Polyprotic Acid-Base Equilibria Introduction 1.)Polyprotic systems Acid or bases that can donate or...

Polyprotic Acid-Base Equilibria

Introduction

1.) Polyprotic systems Acid or bases that can donate or accept more than one proton Proteins are a common example of a polyprotic system

- why activity of proteins are pH dependent

Polymer of amino acids- Some amino acids have acidic or basic substituents

Introduction

1.) Polyprotic systems Amino acids

- Carboxyl group is stronger acid of ammonium group- R is different group for each amino acid

Amino acids are zwitterion – molecule with both positive and negative charge- At low pH, both ammonium and carboxy group are protonated- At high pH, neither group is protonated- Stabilized by interaction with solvent


(acidic)

(basic)

Overall charge is still neutral

Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

2.) Multiple Equilibria Illustration with amino acid leucine (HL)

Equilibrium reactions

low pH high pH

Carboxyl groupLoses H+

ammonium groupLoses H+

Diprotic acid: 11a KK

22a KK


2.) Multiple Equilibria Equilibrium reactions

Diprotic base:1bK

2bK

Relationship between Ka and Kb

w2b1a KKK

w1b2a KKK

pKa of carboxy and ammonium group vary depending on substituents

Largest variations


3.) General Process to Determine pH Three components to the process

Acid Form [H2L+]

Basic Form [L-]

Intermediate Form [HL]

low pH high pH

Carboxyl groupLoses H+

ammonium groupLoses H+


3.) General Process to Determine pH Acid Form (H2L+) Illustration with amino acid leucine

H2L+ is a weak acid and HL is a very weak acid

K1=4.70x10-3 K2=1.80x10-10

21 KK

Assume H2L+ behaves as a monoprotic acid

1a KK


3.) General Process to Determine pH 0.050 M leucine hydrochloride

+ H+

H2L+ HL H+

0.0500 - x x x

K1=4.70x10-3

]H[]HL[M10x32.1xxF

x

LH

HHL107.4K 2

2

2

3a

][

]][[

88.1)M1023.1log(]Hlog[pH 2

M1068.3xF]LH[ 22

Determine [H+] from Ka:

Determine pH from [H+]:

Determine [H2L+]:


3.) General Process to Determine pH Acid Form (H2L+)

What is the concentration of L- in the solution?

[L-] is very small, but non-zero. Calculate from Ka2

][

][][

][

]][[

H

HLKL

HL

LHK 2a

2a

)( 2a10-

2-

2-10-K1080.1

)1032.1(

)1032.1()1080.1(L

][

Approximation [H+] ≈ [HL], reduces Ka2 equation to [L-]=Ka2

]HL[1032.11080.1]L[ 210 Validates assumption


3.) General Process to Determine pH For most diprotic acids, K1 >> K2

- Assumption that diprotic acid behaves as monoprotic is valid- Ka ≈ Ka1

Even if K1 is just 10x larger than K2

- Error in pH is only 4% or 0.01 pH units

Basic Form (L-)

L- is a weak base and HL is an extremely weak base

52aw1b 1055.5K/KK

122aw2b 1013.2K/KK

2b1b KK

Assume L- behaves as a monoprotic base

1bb KK


3.) General Process to Determine pH 0.050 M leucine salt (sodium leucinate)

L- HL OH-

0.0500 - x x x

]OH[]HL[M10x64.1xxF

x

L

OHHL1055.5K 3

2-5

b

][

]][[-

21111010610641

101 123

14.pHM.

.]OH[K]H[ w

M1084.4xF]L[ 2

Determine [OH-] from Kb:

Determine pH and [H+] from Kw:

Determine [L-]:


3.) General Process to Determine pH Basic Form (L-)

What is the concentration of H2L+ in the solution?

[H2L+] is very small, but non-zero. Calculate from Kb2

][][

][

]][[

LHx

xLH

HL

OHLHK 2

222b

]HL[1064.11013.2]LH[ 3122

Validates assumption [OH-] ≈ [HL],

Fully basic form of a diprotic acid can be treated as a monobasic, Kb=Kb1


3.) General Process to Determine pH Intermediate Form (HL)

- More complicated HL is both an acid and base

Amphiprotic – can both donate and accept a proton

Since Ka > Kb, expect solution to be acidic- Can not ignore base equilibrium

Need to use Systematic Treatment of Equilibrium

102aa 1080.1KK

122bb 1013.2KK


Step 1: Pertinent reactions:

Step 2: Charge Balance:

Step 3: Mass Balance:

Step 4: Equilibrium constant expression (one for each reaction):

