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160:341 lecture Note 4
#6. 26.8 ml 0.2 M HCl is mixed with 50 ml 0.2 M Tris, then pure water is added to a final volume
of 200 ml. Tris has a pKa of 8.3 at 20 ˚C and ∆pK/∆T = -0.029 K-1 (a) what is the pH of Tris buffer
at 20˚C? (b) what is the pH of the buffer at 37˚C? (c) what is the buffering capacity of the Tris buffer
at 20˚C?
(a) In the buffer (200 ml), the [Tris-H+] = 26.8 mM and [Tris] = (50 - 26.8) mM = 23.2 mM.
CS
23.2
= 8.3 + log
= 8.2
CA
26.8
pK
pK37 - pK20
pK37 - 8.3
(b)
=
=
= 0.029  pK37 = 7.8
T
37 - 20
37 - 20
CS
23.2
 pH = pKa + log
= 7.8 + log
= 7.7
CA
26.8
 pH = pKa + log
(c) buffering capacity is defined as ∆pH when 1 mmol of HCl is added to 1 liter of the buffer.
23.2
and
26.8
pH1 = - 0.034
pH1 = 8.3 + log
 pH = pH2 -
22.2
27.8
pH2 = 8.3 + log
#7. What is the pH of pure water at 37˚C? The standard molar enthalpy change for water
dissociation is 55.84kJ/mol.
ln
K2
 HÞ  1
1 
-55840  1
1 
=  log K 2 - log 10 -14 =
K1
R T2
T1 
8.314x2.303 310
298 
 K2 = 10 -13.622 = 2.4x10 -14
 pH = 6.81
5. Amino acids: Amino acids are all zwitterions. Each amino acid has an isoelectric point.
#8. Calculate the concentrations of all the histidine species in 0.1 M histidine at pH 7.0.
pK3 = 10.53
H
N
NH3
+
pK3 = 9.16
COOH
+
+
HN
NH3
pK1 = 1.82
COOH
pK2 = 8.95
pK2 = 6
[h] + [hH] + [hH2] + [hH3] = 0.1 M K 1 =
pK1 = 2.18
[H] [hH 2]
= 10-1.82 = 1.51x10 -2
[hH 3]
[H] [h]
-9.16
-10
= 10
= 6.9x10
[hH]
[hH2 ]
[hH]
[h]
5
-3
= 1.51x10
= 10
= 6.9x10 Approximation: 0.1 M >> [h] and 0.1 M >> [hH3 ]
[hH 3 ]
[hH 2]
[hH]
[hH]
-2
-3
 [hH] + [hH 2] = 0.1 and
= 10  [hH] = 9.1x10 M [hH 2] = 9.1x10 M
[hH 2 ]
K2 =
[H] [hH]
-6
= 10
[hH 2 ]
H3N
+
K3 =
[h] = 6.3x10 -4 M and [hH 3] = 6x10 -8 M
1.How to make an unfavorable reaction (positive ∆G) to proceed?
(a) Coupling of the reaction to another one with a huge negative ∆G.
HOH 2 C
O4POH 2C
O
+ PO4 -3
2
O
²GÞ ' = 16.7 kJ/mol
160:341 lecture Note 4
+)
ATP + H2O
ADP + Pi
∆G˚' = -31 kJ/mol
________________________________________________________________________________________________________
HOH 2 C
O 4POH 2C
O
O
+ ATP
+ AD P
²GÞ ' = -14.3 kJ/mol
(b) LeChatelier Principle
F6P + G3P
E4P + X5P
∆G˚' = 6.3 kJ/mol
At F6P (53x10-5 M), G3P (3.2x10-5 M), E4P (2x10-5 M), and X5P (2.1x10-5 M)
-5
 G' =  GÞ' + RT ln
-5
[E] [X]
[2x10 ] [2.1x10 ]
= 6300 + 8.314x298 ln
= -2.86 kJ/mol
[F] [G]
[53x10 -5 ] [3.2x10 -5 ]
2. ∆G is equivalent to maximal useful work (wmax,useful).
G = H - TS  G = H - (TS) = E + (PV) - (TS) = E + PV + V(P) - TS- S(T)
At constant T and P, P = T = 0  G = E + PV - TS = qrev + wrev + PV - TS
qrev = TS G = qrev + wrev + PV - TS = wrev + PV = wmax , useful
Maximal useful work is the reversible work minus PV work (note wPV = -P∆V)
3.Free energy and electromotive force ∆G = wrev, electric= -nE (e.v.) = -nEF (J/mol)
1 volt x 1 coulomb = 1 joule; 1 amp = 1 coulomb/sec; 1 volt x 1amp = 1 watt;
1 watt x 1 sec = 1 joule; 1 mol electrons = 96500 coulomb; 1 F= 96500 coulomb/mol.
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