first, notice that the ph where two species concentrations are the same is around the pka for that...

First, notice that the pH where two species concentrations are the same is around the pKa for that equilibrium. In fact, for polyprotic acids with pKa's that differ by over 3 to 4 units, the pH is equal to the pKa.

Take for example the point where [H3PO4]=[H2PO4-].

The equilibrium equation relating these two species is

If we take the -log10, or "p", of this equation

Since [H3PO4]=[H2PO4-], and log10(1) = 0, pH=pKa1

pKa1

pKa2 pKa3

Second, you might notice that the concentrations of the conjugate bases are maximum half-way between the pKa points.

For example, the point where [H2PO4-] is a maximum lies half-way between

between pKa1 and pKa2. Since H2PO4- is the major species present in

solution, the major equilibrium is the disproportionation reaction.

This equilibrium cannot be used to solve for pH because [H3O+] doesn't occur in the equilibrium equation. We solve the pH problem adding the first two equilibria equations

+

Since the disproportionation reaction predicts [H3PO4]=[HPO42-]

Note that when we add chemical equilibria, we take the product of the Note that when we add chemical equilibria, we take the product of the equilibrium equations. Taking the -log10 of the last equationequilibrium equations. Taking the -log10 of the last equation

Zwitterions – (German for “Double Ion”) – a molecule that both accepts and losses protons at the same time.

EXAMPLES??? How about – AMINO ACIDS

C

H

COOH

NH2

R C

H

COOH

NH3

R C

H

COO

NH3

R C

H

COO

NH2

R

neutral amino-protonated zwitterion

Both groups protonatedcarboxylic-deprotonated

- why activity of proteins are pH dependent

Let’s look at the simplest of the amino acids, glycine

C

H

COOH

NH3

H C

H

COO

NH3

H C

H

COO

NH2

H

H2Gly+

glycinium

HGly Gly-

glycinate

][

]][[10

2

35.21

GlyH

HGlyHK

K1 K2

][

]][[10 78.9

2 HGly

GlyHK

][][][][ 2 GlyOHHGlyH

][][

][][

]][[ 2

1 H

K

H

HGlyKH

K

HGlyH w

1/][

][][

1

2

KHGly

KHGlyKH w

In water the charge balance would be,

Combining the autoprotolysis of water and the K1 and K2 expressions into the charge balance yields:

HGlyGly-

H2Gly+

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 Equilibriums 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

Polyprotic Acid-Base Equilibria Diprotic Acids and Bases

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

Polyprotic Acid-Base Equilibria

Step 1: Pertinent reactions:

Step 2: Charge Balance:

Step 3: Mass Balance:

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



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:


Polyprotic Acids and Bases

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.

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

][



3.) Example: Determine isoelectric and isoionic pH for 0.10 M alanine.

Solution:For isoelectric point:

10.6)87.934.2()pKpK(pH2

1

2

121

Isoelectric and isoionic points for polyprotic acid are almost the same

first, notice that the ph where two species concentrations are the same is around the pka for that...

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