 mL, Cb = 0.2 M):

5

6

7

8

9

24.999 24.9995 25 25.0005 25.001

Vb

pH

OH- neglected

Full equation

H+ neglected

Plotting the titration curve

Strong acid titrated with strong base:

]/[][

]/[][

HKHC

HKHCVV

wb

waab

Titration of strong base with strong acid:

]/[][

]/[][

HKHC

HKHCVV

wa

wbba


Page 29

Example:

Titration curve

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

Vb

pH

Conductometric titration:

HCl titrated with NaOH

][][][][ ClOHNaH ClOHNaH

- overall conductance ( ][ 11 cmk )

X - equivalent conductance.

At 25°C, limiting conductances 0 are:

8.3490 H

10.500 Na

1.1990 OH

35.760 Cl

Procedure:

assume the values of pH;


Page 30

calculate [H+] and [OH

-];

calculate V from the titration curve equation;

calculate )/(][ 0VVVCNa b (from mass balance);

calculate )/(][ 00 VVVCCl a (from mass balance).

50 mL 0.1 M HCl titrated with 0.2 M NaOH:

0

5

10

15

20

25

30

35

0 10 20 30 40 50

Vb

Co

nd

uc

tan

ce


0.006

0.0065

0.007

0.0075

0.008

0.0085

0.009

0.0095

0.01

0.0105

0.011

40 50 60

Vb

Co

nd

uc

tan

ce


Page 31

Titration error

'

'

V

VVerrorTitration

ep

Vep - V at end point

V' - V at equivalence point

Titration of 50 mL 0.1 M HCl with 0.2 M NaOH:

Enlarged section (end point detected with methyl red at pH = 5):


Page 32

Titration error:

%015.025

259963.24100

________________


Titration error:

%2050

5040100

Gran plots

Titration of a strong acid with a strong base:

][][

H

K

VV

VC

VV

VCH w

ba

aa

ba

bb

Before the equivalence point, [OH-] is negligibly small, thus:

bbaaba VCVCHVV ])[(


Page 33

or bbaapH

ba VCVCVVf 10)(1

Ca , Cb and Va are constant, thus a plot of f1 as a function of Vb should be a

straight line with a slope of bC intersecting the X axis at the equivalence

point, V' = CaVa /Cb.

Example: 50 mL 0.0001 M HCl titrated with 0.0002 M NaOH:

0

0.00005

0.0001

0.00015

0.0002

0.00025

0.0003

0.00035

0.0004

0.00045

0.0005

22 23 24 25 26 27

Vb [mL]

f1

In the vicinity of the equivalence point:

0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

0.00007

0.00008

0.00009

0.0001

24.5 24.6 24.7 24.8 24.9 25 25.1 25.2 25.3

Vb

f1


Page 34

Weak monoprotic acids and bases

AHHA

][

]][[

HA

AHKa

OHBHOHB 2

][

]][[

B

OHBHK b

wba KKK

wba pKpKpK


Page 35

NH4+: pKa = 9.24

NH3: pKb = 4.76

Dependence of pKa on ionic strength:

00

0

][

]][[

aa KHA

AHK

00 logloglog aa pKpK

Using Davies equation and setting bI0log (activity coefficient for an

uncharged molecule):

bIIbI

IpKa

'

151.02757.4

where b' is the Davies coefficient (usually 0.2).


Page 36

Best fit: b'=0.3, b = 0

Temperature dependence of apK 0 :


Page 37

Calculating the pH of weak acid

Known: CHA, Ka, Kw

Unknown: [H+], [OH

-], [A

-], [HA]

][

]][[

HA

AHKa

]][[ OHHK w

Mass balance: ][][ HAACHA

Charge balance: ][][][ OHAH

a

a

aa

HAK

KHA

K

HA

K

AHAC

][][

][1][

]][[][

a

aHA

KH

KCA

][][

From mass balance:

a

HA

KH

HCHA

][

][][

From Kw:

][][

H

KOH w

Substituting [A-] and [OH

-] into charge balance:

][][][

H

K

KH

KCH w

a

aHA

Thus:

0)]([][][ 23 wawaHAa KKKKCHKHH


Page 38

Simplifying assumption: [OH-] is negligibly small

a

aHA

KH

KCH

][][

0][][ 2 aHAa KCKHH

When aKH ][ :

