Objectives: Identification of an Amino Acid through Potentiometric Titration Keywords: Amino acid, zwitterions, enderson asselbach, pka, p arry out rough and then an accurate potentiometric titration of two amino acids, one known, one unknown. ompare data for known with literature pka, and combined with molecular weight determined from the data, solve identity of unknown. The amino acid is one of the fundamental building blocks of biology, almost without exception, proteins are made up from some combination of the set of 20 naturally occurring amino acids. All these contain a number of similar structure features a central carbon atom to which are bonded a hydrogen (-), an amine group, (-N 2 ), and a carboxylic acid group, (-OO). In addition there is an "" group that differs for each amino acid. The symbol "" is used here to represent a generalized abbreviation for an organic group. 2 N OO The nature of the group is critical in the structure and function of the protein and the 20 basic amino acids contain a wide range of groups which vary by size, electronegativity and a number of other properties. The electron withdrawing or electron donating nature of the group can affect the pka values of the amino acid and as a result, can affect the structure of the protein under a range of p values. If we just consider the carboxylic acid and the amine group present in every amino acid and ignore the group which can be either an acid or a base depending on what amino acid we are dealing with, there is therefore a p dependence on the state of protonoation of the two groups. In physiological systems where the p is near neutrality, the amino group of an amino acid will be protonated and the carboxylic acid group will be deprotonated. This is called the zwitterion form. 3 N OO hem(bio) Spring 2012 Week 5-1
In strongly acidic solutions the carboxylic acid group will also be protonated, while in strongly basic solutions both the carboxylic acid group and the amino group will both be unprotonated. 3 N OO 2 N OO Acidic Basic The acid-base behavior of amino acids is best described by the Brønsted-Lowry theory of acids and bases. A simple amino acid (that does not have an acid or base group in the "" group) is a diprotic acid in its fully protonated form; it can donate two protons during its complete titration with a base. Therefore, if we start with the fully protonated form, (labeled Acidic above), the titration with NaO will be a two-stage titration represented by the reactions below. + N 3 ()OO + O - + N 3 ()OO - + 2 O + N 3 ()OO - + O - N 2 ()OO - + 2 O The hydrochloride salt of a simple amino acid contains one mole of l for each mole of amino acid such that the amino acid is fully protonated. l - + N 3 ()OO The resulting titration curve will therefore be biphasic (see diagram below). There will be two separate flat portions (called legs) on the titration curve. The midpoint of the first leg (B) is where the amino acid is half in the acidic form and half in the zwitterion form. The point of inflection () occurs when all of the original amino acid is in the zwitterions form (assuming the "" group has no charge). The actual p at which this occurs is called the isoelectric p (or isoelectric point), and is given the symbol pi. During the p titration of an amino acid with a non-ionizable "" group, the equivalence point occurs at the pi of the amino acid. At the midpoint of the second leg (D), half the amino acid is in the zwitterion form and half is in the basic form. The apparent pk values for the two dissociation steps may be extrapolated from the midpoints of each step. This can be shown by the enderson-asselbach equation: p = pk a + log([base]/[acid]) The pk acid (pk a for the carboxylic acid group) is point (B) where half the acid group has been titrated. Therefore the equation becomes: p = pk a hem(bio) Spring 2012 Week 5-2
In the same way, point (D) gives us the pk amine. In this experiment you will titrate first a known amino acid and compare the values obtained with the literature then titrate an unknown amino acid, determine its pi, pk acid and pk amine, and compare your values to literature values to identify it. hem(bio) Spring 2012 Week 5-3
The Experiment You will titrate first one of the known amino acids twice then one of the unknown amino acid solutions two or three times. We will use the dry-run idea where we will do a rough run through using fairly large additions then do a second run where we can add larger additions in the regions of the curve where there is no significant change and smaller additions in the region where the curve changes dramatically. The dry-run tells us these regions. A. Titration of the known amino acid solution 1. alibrate the p meter as per the procedure on the sheet 2. Obtain ~0.2 g of the known amino acid, record exact mass and dissolve in 50.0 ml of deionized water. Mix well. 3. Obtain approximately 100 ml of NaO solution. ecord the concentration. inse and then fill the buret (to the top of the graduated markings) with NaO solution. emove any air bubbles (especially from the tip of the buret) and note the starting volume (which may or may not be 0.0 ml ). 