Amino acids have the general structure: All amino acids have at least two acidic protons. One is the proton

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1 Worksheet rganic Acids and Bases II Amino Acids Acids are proton donors in aqueous solutions: A A - acid base conjugate conjugate base acid The strength of an acid is measured by the [ 3 + ] at equilibrium. The higher the concentration, the stronger the acid. This is given by the K a, the acid equilibrium constant. K a = [ 3 + ] [A - ] [A] The pk a of an acid is defined as the log K a. The smaller the pk a, the stronger the acid and the higher the [ 3 + ] at equilibrium. Amino acids have the general structure: All amino acids have at least two acidic protons. ne is the proton on the -. These generally have a pk a 2.0. The second proton is on the - 3 group. This is the conjugate acid of a R weak base. The pk a for this group is about Some amino acids have a third pk a due to an acidic proton on R, the side chain. ne of these is aspartic acid, shown below, with its pk a values. pka ( -N3 + ) = pka ( -) = 2.10 pka (R-) = 3.86 Rank the three groups from most acid to least acidic > > This will be the order in which the groups deprotonate during a titration. The most acidic group will also determine the initial p of a solution of this amino acid.

2 To make a 1.0 M solution of aspartic acid, 1.0 moles of the amino acid are dissolved in 1.0 L of water. What is the p of the resulting solution? First, assume that only the strongest acid (lowest pk a ) will react with water: The pk a for the strongest acid ( -) is 2.1. What it the acid equilibrium constant, K a, for this reaction? K a = 2. Set up an IE table and solve for the [ 3 + ] equilibrium. initial change equilibrium [A] [ 3 + ] [A - ] K a = [ 3 + ] e [A - ] e [A] e 3. alculate the initial p of the solution (p = - log [ 3 + ] e ). p = 4. What is the total charge on the amino acid in this solution? Remember, this is a weak acid and very little of it is deprotonated. charge = In a titration, a strong base would now be added and the p measured as a function of the amount of base added. Every equivalent of base will deprotonate one acidic proton. 5. ow many moles of base are needed to react completely with1.0 L of 1.0 M aspartic acid? moles of base =

3 3=2p<2.10The enderson-asselbalch equation relates the p of a solution to the pk a values of the species in solution. p = pk a + log [A - ] [A] When p < pk a, the log term is negative, meaning that the ratio [A - ]/[A] is less than one, i.e., there is little dissociation of the acid. When p > pk a, the log term is positive and indicates that the ratio [A - ]/[A] is greater than one, i.e. the acid has been deprotonated to a great extent. When exactly one-half of the acid has been deprotonated, ratio of [A - ]/[A] = 1, the equation reduces to p = pk a. This is called the halfway point, This can be applied to aspartic acid. The initial p of the 1.0 M solution was1.07 and all three acidic functional groups were protonated. 2 pka ( -) = 2.10 pka (R-) = 3.86 pka ( -N3 + ) = +N9.82Both carboxylic acids are neutral (no charge) and the ammonium ion has a charge of +1. The overall charge at this p is +1. At p = 2.10, the pk a of the strongest acid, exactly ½ of the α- group will be deprotonated. In the titration of 1.0 L of 1.0 M aspartic acid, addition of 0.50 moles of - will bring the p to 2.10, and will form a 50%/50% mixture of the two structures shown below: +N32 - p.10250%50%the charge on the amino acid at p = 2.10 is.

4 3+N33+N=2- The next important point in the titration curve is the first equivalence point. This comes after the addition of 1.0 moles of strong base to the 1.0 moles of aspartic acid. All of the α- will be deprotonated. +N2- p.98100%we can approximate the p at this point. All of the α- has been deprotonated (pk a = 2.10) but none of the R- has been deprotonated (pk a = 3.86). The average of these two pk a values is close to the observed p; ( ) / 2 = verall, the charge on the amino acid at this point is zero. This is the isoelectric point of aspartic acid, the p at which it will precipitate out of solution. The next significant point in the titration is the second half-way point, when p = pk a of the R- group %- p=50%addition of 1.5 equivalents (1.5 moles) of - gives this p. The charge on the amino acid is now. Draw the structure of aspartic acid at the second equivalence point and determine its charge and the p of the solution. p = charge =

5 Draw the structures, distribution and charge of aspartic acid at p = p = 9.82 charge = ow many equivalents of base need to be added to bring 1.0 L of 1.0 M aspartic acid to the third equivalence point? equivalents of base = Draw the structure and determine the charge on aspartic acid at the third equivalence point. charge = Summary of values: p charge initial 1 st half-way 1 st equiv. 2 nd half-way 2 nd equiv. 3 rd half-way 3 rd equiv. Use these values to plot the titration curve of 1.0 M aspartic acid on the next page. Also enter the charge at each point in the titration. Indicate which is the p I, isoelectic point. The curve should be relatively flat at the half-way points, since both the acid and its conjugate base are present, creating a buffer.

6 Plot these values in the titration curve below charge 10 8 p equivalents of base