Ch.9 POLYPROTIC ACID-BASE EQUILIBRIA - Yonsei …chem.yonsei.ac.kr/~mhmoon/pdf/AnalChem/Ch9.pdf ·...

9.1

Anal. Chem. by Prof. Myeong Hee Moon

Ch.9 POLYPROTIC ACID-BASE EQUILIBRIA

Polyprotic system : those which can donate or accept more than one proton.

Start with diprotic system, then expands the concept to polyprotic

9.2


Amino acid. Box 11-1 H3N+-CH-COO- R: different group

|R

ammonium carboxylicgroup group

9-1. Diprotic acids and bases.

9.3


at low pH : for the case of leucine, R=isobuthyl -CH2CH(CH3)2 : L

Diprotic acid : H2L+ HL + H+ Ka1 K1

HL L- + H+ Ka2 K2Diprotic base : L- + H2O HL + OH- Kb1

HL + H2O H2L+ + OH- Kb2

Ka1 Kb2 = KwKa2 Kb1 = Kw


9.4


9.5


9.6


* In case of 0.0500 M H2L+, 0.0500 M HL, 0.0500 M L-

What is the pH of each solution ?

1. H2L+, K1 = 4.69 x 10-3 (weak acid)

HL, K2 = 1.79 x 10-10

since K2 is very small, dissociated HL (from K1 reaction) only dissociates very little to L-

H2L+ HL + H+

0.0500-x x x

Ka = K1 = = 4.69 x 10-3

x = [HL] = 1.31 x 10-2 M= [H+] pH = 1.88

[H2L+] = 0.0500 - x = 3.69 x 10-2 M left


9.7


[L-] = ? Ka2 = ]HL[

]L][H[

]H[

]HL[K]L[ a

2

= = Ka22

210

10311

1031110791

x.

)x.)(x.(

(if Ka2 is very small, [L-] is negligible)But this is only when H2L

+ alone exists in solution.


9.8


2. Basic form, L- from sodium leucinate.L- + H2O HL + OH- Kb1=Kw/Ka2=5.59x10-5

HL + H2O H2L+ + OH- Kb2=Kw/Ka1=2.13x10-12

from small Kb1 L- will hardly hydrolyze.Very small Kb2

Kb1 Kb as monobasic.

3. Intermediate form, HL : amphiproticHL H+ + L- Ka = Ka2 = 1.79x10-10

HL + H2O H2L+ + OH- Kb = Kb2 = 2.13x10-12

Both K's are very small and close each other,but we can't ignore.


9.9


* In case of 0.0500 M leucine, in which both reactions happen.Charge balance : [H+] + [H2L

+] = [L-]+[OH-]

[H2L+] - [L-]+[H+] -[OH-] = 0

[H2L+] = [L-] = [OH-] =

plug these eqns back to the balance eqn.

1K

]H][HL[

]H[

]HL[Ka

2]H[

Kw

multiply with [H+] for both sides,

[H+] = ]HL[K

KK]HL[KK w

1

121


9.10


Problem ! - [HL] at equilibrium is not calculablewe assume initial concentration (=F) becomes [HL]

(i.e. 0.0500M leucine, F=[HL]=0.0500M pH=6.06)from this calc. [H2L

+]=9.34x10-6M, [L-]=1.02x10-5MAll leucine remain as HL form

* Simple Approximation if K2F >> Kw , K1<<F, ]HL[K

KK]HL[KK]H[ w

1

121

21KK 21 aa KKor

pH = (pK1 + pK2)/2(i.e. for 0.0500 M leucine, pH = 6.04 by the above

pH = 6.06 from full calculation)Regardless to the concentration, pH depends on K.


9.11


9-2 Diprotic Buffers

Buffer from a diprotic acid H2AH2A HA- + H+ K1HA- A-2 + H+ K2

pH = pK1 + log , pH = pK2 + log]AH[

]HA[

2 ]HA[

]A[

2

ex) Find the pH of a solution prepared by dissolving 1.00 g of KHP (fw=204.223) & 1.20 g of disodium phthalate (fw=210.097) in 50.0 mL of water.