][][

H

KCH aHA aHAKCH 2][ aHAKCH ][

Flood's diagram

From mass and charge balances:

][][

][

H

KH

K

KHC w

a

aHA

Flood's diagram

-8

-6

-4

-2

0

0 2 4 6 8

pH

log C

Strong acid

pKa=4.75

pKa=7.53

pKa=10.72


Page 39

Degree of dissociation

a

aHA

KH

KCA

][][

a

HA

KH

HCHA

][

][][

Degree of dissociation:

a

a

HA KH

K

AHA

A

C

A

][][][

][][

Degree of formation:

aHA KH

H

AHA

HA

C

HA

][

][

][][

][][1

Degree of dissociation and formation

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12 14

pH

degree of dissociationdegree of formation


Page 40

Sillén's diagram (EquiligrapH, equilibrium diagram)

Acetic acid, 0.01 M, pKa = 4.75

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H

+]

[A-]

[HA]

1. [H+] is determined from the definition of pH:

pHH ]log[

2. [OH-] is determined from the same definition:

ww pKpH

H

KOH

][log]log[

3. [A-] is determined from mass balance and Ka:

a

aHA

KH

KCA

][][

4. [HA] is determined from mass balance and Ka:

a

HA

KH

HCHA

][

][][


Page 41

pH of a given system can be determined from the proton condition:

][][][ OHAH

Acidic solution, thus [OH-] can be neglected

][][ AH

Solution for the proton condition can be easily found on equilibrium diagrams

using the pointer function:

][][][log OHAH

Acetic acid, 0.01 M, pKa = 4.75

3.4

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H

+]

[A-]

[HA]

Pointer


Page 42

Acetic acid, 10-7

M, pKa = 4.75

6.8

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

Pointer

[OH-][H

+]

[A-][HA]

Plotting equilibrium diagrams

1. [H+] and [OH

-]: straight lines at 45° angles (slopes of -1 and +1,

respectively)

2. [A-]:

a

aHA

KH

KCA

][][

for pH < pKa, Ka << [H+]

][][

H

KCA aHA

pHpKCA aHA log]log[

1]log[

dpH

Ad

for pH > pKa, Ka >> [H+] HACA ][


Page 43

3. [HA]:

a

HA

KH

HCHA

][

][][

for pH < pKa, Ka << [H+] HACHA ][

for pH > pKa, Ka >> [H+]

a

HA

K

HCHA

][][

aHA pKpHCHA log]log[

1]log[

dpH

HAd

4. When pH = pKa:

2][][ HAC

HAA

3.0log2loglog2

log]log[]log[ HAHA

HA CCC

HAA

What is the pH of 0.001 M NaAc?

PBE:

][][][ OHHAH

][][ OHH , thus PBE:

][][ OHHA


Page 44

0 2 4 6 8 10 12 14

-14

-12

-10

-8

-6

-4

-2

0

pH

log C

once

ntr

atio

n [

M]


Page 45

What happens when the acid concentration is close to Ka?

0.001 M HF, pKa = 3.17

Proton condition: ][][][ OHFH

][][ FH

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H

+]

[F-]

[HF]


Page 46

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

Pointer

pH=3.26

[OH-]

[H+]

[F-]

[HF]

Checking the results:

pH = 3.26

[OH-] = 10

-10.74

[HF] = 10-3.34

)(101010

1010

][

]][[ 17.318.3

34.3

26.326.3

aa KtrueHF

FHK

Mass balance:

)(10101010][][ 3997.234.326.3HFCtrueHFF

Algebraic solution:

0][][ 2 aHAa KCKHH

41051.5][ H

pH = 3.26


Page 47

Low concentrations of very weak acids:

5 x 10-5

M HCN, pKa = 9.32

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-][H

+]

[HCN] [CN-]

[CN-]+[OH

-]

Mixture of acids

Strong acids represented by an unreactive anion.

Strong bases represented by an unreactive cation.

Typically not depicted on EquiligrapHs.

Each weak acid or base represented by the expressions:

a

aHA

KH

KCA

][][

a

BH

KH

HCBH

][

][][

pH found at the point where PBE is fulfilled.