4. Pipet 10.00 ml of the known amino acid solution in a 100 ml beaker and add 25.00 ml of deionized water using a volumetric pipette (for a total volume in the beaker of 35.00 ml). 5. Place the p electrode assembly and a magnetic stirring bar into the beaker. lamp the electrode so that the stirring bar will not hit it as stirring occurs. If the electrode is not properly immersed, the p reading will be erratic. The p should start fairly low since we are dealing with the hydrochloride salt of the acid so we only have to add NaO in other words, we are starting down in the A region of the plot. 6. Titrate the amino acid solution with the NaO from the buret. The first run (called the dry run ) is done by adding the NaO at ~1 ml intervals (note the exact amount dispensed each time) until you are just past the first endpoint; in other words, keep adding 1 ml increments until the p rises abruptly (the part of the ideal graph on the previous page. ecord all of this information in a dry run data table of volume added vs. p. Note that you can calculate the second endpoint s theoretical volume of NaO easily by doubling the volume dispensed in getting to the first endpoint. Allow time to reach a stable p before reading after each addition. 7. efill the buret with NaO solution. Get another clean dry 150 ml beaker and pipet 10.00 ml of your chosen amino acid solution into it; add 25 ml of distilled water. Titrate the amino acid solution again, this time, use ~0.5 ml intervals until just before each endpoint and then dropwise until just after each endpoint. This hem(bio) Spring 2012 Week 5-4
will be the real run that you will graph. ontinue until 2.5 equivalents of NaO have been added or the p reaches about 12.5. B. Titration of the unknown amino acid solution epeat what you did for the known as previously described but for your assigned unknown amino acid. Waste disposal: Solutions must be between p 5 and 12 before being poured down the drain. You can lower the p by adding hydrochloric acid, or raise it by adding solid sodium bicarbonate. inse the burette, the p electrode and all glassware well with plenty of DI water esults In your lab notebook record the following: Dry run titration of known amino acid o Make a table of ml NaO added and p o First endpoint (ml NaO added) o Second endpoint (ml NaO added; this may need to be calculated, rather than measured) Titration 1 of known amino acid In Excel, make a table of ml NaO added and p epeat the above for the unknown amino acid. For the known amino acid, calculate the number of moles of amino acid in the original solution based on the mass used and the formula weight (will be on the bottle I hope ) For the good run on both the known, plot mole equivalents of NaO added vs p (similar to the graph we saw earlier in the handout). Mole equivalents is essentially how many moles of NaO is added per mole of amino acid. This can be calculated by converting your ml of NaO to mole equivalent in Excel and I ll talk about how to do this in lab don t worry about it now. The first endpoint ( on figure) marks the region where 100% of the amino acid is present as the neutral form while the second endpoint marks the region (E) where 100% is present as the deprotonated or basic form. Points B and D represent where we have a 50/50 ratio of the two forms in the equilibrium and as a result, we can, based on the p these occur at, calculate the actual pka values which should be compared to the literature value. Now, for your unknown amino acid, plot the ml of NaO vs p as above. We don t know the formula weight of the unknown acid but we can, from the shape of the line, hopefully determine the p at the 2 points B and D. This allows us to calculate the 2 pka values for the acid. hem(bio) Spring 2012 Week 5-5
Now it gets interesting, we may be able to determine the acid by comparing pka alone but we can take it to the next level. At points B and D, how many moles of acid have been added? Point B corresponds to adding 0.5 mole equivalents of NaO and point D to 1.5 mole equivalents of NaO. So we can at these points know exactly how many ml of NaO was added thus, how many moles of NaO at these points. Now, we can actually solve for how many moles of our unknown amino acid we have because: At point B, # of moles of unknown amino acid x 0.5 = # moles NaO added at point B At point D, # of moles of unknown amino acid x 1.5 = # moles of NaO added at point D So we can solve for both and hopefully get the # of moles of unknown amino acid present in the flask. And since we know how much we weighed in, we can calculate formula weight and hopefully solve the identity of the unknown amino acid based on it s formula weight, the pka values and the fact it can t be an amino acid with an ionizable sidechain! Simple, really Grading of Week 5: Data report not a full blown one as but show me graphs and calculations and convince me you know what your amino acid and why. hem(bio) Spring 2012 Week 5-6