Diprotic Acids and Bases


1K 2bK

][][][][ --2 OHLLHH

][][][ -LLHHLF 2

][

]][[ -

HL

HLK2

][

]][[

L

OHHLK

-

2b

2K 1bK

][

]][[

LH

HHLK

21

][

]][[

HL

OHLHK

-2

1b

Step 6: Solve:



Substitute Acid Equilibrium Equations into charge balance:

0OHHLLH --2 ][][][][

12 K

HHLLH

]][[][

][

][][ -

H

KHLL 2

0H

KH

H

KHL

K

HHL w2

1

][][

][

][]][[

][][ -

H

KOH w

All Terms are related to [H+]

Multiply by [H+]

0KHKHLK

HHLw2

1

2

2][][

]][[



Step 6: Solve:

0KHKHLK

HHLw2

1

2

2][][

]][[

Factor out [H+]2:

w21

KHLK1K

HLH

][

][][ 2

1KHL

KHLKH

1

w2

][][

][ 2

Rearrange:



Step 6: Solve:

1KHL

KHLKH

1

w2

][][

][ 2

Multiply by K1 and take square-root:

][

][][

HLK

KKHLKKH

1

w112

Assume [HL]=F, minimal dissociation:(K1 & K2 are small)

FK

KKFKKH

1

w112

][



Step 6: Solve:

FK

KKFKKH

1

w112

][

Calculate a pH:

06.6pHM10x80.8

0500.010x70.4

)10x0.1)(10x70.4()0500.0)(10x80.1)(10x70.4(H

7

3

143103

][



Step 7: Validate Assumptions

Assume [HL]=F=0.0500M, minimal dissociation (K1 & K2 are small).

Calculate [L-] & [H2L+] from K1 & K2:

63

7

12 10x36.9

10x70.4

)10x80.8)(0500.0(

K

HHLLH

]][[][

57

102 10x02.1

10x80.8

)10x80.1)(0500.0(

H

KHLL

][

][][ -

[HL]=0.0500M >> 9.36x10-6 [H2L+] & 1.02x10-5 [L-] Assumption Valid



Summary of results: [L-] ≈ [H2L+] two equilibriums proceed equally even though Ka>Kb

Nearly all leucine remained as HL

Range of pHs and concentrations for three different forms

Solution pH [H+] (M) [H2L+] (M) [HL] (M) [L-] (M)

Acid form 0.0500 M H2A 1.88 1.32x10-2 3.68x10-2 1.32x10-2 1.80x10-10

Intermediate form 0.0500 M HA- 6.06 8.80x10-7 9.36x10-6 5.00x10-2 1.02x10-5

Basic form 0.0500 M HA2- 11.21 6.08x10-12 2.13x10-12 1.64x10-3 4.84x10-2


3.) General Process to Determine pH Simplified Calculation for the Intermediate Form (HL)

FK

KKFKKH

1

w112

][

FK

FKKH

1

12

][

Assume K2F >> Kw:

Assume K1<< F:

F

FKKH 12 ][


3.) General Process to Determine pH Simplified Calculation for the Intermediate Form (HL)

F

FKKH 12 ][

Cancel F:

12KKH ][

Take the -log:

)KlogKlog(Hlog 212

1 ][-

)pKpK(pH 212

1 pH of intermediate form of a diprotic acid is close to midway between pK1 and pK2

Independent of concentration:


Diprotic Buffers

1.) Same Approach as Monoprotic Buffer Write two Henderson-Hasselbalch equations

Both Equations are always true

Solution only has one pH

Choice of equation is based on what is known- [H2A] and [HA-] known use pK1 equation- [HA-] and [A2-] known use pK2 equation

][

][

AH

HAlogpKpH

21

][

][

HA

AlogpKpH

2

2


Diprotic Buffers

1.) Example 1: How many grams of Na2CO3 (FM 105.99) should be mixed with 5.00 g of

NaHCO3 (FM 84.01) to produce 100 mL of buffer with pH 10.00?

FM = 84.01 FM = 105.99FM = 62.03

35161 .pKa 329102 .pKa


Diprotic Buffers

1.) Example 2: How many milliliters of 0.202 M NaOH should be added to 25.0 mL of 0.0233 M

of salicylic acid (2-hydroxybenzoic acid) to adjust the pH to 3.50?

