AP Chemistry
Buffers – POGIL
Name:
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Period:
In this activity we will develop principles involved in making a buffer solution and then
determine how a buffer minimizes changes in the pH upon addition of strong acid or base.
Buffers consist of a weak acid and its conjugate (acidic buffer) or a weak base and its conjugate
(basic buffer). Let’s prepare a buffer!
The first step involves obtaining a weak acid. Let’s pick 100.0 mL of a 0.200M solution of
benzoic acid, HC7H5O2. The Ka for benzoic acid is 6.4 × 10-5.
Write the equation for the ionization (dissociation) of benzoic acid in water and indicate the
Bronsted-Lowry acid-base conjugate pairs:
Now calculate the pH of this solution and the percent ionization of the acid:
pH = _______
% ionization = _________
The second step in making our acidic buffer is to add a salt of the acid. In this case let’s pick
sodium benzoate, NaC7H5O2. What ions does sodium benzoate dissociate into in solution?
Which of these ions do you predict would be a “spectator ion”?
Which of these ions is “common” to the acidic solution?
Rewrite the reaction for the ionization benzoic acid in water below:
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AP Chemistry
Buffers – POGIL
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Note this is an equilibrium reaction. What do you predict will happen if you added some sodium
benzoate to this equilibrium solution?
Will the equilibrium shift and if so in what direction?
How will this shift affect the concentrations of unionized benzoic acid and hydronium ions in the
solution and the pH?
predict [HC7H5O2] will _______________
predict [H3O+] will _______________
predict pH will _______________
Now let’s add exactly 0.0200 mole worth of solid sodium benzoate to the 100.0 mL of the
benzoic acid solution. Assume the addition of the solid does not affect the volume of the
solution.
Calculate the molar concentration of benzoate ion (C7H5O2-) in the solution:
Rewrite the equation for the ionization of benzoic acid in water. Note the original concentration
of benzoic acid was 0.200M and the concentration of benzoate ion in this solution is now what
you calculated above. Calculate the pH and percent ionization of the acid after the salt has been
added:
pH = _______
% ionization = _________
Do the new pH and percent ionization values match your previous prediction?
We have just created 100.0 mL of an acidic buffer solution in which the concentration of benzoic
acid and benzoate ion are both essentially 0.200M. This is a buffer system because both the
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Buffers – POGIL
AP Chemistry
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benzoic acid and its conjugate base are now in relatively high concentrations. This means that
there is a relatively large amount of weak acid (HC5H7O2) to neutralize any strong base (OH-)
that might be added to it, and there is a relatively large amount of weak base (C5H7O2-) that can
neutralize any strong acid (H3O+) that might be added to it.
Let’s review, in general formulas, how an acidic buffer is made. Again, a buffer consists of a
weak acid and a salt of the acid (or weak base and its salt).
weak acid:
HA + H2O
add salt:
XA
create buffer:
HA + H2O
=
X+
=
+
=
H3O+ + A-
A- (X+ is spectator ion)
H3O+
+
A-
Note the large amounts of HA and A-. The pH is governed by the ratio of [HA]/[A-].
The real utility of buffers comes when we add a strong acid or base. A buffered solution should
resist pH change when a substantial amount of H+ or OH- is added. ("Substantial amount" will
vary from solution to solution, depending on the "buffer capacity".)
In the case of making the benzoic acid buffer:
weak acid:
HC5H7O2
add salt:
NaC5H5O2
create buffer:
HC7H5O2 + H2O
+
H2O
=
= Na+ +
=
H3O+ + C5H7O2-
C7H5O2- (Na+ is spectator ion)
H3O+
+
C7H5O2-
Now that we have created a buffer, let’s see how it actually works. Suppose we add 5.00 mL of
1.00M hydrochloric acid to our buffer. That’s a lot of strong acid! What component of our
buffer will the strong acid react with?
If you said the benzoate ion (weak base) you are correct! The reaction between the strong acid
(HCl) and weak base (C7H5O2-) will be complete and continue until one runs out (hopefully the
strong acid). This part of the problem involves writing the equation for the neutralization
reaction and doing the stoichiometry (math part) to calculate the molar concentrations of benzoic
acid and benzoate ion after the strong acid has been neutralized.
Write the equation for the reaction between the strong acid added and the weak base in our
buffer below: note, recall any strong acid is represented by H3O+ (the Cl- is a spectator ion).
Under each species in the reaction, indicate the initial moles of each species present (mole = M X
V), the change in the moles after the reaction, and the new equilibrium moles after the reaction is
complete. Then convert to molar concentrations by dividing moles by total volume (in liters).
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Just to keep track of the numbers, we are starting with 100.0 mL of our buffer in which both the
benzoic acid and benzoate ion concentrations are 0.200M. We are adding 5.00 mL of 0.100M
HCl.
Neutralization reaction
I
C
E
[HC7H5O2] = __________
[C7H5O2-] = __________
Now go back to the original buffer and put in the new concentrations and solve for x (the
[H3O+]) and pH using the Ka expression.
buffer:
HC7H5O2 + H2O
=
H3O+
+
C7H5O2-
pH after addition of strong acid = __________
pH of original buffer = __________
Did the buffer work well?
To review in general terms, if we add a strong acid to a buffer, the reaction is complete in which
the strong acid reacts with the weak base:
A-(aq) + H3O+(aq)  HA(aq) + H2O(l)
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Note that the [A-] will decrease and the [HA] will increase but the relative [H3O+] will change
only a little. Keep in mind that the pH isn’t a measure of how much acid (HA) is present. pH is
a measure of the concentration of H3O+.
