Transcript Document

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.
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
Blood buffer is made up from the dissolved carbon
dioxide in the plasma.
CO2(g) + H2O <-------> H2CO3 <------> HCO3- + H3O+
When a base is added it reacts with the carbonic
acid.
OH- + H2CO3 <--------> HCO3- + H2O
When an acid is added it reacts with the bicarbonate
ion.
H3O+ + HCO3- <---------> H2CO3 + H2O
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.
HA <--------> H+ + Aie. The value of [HA] at equilibrium = [HA] initially
and the value of [A-] at equilibrium = [A-] initially.
Therefore we can make 2 assumptions:
[HA]equilibrium = [HA]from 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:
-log[H+] = -log Ka + (-log [ acid] )
[anion]
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.
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)
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- <-------> 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 !!!)
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.
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.
A Basic Buffer:
NH3 + H2O <-------> NH4+ + OH-
Taking the -log of both sides
A generalized version of this basic HendersonHasselbach equation is thus
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.
Example 1.
What is the pH of a Ca M acid solution whose acid
dissociation constant is Ka?
Solution:
Let 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
Ka = -----Ca-x
-Ka + (Ka2 + 4 CaKa)1/2
x = --------------------------2
x2 + K a x - C a K a = 0
pH = -log(x)
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 ?
Solution:
After mixing, the concentrations Ca and Cs of the acid HA and its salt NaA respectively are Ca
= C Va / (Va+Vs) and Cs = C Vs / (Va+Vs) 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+ + ACa-x x x
NaA = Na+ + ACs Cs
Common ion [A-] = x+Cs
x(x+Cs)
Ka=
-------Ca-x
x2 + (Ka+Cs)x - Ca Ka = 0
-(Ka+Cs) + ((Ka+Cs)2 + 4 Ca Ka)1/2
x = ----------------------------------------2
pH = -log(x)
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 are:
Cs = 0.1*1.0 M / 10.1 = 0.0099 M
Ca = 9.9*1.0 M / 10.1 = 0.98
HA = H+ + A
Ca-x
x
-
x
-
[A ] = x + 0.0099 (= Cs)
Ka =
x (x + 0.0099)
------------------0.98 - x
= 1e-5
x2 + 0.0099 x = 9.8e-6 - 1e-5 x
x2 + (0.0099 + 1e5) x - 9.8e-6 = 0
x = (-0.0099 + (0.00992 + 4*0.98*1e-5)1/2) / 2
pH = 3.042 .
= 0.000907,
Note the sharp increase in pH when 0.1 mL of basic solution is
added to the solution.
C.
When 1.0 mL NaOH is added, concentration of salt,
Cs = 1.0*1.0 M / 11.0 = 0.0901 M
Ca = 9.0*1.0 M / 11.0 = 0.818
HA =
Ca-x
H+
x
x (x + 0.0901)
Ka = -----------------0.818 - x
+
Ax + 0.0901
= 1e-5
x = (-0.0901 + (0.09012 + 4*0.818*1e-5)1/2) / 2
= 0.0000907 (any approximation to be made?)
pH = 4.042
Using the Henderson Hasselbalch equation yields
0.0901(=[A-])
pH = pKa + log (-------------------) = 5 - 0.958 = 4.042 (same result)
0.818(=[HA])
D. When 5.0 mL NaOH is added, concentration of salt,
Cs = 5.0*1.0 M / 15.0 = 0.333 M
Ca = 5.0*1.0 M / 15.0 = 0.333 M
HA
Ca-x
Ka =
=
H+ +
x
Ax+0.3333
x (x + 0.333)
------------------------- = 1.0e-5
0.333 - x
x 2 + (0.333-1e-5)x - 0.333*1e-5 = 0
x = (-0.333 + (0.3332 + 4*3.33e-6)1/2) / 2
= 0.0000010
pH = 5.000
Note: Using the Henderson Hasselbalch equation yields the same result
0.333(=[A-]
pH = pKa + log (---------------------) = 5 + 0.000 = 5.000
0.333(=[HA])
E.
When 10.0 mL NaOH is added, the concentration of salt, Cs =
10.0*1.0 M / 20.0 = 0.500 M and Ca = 0.0*1.0 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:
Kb =
[HA] [OH-] [H+]
--------------- ---[A-] [H+]
A- + H2O = HA + OH Cs-x
x
x
Kw 1e-14
x2
= --- = ------ = 1e-9 = -----Ka
1e-5
Cs-x
x = (0.500*1.0e-9)1/2 = 2.26e-5
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 ?
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
pH 14
NH2 – CH2 – COO–
net charge –1
zwitterion
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
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<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
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
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 = 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:
1.
Salting in: increasing ionic strength up to a point (relatively low I), proteins go
into solution
2.
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