Page 48

0.01 M HAc (Ka = 10-4.75

) and 0.001 M HFo (Ka = 10-3.75

)

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H

+]

[Ac-][HAc]

[HFo] [Fo-]

[Ac-]+[Fo

-]+[OH

-]

Proton condition: ][][][][ OHFoAcH

Pointer function:

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

Pointer

pH=3.3

[OH-] [H

+]

[Ac-][HAc]

[HFo] [Fo-]


Page 49

Assumptions:

[OH-] negligibly small

[H+] >> Ka

][][][][

H

K

KH

KC

KH

KCH w

f

ff

a

aa

f

ffaa

KH

KC

H

KCH

][][][

Iteration: f

ff

aaKH

KCHKCH

][

][][ 2

Circular reference: f

ff

aaKH

KCHKCH

][

][][

Mixture of strong and weak acid:

0.001 M HCl and 0.01 M HAc

][][][][ OHClAcH

][][

][][

H

KCl

KH

KCH w

a

aa

[H+] >> [OH

-] and Ka

]][[][ 2 ClHKCH aa

Solution: pH = 2.94


Page 50

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H

+]

[Ac-]

[HAc]

[Cl-]

[Ac-]+[Cl

-]+[OH

-]

Salt of a weak acid and a weak base

Two independent weak acid systems linked by the condition that they have the

same total concentration.

][]][[ 1 HAKAH a

][]][[ 2 BHKBH a

wKOHH ]][[

mass balances: ][][][][ BBHAHAC

charge balance: ][][][][ OHAHBH

Proton condition: ][][][][ BOHHHA


Page 51

If [H+] and [OH

-] are small:

2

2

1 ][][

][

a

a

a KH

CK

KH

HC

212][ aa KKH

21][ aa KKH

pH does not depend on C (provided the assumption above is fulfilled)!

Example: pH of 0.01 M NH4Ac (pKa1 = 4.75, pKa2 = 9.25)

00.7)(2/1 21 aa pKpKpH

The value of 7 is coincidental.

Equilibrium diagram:

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

[OH-] [H

+]

[Ac-]

[HAc][NH3]

[NH4+]


Page 52

Pointer function:

-14

-12

-10

-8

-6

-4

-2

0

0 2 4 6 8 10 12 14

pH

log

C

Pointer

pH=7

[OH-] [H

+]

[Ac-]

[HAc][NH3]

[NH4+]

Full solution:

][][][

][

][

2

2

1

H

K

KH

CKH

KH

HC w

a

a

a

Quartic equation in [H+].

The equation is linear in C, thus:

2

2

1 ][1

][

1][

][][

a

a

a

w

KH

K

KHH

HH

K

C


Page 53

Example: Dimethylammonium acetate

pKa1 = 4.75 (acetic acid), pKa2 = 10.76 (dimethylamine)

7

7.1

7.2

7.3

7.4

7.5

7.6

7.7

7.8

0 2 4 6 8

-log C

pH

pH = 7.755

General equation for the titration curve

Mass balance:

ba

aa

VV

VCAHA

][][

Mass balance:

ba

bb

VV

VCNa

][

][

]][[

HA

HAKa

Charge balance: ][][][][ OHANaH


Page 54

a

a

ba

aa

KH

K

VV

VCA

][][

][][][

H

K

KH

K

VV

VC

VV

VCH w

a

a

ba

aa

ba

bb

or ][

][

H

K

VV

VC

VV

VCH w

ba

aaHA

ba

bb

][][

][][

][

H

KHC

H

KH

KH

KC

VVw

b

w

a

aa

ab

][][

][][

H

KHC

H

KHC

VVw

b

wHAa

ab

Titration of a strong acid with a strong base:

][][

][][

H

KHC

H

KHC

VVw

b

wa

ab

Simplifying assumptions: before the equivalence point [H+]>>[OH

-]; after the

equivalence point [H+]<<[OH

-]


Page 55

Example: titration of 10 mL 0.1 M HAc with 0.1 M NaOH

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14Vb

pH

Derivative curve

)(

)(

)(

)(

bb V

pH

V

pH

End point determined by the maximum of the derivative curve. If necessary,

second derivative can be obtained in the same manner.

