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Buffer Know-how

Buffer Know-how.

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Buffer Know-how

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  1. 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 7.5 buffer”. 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 to understand their make up and mechanism of action. We will also give some insights into how to solve buffer problems.

  2. HA + A- + H2O HA + A- + H2O 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). Adding NaOH or HCl to this buffer would shift the reaction toward the right, but with different results. This is because of Rule 2 (click 1). Both components in the buffer must change any time acid or base is added to the buffer. This is because of Rule 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 of Rule 4 (click 1). Hence, Rule 5 summarizes an important principle that we must know (click 1). Rules 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. 5. Only the ratio, never the total, of HA to A- changes as a result of buffer action.

  3. 10 moles, ratio A-/HA = 2.43/1.0 OH- A- + H2O + HA 10 moles, ratio = 1.0 Putting the Rules to Work Let’s 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) 2 moles 5 moles 5 moles HA + A- + H2O 3 moles 7 moles We see that HA decreases and A- increases, both to the same increment. This illustrates Rule 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). 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. 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.

  4. HA A- HA A- A- HA HA A- HA A- A- HA HA A- HA A- HA A- A- HA HA A- Reserve salt 11 moles Reserve acid 11 moles 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 buffer’s 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 acid’s 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. OH- Neutralized A- OH- A- A- A- A- 6 moles 16 moles

  5. [HA] [A-] [HA] [A-] [H+] = Keq Log [H+] = Log Keq + Log [HA] [A-] –Log [H+] = –Log Keq – Log variable constant [A-] [HA] pH = pK + Log 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), Mutiplying components on both sides of the equation by -1 gives (click 1) 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) 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.

  6. [A-] [HA] [A-] [HA] Solving for Log pH = pK + Log [A-] [HA] Log = pH – pK = 5.2 – 4.8 = 0.4 [A-] [HA] = 2.5 / 1.0 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 to know 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) 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

  7. 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 to conjugate 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.

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