# Buffers and the Henderson

```Buffers and the Henderson-Hasselbalch Equation
-many biological processes generate or use H+
- the pH of the medium would change dramatically if it were
not controlled (leading to unwanted effects)
--biological reactions occur in a buffered medium where
pH changes slightly upon addition of acid or base
-most biologically relevant experiments are run in buffers
how do buffered solutions maintain pH under varying
conditions?
to calculate the pH of a solution when acid/base ratio of
weak acid is varied: Henderson-Hasselbalch equation
comes from:
Ka = [H+] [A–] / [HA]
take (– log) of each side and rearrange, yields:
pH = pKa + log ( [A–] / [HA] )
some examples using HH equation:
what is the pH of a buffer that contains the following?
1 M acetic acid and 0.5 M sodium acetate
Titration example (similar one in text:)
Consider the titration of a 2 M formic acid solution with
NaOH.
1. What is the pH of a 2 M formic acid solution?
Ka = [H+] [A–] / [HA]
use
HCOOH
H+ + HCOO–
let x = [H+] = [HCOO–]
then
Ka = 1.78 x 10 –4 = x2 / (2 – x)
formic acid is a weak acid (Ka is small),
x &lt;&lt;&lt; [HCOOH]
and equation becomes Ka = 1.78 x 10 –4 = x2 / 2
so
x = [H+] = [HCOO–] = 0.019
and pH = 1.7
2. Now start the titration. As NaOH is added, what
happens?
•NaOH is a strong base --- completely dissociates
•OH– is in equilibrium with H+ , Kw = [H+] [OH–] = 10–14 ,
•Kw is a very small number so virtually all [OH–] added
reacts with [H+] to form water
Titration continued:
- to satisfy the equilibrium relationship given by Ka
Ka = [H+] [HCOO–] / [HCOOH] = 1.78 x 10 -4
more HCOOH dissociates to replace the reacted [H+] and
-applying HH, see that [HCOO–] / [HCOOH] will increase
pH = pKa + log ( [HCOO–] / [HCOOH] )
-leading to a slow increase in pH as the titration proceeds
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consider midpoint of titration where half of the HCOOH
has been neutralized by the NaOH
[HCOO–] / [HCOOH] = 1
HH becomes: pH = pKa + log 1 = pKa = 3.75 for HCOOH
Titration curve:
- within 1 pH unit of pKa
over most of curve
- so pKa defines the
range where buffering
capacity is maximum
- curve is reversible
Simple problem:
-have one liter of a weak acid (pKa = 5.00) at 0.1 M
-measure the initial pH of the solution, pH = 5.00
-so it follows that initially,
[A–] = [HA] where pH = pKa
-add 100mL of 0.1M NaOH, following occurs
HA + OH– = A–
0.01moles
+
H2 O
-so, 0.01 moles of HA reacted and
new [HA] = 0.1 – 0.01 = 0.09
new [A–] = 0.11
-use HH to get new pH = 5 + log (0.11 / 0.09) = 5.087
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now consider,100mL of 0.1 M NaOH added to 1 L without
the weak acid to see how well the weak acid buffers
0.01 moles OH– / 1.1L = 9.09 x 10 -3 = [OH– ]
use Kw = [OH–] [H+] = 1 x 10 -14
to get pH = 11.96
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what happens when 0.1 moles of base have been added?
what happens when the next 1 mL of base is added?
Known as overrunning the buffer
Sample Buffer Calculation (in text)
-want to study a reaction at pH 4.00
-so to prevent the pH from drifting during the reaction, use
weak acid with pKa close to 4.00 -- formic acid (3.75)
-can use a solution of weak acid and its conjugate base
-ratio of formate ion to formic acid required can be
calculated from the Henderson - Hasselbalch equation:
4.00 = 3.75 + log [HCOO–] / [HCOOH]
[HCOO–] / [HCOOH] = 10 0.25 = 1.78
-so can make a formate buffer at pH 4.0 by using equal
volumes of 0.1 M formic acid and 0.178 M sodium formate
-Alternatively, exactly the same solution could be prepared
by titrating a 0.1 M solution of formic acid to pH 4.00 with
sodium hydroxide.
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some buffer systems controlling biological pH:
1. dihydrogen phosphate-monohydrogen phosphate
pKa = 6.86 - involved in intracellular pH
control where phosphate is abundant
2. carbonic acid-bicarbonate pKa = 6.37, blood pH control
3. Protein amino acid side chains with pKa near 7.0
Example of an ampholyte - molecule with both acidic
and basic groups
NH3+ – CH2 – COOH
net charge +1
pH 6
NH3+ – CH2 – COO–
zwitterion
net charge 0
pH 14
NH2 – CH2 – COO–
net charge –1
glycine: pH 1
pKa values
carboxylate group
amino group
2.3
9.6
can serve as good buffer in 2 different pH ranges
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use glycine to define an important property
isoelectric point (pI) - pH at which an ampholyte or
polyampholyte has a net charge of zero.
for glycine, pI is where:
[NH3+ – CH2 – COOH] = [NH2 – CH2 – COO– ]
can calculate pI by applying HH to both ionizing groups
and summing (see text) yields:
pI = {pK COOH + pK NH 3+ } / 2 = {2.3 + 9.6} / 2 = 5.95
pI is the simple average for two ionizable groups
polyampholytes are molecules that have more than 2
ionizable groups
lysine
NH3+- C- (CH2)4 - NH3+
COOH
titration of lysine shows 3 pKa’s:
• pH&lt;2, exists in above form
•first pKa = 2.18, loss of carboxyl proton
•second at pH = 8.9
•third at pH = 10.28
•need model compounds to decide which amino group
loses a proton first
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to determine pI experimentally use electrophoresis
(see end of Chapter 2)
1. Gel electrophoresis-electric field is applied to solution
of ions, positively charged ions migrate to cathode and
negatively charged to anode, at it’s pI an ampholyte does
not move because net charge = zero
2. Isoelectric focusing- charged species move through a
pH gradient, each resting at it’s own isoelectric point
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Macromolecules with multiples of either only negatively or
only positively charged groups are called polyelectrolytes
polylysine is a weak polyelectrolyte - pKa of each group
influenced by ionization state of other groups
Solubility of macroions (polyelectrolytes and
polyampholytes, including nucleic acids and proteins)
depends on pH.
For polyampholytes:
•high or low pH leads to greater solubility (due to – or +
charges on proteins, respectively)
•At the isoelectric pH although net charge is zero, there are
+ and – charges and precipitation occurs due to:
- charge-charge intermolecular interaction
- van der Waals interaction
•to minimize the electrostatic interaction, small ions (salts)
are added to serve as counterions, they screen the
macroions from one another
Ionic Strength = I = &frac12;  (Mi Zi2)
(sum over all small ions)
M is molarity
Z is charge
Consider the following 2 processes that can take place for
protein solutions:
1. Salting in: increasing ionic strength up to a point
(relatively low I), proteins go into solution
2. Salting out: at high salt, water that would normally
solvate the protein goes to solvate the ions and protein
solubility decreases.
Most experiments use buffers with NaCl or KCl
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