First derivative:

0

2

4

6

8

10

12

14

0 2 4 6 8 10 12 14Vb

pH

d(pH)/dVb


Page 56

First and second derivative:

-8

-6

-4

-2

0

2

4

6

8

10

12

14

9.90 9.95 10.00 10.05 10.10

Vb

Another way to plot the titration curve: through fraction titrated :

aa

bb

VC

VC

Substitution to the general equation:

][

][

)(

][H

H

K

VC

VV

KH

K w

aa

ba

a

a

Iteration – set baab CVCV / and get the fraction titrated


Page 57

0

2

4

6

8

10

12

14

0 1 2Fraction titrated

pH

Non-iterative solution:

b

a

aa

bb

C

OHH

C

OHH

VC

VC

][][1

][][

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH


Page 58

The plot of pH vs. fraction titrated enables easy comparisons of different

titration curves.

Examples:

Different volumes of the acid

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH

Different Ka values

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH

Ka = 1e-6

Ka = 1e-5

Ka = 1e-4

Ca = Cb = 0.01 M

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40Vb

pH

Va = 10 mL

Va = 20 mL

Va = 30 mL

0

2

4

6

8

10

12

14

0 5 10 15 20Vb

pH

Ka = 1e-6

Ka = 1e-5

Ka = 1e-4

Ca = Cb = 0.01 M


Page 59

Different concentrations of the acid:

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH

Ca = 0.0001 M

Ca = 0.001 M

Ca = 0.01M

pKa = 4.75

Cb = 0.01 M

Different concentrations of the base:

0

2

4

6

8

10

12

14

0 1 2

Fraction titrated

pH

Cb = 0.0001 M

Cb = 0.001 M

Cb = 0.01 M

pKa = 4.75

Ca = 0.01 M

0

2

4

6

8

10

12

14

0 5 10 15

Vb

pH

Ca = 0.0001 M

Ca = 0.001 M

Ca = 0.01 M

pKa = 4.75

Cb = 0.01 M

0

2

4

6

8

10

12

14

0 500 1000 1500 2000

Vb

pH

Cb = 0.0001 M

Cb = 0.001 M

Cb = 0.01 M

pKa = 4.75

Ca = 0.01 M


Page 60

Titration of a weak acid with a weak base:

][][

H

K

VV

VC

VV

VCH w

ba

aa

A

ba

bb

BH

b

BH

a

A

aa

bb

C

OHH

C

OHH

VC

VC

][][

][][

where

BHB

BH KH

H

C

BH

][

][][ ;

HA

HA

HAA KH

K

C

A

][

][

0

2

4

6

8

10

12

14

0 1 2Fraction titrated

pH

Strong base

Kb = 5

pKa = 4.75

pKb = 5

Ca = 0.01 M

0

2

4

6

8

10

12

14

0 5 10 15 20

Vb

pH

Strong base

Kb = 5

pKa = 4.75

pKb = 5

Ca = 0.01 M

Va = 10 mL


Page 61

Titration error

At the equivalence point 1

aabb VCVC '

where Vb’ is Vb at the equivalence point

b

aab

C

VCV '

Titration error:

1111''

'..

ep

aa

epb

b

aa

ep

b

ep

b

bep

VC

VC

C

VC

V

V

V

V

VVeT

At the equivalence point and its vicinity:

b

ab

a

ba

C

CC

V

VV

Also, near the equivalence point, pH >> pKa , thus [H+] << Ka

aaa

a

K

H

KH

H

KH

K ][

][

][1

][1

][

][

)(11 H

H

K

VC

VV w

aa

baep

a

w

aa

baep

K

HH

H

K

VC

VV ][][

][

)(1

a

ep

ep

ep

w

ba

baep

K

HH

H

K

CC

CC ][][

][

)(1


Page 62

Example: 0.1 M HAc, 0.1 M NaOH, pHep = 8 instead of pH = 8.92 at the

equivalence point

%)054.0(104.510

10)1010)(20(1 4

375.4

886

ep

Titration of a weak base with a strong acid:

ep

a

ep

wep

ba

baep

H

K

H

KH

CC

CC

][][][

)(1

tadeusz górecki ionic equilibria acid-base equilibria · acid-base equilibria brønsted-lowry: an...

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