First, notice that the pH where two species concentrations are the same is around the pKa for that...
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Transcript of 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
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