9.12


ex) Calculate the volume of 0.800 M KOH needed to make a buffer with 3.38g oxalic acid (fw=92.036) to give pH of 4.40 when diluted to 500 mL. pK1 = 1.252, pK2=4.266

9.13


9-3. Polyprotic acids & bases

for a triprotic system,

H3A H2A- + H+ Ka1 = K1

H2A HA- + H+ Ka2 = K2

HA-2 A- + H+ Ka3 = K3

A-3 + H2O HA-2 + OH- Kb1 = Kw/Ka3

HA-2 + H2O H2A-1 + OH- Kb2 = Kw/Ka2

H2A- + H2O H3A + OH- Kb3 = Kw/Ka1

9.14


SUMMARY---------------------------------------------------------------------1. H3A - treated as monoprotic acid Ka = K12. H2A

- - treated as diprotic acid

[H+] =

3. HA-2 - similar to the above [H+] =

4. A-3 - treated as monobasicKb = Kb1 = Kw/Ka3

FK

KKFKK w

1

121

FK

KKFKK w

2

232


9.15


ex) Find the pH of 0.10 M H3His+2, 0.10M H2His+, 0.10M HHis, 0.10M His-, (His: histidine pK1 = 1.7, pK2=6.02, pK3=9.08).


9.16


9-4. Which is the principal species ?

benzoic acid in an aqueous solution pH=8,

COOH pH=pKa+log

pKa=4.20

]acid benzoic[

]benzoate[

at pH=4.20, [benzoic acid] = [benzoate]at pH>4.20, [benzoic acid] < [benzoate]

predominant form

9.17


5-5. Fractional Composition Equations

How many of each component does exist ?

1. monoprotic system

HA H+ + A- Ka = mass balance:F = [HA] + [A-] = [HA]original

Ka =

[HA] = (10-15)

]HA[

])HA[F](H[

aK]H[

F]H[

F

]A[A

a

HAK]H[

]H[

F

]HA[

9.18


from eq 11-15

aHA

K]H[

]H[

F

]HA[

a

aaA K]H[

K

F

K]H[

F]H[F

F

]A[

In case of pKa = 5.00


9.19


2. Diprotic systems

H2A H+ + HA- K1HA- H+ + A-2 K2

[HA] =

[A-2] = [HA-]

]H[

K]AH[

12

221

22

]H[

KK]AH[

]H[

K

mass balanceF = [H2A] + [HA-] + [A-2]

= [H2A]

22111]H[

KK

]H[

K


9.20


2112

22

2 KK]H[K]H[

]H[

F

]AH[AH

2112

112

KK]H[K]H[

]H[K

]H[

K

F

]AH[

F

]HA[HA

2112

212212

2

2KK]H[K]H[

KK

]H[

KK

F

]AH[

F

]A[A


9.21


In case of fumaric acid pK1 = 3.053, pK2 = 4.494


9.22


9-6 Isoelectric & Isoionic pH

Isoionic point (or iosoionic pH): when pure, neutral polyprotic acid HA is dissolved in water

H2A+ HA A-

alanine alanine alaninecation anion

pKa1=2.35 pKa2=9.87

Isoelectric point (iselectric pH): pH at which the average charge of the polyprotic acid is 0.

[H+] = FK

KKFKK w

1

121

9.23


pH at [H2A] = [A-]

[H+] =

Isoelectric point pH = (pK1 + pK2) / 2

]H[

]HA[K

K

]H][HA[

2

1

21KK

similar to polyprotic acid.

Isoelectric pH

9.24


Homework

5, 12, 18, 20, 23, 28

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Ch.9 POLYPROTIC ACID-BASE EQUILIBRIA - Yonsei …chem.yonsei.ac.kr/~mhmoon/pdf/AnalChem/Ch9.pdf ·...

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