97221 .pKa 7132 .pKa


Polyprotic Acids and Bases

1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems Equilibria for triprotic system

For a polyprotic system, would simply contain n such equilibria

Acid equilibria:

Base equilibria:



1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems Rules for triprotic system

1. H3A is treated as a monoprotic acid, Ka = K1

2. H2A- is treated similarly as an intermediate form of a diprotic acid

3. HA2- is also treated similarly as an intermediate form of a diprotic acida. Surrounded by H2A- and A3-

b. Use K2 & K3, instead of K1 & K2

4. A3- is treated as monobasic, with Kb=Kb1=Kw/Ka3

FK

KKFKKH

1

w112

][

FK

KKFKKH

2

w232

][

“End Forms” of Equilibria that Bracket Reactions are Treated as Monoprotic Use Kas that “bracket” or contain form, K1 & K2Use Kas that “bracket” or contain form, K2 & K3



1.) Extend Treatment of Diprotic Acids and Bases to Polyprotic Systems Rules for triprotic system

Treat asMonoprotic acid:

Treat asMonoprotic base:

Treat asIntermediate Forms

Treat asIntermediate Forms

For more complex system, just have additional intermediate forms in-between the two monoprotic acid and base forms at “ends”



3.) Which is the Principal Species? Depends on the pH of the Sample and the pKa values

- At pKa, 1:1 mixture of HA and A-

- For monoprotic, A- is predominant when pH > pKa

- For monoprotic, HA is predominant when pH < pKa

Similar for polyprotic, but several pKa values

pH Major Species

pH < pK1 H2A

pK1 < pH < pK2 HA-

pH > pK2 A2-

Diprotic acidTriprotic acid

Determine Principal Species by Comparing the pH of the Solution with the pKa Values



4.) Fractional Composition Equations Fraction of Each Species at a Given pH Useful for:

- Acid-base titrations- EDTA titrations- Electrochemical equilibria

Combine Mass Balance and Equilibrium Constant

][

]][[

HA

AHKa

][][][][ HAFAAHAF --

][

][

HA

F-[HA]HKa

)(

Rearrange:

aKH

FHHA

][

][][



4.) Fractional Composition Equations Combine Mass Balance and Equilibrium Constant

aKH

FHHA

][

][][

F

HA

AHA

HAHA

][

][][

][

Recall: fraction of molecule in the form HA is:

Divide by F:

aHA

KH

H

F

HA

][

][][Fraction in the form HA:

Fraction in the form A-:a

aA KH

K

F

A

][

][



4.) Fractional Composition Equations Diprotic Systems Follows same process as monoprotic systems

211

2AH

KKKHH

H

F

AH2

][][

][][2

2

Fraction in the form H2A:

Fraction in the form HA-:

211

1HA KKKHH

HK

F

HA

][][

][][2

Fraction in the form A2-:

211

212

A KKKHH

KK

F

A2

][][

][2


Isoelectric and Isoionic pH

1.) Isoionic point – is the pH obtained when the pure, neutral polyprotic acid HA is dissolved in water Neutral zwitterion Only ions are H2A+, A-, H+ and OH-

- Concentrations are not equal to each other

FK

KKFKKH

1

w121

][Isoionic point: pH obtained by simply dissolving alanine

Remember: Net Charge of Solution is Always Zero!



2.) Isoelectric point – is the pH at which the average charge of the polyprotic acid is 0 pH at which [H2A+] = [A-]

- Always some A- and H2A+ in equilibrium with HA Most of molecule is in uncharged HA form

To go from isoionic point (all HA) to isoelectric point, add acid to decrease [A -] and increase [H2A+] until equal- pK1 < pK2 isoionic point is acidic excess [A-]

Remember: Net Charge of Solution is Always Zero!



2.) Isoelectric point – is the pH at which the average charge of the polyprotic acid is 0 isoelectric point: [A-] = [H2A+]

Isoelectric point: )pKpK(pH 212

1

12 K

HHAAH

]][[][

][

][][ -

H

HAKA 2

212

1KKH

H

HAK

K

HHA

][

][

][]][[



3.) Example: Determine isoelectric and isoionic pH for 0.10 M alanine (pK1 = 2.34, pK2=9.87).

Solution:For isoionic point:

11.6pHM107.7

10.010

)101)(10()10.0)(10)(10(

FK

KKFKKH

7

34.2

1434.287.934.2

1

w121

][

For isoelectric point:

1068793422

1

2

121 .)..()pKpK(pH

Isoelectric and isoionic points for polyprotic acid are almost the same

Polyprotic Acid-Base Equilibria Introduction 1.)Polyprotic systems Acid or bases that can donate or...

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Transcript of Polyprotic Acid-Base Equilibria Introduction 1.)Polyprotic systems Acid or bases that can donate or...