Lesson 17: Buffers
Buffer solution: - Solutions in equilibrium that will maintain a RELATIVELY constant
pH when small amounts of acid or base are added to them.
Q: How do you make a buffer?
A: You need two ingredients:
1. Either a weak acid or a weak base
2. A weaker CONJUGATE substance in the form of a salt.
e.g.,
Let’s prepare a buffer solution using Acetic acid (a weak acid)
Step #1: Write the equilibrium equation for the weak acid or base you are using.
(show the relative concentrations of each species)
CH3COOH ====== H+(aq) + CH3COO-(aq)
[HIGH]
[LOW]
[LOW]
Step #2: Identify the conjugate species and add a salt that contains this species.
(show the relative concentrations of each species)
ADD THE SALT NaCH3COO (Sodium ion is a spectator b/c it’s from group #1
CH3COOH ====== H+(aq) + CH3COO-(aq)
[HIGH]
[LOW]
[HIGH]
Q: Why add a salt?
A:
1. A weak acid without a salt is not effective as a buffer when acid is added.
2. A weak base without a salt is not effective as a buffer when base is added.
Q: Why is the buffer ineffective without the salt??!
A: There is only a small concentration of the conjugate species available in a weak acid
or weak base equilibrium. This means that the equilibrium will only be able to shift a
VERY SMALL amount to the left.
Q: How does adding the appropriate salt convert our solution into an effective buffer
solution?
A: Addition of the appropriate salt will increase the concentration of the CONJUGATE
SPECIES in equilibrium. This allows the equilibrium to shift a SIGNIFICANT
amount of material to the left and the right. i.e., it allows the equilibrium to consume
both small amounts of acid and base!
Q: What happens to the ion that is not part of the equilibrium?
A: It will be a SPECTATOR ION.
Q: How do I ensure that the spectator ions in the salt will really be spectators?
A: For cation spectators select from group #1 or #2 from the periodic table.
For anion spectators select from strong CON ACID.
** Recall the hydrolysis lesson. These are the ions that do not react with water!
Exercises:
Prepare buffers for the following substances:
1. HCN ===== H+(aq) + CN-(aq)
[High]
[Low]
[Low]
Add a salt with CN(e.g., NaCN)
HCN ===== H+(aq) + CN-(aq)
[High]
[Low]
[High]
2. NH3 + H2O(l) ==== OH-(aq) + NH4+(aq) Add a salt with NH4+
[High]
[Low]
[Low]
(e.g., NH4Cl)
NH3 + H2O(l) ==== OH-(aq) + NH4+(aq) Add a salt with NH4+
[High]
[Low]
[Low]
Q: What will happen to the overall pH of both buffers (1 and 2) if a small amount of acid
is added?
A: The HCN equilibrium will shift to the LEFT.
The NH3 equilibrium will shift to the RIGHT.
The overall pH of both buffer solutions will DECREASE slightly.
Q: Can HCl be used to make a buffer? Explain.
A: NO. HCl is strong so it does not form an equilibrium system.
Q: Which of the following salts would be the best choice for preparing a buffer with
NaHC2O4?
NaF, NaCN, Na2C2O4, NH4HC2O4
A: Na2C2O4. C2O4-2 is the only conjugate species available. Also
HC2O4- is amphiprotic with Ka > Kb, so the original substance is a
weak acid.
Calculations with Buffers
Since buffers are equilibrium systems, the calculations that govern them are the same for
any ICE box.
1. Calculate the approximate pH for a buffer prepared using 1.0M acetic acid and 1.0M
sodium acetate.
Step #1: Write the dissociation equation for the weak acid.
Step #2: Prepare an ICE box.
CH3COOH(aq) === H+(aq) + CH3COO-(aq)
Initial
1.0M
-x
1.0M - x
Change
Equilibrium
0.0M
+x
x
1.0M
+x
1.0 + x
Insignificant. Also, the system shifts right b/c there is no H+
Step #3: Use Ka (for weak acid equilibrium) or Kb (for weak base equilibrium) to
calculate for the “ion of interest” (i.e., H+ for acid or OH- for base).
Ka =
[𝑯+][𝑪𝑯𝟑𝑪𝑶𝑶−]
[𝑪𝑯𝟑𝑪𝑶𝑶𝑯]
Ka =
(𝒙)(𝟏.𝟎)
(𝟏.𝟎)
1.8 x 10-5 M = x = [H+]
Sub in from pg.6 of data booklet.
Step #4: Roadmap.
pH = -log[H+]
pH = -log(1.8 x 10-5 M)
**** pH = pKa ****
pH = 4.74
2. Calculate the approximate pH of the buffer prepared using 2.0M NH3 and 2.0M
NH4NO3.
NH3(aq) + H2O(l) ===== OH-(aq) + NH4+(aq)
Kb =
𝑲𝒘
𝑲𝒂 𝒐𝒇 𝑵𝑯𝟒+
=
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒
𝟓.𝟔 𝒙 𝟏𝟎−𝟏𝟎
= 1.7857 x 10-5
pOH = pKb
pOH = -log[OH-] = -log(1.7857 x 10-5) = 4.748188
Therefore, the pH = 9.25
3. How would you create a buffered solution with a pH around 4.5. List the acid/base
needed as well as the conjugate species.
pH = pKa
[H+] = Ka
[H+] = 2nd log –pH
[H+] = 3.16 x 10-5 = Ka
Use Acetic Acid as it has a Ka of 1.8 x10-5 and use NaCH3COO
as your conjugate base.