UCLA CHEM 14BL - Assignment: “Analysis of an Unknown Amino Acid”

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Milena Asatryan605204517Lab 2HAssignment: “Analysis of an Unknown Amino Acid”(I) Statement of PurposeIn the following experiment, we are expected to use an acid-base titration in order toexamine an amino acid. Based on the titration data, we will find the pKa values (includingisoelectric pH value) and the molarity of the amino acid. With the three point method withbuffers of pH 4,7, and 10, we will calibrate the pH meter. After we shall begin the titrationprocess of the amino acid by rinsing the buret with NaOH. we will then set and initial volume ofthe NaOH prior to doing the titration simply by filling up the buret with the solution between a0.00-to 1.00 mL marking. Using a pipet, we are expected to Fill a 125-mL beaker with a10.00-mL aliquot of the amino acid. Then make sure that the electrode tip is covered by adding20mL of distilled water. We are expected to record the pH of original amino acid solution and thepH of the final amino acid solution. Titrate the amino acid solution with the standardized NaOHsolution every 0.2 - 0.3 pH units, taking pH and volume data. Continue titrating until the pHreading falls between 12 and 13.Repeat the experiemnt one more time to ensure accuracy.(II) Data Table Set Up for Titration GraphTitration of Unknown Amino Acidwith NaOHTitration of Unknown Amino Acidwith NaOHTrial 1Trial 2NaOH Volume(mL)pHNaOH Volume(mL)pH0. Plotting Titration GraphIV) Data Analysis – Unknown Acid ConcentrationQuestion 1.Trial #1● Equivalence point #1: 6.17○ Volume: 7.30 mL● Equivalence point #2: 11.15○ Volume: 15.44 mLTrial #2● Equivalence point #1:6.28○ Volume:7.35 mL● Equivalence point #2:11.45○ Volume:16.01 mLQuestion 2NaOH M=0.2153 M 10-mL volumetric pipetMacidic protons x Vunknown acid=MNaOH x VNaOHTrial 1:M1acid x 10.00 mL=0.2153M x7.30 mLM1acid=0.157MTrial 2:M2acid x 10.00 mL=0.2153Mx 7.35mLM2acid = 0.158MQuestion 3a) The Average Macid= M1acid+M2acid/2 =0.1577= 0.158Mb) HA++OH-→ HA+H2Oi) I 0.158M 0.2153M ___ii) C -0.158M -0.158M +0.158Miii) E 0 .0573M .158MMHA=.158MQuestion 4Trial 1: pKa1= pH at Volume of 3.65mLPKa1=pH 2.54Pka2=pH at volume of 7.72mLPka2=pH=8.12Trial 2: pka1=pH at volume of 3.68 mLPka1=pH=2.65Pka2=pH at volume of 8.00 mLPka2=pH=8.57Question 5The isoelectric point pH (IpH):Trial 1:IpH=pka1+pka2/2=2.54+8.12/2=5.33Trial2: IpH=pka1+pka2/2=2.65+8.57/2=5.61Question 6The average IpH for the amino acid based on the results from Q5 for the 2 trials.Avg IpH: 5.33+5.61/2 =5.47Question 7Trial 1:IpH is 6.18Trial2: IpH is 6.28Avg: 6.18+6.28/2=6.23Question 8Yes there is a significant difference more than 5% between the average IpH calculated from question 6.The significant percent difference is calculated by first subtracting 6.24 and 5.47 then dividing by 6.24,times 100% which equals to 12.2%.So, 12.2% is greater than

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UCLA CHEM 14BL - Assignment: “Analysis of an Unknown Amino Acid”

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