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INTRODUCTION TO BUFFERS
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
WHAT IS A BUFFER COMPOSED OF?
To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired pH when preparing the buffer. For example, the following could function as buffers when together in solution:
Acetic acid (weak organic acid w/ formula CH COOH) and a salt containing its conjugate base, the acetate anion (CH COO ), such as sodium acetate (CH COONa) Pyridine (weak base w/ formula C H N) and a salt containing its conjugate acid, the pyridinium cation (C H NH ), such as Pyridinium Chloride. Ammonia (weak base w/ formula NH ) and a salt containing its conjugate acid, the ammonium cation, such as Ammonium Hydroxide (NH OH)
HOW DOES A BUFFER WORK?
A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H O and OH ) when the are added to the solution. To clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F ion and solvated protons (H O ), which does not allow it to dissociate completely in water. Therefore, if we obtain HF in an aqueous solution, we establish the following equilibrium with only slight dissociation (K (HF) = 6.6x10 , strongly favors reactants):
We can then add and dissolve sodium fluoride into the solution and mix the two until we reach the desired volume and pH at which we want to buffer. When Sodium Fluoride dissolves in water, the reaction goes to completion, thus we obtain:
Since Na is the conjugate of a strong base, it will have no effect on the pH or reactivity of the buffer. The addition of to the solution will, however, increase the concentration of F in the buffer solution, and, consequently, by Le Châtelier’s Principle, lead to slightly less dissociation of the HF in the previous equilibrium, as well. The presence of significant amounts of both the conjugate acid, , and the conjugate base, F , allows the solution to function as a buffer. This buffering action can be seen in the titration curve of a buffer solution.
3 3
3 5 5 5 5 +
3 4
3
3
a
H F ( aq ) + H 2 O ( l ) ⇌ F (^) (− aq ) + H 3 O + ( aq ) (1)
Na F ( aq ) + H 2 O ( l ) → N a + ( aq ) + F (^) (− aq ) (2)
HF -
As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted through neutralization. This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity.
If acid were added:
In this reaction, the conjugate base, F , will neutralize the added acid, H O , and this reaction goes to completion, because the reaction of F with H O has an equilibrium constant much greater than one. (In fact, the equilibrium constant the reaction as written is just the inverse of the K for HF: 1/K (HF) = 1/(6.6x10 ) = 1.5x10 .) So long as there is more F than H O , almost all of the H O will be consumed and the equilibrium will shift to the right, slightly increasing the concentration of HF and slightly decreasing the concentration of F , but resulting in hardly any change in the amount of H O present once equilibrium is re-established.
If base were added:
In this reaction, the conjugate acid, HF, will neutralize added amounts of base, OH , and the equilibrium will again shift to the right, slightly increasing the concentration of F in the solution and decreasing the amount of HF slightly. Again, since most of the OH is neutralized, little pH change will occur.
These two reactions can continue to alternate back and forth with little pH change.
SELECTING PROPER COMPONENTS FOR DESIRED PH
Buffers function best when the pK of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the K for hydroflouric acid is 6.6 x 10 so its pK = -log(6.6 x 10 ) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18.
For the weak base ammonia (NH ), the value of K is 1.8x10 , implying that the K for the dissociation of its conjugate acid, NH , is K /K =10 /1.8x10 = 5.6x10. Thus, the pK for NH = 9.25, so buffers using NH /NH will work best around a pH of 9.25. (It's always the pK of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pK of the conjugate base, obviously.)
When the desired pH of a buffer solution is near the pK of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a simple approximation of the solution pH, as we will see in the next section.
EXAMPLE 1: HF BUFFER
In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of 3.0. We can use the Henderson-Hasselbalch approximation to calculate the necessary ratio of F and HF.
F (^) (− aq ) + H 3 O + ( aq ) ⇌ H F ( aq ) + H 2 O ( l ) (3)
- 3 +
- 3 + a a -4^ +3^ -^3 +^3 +
- 3 +
H F ( aq ) + O H (^) (− aq ) ⇌ F (^) (− aq ) + H 2 O ( l ) (4)
a
a -4^ a -
3 b -5^ a 4 + w b -14^ -5^ -10^ a 4 +^4 +^3 a b
a
pH = pKa + log
[ Base ] [ Acid ]
3.0 = 3.18 + log
[ Base ] [ Acid ]
log = −0.
[ Base ] [ Acid ]
[ Base ] [ Acid ]
10 −0.18^ (8)
0.066 initial moles F - 0.010 moles reacted with H O = 0.056 moles F remaining
Also during this process, more HF is formed by the reaction:
0.10 initial moles HF + 0.010 moles from reaction of F with H O = 0.11 moles HF after reaction
Plugging these new values into Henderson-Hasselbalch gives:
pH = pK + log (base/acid) = 3.18 + log (0.056 moles F /0.11 moles HF) = 2.
Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of strong acid.
REFERENCES
- Brown, et al. Chemistry:The Central Science. 11th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2008.
- Chang, Raymond. General Chemistry:The Essential Concepts. 3rd ed. New York: Mcgraw Hill, 2003
- Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007.
OUTSIDE LINKS
Urbansky, Edward T.; Schock, Michael R. "Understanding, Deriving, and Computing Buffer Capacity." J. Chem. Educ. 2000 1640..
CONTRIBUTORS
Jose Pietri (UCD), Donald Land (UCD)
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