http://en.wikipedia.org/wiki/Acid_Queen
Bio 98 - Lecture 2
Acid-Base Equilibria,
pH and Buffers
pH profiles of enzymatic reactions
Amylase
Pepsin
UCI Bio199 Independent Research
Pure water is only slightly ionized
H2O
H+ + OH-
H+ ions (protons) do not persist free in solution, they are
immediately hydrated to hydronium ions (H+ + H2O
H3O+).
Proton wire
Achieving equilibrium
H2O
H+ + OH-
[H+] [OH-]
Keq =
= 1.8x10-16 M
[H2O]
Concentration of “water in water” ([H2O]) is 55.6 M [next slide], thus
Kw = [H+] [OH-] = 10-14 M2
Constant ion product!
Pure water has equal quantities of H+ and OH- ions, or, put differently,
pure water has equal [H+] and [OH-].
[H+] = [OH-] = 10-7 M = 0.1 mM
[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.
Base: proton acceptor
The pH scale
pH = log
1
+]
=
-log
[H
[H+]
neutral/pure water has
[H+] = [OH-] = 10-7 M, so
pH = -log(10-7) = -(-7) = 7
Logarithm (base 10) refresher:
if log(x)=y then x=10y
Acid: proton donor
Strong acids and bases
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.
Dissociation of a weak acid or weak base
O
R-C-OH
O
R-C-O- + H+
(C-term/Asp/Glu)
R-NH3+
R-NH2 + H+
(N-term/Lys)
HA
[H+] [A-]
Ka = –––––––––
[HA]
A-
+ H+
(general)
= acid dissociation constant
Weak acids and weak bases
Acetic acid is a weak acid as it does not completely dissociate in water.
O
O
H3C
C OH
C O-- + H+
Ac-
HAc
Keq = Ka =
H3C
[H+] [Ac-]
[HAc]
= 1.7x10-5 M
[H+] [OH-]
Recall for water: K =
= 1.8x10-16 M with [H2O] = 55.6 M!
eq
[H2O]
pKa = -log(Ka) = -log(1.7x10-5 M) = 4.8
pKa and pH
[Ac-][H+]
Ka =
[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.
Titration curves
Ka =
[H+] [Ac-]
[HAc]
-]
[Ac
pH = pKa + log
[HAc]
2
Start at low pH and begin to add HO-.
The product of [H+] [HO-] must
remain constant, so adding HOmeans [H+] must decrease and thus
pH increases. At the pKa, [Ac-] and
[HAc] are equal, so adding more HOdoes 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).
Measuring pKa values
NH4+
Ka =
H+ + NH3
[H+] [NH3]
[NH4+ ]
pKa = pH when [NH4+] = [NH3]
The Henderson-Hasselbalch Equation
HAc
H+
+
Ac-
take the -log on both sides
apply p(x) = -log(x)
Ka =
[H+] [Ac-]
[HAc]
-log Ka = -log
pKa =
[H+]
pH
[Ac-]
-log
[HAc]
-]
[Ac
-log
[HAc]
and finally solve for pH…
-]
[Ac
pH = pKa + log
[HAc]
[Proton acceptor]
= pKa + log
[Proton donor]
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]
[Acetic acid]
10x then *[Acetic acid]
on both sides
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
How does a buffer work?
-
At the pKa, [HAc] = [Ac-] so the system is able to absorb the addition of HOor H+. If we add HO- near the pH where [Hac] = [Ac-] (ie pH ~= pKa) then
HAc can release H+ to offset the HO- added but the ratio of HAc to Ac- does
not change much. If we add H+ then Ac- can absorb H+ to form HAc.
Hence, the pH does not change much.
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+ + HPO42-
pKa2 = 7.2
HPO42-
H+ + PO43-
pKa3 = 12.7
Carbonate
CO2 + H2O
H2CO3
H2CO3
H+ + HCO3-
pKa1 = 6.4
HCO3-
H+ + CO32-
pKa2 = 10.2
Phosphate buffering
Carbon dioxide - carbonic acid - bicarbonate buffer
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
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.
What happens to blood pH when you hyperventilate?
What happens to blood pH when you hypoventilate?
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.
O
R-C-OH
O
R-C-O- + H+
[H+] [A-]
Ka =
_________
[HA]
and
[H+]=[A-]
[HA]=0.15-[H+]
[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)
ax2 +
b
x +
c
= 0 Quadratic Formula
[H+] = 1.53x10-3 M and thus pH = 2.8
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.
O
R-C-OH
O
R-C-O- + H+
[H+] [A-]
Ka =
_________
[HA]
and
[H+]=[A-]
[HA]=0.15-[H+]
[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+]2 = 2.37x10-6 M2 Assumption: [H+] << 0.15 M!
[H+] = 1.54x10-3 M and thus pH = 2.8
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?
KH2PO4
H2PO4- + K+
H2PO4-
and
K2HPO4
HPO42- + H+
[HPO42-]
pH = 7 = pKa + log
[H2PO4-]
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]
HPO42- + 2 K+
pKa = 6.86
Another HH calculation
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-.
(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?
-]
[Ac
pH = pKa + log
HH equation
[HAc]
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
(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).