Now let’s add some strong base to our original benzoic acid buffer. Suppose we add 10.00 mL
of 1.00M sodium hydroxide (that’s a lot of strong base). What component of our buffer will the
strong base react with?
If you said the benzoic acid (weak acid) you are correct! The reaction between the strong base
(NaOH) and weak acid (HC7H5O2) will be complete and continue until one runs out (hopefully
the strong base). This part of the problem involves writing the equation for the neutralization
reaction and doing the stoichiometry (math part) to calculate the molar concentrations of benzoic
acid and benzoate ion after the strong base has been neutralized.
Write the equation for the reaction between the strong base added and the weak acid in our
buffer below: note, recall any strong base is represented by OH- (the Na+ is a spectator ion).
Under each species in the reaction, indicate the initial moles of each species present (mole = M X
V), the change in the moles after the reaction, and the new equilibrium moles after the reaction is
complete. Then convert to molar concentrations by dividing moles by total volume (in liters).
Just to keep track of the numbers, we are starting with 100.0 mL of our buffer in which both the
benzoic acid and benzoate ion concentrations are 0.200M. We are adding 10.00 mL of 1.00M
NaOH.
Neutralization reaction
I
C
E
[HC7H5O2] = __________
[C7H5O2-] = __________
Now go back to the original buffer and put in the new concentrations and solve for x (the [H3O+]
and pH using the Ka expression.
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Buffers – POGIL
AP Chemistry
buffer:
HC7H5O2 + H2O
=
H3O+
+
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C7H5O2-
pH after addition of strong base = __________
pH of original buffer = __________
Did the buffer work well?
If we add a strong base to a solution that contains a weak acid, the reaction is complete in which
the strong base reacts with the weak acid):
HA(aq) + OH-(aq)  A-(aq) + H2O(l)
Again, the [HA] will decrease and the [A-] will increase, but the [OH-] will change only a little,
and consequently the [H3O+] and pH will change only slightly
Notice that in these equations, unless excess strong acid or base is added, no strong acid or base
builds up. Only weak species remain. This is why a buffer works so well.
Your textbook talks about using the Henderson-Hasselbach equation in solving the pH of buffer
problems. You can review this on the last page of this worksheet.
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AP Chemistry
Buffers – POGIL
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Buffer Problems
1. A solution consists of 200.0 mL of 0.800M nitrous acid and 0.500M sodium nitrite. The
Ka for nitrous acid = 4.0 X 10-4.
a) a) write the equation for the ionization of nitric acid in water and indicate the
acid-base conjugate pairs:
b) calculate the pH of this acidic buffer solution:
c) Calculate the pH of the solution after the addition of 5.00 mL of 1.00M nitric
acid:
Neutralization reaction & math part (strong acid reacts completely with weak base in buffer):
Now back to buffer equilibrium with new concentrations:
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Buffers – POGIL
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Now lets try a basic buffer!
2. A solution consists of 250.0 mL of 0.500M ammonia (NH3) and 0.500 M ammonium
chloride. The Kb for ammonia = 1.8 X 10-5.
a) write the equation for this reaction and indicate the acid-base conjugate pairs:
b) calculate the pH of this basic buffer solution:
c) Calculate the pH of the solution after the addition of 10.0 mL of 1.00M nitric
acid:
Neutralization reaction & math part (strong acid reacts completely with weak base in buffer):
Now back to buffer equilibrium with new concentrations:
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AP Chemistry
Buffers – POGIL
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3. A buffered solution is made by adding 7.50 g of methylamine chloride, CH3NH3Cl, to
500.0 mL of a 0.100 M solution of methylamine (CH3NH2). Assume no change in
volume on the addition of the salt. The Kb for methylamine is 4.38 X 10-4.
a) Determine the pH of this buffered solution. Show relevant equation.
b) Exactly 10.0 mL of a 2.00 M hydrochloric acid solution is added to this buffer.
Calculate the resulting pH. Show relevant equations.
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AP Chemistry
Buffers – POGIL
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4. Exactly 15.0 g of sodium lactate, NaC3H5O3, is added to 200.0mL of a 0.400M lactic
acid HC3H5O3, solution. Assume no change in volume upon the addition of the salt. The
Ka for lactic acid is 1.38 X 10-4.
a) Determine the pH of this buffered solution. Show relevant equation.
b) Exactly 10.0 mL of a 1.50 M potassium hydroxide solution is added to this buffer.
Calculate the resulting pH. Show relevant equations.
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All other things being equal, the closer the [HA]/[A-] ratio is to 1, the better will be the buffer
capacity of the solution. This will happen when [H+] = Ka, or pH = pKa. This means that when
selecting a buffer that will be used to maintain the pH of a solution in a particular small range,
The pKa of the buffer should be as close as
possible to the desired pH of the solution
Problem: We wish to buffer a solution at pH = 10.07. Which one of the following bases (and
conjugate acid salts) would be most useful?
a) NH3 (Kb = 1.8 X 10-5)
b) C6H5NH2 (Kb = 4.2 X 10-10)
c) N2H4 (Kb = 9.6 X 10-7)
pH and pKa/ the Henderson-Hasselbach Equation
H+ + A-  HA
[H+] = Ka
[ HA]
[ A ]
-log[H+] = -log(Ka) - log
pH = pKa – log
[ HA]
[ A ]
[ HA]
[ A ]
or, where inverting the log term reverses the sign:
[ A ]
pH = pKa + log [ HA]
Note that if [HA] = [A-], then pH = pKa (and pOH = pKb)
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