la. In an experiment to determine the amount of acetic acid present in the given volume of vinegar, following data was obtained.The chemical reaction between acetic acid(in vinegar) and sodium hydroxide(base) is given below: CH,COOH(aq) + NaOH(aq) → CH,COONA(aq) + H,0(1) At equivalence point: moles of NaOH = moles of CH;COOH Molarity of NAOH solution = MN4OH = 0.100 M Complete calculations using the following data: Trial-1 Volume of Vinegar in Flask 2.00 mL Initial reading of NaOH in the buret in mL 2.05 Final reading of NaOH in the buret in mL 12.34 Total volume of NaOH used (Show calculations) in mL. Trial-1 Total Volume of NaOH solution used (in mL) =VFinal VInitial = Total Volume of NaOH solution used(in Liters) = V NaOH Moles of NaOH = ( MNAOH VNAOH ) Moles of Acetic acid = Mass of acetic acid in vinegar in grams (Use molar mass of acetic acid) = Volume of Vinegar used: 2.00 mL Weight/volume percent of acetic acid in vinegar %(w/v) =
la. In an experiment to determine the amount of acetic acid present in the given volume of vinegar, following data was obtained.The chemical reaction between acetic acid(in vinegar) and sodium hydroxide(base) is given below: CH,COOH(aq) + NaOH(aq) → CH,COONA(aq) + H,0(1) At equivalence point: moles of NaOH = moles of CH;COOH Molarity of NAOH solution = MN4OH = 0.100 M Complete calculations using the following data: Trial-1 Volume of Vinegar in Flask 2.00 mL Initial reading of NaOH in the buret in mL 2.05 Final reading of NaOH in the buret in mL 12.34 Total volume of NaOH used (Show calculations) in mL. Trial-1 Total Volume of NaOH solution used (in mL) =VFinal VInitial = Total Volume of NaOH solution used(in Liters) = V NaOH Moles of NaOH = ( MNAOH VNAOH ) Moles of Acetic acid = Mass of acetic acid in vinegar in grams (Use molar mass of acetic acid) = Volume of Vinegar used: 2.00 mL Weight/volume percent of acetic acid in vinegar %(w/v) =
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![la. In an experiment to determine the amount of acetic acid present in the given volume of
vinegar, following data was obtained.The chemical reaction between acetic acid(in vinegar) and
sodium hydroxide(base) is given below:
CH;COOH(aq) + NaOH(aq) → CH,COONA(aq) + H,O(1)
At equivalence point: moles of NaOH = moles of CH;COOH
Molarity of NAOH solution = MN2OH = 0.100 M
Complete calculations using the following data:
Trial-1
Volume of Vinegar in Flask
2.00 mL
Initial reading of NaOH in the
2.05
buret in mL
Final reading of NaOH in the
buret in mL
12.34
Total volume of NaOH used
(Show calculations) in mL.
Trial-1
Total Volume of NaOH solution used (in mL) = VFinal
- VInitial
Total Volume of NaOH solution used(in Liters) = V NaoH =
Moles of NaOH=( MNaOH VNAOH )
Moles of Acetic acid =
Mass of acetic acid in vinegar in grams (Use molar mass of acetic acid)
Volume of Vinegar used: 2.00 mL
Weight/volume percent of acetic acid in vinegar %(w/v) =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2c00dd4-5472-4bd9-89db-0730954ee662%2F2e7ba8f4-39e9-4ff3-90d2-5be66cc50e45%2Fh6ijxgu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:la. In an experiment to determine the amount of acetic acid present in the given volume of
vinegar, following data was obtained.The chemical reaction between acetic acid(in vinegar) and
sodium hydroxide(base) is given below:
CH;COOH(aq) + NaOH(aq) → CH,COONA(aq) + H,O(1)
At equivalence point: moles of NaOH = moles of CH;COOH
Molarity of NAOH solution = MN2OH = 0.100 M
Complete calculations using the following data:
Trial-1
Volume of Vinegar in Flask
2.00 mL
Initial reading of NaOH in the
2.05
buret in mL
Final reading of NaOH in the
buret in mL
12.34
Total volume of NaOH used
(Show calculations) in mL.
Trial-1
Total Volume of NaOH solution used (in mL) = VFinal
- VInitial
Total Volume of NaOH solution used(in Liters) = V NaoH =
Moles of NaOH=( MNaOH VNAOH )
Moles of Acetic acid =
Mass of acetic acid in vinegar in grams (Use molar mass of acetic acid)
Volume of Vinegar used: 2.00 mL
Weight/volume percent of acetic acid in vinegar %(w/v) =
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