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Buffer Know-how

Buffers are important in biochemical processes. Whether they occur

naturally in plasma or in the cytosol of cells, buffers assure biological reactions

occur under conditions of optimal pH. They do this by controlling the hydrogen

ion concentration of solutions. The word buffer is so common in biochemistry

its replaces the word water in experimental protocols. For example, typical

of the statement seen in publications is the pellet was dissolved in pH 7.5buffer. All this should alert you to the importance of thoroughly understanding

buffers and buffering agents. Words such as pH, pKa, conjugate acid,

conjugate base, Henderson-Hasselbalch equation are used frequently in

biochemical language and every publication that describes an experiment

performed in vitro (Lat., in glass), must include a clear description of the

buffer that was used. In this tutorial we will revisit buffers and attempt tounderstand their make up and mechanism of action. We will also give some

insights into how to solve buffer problems.

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Rules Governing Buffer Reactions

To help you understand buffer action, consider the equation that describes a

buffer reaction (click 1). The HA and A- represent the two components of any buffer: the

conjugate acid and the conjugate base, respectively. Note that the right side and left

side of the equation are the same.Rule1

gives the meaning of the reaction (click 1).

Rules

HA + A- + H2O HA + A- + H2O

1. Buffer reactions never go to completion.

2. H+ can only react with A-, OH- can only react with HA.

3. The conjugate acid and conjugate base must change proportionately and in opposite

directions.

4. The sum of the concentrations of base and acid components stays constant, i.e, HA + A-

before = HA + A- after.

Adding NaOH or HCl to this buffer would shift the reaction toward the right, but with different

results. This is because ofRule 2 (click 1). Both components in the buffer must change

any time acid or base is added to the buffer. This is because ofRule 3 (click 1). Finally, we

take into account the HA and A- components, expecting to see a decline in the overall buffer.Such is not found because ofRule 4 (click 1). Hence, Rule 5 summarizes an important

principle that we must know (click 1).

5. Only the ratio, never the total, of HA to A- changes as a result of buffer action.

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Putting the Rules to Work

Lets now see how the 5 rules apply to a buffer reaction. Lets start with the buffer

(click 1). Suppose we add OH- to the buffer(click 1). Rule 2 tells us the following would occur

(click 1)

HA + A- + H2O

OH-

HA + A- + H2O

We see that HA decreases and A-

increases, both to the same increment. This illustratesRule 3. H2O would also increase, but we can ignore H2O because the increase would be

insignificant compared to the amount of H2O present (click 1).

2 moles

5 moles 5 moles

Finally, if we calculate HA and A- before and after the addition of OH-, we see that both

add up to 10 moles in either case (click 1). This verifies Rule 4. Thus, we started with a

ratio of A- to HA = 5/5 = 1.0 before the reaction and after the reaction the ratio of A to HA-

= 7/3= 2.3 to 1.0, but the total did not change. Click to go on.

10 moles, ratio = 1.0

10 moles, ratio A-/HA = 2.43/1.0

3 moles 7 moles

If we applied numbers to the concentrations of OH-, HA and A we can test the other rules

(click 1). Here we see that we are adding 2 moles of OH- to 5 moles each of HA and A-.

The reaction of OH- with HA lowers the HA from 5 to 3 and raises the A- from 5 to 7

(click 1). The decrease in HA was matched by an equal increase in A-, which is Rule 3.

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Principle Behind Buffer Action

Buffers are composed of weak acids and their salts. A salt is the acid minus its

proton. Weak acids and their salts have two properties that are important for buffering

action. First, weak acids are a reserve of the protons that neutralize OH- and prevent the

solution from becoming alkaline. Salts of weak acids are strong bases and prevent the

solution from becoming acidic. Both components are needed and both are

interchangeable through the loss (or gain) of a single proton.

A buffers power lies in its reserves (click 1). A buffer is at optimal strength

when there is an equal amount of HA and A- in solution as shown. This will only occur

when the pH of the solution equals the pKa of the acids group. Adding OH- causes the

buffer to respond by calling on the reserve pool of HA. A-

is formed at the expense of HA(click 1). This continues until all the excess OH- is neutralized. At the end the salt pool

has increased (and the acid pool has decreased) by the same number of moles of base

that were added. Click to go on.

