1. Acid-Base Equilibria,
pH and Buffers

2. pH profiles of enzymatic reactions
UCI Bio199 Independent Research
Pepsin
Amylase

3. H2O H+
+ OH-
Pure water is only slightly ionized
H+
ions (protons) do not persist free in solution, they are
immediately hydrated to hydronium ions (H+
+ H2O H3O+
).
Proton wire

4. Achieving equilibrium
Kw = [H+
] [OH-
] = 10-14
M2
[H+
] = [OH-
] = 10-7
M = 0.1 µM
[H+
] [OH-
]
[H2O]
Concentration of “water in water” ([H2O]) is 55.6 M [next slide], thus
Pure water has equal quantities of H+
and OH-
ions, or, put differently,
pure water has equal [H+
] and [OH-
].
Constant ion product!
= 1.8x10-16
MKeq =
H2O H+
+ OH-

5. [H2O]
Concentration is measured in moles per liter (mol/l) or simply M.
1 l = 1,000 ml of water has a mass of 1,000 gr.
1 mole of water has a mass of 18 gr (hydrogen 1 Da, oxygen 16 Da).
Thus 1 liter of water (1,000 gr) contains 1,000 gr / 18 gr moles of water.
[H2O] = (1,000 gr / 18 gr) M = 55.6 M.

6. pH = log
[H+
]
1
= -log [H+
]
pH = -log(10-7
) = -(-7) = 7
The pH scale
neutral/pure water has
[H+
] = [OH-
] = 10-7
M, so
Logarithm (base 10) refresher:
if log(x)=y then x=10y
Base: proton acceptor

7. pH = -log [H+
]
HCl H+
+ Cl-
HCl is a strong acid that completely dissociates in
water. 1 M HCl will thus yield 1 M [H+
] and the pH will be
pH = -log [H+
] = -log(1) = 0
NaOH is a strong base that completely dissociates in
water. 1 M NaOH will thus yield 1 M [OH-
]. Since
[H+
] [OH-
] = 10-14
M and must remain constant
[H+
] = 10-14
M and the pH will be
pH = -log [H+
] = -log(10-14
) = 14
Life is compatible only in a narrow pH range around pH 7.
Strong acids and bases

8. Dissociation of a weak acid or weak base
[H+
] [A-
]
Ka = ––––––––– = acid dissociation constant
[HA]
R-C-OH R-C-O-
+ H+
(C-term/Asp/Glu)
R-NH3
+
R-NH2 + H+
(N-term/Lys)
HA A-
+ H+
(general)
O O

9. Weak acids and weak bases
H3C C
O
OH H3C C
O
O- + H+
HAc Ac-
Keq = Ka =
[H+
] [Ac-
]
[HAc]
= 1.7x10-5
M
pKa = -log(Ka) = -log(1.7x10-5
M) = 4.8
Acetic acid is a weak acid as it does not completely dissociate in water.
-
[H+
] [OH-
]
[H2O]
= 1.8x10-16
M with [H2O] = 55.6 M!Keq =Recall for water:

10. Ka =
[Ac-
][H+]
[HAc]
when [Ac-
] and [HAc] are equal then Ka = [H+
]
pKa = 4.8 = -log(Ka) = -log [H+
]
But since pH is defined as -log [H+
]
The pKa = pH when the concentrations of Ac-
and HAc
are equal.
pKa and pH

11. pH = pKa + log
[Ac-
]
[HAc]
Ka =
[H+
] [Ac-
]
[HAc]
Start at low pH and begin to add HO-
.
The product of [H+
] [HO-
] must
remain constant, so adding HO-
means [H+
] must decrease and thus
pH increases. At the pKa, [Ac-
] and
[HAc] are equal, so adding more HO-
does not change the ratio of [Ac-
] to
[HAc] very much and thus the pH
does not change very much (shallow
slope of titration curve from ~1 pH
unit below pKa to ~1 pH unit above).
Titration curves
2

12. Measuring pKa values
NH4
+
H+
+ NH3
[NH4
+
]
[H+
] [NH3]
Ka =
pKa = pH when [NH4
+
] = [NH3]

13. Ka =
[H+
] [Ac-
]
[HAc]
take the -log on both sides
The Henderson-Hasselbalch Equation
-log Ka = -log [H+
] -log
[Ac-
]
[HAc]
pH = pKa + log [Ac-
]
[HAc]
= pKa + log
[Proton acceptor]
[Proton donor]
HAc H+
+ Ac-
pKa = pH -log [Ac-
]
[HAc]
apply p(x) = -log(x)
and finally solve for pH…

14. Acetic acid has a pKa of 4.8. How many ml of 0.1 M
acetic acid and 0.1 M sodium acetate are required to
prepare 1 liter of 0.1 M buffer with a pH of 5.8?
Substitute the values for the pKa and pH into the Henderson-Hasselbalch
equation:
5.8 = 4.8 + log [Acetate]
[Acetic acid]
1.0 = log [Acetate] 10x
then *[Acetic acid]
[Acetic acid]
10 [Acetic acid] = [Acetate]
For each volume of acetic acid, 10 volumes of acetate must be added (total
of 11 volumes).
Acetic acid needed: 1/11 x 1,000 ml = 91 ml
Acetate needed: 10/11 x 1,000 ml = 909 ml
on both sides

