titrating polyfunctional acids and bases

Titrating Polyfunctional Acids and Bases

920207 1http:\\asadipour.kmu.ac.ir 40 slides

1. Treating Complex Acid-Base Systems

• Complex systems are defined as solutions made up of:

(1) An acid or base that has two or more acidic protons or basic functional groups

H3PO4

Ca(OH)2

(2) Two acids or bases of different strengths

HCl + CH3COOH

NaOH + CH3COO-920207 2http:\\asadipour.kmu.ac.ir 40

slides

)3) An amphiprotic substance that is capable of acting as both acid and base

HCO3- + H2O CO3

2- + H3O+

HCO3- + H2O H2CO3 + OH-

NH3+CH2COO- + H2O NH2CH2COO- + H3O+

NH3+CH2COO- + H2O NH3

+CH2COOH + OH-


•


34

324

24243

1 POHPOPOHPOH KaKaK a

Kb3Kb2 Kb1

Ka1×Kb3=Kw Ka2×Kb2=Kw Ka3×Kb1=Kw

Ka1=1×10-2 >Ka2=1×10-7> Ka3=1×10-12

Ktotal=Ka1×Ka2× Ka3=1×10-21

pH of H3PO4

1. Calculate the pH of 0.100M H3PO4 solution.


negligibleisKK

Ka

a

a ..1000 22

1

H

POHHKa 100.0

]].[[101 422

1

H+ is not negligible

pH of HA-

pH of HA- solution

HA- A2- + H+

HA- H2A + OH-


Ka2

Kb2

Ka1 1

2

1][

a

HA

waHA

K

CKKC

H

pH of HA-

Calculate the pH of 0.100M NaHCO3 solution.

Ka2×CHA- =1×10-10 ×1.00>>Kw…….Kw is negligible


Ka1=1×10-6 >Ka2=1×10-10

66

a1

HA 101101

00.1

K

C

82

..........1 21

aa pKpKpHleisnegligib

1

2

1][

a

HA

waHA

K

CKKC

H

pH of HA-

Calculate the pH of 0.0100M NaH2PO4 solution.


Ka1=1×10-2 >Ka2=1×10-7> Ka3=1×10-12

Ka2×CHA- =1×10-7 ×0.01>>Kw…….Kw is negligible

1101

01.0

K

C2

a1

HA

negligiblenotis ......1

1

2

1][

a

HA

waHA

K

CKKC

H

1

2

1][

a

HA

aHA

K

CKC

H

pH of HA-

Calculate the pH of 1.00×10-3M Na2HPO4 solution.

Ka2×CHA- =1×10-10 ×0.001=1×10-13 Kw isnot negligible


Ka1=1×10-2 >Ka2=1×10-7> Ka3=1×10-12

4

a1

HA 101101

001.0

K

C7

negligibleis.....1

1

2

1][

a

HA

waHA

K

CKKC

H

1

2][

a

HA

waHA

K

CKKC

H

Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and the other as a weak acid (Ka2 = 1.02 X 10-2). Let us consider how the hydronium ion concentration of sulfuric acid solutions is computed using a 0.0400M solution as an example.

H2SO4 →H+ +HSO4- SO42- + H+

We will first assume that the dissociation of HSO4 is negligible because of the large excess of H30+ resulting from the complete dissociation of H2SO4. Therefore,

This result shows that [SO4- ] is not small relative to [HSO4 ], and a more rigorous so lution is required.

From stoichiometric considerations, it is necessary that

Mixture of weak and strong acids


[SO[SO44] = [H] = [H++] - 0.0400 ] - 0.0400 [H[H++] = 0.0400 + ] = 0.0400 + [SO[SO442-2-]] CCHH22SOSO44, = 0.0400 = [HS0, = 0.0400 = [HS044

-- ] + [SO ] + [SO442-2-] ]

[HSO[HSO44--] = 0.0800 - [H] = 0.0800 - [H33OO++]]

[H[H++] ≈ [HSO] ≈ [HSO44 ] ≈ 0.0400 M ] ≈ 0.0400 M 04.0

][04.0 24

SOKa

Sulfuric acid is unusual in that one of its protons behaves as a strong acid in water and the other as a weak acid (Ka2 = 1.02 X 10-2). Let us consider how the hydronium ion concentration of sulfuric acid solutions is computed using a 0.0400M solution as an example.

H2SO4 →H+ +HSO4- SO42- + H+

We will first assume that the dissociation of HSO4 is negligible because of the large excess of H30+ resulting from the complete dissociation of H2SO4. Therefore,

Mixture of weak and strong acids


])[04.0(

][])[04.0(

][

][][2

4

24

24

4

24

SO

SOSO

HSO

SOHKa

[H[H++] = 0.0400 + ] = 0.0400 + [SO[SO442-2-]] [HSO[HSO44

--] = 0.0400 - ] = 0.0400 - [SO[SO442-2-]]

Curves for the titration of strong acid / weak acid mixture with 0.1000 M NaOH. Each titration is on 25.00 ml of a solution that is 0.1200 M in HCl and 0.0800 M in HA.920207 27http:\\asadipour.kmu.ac.ir 40

slides

Curves for the titration of 25.00 ml of polyprotic acid with 0.1000M NaOH solution .

