Presentation on theme: "Lecture 1 Determination of the Concentration and THE Acid Dissociation Constants of an Unknown Amino Acid."— Presentation transcript:
1Lecture 1Determination of the Concentration and THE Acid Dissociation Constants of an Unknown Amino Acid
2Scheduling This week in lab: Tuesday/Wednesday: Check-in and Pipette calibrationThursday/Friday: Start of experiment 8: “The determination of the concentration and the acid dissociation constants of an amino acid”Experiment 8 takes a total of two lab periods (4/3-4/9)Today’s lecture: Titration of an unknown amino acidsHint: You may think of the unknown amino acid containing both HA+/- and H2A+ forms of the amino acid
3Polyprotic AcidsDefinition: Polyprotic acids, also known as polybasic acids, are able to donate more than one proton per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule. The protons are usually released one at a time.Examples: sulfuric acid (H2SO4=O2S(OH)2), phosphoric acid (H3PO4= (OP(OH)3), carbonic acid (H2CO3=OC(OH)2), oxalic acid ((COOH)2), all amino acids (H2N-CHR-COOH)AcidpKa1pKa2pKa3Ascorbic acid4.1011.60Carbonic acid6.3710.32Malic acid3.405.20Oxalic acid1.274.27Phthalic acid2.985.28Phosphoric acid2.157.2012.35Sulfuric acidstrong1.92
4Amino AcidsAmino acids are the building blocks of proteins and enzymesAmino acids have the form H2N-CHR-COOH where R is a side chainProteins dominantly contain the (S)-enantiomer (exception: (R)-cysteine, glycine (achiral))NutraSweet (aspartame, artificial sweetener) is a famous dipeptide composed of phenylalanine and aspartic acidPenicillins are tripeptides (L-Cysteine, D-Valine, L-Aminoadipic acid)The isoelectric point is the pH value at which the molecule carries no net electrical charge (HL). At this point, the amino acid displays its lowest solubility in polar solvents (i.e., water, salt solutions) and does not migrate in the electrical field either.
5Diprotic Acids I Diprotic acids undergo the following equilibria: H2L+ HL + H+ Ka1HL L- + H+ Ka2Three possible forms in solution: H2L+, HL, L-The solution contains all three species at any given time. The individual concentration depends on the pH-value.
7Example ILeucineSince Ka1 >>> Ka2, only the first equilibrium has to be considered at low pH-valuespKa1=2.33pKa2=9.75
8Example II What is the pH-value of a 0.05 M H2L+ solution? The 5 % rule fails in this case. Thus, the quadratic formula has to be used here. x =1.31 * 10-2 M = [H+] (=26.2 %>>5 %)pH= -log([H+])=1.88For the calculation above, we assumed that the second equilibrium was unimportant (L- ≈ 0).Ka1 =
9Example IIIHowever, using the number above we can find the true concentration.With [HL] = [H+] =1.31 * 10-2 M.The calculation shows that the concentration of L- is indeed very low compared to the other concentrations.Ka2 =L- == x10-10
10Titration I Titration of diprotic acid has six points of interest P1: Initial pH-valueP2: pH-value at halfway to first equivalence point (pH=pKa1)P3: pH-value at first equivalence pointP4: pH-value at halfway to second equivalence point (pH=pKa2)P5: pH-value at first equivalence pointP6: pH-value after adding excess of baseV= Veq/ Veq Veq 2 Veq Veq
11Titration IIExample: Titration of 10 mL of M H2L+ with M NaOHTwo reactions have to be consideredH2L OH HL + H2O (1)HL OH L- + H2O (2)Step 1: Initial pH-value (see previous calculation)Step 2: After 5.0 mL of base have been added, [H2L+]=[HL] pH=pKa1=2.33Step 3: After 10.0 mL of base were added, the first equivalence point is reached (=isoelectric point) pH=6.04
12Titration IIIStep 4: After 15.0 mL of base have been added, [HL]=[L-] pH=pKa2= 9.75Step 5: After 20.0 mL of base were added, the second equivalence point is reached. Since all of HL is converted to L-, the hydrolysis of L- has to be considered (ICE) L- + H2O HL + OH-L-HLOH-Initial5.0*10-4 moles (=0.010 L *0.050 M)~0Change-x+xEquilibrium
13Titration IV Step 5 (continued): Determine pKb of L- Determine [OH-] Using the quadratic equation, one obtains y = [OH-] = 9.38*10-4 M (= 5.5 % of M) pOH = pH=10.97
14Titration VStep 6: After 25.0 mL of base have been added, all H2L+ has been converted to L-. This required 20.0 mL of base to accomplish.There is an excess of 5.0 mL of base in the solutionFind number of moles of basen = L * M = 2.5*10-4 molesFind concentration of basec = 2.5*10-4 moles/ L = 7.14*10-3 MFind pOH and pHpOH = -log([OH-]) = 2.15pH = 14 – pOH = 11.85
15Summary for LeucineThe six points of interest in the titration of M leucine with M NaOH arePointBase addedEquivalencepH-valueCommentsP10.0 mL0.01.88P25.0 mL0.52.33=pKa1P310.0 mL1.06.04=(pKa1+pKa2)/2P415.0 mL1.59.75=pKa2P520.0 mL2.010.97P625.0 mL2.511.85
16Individual Work In lab, this week on Thursday and Friday The student obtains a standardized NaOH solution.The students have used pH meters before (Chem 14BL) so the calibration should go rather smoothly. If you do not remember how to do it anymore, please review it in the lab manual (page 12).Make sure to keep the standardized sodium hydroxide and the unknown amino acid solution. DO NOT store your standard solution (NaOH) in volumetric flasks. Use other glassware to store the solutions (ask your TA).The student has to perform three titrations of the unknown amino acid solution (until pH=12)Clean-upNeutralize all titrant solutions with citric acid until the pH paper turns light green or orange before discarding in the drain. Pour the small amount of waste NaOH used to rinse the burette into the labeled waste container. Do not pour un-neutralized NaOH solutions down the drain.At the end of the assignment, place the capped bottles of unused NaOH and amino acid on the lab cart for return to the lab support
17ReportUse Excel for plotting titration curves and first-derivative graphs.The pKa’s of the amino acid are determined from the full titration graphTo determine pKa1 and pKa2, locate the volume on the graphs half way between the two equivalence point volumes determined from the expanded derivative curves. The pH at this point is in the titration is equal to pKa2.Next, measure an equal distance on the graph to the left of Vep1. The pH at this point is equal to pKa1.Error Analysis:Calculate the relative average deviation in the concentrations of your amino acid.Compare the relative average deviation with the inherent error calculated by propagating the errors in measurements of the pipet, the volumes determined from the graphs, and the standard base solution.Estimate the absolute error in your pKa’s by considering the variability you had in the pH’s of the solutions at the |DVep/2| points in the three titrations. Report the range for each of the pKa’s.The report is due on April 15, 2014 or April 16, 2014 at the beginning of the lab section.