Transcript Buffers and Titrations

Buffers
A Buffer in Action
Definition of a Buffer


A buffer solution is one which resists changes in
pH when small quantities of an acid or a base are
added to it.
The most important way that the pH of the blood is
kept relatively constant is by buffers dissolved in
the blood.
H2CO3 + H2O HCO3- + OHHCO3- + H+  H2CO3
HCO3- + 3OH  H2O + H2CO3

When a substance acts as both an acid and a base
such as HCO3- does, it is called amphoteric.
How do buffer solutions work?
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A buffer solution contains a weak acid and
base, which removes any hydrogen ions or
hydroxide ions that are add – keeping the pH
constant. (not neutral)
The Bicarbonate Buffer:
weak acid
weak base
H2CO3 + H2O  HCO3- + OH-
Another Buffer solution
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Ammonia is a weak base, and the position of
this equilibrium will be well to the left:

Any added hydrogen ions will react with the
ammonia to make more of the conjugate
acid, maintaining the pH.
Another Buffer solutions

Any added OH- ions will react with the weak
acid to make the conjugate base and again
maintain a constant pH.
Determination of the pH of a Buffer
solution
Example: What is the pH of a solution of 0.11 M
NaC2H3O2 and 0.090 M HC2H3O2?
(ka = 1.8 x 10-5)
HC2H3O2  C2H3O2- +
H+
Initial
0.09
0.11
0
Change -x
+X
+x
Equil.
.09-x
0.11 +X
x
ka = x (0.11) /.09-x
X = 1.47 x 10-5, pH = 4.83

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(buffered solution will have a very small dissociation so + or – x can
be ignored)
Preparing Buffer Solutions
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The Henderson-Hasselbalch Equation gives
the ratio of weak acid to base needed to
maintain a constant pH.
pH = pka + log [A-]/[HA]
(The pH can be approximated: pH = pka +1

Preparing Buffers
1.
Example. What is the mole ratio of acetic acid to
acetate ion needed to prepare a buffer solution at
a pH = 5.00?
ka = 1.8 x 10-5
ka = [H+]
[A-] =
[HA]
pH = 5 so,
mol C2H3O2- = [H+]
mol HC2H3O
ka
[H+] mol C2H3O2mol HC2H3O
[H+] = 1 x 10-5
= 1x 10-5
1.8 x 10-5
= .56
Preparing Buffers
2.
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What is the pH of a solution containing 0.11
M of sodium acetate and 0.090 M acetic
acid? Ka = 1.8 x 10-5
pH = pka + log [A-]/[HA]
pH = 4.74 + log (0.11mol A-)
(0.090 mol HA)
pH = 4.83
Titrations
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Titration of a strong acid by a strong base
The equivalence point occurs at 25.00 mL added base
with a pH of 7.0.
Titrations
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Titration of a weak acid by a strong base

This can be divided into four regions
1)
2)
3)
4)
Before the titration begins: this is simple a
solution of weak acid
During the titration, but before the
equivalence point: the solution is a buffer
At the equivalence point: the solution contains
a salt of the weak acid, and hydrolysis can
occur
Past the equivalence point: the excess added
OH- is used to determine the pH of the
solution
Buffered
Region
The titration curve for the titration of 25.00 mL of 0.200 M
acetic acid with 0.200 M sodium hydroxide. Due to hydrolysis,
the pH at the equivalence point higher than 7.00.
Titrations
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Titration of a weak base by a strong acid
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This is similar to the titration of a weak acid
by strong base
Again dividing into four regions
1)
2)
3)
4)
Before the titration begins: this is a solution of a
weak base in water
During the titration, but before the equivalence
point: the solution is a buffer
At the equivalence point: the solution contains the
salt of the weak base, and hydrolysis can occur
Past the equivalence point: excess added H+
determines the pH of the solution
Buffered
Region
Titration curve for the titration of 25.00 mL of 0.200 M NH3
with 0.200 M HCl. The pH at the equivalence point is below
7.00 because of the hydrolysis of NH4+.
Titrations
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Titration curves for diprotic acids
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The features are similar to those for
monoprotic acids, but two equivalence points
are reached
The titration of
the diprotic acid
H2A by a strong
base. As each
equivalence
point is reached,
the pH rises
sharply.
Titrations
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A few general comments about indicators
can be made
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Most dyes that are acid-base indicators are
weak acids, which can be represented as HIn
The color change can be represented as:

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
HIn ( aq )  H ( aq )  In ( aq )
acid form
base form
(one color)
(another
K HIn 
color)
[ H ][ In  ]
[ HIn ]
Titrations
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The color change will “appear” to the human
eye near the equivalence point of the indicator
At the equivalence point, the concentration of
the acid and base form are equal, so that
pH at the

equivalenc e point
 p K HIn
The best indicators have intense color(s) so
only a small amount will produce an intense
color change that is “easy” to see and won’t
consume too much of the titrant.