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10 M HAc Buffer Intensity - UMass AmherstAcids/Bases and Buffers: Fundamentals & Buffer Intensity...
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CEE 680 Lecture #17 2/24/2020 1 Lecture #17 Acids/Bases and Buffers: Fundamentals & Buffer Intensity (Benjamin, Chapter 5) (Stumm & Morgan, Chapt. 3 ) David Reckhow CEE 680 #17 1 Updated: 24 February 2020 Print version Buffer Intensity Amount of strong acid or base required to cause a specific small shift in pH David Reckhow CEE 680 #17 f -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 pH 2 3 4 5 6 7 8 9 10 11 12 g -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 pH 3.35 pH 4.7 pH 8.35 StartingPoint Mid-point EndPoint dpH dC dpH dC A B 10 -2 M HAc B C pH B C pH Slope = 1/
Transcript of 10 M HAc Buffer Intensity - UMass AmherstAcids/Bases and Buffers: Fundamentals & Buffer Intensity...
CEE 680 Lecture #17 2/24/2020
1
Lecture #17
Acids/Bases and Buffers: Fundamentals & Buffer Intensity
(Benjamin, Chapter 5)
(Stumm & Morgan, Chapt. 3 )
David Reckhow CEE 680 #17 1
Updated: 24 February 2020 Print version
Buffer Intensity Amount of strong acid or base required to cause a specific small shift in pH
David Reckhow CEE 680 #17 2f
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2
pH
2
3
4
5
6
7
8
9
10
11
12
g
-0.20.00.20.40.60.81.01.2
pH 3.35
pH 4.7
pH 8.35
Starting Point
Mid-point
End Point
dpH
dC
dpH
dC AB
10-2M HAc
BC
pH
BC
pH
Slope = 1/
CEE 680 Lecture #17 2/24/2020
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Buffers: Acetic Acid with Acid/Base Addition 1. List all species present
(use NaOH and HCl as acid/base)
H+, OH‐, HAc, Ac‐, Na+, Cl‐
2. List all independent equations equilibria
Ka = [H+][Ac‐]/[HAc] = 10‐4.77
Kw = [H+][OH‐] = 10‐14
mass balances CT = [HAc]+[Ac‐]
electroneutrality: (positive charges) = (negative charges) Note: we can’t use the PBE because we’re essentially adding an acid and its conjugate base
[Na+] + [H+] = [OH‐] + [Ac‐] + [Cl‐]
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12
3
4
Six total
CA = [Cl-]CB = [Na+]
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Acetic Acid with Acid/Base Addition (cont.) 3. Use ENE, substitute & solve for CB‐CA
[Na+] + [H+] = [OH‐] + [Ac‐] + [Cl‐]
CB + [H+] = Kw/[H+] + KaCT/{Ka+[H
+]} + CA
CB ‐ CA = Kw/[H+] ‐ [H+] + KaCT/{Ka+[H
+]}
4. Take derivative with respect to [H+]
David Reckhow CEE 680 #17 4
4
Kw = [H+][OH-][OH-] = Kw/[H+]
2
3
1,2,
3,4,
5,6
CT = [HAc]+[Ac-][HAc]= CT - [Ac-]
CA = [Cl-]CB = [Na+]
56
1 Ka = [H+][Ac-]/[HAc] Ka = [H+][Ac-]/ {CT-[Ac-]}KaC-Ka[Ac-]= [H+][Ac-]KaC=[Ac-]{Ka+[H+]}[Ac-]=KaCT/{Ka+[H+]}
1+3
CEE 680 Lecture #17 2/24/2020
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Acetic Acid with Acid/Base Addition (cont.) Take the derivative with respect to [H+] of:
CB = CA + Kw/[H+] ‐ [H+] + KaCT/{Ka+[H
+]}
But this is not exactly what we want
Factor out equation
and recall:
David Reckhow CEE 680 #17 5
22][
1][][
HK
KC
H
K
Hd
dC
a
aTwB
dpH
Hd
Hd
dC
dpH
dC BB ][*
][
][303.2][
][303.2
][
303.2
]ln[
303.2
]ln[]log[
HdpH
Hd
H
HdHddpH
HHpH
Acetic Acid with Acid/Base Addition (cont.) so:
and combining:
David Reckhow CEE 680 #17 6
][][303.2
Hd
dCH B
2
22
][
][][
][303.2
][1
][][303.2
HK
HKCH
H
K
HK
KC
H
KH
a
aTw
a
aTw
10][][303.2 TCHOH
2][][
]][[][][303.2
AHA
AHACHOH T
][
][][0
HK
H
C
HA
aT
][
][1
HK
K
C
A
a
a
T
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Example
Trichlorophenol
pKa = 6.00
CT = 10‐2
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f
0.0
0.2
0.4
0.6
0.8
1.0
pH
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Lo
g C
-7
-6
-5
-4
-3
-2
-1
0
H+ OH-
Trichlorophenol Trichlorophenate ion
pH
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
g
0.0
0.2
0.4
0.6
0.8
1.0
pH 4
pH 6.0
pH 9
Starting Point
Mid-point
End Point
See also S&M fig 3.10
David Reckhow CEE 680 #17 8
pH
2 3 4 5 6 7 8 9 10 11 12
Bu
ffer
Inte
nsity
, B (
M/p
H)
0.000
0.001
0.002
0.003
0.004
0.005
0.006
Local Maximimum@ g=0.5
Local Min.@ g=0
Local Min.@ g=1
CEE 680 Lecture #17 2/24/2020
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Equations for polyprotic acids Analogous to the monoprotic systems
monoprotic
diprotic
triprotic
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10][][303.2 TCHOH
2110][][303.2 TT CCHOH
322110][][303.2 TTT CCCHOH
Buffer example
Design a buffer using phosphate that will hold its pH at 7.0 0.05 even when adding 10‐3 moles per liter of a strong acid or base
first determine the required buffer intensity
Next look at the buffer equation and try to simplify based on pH range of interest
David Reckhow CEE 680 #17 10
322110][][303.2 TTT CCCHOH
02.005.0
10 3
dpH
dCB
0 0 0 0
CEE 680 Lecture #17 2/24/2020
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Buffer example (cont.) This gives us the simplified version that can be further simplified
David Reckhow CEE 680 #17 11
M
C
KH
H
K
H
KKH
KKH
H
KK
H
KKH
T
037.0
22.4303.202.0
11303.202.0
11303.202.0
303.2
1
1][1
][
1
][
][][1
][][
][
21
2
2
3
221
2
2322
1
0 0 0 0
Acid Neutralizing Capacity
Net deficiency of protons
with respect to a proton reference level
when the reference level is H2CO3, the ANC=Alkalinity
conservative, not affected by T or P
In a monoprotic system:
[ANC] = [A‐] + [OH‐] ‐ [H+]
= CT1 + [OH‐] ‐ [H+]
David Reckhow CEE 680 #17 12
xf
nf
dpHANC
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CEE 680 Lecture #17 2/24/2020
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David Reckhow CEE 680 #17 15