HA

HA

HAHA HA

HAHA

HAHA

HA

HA

Reserve acid

11 moles

Reserve salt

11 moles

A-A-

A-

A-

A-A-

A-

A-

A-

A-

A-

OH-

A-

A-

A-

A-

A-

6 moles 16 moles

OH-

Neutralized

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Focus on the Ratio of [A-]/[HA]

In the previous illustration you saw the importance of knowing the ratio of HA and

A-. Now you will see that it is the ratio that determines the pH of the solution, and vice

versa, the pH allows you to determine the ratio. It all begins with an equilibrium expression(click 1). If we take the log of all components we derive a logarimic expression of the same

equation (click 1),

[H+] = Keq[HA]

[A-]Log [H+] = Log Keq + Log

[HA]

[A-]

Mutiplying components on both sides of the equation by -1 gives (click 1)

Log [H+] = Log Keq Log[HA]

[A-]

Substituting pH and pK for the appropriate terms in the equation and making the log of HA/A-

positive by reversing numerator and denominator gives (click 1)

pH = pK + Log[A-]

[HA]

Note, in the equation, pK is a

constant and A/HA is the only

variable (click 1). This is the

Henderson-Hasselbalch equation.

Click to go on.

constant variable

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Putting Henderson-Hasselbalch to use

Knowing the ratio of [A-]/[HA] allows you to calculate pH. Always treat the ratio as a

whole number, i.e., do not separate numerator from denominator. As an example, assume 2

moles of NaOH are added to 10 moles of a pH 5.2, pK = 4.8 buffer (HA + A-). You want toknow the moles of HA after neutralization and the new pH. Follow these steps to the solution

(click 1)

First determine the moles of HA at the start (click 1)

pH = pK + Log[A-]

[HA]Solving for Log [A-]

[HA]

[A-]

[HA]Log = pH pK

= 5.2 4.8

= 0.4

[A-]

[HA] = 2.5 / 1.0

The ratio of A- to HA is 2.5 parts to 1 part. This means the 10

moles are represented by 3.5 parts. If 2.5 parts of the 10 are

moles of A- and 1.0 part is HA, then before OH- was added

there were 7.1 moles of A- and 2.9 moles of HA. Together the

two add up to 10 and their ratio is 2.5:1.0 (click 1).

When OH- is added, 2.0 moles of NaOH react with 2.9 moles

of HA. As a consequence, HA goes from 2.9 to 0.9 and A-

goes up from 7.1 to 9.1 moles. The new pH is determined

from the ratio 9.1 to 0.9 or 10.1. This computes to pH = 5.8

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Test and Enhance your Understanding

Q: Why must buffers always have two components?

A: Several reasons. First, Rule 2 say that a base will only react with an acid and an acid with a base.

The two components assure the buffer protects against both. Second, it is the ratio of conjugate base toconjugate acid that determines pH. Neither one alone would suffice.

Q: How does one select a buffer in a particular biochemical experiment?

A: The decision is based on the desired pH that must be maintained. An acetate buffer, for example, is

useless at pH 8.0 because the buffer (pK = 4.8) at this pH is almost all conjugate base. In contrast a

buffer with a pK around 8.0, Tris buffer, for example, would be more suitable.

Q: Is there a way of telling when a buffer will be most effective at a given pH?

A: Yes. The pK of the buffer (sometimes referred to as pKa for an acid) immediately tells you the

pH that will result in equal amounts of conjugate acid and base in solution, which is the optimal

condition for any buffer.

Q: Suppose you added 2 moles of acid to 10 moles of buffer in a solution where pH and

the pK have the same value. What would be the result?

A: Since the pH = pK, you know that there are equal moles of conjugate acid and conjugate base, 5

moles of each. The 2 moles of acid would convert 2 moles of the base component to the acid changing

the ratio of A-/HA from 1.0 to 0.43, which would lower the pH of the solution by 0.37 units.