Question

I've completed the data table for part one. I am just missing the data table for part two.

Procedure IMPORTANT SAFETY CONSIDERATIONS: Avoid getting NaOH on your skin. Should you come in contact with this chemical, immediately wash with water. Review all MSDS associated with this lab. Any broken or chipped glassware should be replaced. Clean up spills immediately. Download the following tables and record all data gathered throughout Part 1 and Part 2 of the procedure: Data Analysis Tables. You can also recreate these tables and record the data in your lab notebook PART 1 - Potentiometric Titration of Acetic Acid with NaOH 1. Secure the syringe stopcock onto the syringe (if this is not already in place). 2. Quantitatively place 5.0 ml of 5% white distilled vinegar (-0.9 M Acetic Acid) in a 250 ml beaker 3. Using a small funnel, fill the syringe with the 1.0 M NaOH solution and adjust the volume so that the meniscus is at 30.0 ml. 4. Place the pH probe in the beaker and add enough water to the beaker to cover the end of the pH probe. 5. Add 4 drops of phenolphthalein indicator to the solution. 6. Hold the pH probe in middle of the beaker and collect an initial pH reading. (Take a photo) 7. Add 1.0 mL of NaOH solution; mix by swirling the mixture gently, 8. Hold the pH probe in middle of the beaker and collect a pH reading. 9. Repeat steps 1.6 and 1.7 until the downloaded table is completely filled out. You can also recreate the table in your lab notebook. (Take a photo of the last pH reading) 10. Rinse the beaker and the pH probe with tap water. The contents of the beaker may be poured down the drain. 11. Enter the following data into the calculator and graph pH versus mL of NaOH added. 12. At the endpoint you should see a rapid change in pH. Identify the two volumes on each side of the endpoint, and determine the color change of the indicator for the end point.

Answer 1

In Part-1 point 2 ---   5 mL of 0.9 M acetic acid is used

In Part-2 point 1 --- 15 mL of vinegar (nothing but acetic acid) is used

1 mol acetic acid reacts with 1 mol NaOH

from Part-1 we found volume of NaOH (sudden change in pH) = 3 mL for 5 mL of acid

hence for 15 mL of acid, NaOH will be 3 x 3 = 9 mL

total volume = 15 mL + 9 mL = 24 mL = 0.024 L

moles vinegar in 15 mL = 15 mL x 10^-3 L x 0.9 M = 0.0135 moles

Molarity = moles / liter = 0.0135 moles / 0.024 L = 0.5625 M

Answers:

Molarity of vinegar = 0.5625 M

Volume of NaOH added = 9 mL

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