Buffer Solutions

1
Buffer Solutions Buffers are solutions with the
ability to resist the addition of strong acids or
strong bases, within limits. They play an
important role in chemical processes where it is
essential that a fairly constant pH is
maintained. For example, the pH of your blood
lies at about 7.35. If this value drops below 7.0
(acidosis) the results are fatal. Also if it
rises above 7.7 (alkalosis) the results are also
fatal. Fortunately our blood contains a buffering
system which maintains the acidity at the proper
level. If it were not for the protection of the
buffering system, we could not eat and adsorb
many of the acidic fruit juices and foods in our
diet.
2
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
3
Blood buffer is made up from the dissolved carbon
dioxide in the plasma.  CO2(g) H2O
lt-------gt    H2CO3  lt------gt  HCO3- H3O
  When a base is added it reacts with the
carbonic acid.                        OH-
H2CO3 lt--------gt  HCO3- H2O   When an acid is
added it reacts with the bicarbonate ion.
                     H3O HCO3-   
lt---------gt  H2CO3 H2O
4
There are 3 basic types of calculations that can
be done with buffer systems and these will be
covered next.    1. The pH of a Buffered
Solution Because HA is a weak acid, very little
of it is dissociated at equilibrium. Even if HA
were the only solute. But the solution also
contains A- from the dissolved salt. The presence
of A- suppresses the already slight ionization of
HA by shifting the following equilibrium to the
left. You should recognize this for the common
ion effect , which it is.
5
HA   lt--------gt    H A- ie. The value of HA
at equilibrium HA initially and the
value of A- at equilibrium A- initially.
Therefore we can make 2 assumptions
  HAequilibrium HAfrom the initial acid
acid   A-equilibrium A-from the
initial salt anion   Therefore we can do
this H Ka x acid                   
anion If we take the -log of both sides
-logH -log Ka (-log  acid )
                                         
anion
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Two factors govern the pH in a buffered solution.
(1) pKa of the weak acid (2) ratio of the
initial molar 's of the acid and it's salt.
If a solution has anion acid then the log
anion log 1 0       acid Therefore
the pH of the solution will turn out equal to the
pKa of the weak acid. What mostly determines
where on a pH scale a buffer can work best is the
pKa of the weak acid. Then by adjusting the ratio
of anion to acid, we can cause shifts so that
the pH of the buffered solution comes out on one
side or the other of this value of pH. 
7
Easy Problem Calculating the pH of a Buffered
Solution To study the effect of a weakly acidic
culture medium on the growth of a certain strain
of bacterium, a microbiologist prepared a buffer
solution by making it with 0.11 mol/L NaCH3COO,
sodium acetate, and also 0.09 mol/L CH3COOH,
acetic acid, What is the final pH? Ka for acetic
acid 1.8 x 10-5 Therefore pKa 4.74 acid
0.090 mol/L      anion 0.11 mol/L pH 4.74
log (0.11) 4.74 log 1.2 4.74 0.079
4.82                            (0.9)
8
The Effectiveness of a Buffer (Important
calculation) Suppose we drop 0.01 moles of strong
base into our buffer from the last example. What
will be the measured effects?             HA    
   OH- lt-------gt  A-           H2O start  
0.09 M      -----               0.11 M finish
-0.01 M    -0.1 M           0.01 M (These are
changes)          0.08 M      0.0 M           
0.12 M pH pKa log anion                   
        acid       4.74 log (0.12)
                          (0.08) 4.74
0.18       4.92     The change in pH is small
compared to what it would have been in pure
water!  (Do the calculation !!!)
9
Buffer Capacity No buffer has an unlimited
capacity. ie buffers can only absorb so much
abuse before they are destroyed. The capacity of
a buffer is the amount of acid or base it can
handle before the pH of the solution changes
drastically. If you add enough strong acid to
neutralize all of the buffers basic component,
then additional strong acid will make the pH drop
rapidly. The same applies for a strong base and
the buffers acidic component. The buffer's pH is
a function of it's pKa and the ratio of
concentrations of anion and acid, but the
buffer's capacity depends upon actual
concentrations. 
10
Preparation of A Buffer To prepare a buffer
first choose an acid with a pKa within a 1 pH
unit of the desired value. Then manipulate the
ratios to get the desired pH. A solution buffered
at pH 5.00 is needed in a chemistry experiment.
Can we use acetic acid and sodium acetate to make
it? If so, what ratio of acetate ion to acetic
acid is needed? Ka of acetic acid 1.8 x 10-5
pKa 4.74 the pKa is within a range of 5
1      i.e. between the values of  4 - 6.
therefore, acetic acid is okay to use. next
use  pH pKa log anion                      
                   acid so
5.00 4.74 log anion                        
                 
acid          
100.26 anion 1.8                       
acid                Therefore the ratio of
anion to acid is 1.8 to 1. Therefore, use a ratio
of 1.8 moles of acetate ion to 1.0 mole of acetic
acid. The solution will then be buffered at a pH
of 5.00 OR We could use a ratio of 0.18 moles
to 0.100 moles if we wanted a smaller buffering
capacity.
11
A Basic Buffer NH3 H2O lt-------gt  NH4
OH- Taking the -log of both sides A
generalized version of this basic
Henderson-Hasselbach equation is thus