15. Buffers are vitally important in biochemical systems
since pH needs to be controlled. Living systems must
be “buffered” to resist large variations in pH.
Phosphate
H3PO4 H+
+ H2PO4
-
pKa1 = 2.2
H2PO4
-
H+
+ HPO4
2-
pKa2 = 7.2
HPO4
2-
H+
+ PO4
3-
pKa3 = 12.7
Carbonate
CO2 + H2O H2CO3
H2CO3 H+
+ HCO3
-
pKa1 = 6.4
HCO -
H+
+ CO 2-
pK = 10.2

16. Phosphate buffering

17. CO2 + H2O H2CO3
H2CO3 H+
+ HCO3
-
pKa = 6.4
HCO3
-
CO3
-2
+ H+
pKa = 10.2 (not relevant, far from pH 7.4)
Carbon dioxide - carbonic acid - bicarbonate buffer

18. Carbon dioxide - carbonic acid - bicarbonate buffer
What happens to blood pH when you hyperventilate?
What happens to blood pH when you hypoventilate?
If blood pH drops due to metabolic
production of H+
then [H2CO3] increases
by protonation of HCO3
-
, H2CO3 rapidly
loses water to form CO2(aq), which is
expelled as CO2(g).
If the blood pH rises, [HCO3
-
] increases
by deprotonation of H2CO3, then
breathing rate changes and CO2(g) is
converted to CO2(aq) and then to H2CO3
in the capillaries in the lungs.

19. What is the pH of 0.15 M acetic acid?
The pKa of acetate is 4.8, so the Ka = 10-4.8
M = 1.58x10-5
M.
[H+
] [A-
]
Ka = _________
[HA]
[H+
]2
[H+
]2
Ka = ––––– = ––––––––– = 1.58x10-5
M
[HA] 0.15 M - [H+
]
[H+
]2
+1.58x10-5
M [H+
] + (-2.37x10-6
M2
) = 0 (ax2
+bx+c = 0)
[H+
] = 1.53x10-3
M and thus pH = 2.8
R-C-OH R-C-O-
+ H+
OO and
[H+
]=[A-
]
[HA]=0.15-[H+
]
ax2
+ b x + c = 0 Quadratic Formula

20. What is the pH of 0.15 M acetic acid?
The pKa of acetate is 4.8, so the Ka = 10-4.8
M = 1.58x10-5
M.
[H+
] [A-
]
Ka = _________
[HA]
[H+
]2
[H+
]2
Ka = ––––– = ––––––––– = 1.58x10-5
M
[HA] 0.15 M - [H+
] Assumption: [H+] << 0.15 M!
[H+
]2
= 0.15 M * 1.58x10-5
M
[H+
] = 1.54x10-3
M and thus pH = 2.8
R-C-OH R-C-O-
+ H+
OO and
[H+
]=[A-
]
[HA]=0.15-[H+
]
[H+
]2
= 2.37x10-6
M2
Assumption: [H+] << 0.15 M!

21. Your 199 prof asks you to make a pH 7 buffer. You
already have 0.1 M KH2PO4. What concentration of K2HPO4
do you need?
pH = 7 = pKa + log
7 = 6.86 + log(x / 0.1 M)
0.14 = log(x / 0.1 M)
100.14
= x / 0.1 M
x = 0.138 M = [K2HPO4]
[HPO4
2-
]
[H2PO4
-
]
H2PO4
-
HPO4
2-
+ H+
pKa = 6.86
KH2PO4 H2PO4
-
+ K+
and K2HPO4 HPO4
2-
+ 2 K+

22. Make 200 ml of 0.1 M Na acetate buffer pH 5.1, starting
with 5.0 M acetic acid and 1.0 M NaOH.
Strategy
1. Calculate the total amount of acetic acid needed.
2. Calculate the ratio of the two forms of acetate (A-
and
HA) that will exist when the pH is 5.1.
3. Use this ratio to calculate the % of acetate that will be in
the A-
form.
4. Assume that each NaOH will convert one HAc to an Ac-
.
Use this plus the % A-
to calculate the amount of NaOH
needed to convert the correct amount of HAc to Ac-
.
Another HH calculation

23. (1) How much acetic acid is needed?
200 ml x 0.1 mol/l = 200 ml x 0.1 mmol/ml = 20 mmol
5.0 mol/l x x ml = 5.0 mmol/ml x x ml = 20 mmol
x = 4.0 ml of 5.0 M acetic acid are 20 mmol
(2) What is the ratio of Ac-
to HAc at pH 5.1?
5.10 - 4.76 = log[A-
]/[HAc], thus [Ac-
]/[HAc] = 2.19 / 1
(3) What fraction of total acetate is Ac-
at pH 5.1?
[Ac-
] [Ac-
] 2.19
–––– = 2.19; –––––––––– = ––––––– = 0.687 or 68.7%
[HAc] [HAc] + [Ac-
] 1 + 2.19
pH = pKa + log HH equation
[Ac-
]
[HAc]

24. (4) How much OH-
is needed to obtain 68.7% Ac-
?
Na+
+ OH-
+ HAc ⇒ Na+
+ Ac-
+ H2O
mmol NaOH = 0.687 x 20 mmol = 13.7 mmol
1.0 mol/l x x ml = 1.0 mmol/ml x x ml = 13.7 mmol
x = 13.7 ml of 1.0 M NaOH
(5) Final answer (Jeopardy…)
• 4.0 ml of 5.0 M acetic acid
• 13.7 ml of 1.0 M NaOH
• Bring to final volume of 200 ml with water
(ie add about 182.3 ml of H2O).