Question

Please Show all work . Use attached Data and graph to answer questions

Titration of Acetic Acid: 0. IM 1) Check to make sure that the volume on If it does not read 0.00 mL, adjust it to do so before you proceed any further. your sodium hydroxide buret reads 0.00 mL 2) Use the buret that is located near the acetic acid container to dispense 25.00 mL of acetic acid solution into a clean dry 100 mL beaker. Record the molarity of this solution on your data sheet. 3) Bring the solution back to your work area. Set up the beaker so that the pH electrode can be immersed and the buret can be over it with its tip slightly below the lip of the beaker. 4) Remove the pH electrode from its storage solution, rinse and blot it and then clamp it so that the electrode is immersed in the acid solution. It's important that the pH electrode does not rest directly on the bottom of the beaker, but that it's immersed in the acid solution. This will allow you to swirl the beaker gently 5) Click the Red Circle (Preview) button in Capstone. Now double click on the first cell in the first column of the table (under V NaOH). Enter 0.00, and then click the General Chemistry II Laboratory Manual, 2015 Revision 111 Keep button at the bottom of the Capstone window. The computer will record the pH of your acid solution before you added any base. The second cell in the table under V NaOHwill now be highlighted, and a point should appear on your graph.. 8) Add about 1 mL of base to the acid and allow the mixture to stir for about 30 seconds. Double click in the highlighted cell under V NaOH, read the volume to the nearest 0.1 mL, type it into the field, and then click Keep. 9) Add base in 1 mL increments until you reach 24 mL. Don't forget to record each volume in Capstone as you've done before and then using the Keep button. 10) Now add base in 0.2 mL increments until you reach 26 mL. This helps to define the steepest part of the titration curve better. Again, make sure each volume is recorded in Capstone. 10) Continue to add base in 1 mL increments remembering to record each volume in Capstone. Do this until you see that the titration curve that is being plotted levels off and you've got 5 points in the level region of the curve after the endpoint. 11) Click the Red Square (Stop) button. Print the plot of the titration curve, and save your data. Also export the data to Excel by clicking on Export under the File menu. Remember that any data you save to the hard drive on the computer will be lost once the laptop has been shut down. You would do well to save your data to a flash drive or to an online storage location. 12) Find the Export Data command in the file menu in Capsotone and select it. Select pH vs. mL Base and click OK. Name this file and make sure that it's saved. Once again, you may want to save your data to a flash drive or an online location. 13) Wait to clean up and shut down Capstone until you're certain that your data is transferred to Excel cleanly, and that your titration curves and other plots look good. You'd do well to consult your instructor about this. Titration of Acetic Acid: mL base needed to reach the equivalence point: determined this from your titration curve Show how you Textbook Kg value: Experimental Kg value: Make sure you reference where you got the textbook Ka value. Show how you determined your experimental Kg value. Turn in your computer generated plot of pH vs. Volume from DataStudio and your Excel generated plot showing the endpoint along with your Excel spreadsheet. General Chemistry II Laboratory Manual, 20155 Revision 117 For each point shown below in the titration, record your measured pH from your data, and then calculate the pH of the solution. Show vour work for all calculations in the space provided. the pH of the acetic acid before any base was added? Calculated pH- 1) What was Measured plH Calculations: 2) What is the pH after 10.00 mL of sodium hydroxide solution are added? Measured pH=_ Calculated pH Calculations: Titration Curves 118 3) What is the pH at the equivalence point? Calculated pH Measured pH= Calculations: econd 4) What is the pH after 30.0 mL of sodium hydroxide solution are added? Calculated pH Measured pH= Calculations: mL base pH d2pH/dmL2 mL(av) dpH/dmL mL'(av) 0 1.1 2.1 0.55 0.454545 1.65 0.272727 2.65 0.444444 2.6 2.2 1.1 -0.16529 2.15 0.171717 3.1 -0.49383 4.025 0210526 -0.1 2.9 3.1 3.3 3.3 3.5 3.55 0.2 4.5 5.5 0.1 5 6 3.6 3.7 3.8 0 6.5 0.1 7.55 0.090909 8.55 0.111111 0.1 0.1 0.1 7.025 -0.00866 8.05 0.020202 9.025 8.1 3.9 9 10 -0.0117 9.5 4 4.4E-16 11 8.88E-16 -0.1 10 10.5 4.1 4.2 12 11.5 12 13 4.2 12.5 0.1 13 0.1 4.3 13.5 14 14 8.88E-16 -8.9E-16 0 17 8.88E-16 0.1 -0.1 0.1 0 0.1 0.1 4.4 14.5 15 15 16 4.5 16 15.5 0.1 4.6 16.5 17.5 17 0.1 0.2 0.1 0.2 0.2 0.3 4.7 18 4.9 18.5 19 19.5 20.5 21.5 5 20 5.2 21 5.4 5.7 22 23 23.8 23.33333 24.2 0.1 22.5 23 1.7 23.5 2 7.7 24 16 24.1 10.9 11.2 11.4 11.5 24.2 24.4 -72.5 -2.5 24.3 24.5 24.7 24.9 25.1 1.5 24.4 24.6 24.8 25 24.6 -2.5 0 0 0.5 24.8 0.5 0.5 11.6 11.7 25 25.2 25.4 25.6 25.8 25.2 4.44E-14 -2.5 0.5 11.8 11.8 11.8 11.9 12 12.1 25.3 25.4 25.6 25.8 26.2 -0.66667 0 0 0.5 0.1 0.1 0.1 25.5 25.7 25.9 0 2.5 26 26.5 27.5 28.5 29.5 30.5 27 27 28 28 0 -0.1 0.1 -0.1 12.2 12.2 12.3 29 29 30 30 0.1 31 31 31.5 12.3 32 の寸r 2 5671 F 89872&& Acelic Acid Titration curve of Sodium Hydroxide and Acetic acid 40 20 26 25.5 25 24 24.5 23 23.5 품-20 -40 -60 mLs of Sodium Hydroxide -80

Answer 1

At 24 mL, the pH of solution increases suddenly after addition of further base. Hence

mL of base needed to reach equivalence point = 24 mL

pH Versus mL NAOH 14 12 10 Equivalence point pH 4 2 0 24 25 mL 0 5 10 15 20 35 (base) mL NaOH-> 30 LO LO

Hence experimental Ka value = 6.31*10^-5

Calculation:

Half-equivalence point = 24 mL / 2 = 12 mL

pH at 12 mL = 4.2

At half-equivalence point, pH = pKa

=> 4.2 = pKa

=> Ka = 10^-4.2 = 6.31*10^-5

Hence experimental Ka value = 6.31*10^-5

#1: pH of acetic acid before any base was added:

Measured pH = 2.1

Calculated pH: Given concentration of CH₃COOH = 0.1 M

Volume of CH₃COOH = 25.00 mL

--CH₃COOH <-----> CH₃COO^-(aq) + H⁺(aq) : Ka = 1.76*10^-5 (Literature value)

I: 0.1 M ---------------- 0 ------------------ 0

C: - X ------------------- +X ---------------- +X

E: (0.1-X) --------------- X ----------------- X

Ka = 1.76*10^-5 = [CH₃COO^-(aq)]*[H⁺(aq)] / [CH₃COOH]

=> 1.76*10^-5 = X*X / (0.1-X)

Since X <<0.1, we can neglect X in denominator

=> 1.76*10^-5 = X² / 0.1

=> X = square root (0.1*1.76*10^-5)

=> X = [H⁺(aq)] = 1.33*10^-3 M

=> pH = - log(1.33*10^-3) = 2.88 (answer)