The shortened version is the more commonly used form (besides, water is a pure liquid in the reaction) and the expression for the equilibrium is:

Transcription

1 Biochemistry I Buffers Acids and Bases 4,5 Substances that dissolve in water with the subsequent production of hydronium ion (H 3 O +, long hand for hydrogen ion or protons, H + ) are called acids. A more literal term for these substances is Brønsted acid; so called proton donors because they are able to release protons that can be picked up by other substances; bases. Acids are generally represented by the formula, HA, where H is the proton that the acid can release and A is the anion that results from the acids losing the proton. The reaction that acids undergo is: HA(aq) + H 2 O(l) º H 3 O + (aq) + A - (aq) HA(aq) º H + (aq) + A - (aq) or shortened The shortened version is the more commonly used form (besides, water is a pure liquid in the reaction) and the expression for the equilibrium is: K eq = [A - ][H + ] [HA] Because the reaction is the loss of hydrogen by the acid, we call this a dissociation reaction and the equilibrium constant is commonly represented as K A to reflect this. Note that a substance enclosed in brackets means the concentration of the substance (i.e. [H + ] is the hydrogen ion concentration). K eq = K A = [A - ][H + ] Note: this is K B for weak bases. [HA] When acids dissolve in water, the concentration of hydrogen ion, [H + ], increases. We typically indicate the level of hydrogen ion in solution using p-values (negative logs). For acid/base solutions, we use ph, the negative log of the hydrogen ion concentration: ph = -log([h + ]) hence, the greater the hydrogen ion concentration, the lower the ph and vice versa. Solutions that have a ph of 7 are called neutral because the amount of hydrogen ion is equal to the amount of hydroxide ion, OH -, in solution. Solutions with a ph less than 7 are called acidic; and the lower the ph the more acidic. Solutions with ph greater than 7 are called basic; and the greater the ph, the more basic. If the bond between the hydrogen to be lost and the anion is weak, virtually all of the molecules of acid in a sample of acid will lose the hydrogen when introduced to an aqueous environment (dissolved in water). We call such an acid a strong electrolyte; a strong acid. This is the case when the electron withdrawing capability of the anion is strong; most commonly when the attachment point for the hydrogen is highly electronegative. In those cases, there is no need to concern oneself with the equilibrium for the system as almost 100% of the acid molecules will lose there hydrogens. The amount of hydrogen ion produced will be equal to the amount of acid you start with. The situation is similar for strong bases. HA(aq) H + (aq) + A - (aq) strong acid base

2 H + (aq) + A(aq) HA + (aq) strong base acid However, in the case that the anion holds the hydrogen more tightly, there is less probability that the hydrogen will dissociate from the anion and fewer hydrogen ions will populate the solution. These acids are called weak electrolytes; weak acids. By the same token, substances that gain protons weakly (only a portion of an aqueous population gains protons) are also weak electrolytes called weak bases. When a weak acids loses a hydrogen, the resulting anion is a basic substance called the conjugate base. Likewise, when a weak base gains a proton, the resulting substance is called the conjugate acid. HA(aq) º H + (aq) + A - (aq) weak acid conjugate base H + (aq) + A(aq) º HA + (aq) weak base conjugate acid We will concentrate on the situation for weak acids for simplicity and because you can easily extrapolate the condition of weak bases from the information for weak acids. Buffering 3,4,5 There is an interesting property of solutions that are made of weak acids and their conjugates: if the proportions are close (i.e. around 1:1), the solution resists changes in hydrogen ion concentration ([H + ]) when small amounts of acid or base are added to them. The presence of approximately equal amounts of acid and conjugate buffers the solution against changes in ph! These solutions are called buffers (specifically ph buffers). Weak acids, by definition, dissociate incompletely in water; at what point does the dissociation take place? As it turns out the dissociation is ph dependent (it is a function of [H + ]. The equilibrium expression for the dissociation of a weak acid is: K A = [A - ][H + ] [HA] This can be rearranged to: ph = pk A + log([a - ]/[HA]) the so called the Henderson-Hasselbach equation. pk A is frequently shortened to just pk. Now, if [A - ] = [HA] then ph = pk. It is around this point that weak acids are most likely to lose their hydrogens. So if you have a solution of almost equal amounts of a weak acid and its conjugate ([A - ] ~ [HA]), and you add OH - to the solution, the OH - reacts to remove H + from solution, but at that point you have approximately equal amounts of weak acid and conjugate base, so the acid readily dissociates to make more H +. So when you remove acid by adding OH - it is replenished by

