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
Diprotic Acids and Bases
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+
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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+]:
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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-]:
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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):
Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
1K 2bK
][][][][ --2 OHLLHH
][][][ -LLHHLF 2
][
]][[ -
HL
HLK2
][
]][[
L
OHHLK
-
2b
2K 1bK
][
]][[
LH
HHLK
21
][
]][[
HL
OHLHK
-2
1b
Step 6: Solve:
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
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][][
]][[
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
Step 6: Solve:
0KHKHLK
HHLw2
1
2
2][][
]][[
Factor out [H+]2:
w21
KHLK1K
HLH
][
][][ 2
1KHL
KHLKH
1
w2
][][
][ 2
Rearrange:
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
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
][
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
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
][
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
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
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
3.) General Process to Determine pH Intermediate Form (HL)
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
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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 ][
Polyprotic Acid-Base Equilibria Diprotic Acids and Bases
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 Acid-Base Equilibria
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
Polyprotic Acid-Base Equilibria
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!
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
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!
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
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
][
][
][]][[
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
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
][
Polyprotic Acid-Base Equilibria
Isoelectric and Isoionic pH
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
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