inegar? required to neutralize the acetic acid in each sample of vinegar. 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. 15. Subtract the initial volume from the final volume to calculate the volume of NaOH Average volume of NaOH required to reach endpoint in mL Convert the average volume of NaOH required to reach endpoint from mL to L I DATA 269 12.55 TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH M TRIAL 1 TRIAL 2 Initial base buret reading Final base buret reading Volume of sodium hydroxide (N2OH) required to reach endpoint in mL Volume of vinegar used O15 12.55 TRIAL 3 12:55 24.11 EXTRA TRIAL 24.11 35.89 11.78 12.05 11.56 15.00 mL 15.00 mL 15.00 mL 15.00 mL (Show calculations here.) 12.05711-56+1178 3) 27.54ML -27.54 a7.54ML X L (Show calculations here.) :0-02754L 0:087541 d ba b00 Molarity of NAOH (recorded from the standard solution bottle) 0.1 I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters Ol= required to reach the endpoint, calculate the moles of NaOH used in the titration. Show your calculations using dimensional analysis or the following equation: moles of NaOH a M = Liters of solution 0.02754L X 0.1 6-002754 moles of NaOH used to reach endpoint of titration 2. Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of acetic acid (HC,H,O,) are present in a 15.00 mL sample of diluted vinegar? moles of HC,H,O, in diluted vinegar sample 0:002754 K-11
inegar? required to neutralize the acetic acid in each sample of vinegar. 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. 15. Subtract the initial volume from the final volume to calculate the volume of NaOH Average volume of NaOH required to reach endpoint in mL Convert the average volume of NaOH required to reach endpoint from mL to L I DATA 269 12.55 TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH M TRIAL 1 TRIAL 2 Initial base buret reading Final base buret reading Volume of sodium hydroxide (N2OH) required to reach endpoint in mL Volume of vinegar used O15 12.55 TRIAL 3 12:55 24.11 EXTRA TRIAL 24.11 35.89 11.78 12.05 11.56 15.00 mL 15.00 mL 15.00 mL 15.00 mL (Show calculations here.) 12.05711-56+1178 3) 27.54ML -27.54 a7.54ML X L (Show calculations here.) :0-02754L 0:087541 d ba b00 Molarity of NAOH (recorded from the standard solution bottle) 0.1 I CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters Ol= required to reach the endpoint, calculate the moles of NaOH used in the titration. Show your calculations using dimensional analysis or the following equation: moles of NaOH a M = Liters of solution 0.02754L X 0.1 6-002754 moles of NaOH used to reach endpoint of titration 2. Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of acetic acid (HC,H,O,) are present in a 15.00 mL sample of diluted vinegar? moles of HC,H,O, in diluted vinegar sample 0:002754 K-11
World of Chemistry, 3rd edition
3rd Edition
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 63A
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