3 dissociation of HA. Hence, the ph hardly changes; the solution buffers against ph changes! Notice that this is not a 1:1 process (removing one H + doesn t result in regaining one H + ) so the ph never flat lines, it just changes more slowly. When you use a buffer, you must also pay attention to the concentration of the buffer. Because buffers work by replenishing hydrogen ion to solution, the more concentrated the buffer (the more buffer molecules there are in solution), the easier it is to replenish hydrogen ion (i.e. the better it buffers and hence holds ph easier). Buffers with low concentrations are easier to perturb than buffers with high concentrations! However, you do not want the concentration to be so high as to drastically alter the physical and/or chemical properties of the solution. That could be detrimental to whatever you re working on. A titration is a process where an unknown amount of some entity is determined using an interaction of precisely known consequences with an entity of known amount. In the case of solutions, this is most commonly done to determine hydrogen ion concentrations in solutions where this is not known. The idea is to react the unknown [H + ] with a known amount of [OH - ]. Since the reaction is 1 to 1 for H + and OH -, knowing how much OH - reacts means you know how much H + reacts. If we plot the titration of a weak acid with a strong base (i.e. OH - ), we see something interesting; a sigmoidal curve. The basic idea behind a titration actually reveals a lot about the true meaning of K a. In a very acidic environment, [H + ] is high and the probability that HA is able to permanently lose H + is low. As OH - is added, [H + ] decreases (i.e. ph goes up) and this probability increases. Since HA is losing H + the act of OH - removing H + from solution is negated and the ph seems to stay the same (or at least changes very little). Eventually, all of the HA is gone and the added OH - only reduces H +. So what is K a?

4 Making Buffers Producing a buffer is not a trivial matter and there are numerous reported ways of accomplishing the task. The substances used to make a buffer must be compatible with the system as far as reactivity and toxicity and should have a pk close (within ±1.0) to the target ph of the buffer. Further, you must make a solution with the desired ph at an appropriate concentration. The ratio of acid to conjugate can be determined using the Henderson-Hasselbach equation: [A - ]/[HA] = 10 (ph-pk) where [A - ] = [HA]*10 (ph-pk) Given that the concentration of a buffer, B c, is defined as the sum of the concentration of the conjugates B c = [HA] + [A - ] We can use these two equations to solve for [HA] and [A - ] since B c, ph, and pk are all known. Consequently the concentration of the acid component of the buffer is [HA] = B c /( pH-pK0 ) and the concentration of the conjugate base component of the buffer is [A - ] = B c (10 9pH-pK0 )/( pH-pK0 ) Now, when you make the buffer you produce it from any combination of two or more solutions and/or solutes. When you mix solutions, the concentrations of the solutes change as a result of the volume change and the correct proportions are not easily calculated unless the ratios are about 1:1 (then each solution should just be made about twice as concentrated as the target). When you mix two or more solutes, the resulting solution must keep the same volume, or the concentrations will be changed. This, of course, means that you can t adjust the ph of the prepared solution; you must half-way make the solution, adjust the ph, then dilute to the proper volume. The Long Way 2 Using the relations for mass of solute and molar mass of solute and solution concentration, we arrive at m HA = M*L*M HA (10 (ph-pk) +1) m A- = 10 (ph-pk) *M*L*M A- (10 (ph-pk) +1) and where m HA is mass of acid required, m A- is the mass of conjugate base required, pk is the pk of the system, ph is the target ph, M is the target molarity, L is the target volume in liters, and M HA and M A- are the molar masses of the acid and conjugate respectively. This gives you the mass of solute required to produce the desired volume and molarity of solution from solid components. If the solution must be made using a more concentrated stock solution, divide the appropriate mass from above, by the product of the molar mass of that buffer component and the stock solution s molarity, and you ll have the volume, in liters, of stock solution required to produce the buffer:

