online CHEM 1412 Acid Base Potentiometric Titration with excel 6-24-20

Online CHEM 1412 Acid-Base Potentiometric Titration Report Sheet for Acid Base Potentiometric Titration Name and date: tonic water Partner (if applicable): Part I: Standardization of NaOH using Anhydrous Citric Acid Trial 1 Trial 2 Trial 3 Mass of citric acid 0.15 g 0.16 g 0.16 g Mol of citric acid 8.3 x 10 -4 mol 8.3 x 10 -4 mol 8.3 x 10 -4 mol Moles of NaOH 2.5 x 10 -3 mol 2.5 x 10 -3 mol 2.5 x 10 -3 mol Initial graduated cylinder volume 10.10 mL 10.13 mL 10.14 mL Final graduated cylinder volume 4.10 mL 4.20 mL 4.47 mL Volume of NaOH used in mL 6.00 mL 5.92 mL 5.67 mL Volume of NaOH used in L 0.006 L 0.00592 L 0.00567 L NaOH concentration 0.416 M 0.4213 M 0.4406 M Average Molarity of NaOH 0.4261 M Show your work for one trial . - Moles citric acid: (0.16g x 1)/192.12g = 8.3 x 10 -4 - Citric acid to NaOH ratio: 1:3 - 3.3x10 -4 x (3/1) = 2.5x10 -3 Trial 1: - M = n/L : 2.5x10 -3 mol/0.006L = 0.4164 M Trial 2: - M = n/L : 2.5x10 -3 mol/0.00593L = 0.4213 M Trial 3: - M: n/L : 2.5x10 -3 mol/0.00567L = 0.4406 M Average: ( 0.4164 + 0.4213 + 0.4406 )/3 = 0.4261

Online CHEM 1412 Acid-Base Potentiometric Titration *I had to upload this as a screen shot of my excel graph, my word document would not save my inserted excel graph. *

10.0 mL 3.1 23.2 mL 6.3 36.0 mL 10.9 12.0 mL 3.3 23.4 mL 6.6 37.7 mL 10.9 14.0 mL 3.5 23.6 mL 7.0 38.4 mL 10.8 16.0 mL 3.8 23.8 mL 7.8 39.1 mL 10.8 18.0 mL 4.4 24.0 mL 8.3 40.4 mL 10.9 20.0 mL 4.8 24.2 mL 8.7 mL 22.0 mL 5.1 24.4 mL 9.1 mL Use the titration data to construct a graph. Label the equivalence point. Fill in the following table using your graph. Titration graph Volume of NaOH pH before titration begins 0 mL 2.4 At half equivalence point 23.0 ml 6.0 At equivalence point 24.6 ml 9.1 Fill in the following table: 1. Equivalent volume of NaOH (from equivalence point on graph) 24.6 mL Molarity of NaOH (from Part I) 0.4261 M moles of NaOH (in the equivalent volume) 0.0098 mol moles of acetic acid (at equivalence pt) 0.0098 mol mass of acetic acid 0.588g mass of initial 5.0 mL vinegar 5.00 g % acetic acid in vinegar 11.7 % North Lake College CHEM 1412 online Potentiometric Titration p. 4

pK a 6.0 K a of acetic acid 1.0x10 - 7.0 Moles NaOH: molarity x volume L = (0.4261M)*(2.46x10^- 3L) = 0.010 mol Moles acetic acid: at equivalence point, = moles NaOH = 0.010 mol Mass Acetic acid: Mass AA = moles AA x molar mass AA = - 0.010 mol x 60.0 g/mol = 0.629 g AA Mass 5.0 mL vinegar: volume of vinegar x density (1.00 g/mL) - 5.0 ml * 1.00 g/mL = 5.0 g % acetic acid vinegar: (mass acetic acid/mass vinegar) x 100 = - (0.629/5.00)x100 = 12.6 % AA in vinegar pKa = pH at half equivalence point = 6.0 - pH = p K a + log [base][acid] = pH = pK a pKa = -logKa = 10 -6.0 = 10 -6.0 = 1.0x10 -7.0 Conclusion The purpose of this lab was to familiarize the student with potentiometric analytical techniques using a pH meter to monitor change in pH as bases are added to solution, to determine the % acetic acid present in performed vinegar solution, and to determine the Ka of acetic acid. I determined the equivalence point using two techniques of titration, volumetric and potentiometric. Part I of the assignment seemed to run smoothly, I was careful to adhere to the instructions to avoid over-titrating my solution. The reliable acid in my NaOH solution was citric acid, and the weak acid Acetic acid was calculated in part II through the use of its dilute solution, vinegar. Moving into part II I faced immediate difficulties trying to calibrate my pH meter, I reached out to my lab partner and later had to find an online video that explained the steps a bit more clearly than those described in the manual. With an incorrect calibration, I noticed my pH meter values would jump around from 1.1 to 11 and such. After correctly calibrating, I was able to carry out my experiment. I had to patiently test the pH following every addition of the base solution to my vinegar solution. I realized, when comparing results with my lab partner, that we reached pH of 5.0 at different volumes. This I am attributing to human error of measurements using the pipets or reading the cylinder. Once the experiment was complete, I found the calculations a bit smooth, following along the example given in the handout. I did also note that my graph did not give a clear equivalence point, this point should have been a bit more obvious to identify as it is seen where a drastic change in pH occurs. I believe this is also subject to human error, my solution may have been over-diluted with either the vinegar or basic solution. Overall, I was able to identify the percentage of acetic acid present in vinegar to be 12.6 % from my calculations, which supports the statement that vinegar is a dilute solution of acetic acid. North Lake College CHEM 1412 online Potentiometric Titration p. 5