AP Buffers CHapter 17 powerpoint

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Bicarbonate buffer system
Aqueous
Equilibria
Buffers:
• What are
they?????
• Solutions of a
weak acid and
its conjugate Or
a weak
base/conjugate
Aqueous
Equilibria
Buffers:
• Why are they
important????
• They are
resistant to a
change in pH
when small
quantities of acid
and bases are
added
Aqueous
Equilibria
How do buffers work???
• Since a buffer consists of both an weak
acid or base and its conjugate, an acid
and a base are present in all buffer
solutions. If a small amount of strong
acid is added to the buffer, there is a
base component ready and waiting to
neutralize the “invader”.
Aqueous
Equilibria
Things to remember…
• HA has A- as its conjugate
• NaA is not only the salt, but that
any soluble salt releases the
conjugate
• Ka × Kb = Kw
• Any titration question involving a
weak A/B is really a buffer problem.
Aqueous
Equilibria
Buffers
If a small amount of hydroxide is added to an
equimolar solution of HF in NaF, for example, the HF
Aqueous
reacts with the OH− to make F− and water.
Equilibria
Buffers
If acid is added, the F− reacts to form HF and water.
Aqueous
Equilibria
Buffer Calculations
Consider the equilibrium constant
expression for the dissociation of a
generic acid, HA:
HA + H2O
H 3O + + A −
[H3O+] [A−]
Ka =
[HA]
Aqueous
Equilibria
Buffer Calculations
Rearranging slightly, this becomes
−]
[A
Ka = [H3O+]
[HA]
[HA]
+
[H3O ] = Ka [A-]
acid
base
Aqueous
Equilibria
•If acid is added to the buffer, simply add
acid to the numerator AND subtract the
same quantity from the base since it was
self-sacrificing and neutralized the acid.
•If base is added, simply add the base to
the denominator and subtract from the
numerator.
• Add or subtract in moles NOT
molarity!!!!
Aqueous
Equilibria
When equal concentrations (or
moles) of Acid and Base are
present [which occurs at the ½
equivalence point of a titration]
the ratio of acid to base equals
ONE and therefore,
the pH = pKa.
Aqueous
Equilibria
Buffer Calculations
or:
[base]
pH = pKa + log
[acid]
• This is the Henderson–Hasselbalch equation.
Aqueous
Equilibria
Calculating pH Changes in
Buffers Ex 1
What is the pH of a buffer solution
composed of 0.50 M formic acid and
0.70 M sodium formate before and after
adding 10.0 mL of 1.00 M HCl. The Ka
of formic acid is 1.8 × 10–4
Aqueous
Equilibria
Before:
H+
= Ka
[acid]
[base]
pH = (1.8 
10−4)
(0.50)
(0.70)
pH H+ = 1.286 x 10-4
[pp pH =3.89
Aqueous
Equilibria
After: I added ACID!!!New
Molarities!!!
H+
= (1.8 
= Ka
[acid]
[base]
(0.50L X .50 mole+.01Lx1M/.51 L)
(0.50L x .70 M -.01Lx1M/.51L )
10−4)
H+ = (1.8  10−4) [.26 moles]
[.34 moles]
pH H+ = 1.38 x 10-4
[pp pH =3.86
Aqueous
Equilibria
Calculating pH Changes in Buffers
EX 2
A buffer is made by adding 0.300 mol
HC2H3O2 and 0.300 mol NaC2H3O2 to
enough water to make 1.00 L of
solution. The pH of the buffer is 4.74.
Calculate the pH of this solution after
0.020 mol of NaOH is added. The Ka
of acetic acid is 1.8x 10-5
Aqueous
Equilibria
Before:
H+
= Ka
[acid]
[base]
pH = (1.8 
10−5)
(0.30)
(0.30)
pH pppH =pKa
[pp pH =4.74
Aqueous
Equilibria
After: I added BASE!!!New
Molarities!!!
H+
= (1.8  10−5)
= Ka
[acid]
[base]
(.30 mole-.2mol)
(.30 mole+ .2 mol)
H+ = (1.8  10−5) [.10 moles]
[.50 moles]
pH H+ = 3.6 x 10-5
[pp pH =4.80 (not much of a change!
Aqueous
Equilibria
Example 3
•
•
•
•
•
•
•
•
•
•
A buffered solution contains 0.25 M NH3 ( Kb = 1.8 × 10–5) and 0.40
M NH4Cl.
Calculate the pH of this solution.
b) Calculate the pH when 0.10 mol of gaseous HCl is added to 1.0 L of
the buffered solution.
Aqueous
Equilibria
pH Range
• The pH range is the range of pH values over
which a buffer system works effectively.
• It is best to choose an acid with a pKa close to
the desired pH, the pH range of a buffer is
generally within one pH unit of the pKa
• The greater the concentrations of the
conjugate aced/base pair, the greater the
buffering capacity
Aqueous
Equilibria
Titration
A known concentration
of base (or acid) is
slowly added to a
solution of acid (or
base).
Aqueous
Equilibria
Titration
A pH meter or
indicators are used to
determine when the
solution has reached
the equivalence point,
at which the
stoichiometric amount
of acid equals that of
base.
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Aqueous
Equilibria
Titration of a Strong Acid with
a Strong Base
Just before and after
the equivalence point,
the pH increases
rapidly.
Aqueous
Equilibria
Titration of a Strong Acid with
a Strong Base
At the equivalence
point, moles acid =
moles base, and the
solution contains only
water and the salt from
the cation of the base
and the anion of the
acid.
Aqueous
Equilibria
Titration of a Strong Acid with
a Strong Base
As more base is
added, the increase
in pH again levels
off.
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
• Unlike in the previous
case, the conjugate base
of the acid affects the pH
when it is formed.
• The pH at the equivalence
point will be >7.
• Phenolphthalein is
commonly used as an
indicator in these
titrations.
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
At each point below the equivalence point, the
pH of the solution during titration is determined
from the amounts of the acid and its conjugate
base present at that particular time.
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
With weaker acids,
the initial pH is
higher and pH
changes near the
equivalence point
are more subtle.
Aqueous
Equilibria
Titration of a Weak Base with
a Strong Acid
• The pH at the
equivalence point in
these titrations is < 7.
• Methyl red is the
indicator of choice.
Aqueous
Equilibria
Titrations of Polyprotic Acids
In these cases
there is an
equivalence
point for each
dissociation.
Aqueous
Equilibria
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