   
12
Base buffer problem A chemistry student needs
250 mL of a solution buffered at a pH of 9.00.
How many grams of ammonium chloride have to be
added to 250 mL of 0.2 mol/L NH3 to make such a
buffer? (Volume is assumed not to change.) pH
9.0      pOH 5.0 base 0.2        
pKb of ammonia 4.74 (look up ) pOH pKb log
cation                              base
5.00 4.74 log cation                      
         (0.2) cation 100.26 X 0.2
            1.8 0.2             0.36
mol/L of the NH4. But we only need enough for
250 mL  so   0.36 mol  __x___
                                                 
                1000 mL     250 mL so
x 0.09 moles of NH4 ions needed Since the
NH4 comes from NH4Cl then we also need 0.09
moles of NH4Cl. g n mm    0.09 moles
53.5 grams/mole   4.8 grams of the salt are
required. 
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Example 1. What is the pH of a Ca M acid
solution whose acid dissociation constant is Ka?
SolutionLet HA represent the weak acid, and
assume x M of it is ionized. Then, the
ionization and equilibrium concentration is
HA ? H A
Ca-x x x
x2 Kax - CaKa 0
x2
Ka ------
Ca-x
 
-Ka (Ka2 4 CaKa)1/2
 
x ---------------------------
pH -log(x)
2
14
Example 2. Let us make a buffer solution by
mixing Va mL of acid HA and Vs mL of its salt
NaA. let us assume both the acid and the salt
solutions have the same concentration C M. What
is the pH of the so-prepared buffer solution ?
  SolutionAfter mixing, the concentrations Ca
and Cs of the acid HA and its salt NaA
respectively are Ca C Va / (VaVs) and Cs C
Vs / (VaVs) Assume x M of the acid is ionized.
Then, the ionization and equilibrium of the acid
is shown below, but the salt is completely
dissociated.
HA H A-
NaA Na A-
Ca-x x x
Cs Cs
Common ion A- xCs
x(xCs)
Ka --------
Ca-x
x2 (KaCs)x - Ca Ka 0
-(KaCs) ((KaCs)2 4 Ca Ka)1/2
x ----------------------------------------- pH
-log(x)
2
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Example 3 Plot the titration curve when a 10.00
mL sample of 1.00 M weak acid HA (Ka 1.0e-5) is
titrated with 1.00 M NaOH. Solution A. Because
the concentration is high, we use the
approximation H (CaKa)1/2 0.00316
therefore, pH 2.500 B. When 0.1 mL NaOH is
added, the concentration of salt (Cs), and
concentration of acid Ca areCs 0.11.0 M /
10.1 0.0099 MCa 9.91.0 M / 10.1 0.98
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HA H A
-
C
-
x x x
a
A
-
x 0.0099 (
C
)
s
x (x 0.0099)
K

-------------------
1e
-
5
a
0.98
-
x
x
2
0.0099 x 9.8e
-
6
-
1e
-
5 x
x
2
(0.0099 1e5) x
-
9.8e
-
6 0
x
(
-
0.0099 (0.0099
40.981e
-
5)
) / 2 0.000907,
2
1/2
pH 3.042 .
Note the sharp increase in pH when 0.1
mL
of basic solution is
added to the solution.
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19
E.
When 10.0
mL NaOH
is added, the concentration of salt, Cs
10.01.0 M / 20.0 0.500 M and Ca 0.01.0 M /
20.0 0.000.
At
the equivalence point, the solution contains
0.500 M of
the salt
NaA
, and the following equilibrium must be
considered
A
H
O HA OH
-
-
2
HA OH
H

-

C
-
x x x
s
K

---------------
----
b
A
-
H


K
1e
-
14 x
2
w

---

------
1e
-
9
------
K
1e
-
5 C
-
x
a
s
x (0.5001.0e
-
9)
2.26e
-
5
1/2
pOH

-
log x 2.651 pH 14
-
2.651 11.349
Note The calculation here illustrates hydrolysis
of
the basic salt
NaA
.
Follow

up
Sketch the titration curve based on the estimates
given in this
example, and
notice the points made along the way. What
happens when 1 ml of
the
NaOH
is
added beyond the equivalence point ?
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Example of an ampholyte - molecule with both
acidic and basic groups glycine pH 1 NH3 CH2
COOH net charge 1 pH 6 NH3 CH2 COO
net charge 0 zwitterion pH 14 NH2 CH2 COO
net charge 1 pKa values carboxylate
group 2.3 amino group 9.6 can serve as good
buffer in 2 different pH ranges. 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
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• polyampholytes are molecules that have more than
  2 ionizable groups
• lysine NH3- C- (CH2)4 - NH3
• ?
• COOH
• titration of lysine shows 3 pKas pHlt2,
  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
• to determine pI experimentally use
  electrophoresis (see end of Chapter 2)
• Gel electrophoresis-electric field is applied to
  solution of ions, positively charged ions migrate
  to cathode and negatively charged to anode, at
  its pI an ampholyte does not move because net
  charge zero
• Isoelectric focusing- charged species move
  through a pH gradient, each resting at its own
  isoelectric point
• 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

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• 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 ½ ? (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
• 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