5 L = m/(m*m) Combine the appropriate amounts of each solute and mix to homogeneity. Once the buffer is made, check the ph with a calibrated ph meter. The meter should be calibrated with a calibration buffer that is close to the required ph for the buffer. Rinse the electrode after calibrating the meter and remove excessive adhering water with a clean Kim-Wipe. Transfer the buffer to a clean beaker at least twice the volume of the buffer but small enough that the height of the buffer in the beaker is at least two inches. Add a small, magnetic stirring bar and commence stirring on a stirring platform at a medium speed. Immerse the meter s electrode in enough buffer so as to cover the tip by at least an inch and read the ph from the meter. If the ph is too low (too acidic) add enough of the conjugate base solution (solution of the conjugate base at the target molarity) to adjust the buffer to the proper ph. If the ph is too high (too basic) add enough of the acid solution (solution of the acid at the target molarity) to adjust the buffer to the proper ph. The Volume Way 2 Produce separate solutions of the acid/base and the conjugate by either 1) determining the mass of solute required to make solutions of the target concentration by multiplying the molar mass of that solute by the target concentration and volume, then diluting that amount to the target volume with pure water, or 2) determining the volume of stock solution required to make solutions of target concentration by dividing the product of the target concentration and volume by the stock solution concentration and diluting that amount to the target volume with pure water. Once the two solutions are made, mix the volume of each indicated by solving the ratio of [A - ]/[HA] using 10 ph-pk ; where 10 ph-pk is the volume of target molar conjugate base, and 1.0 is the volume of the target molar acid (ratio can be multiplied by a factor as needed). Combine the appropriate amounts of each solute solution then check and adjust the ph with the prepared solutions as described above (The Long Way). The Concentration Way 1 Determine the concentration ratio of [A - ]/[HA] using 10 ph-pk ; where 10 ph-pk M will be the concentration of conjugate base, and 1.0M the concentration of the acid. Now the sum of the concentrations, [A - ]+[HA] should equal the target concentration (TC) so that [A - ] = TC - [HA] but [A - ]/[HA] = 10 ph-pk so TC-[HA]/[HA] = 10 ph-pk TC-[HA] = [HA]*10 ph-pk TC = [HA]*10 ph-pk + [HA] = [HA](10 ph-pk + 1) [HA] = TC/(10 ph-pk + 1) and [A - ] = TC*10 ph-pk /(10 ph-pk + 1) so is the actual molar concentration of each solute in the buffer to be made. To determine the amount (mass) of each solute to use, multiply the actual molar concentration by the target volume in liters and the molar mass of the solute, combine the amount of solute determined for each and dilute the mixture to the target volume with pure water.

6 If the buffer must be made using a more concentrated stock solution of a solute, divide the appropriate mass determined for that solute from above, by the product of the molar mass of that solute and the stock molarity, and you ll have the volume of that solute stock solution in liters required to produce the buffer: L = m/(m*m) Combine the appropriate amounts of each solute and mix to homogeneity. Once the buffer is made, check and adjust the ph with solutions of the target concentration as described above (The Long Way). REFERENCES (1) How to Make a Phosphate Buffer Solution, Anne Marie Helmenstine, About.com, v i e w e d : 1 0 / 2 7 / , < (2) How to Make A Buffer, Reginald Stanton, 2010, Personal Communication. (3) Garrett and Grisham, Biochemistry, 5 th ed., 2005, WH Freeman, pp (4) Munowitz, M., 2000, Principles of Chemistry, 1 st ed., WW Norton, pp (5) Skoog, D.A., West, D.M., and Holler, F.J Analytical Chemistry: An Introduction, 5 th ed. Saunders College Publishing. pp , CONCEPT APPLICATION 1. What happens when you add 0.10mol of NaOH to 0.50mol of aqueous formic acid? Formic acid is a weak acid and subsequently doesn t have a strong tendency to dissociate in water (K a = 1.78x10-4 ) so most of the 0.5mol will be HCOOH. Adding OH - will reduce HCOOH by 0.1mol and increase HCOO - by 0.1mol. ph = log(0.10/0.40) ~ 3.15 ~ What happens the ph of the blood when the levels of dissolved CO 2 decrease? That technically will depend upon how the CO 2 (aq) disappears. However, in all likely hood the ph won t be effected since gaseous carbon dioxide is in equilibrium with dissolved carbon dioxide and a reduction in CO 2 (aq) will be compensated by more CO 2 (g) dissolving to make up the difference. 3. How would you make a ph 7.0 buffer? First you need to choose a system that has a pk value close to 7.0. A suitable system is the second phosphoric acid equilibria (phosphate): H 2 PO 4- (aq) + H 2 O(l) º HPO 4 2- (aq) + H 3 O + (aq) K a = 6.31x10-8 [HPO 4 2- ]/[H 2 PO 4- ] = = ~ 0.63 So you would add 0.631mol of dibasic phosphate (HPO 4 2- ) to 1.000mol of monobasic phosphate (H 2 PO 4- ). At least that s the molar ratio to use. You get the same thing by adding equal volumes of 0.631M dibasic to 1M monobasic.