A)0.1000 M H3PO4,

B) 0.1000M oxalic acid,

C) 0.1000 M H2SO4


Ka1 =5.6 × 10-2 and Ka2 = 5.4 x 10-5

Ka1=1×10-2 >Ka2=1×10-7> Ka3=1×10-12

Ka2 = 1.02 × 10-2

Titration of 20.00 ml of 0.1000 M H2A with 0.1000 M NaOH. For H2A, Ka1= 1.00 × 10–3 and Ka2 = 1.00 × 10–7 .

Titration curves for polyfunctional acids


Titration of 25.00 ml of 0.1000M maleic acid with 0.1000M NaOH.


HOOC-C=C-COOHpKa1=1.89 ,pKa2=6.23

Fractional composition diagram for fumaric acid (trans-butenedioic acid).

Fractional composition diagram for maleic acid (Cis-butenedioic acid).


Z-HOOC-C=C-COOHE-HOOC-C=C-COOH

pKa1=3.05 ,pKa2=4.49 pKa1=1.89 ,pKa2=6.23

amino acids

alanine

The amine group behaves as a base, while the carboxyl group acts as an acid.

Aspartic acid920207 32http:\\asadipour.kmu.ac.ir 40

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1-Determining the pK values for amino acidsAmino acids contain both an acidic and a basic group.


NH2-CH2-COOH +NH3-CH2-COO- Zwitterion formation

+NH3-CH2-COO- + H2O NH2-CH2-COO- + H3O+

+NH3-CH2-COO- + H2O +NH3-CH2-COOH + OH-

]COO-CH2-NH3[

]H3O [] COO-CH2-[NH2-

-

Ka

]COO-CH-NH[

]OH []COOH-CH-NH[-

23

23

Kb

Ka×Kb= ??!!!!

2-Determining the pK values for amino acidsAmino acids contain both an acidic and a basic group.


NH2-CH2-COOH +NH3-CH2-COO- Zwitterion formation

+NH3-CH2-COOH +NH3-CH2-COO- NH2-CH2-COO-Kb Ka

Ka1=5×10-3 Ka2=2×10-10

Curves for the titration of 20.00ml of 0.1000M alanine with

A) 0.1000 M NaOH

B) 0.1000M HCl.


AB

• The zwitterion of an amino acid, containing as it does a positive and a negative charge, has no tendency to migrate in an electric field,

• whereas the singly charged anionic and cationic species are attracted to electrodes of opposite charge.

• NH2-CH2-COO- +NH3-CH2-COOH

• No net migration of the amino acid occurs in an electric field when the pH of the solvent is such that [anionic] = [cationic], which is pH dependent.

• The pH at which no net migration occurs is called the isoelectric point; this point is an important physical constant for characterizing amino acids. The isoelectric point is readily related to the ionization constants for the species. Thus, for glycine,

Iso electric point:The pH at which the average charge of the polyprotic acid is zero


+NH3-CH2-COO-

1-Determining iso electric point for amino acids


+NH3-CH2-COO- Zwitterion formation

]COO-CH-NH[

]OH [] COO-CH-[NH-

23

3-

22

Ka ]COO-CH-NH[

]OH []COOH-CH-NH[-

23

23

Kb

]COOH-CH-NH[] COO-CH-[NHpoint electric isoIn 23-

22

][

][ 3

OH

OH

K

K

b

a

b

a

K

OHKOH

][][ 3

b

wa

K

KOHKOH

][][ 3

3

b

wa

K

KKOH

23 ][

b

wa

K

KKOH

][ 3

2-Determining iso electric point for amino acids


+NH3-CH2-COOH +NH3-CH2-COO- NH2-CH2-COO-

pKa1=2.35 Ka2=9.87

221 aa pKpK

pH

Method1=method2

AHAAH bb KK 12

22aK1aK

HAHAAHOH ab KK2

Kb

KwKaOH 3

2

21213

pKapKaKaKaOH

Ka=Ka2,,,,,,,,,,,,Kb=Kb2 1b2

w

K

Ka

b

w KK

K

213 KaKaKb

KwKaOH

For simple amino acids, Ka and Kb are generally so small that their quantitative determination by neutralization titrations is impos sible.

Amino acids that contain more than one carboxyl or amine group can sometimes be determined. If the Ka values are different enough (104 or more), stepwise end points can be obtained just like other polyfunctional acids or bases as long as the Ka values

Formol titration


+NH3-CH2-COO- + OH- Product

+NH3-CH2-COO- + HCOH CH2=NCH2COOH

titrating polyfunctional acids and bases

Documents

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solution ha a2

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h3po4 solution

strong acid

slidessulfuric acid

h ha h2a oh

nahco3 solution