Volume vinegar used (mL)5.00 mL% from label5 %Molarity of NaOH0.0985 MTable viewList viewTitration volume measurementsTrial 1Trial 2Trial 3Inital volume NaOH (mL)0.240.840.07Final volume NaOH (mL)44.9443.0644.6244.70 mL42.22 mL44.55 mLVolume NaOH used (mL) Average volume NaOH used (mL)Average volume NaOH used (L)Moles NaOH used (from avg volume)Moles CH3COOH

The equation involved in this reaction is as follows. NaOH+CH3COOH→CH3COONa+H2O That means NaOH and CH3COOH reacts in 1:1 ratio.Now we can do the calculation part. Average volume of NaOH used mL=44.7+42.22+44.53=43.8 mLAverage volume of NaOH used L=0.0438 LMoles of NaOH used=average volume×molarity=0.0438 L×0.0985 M=0.0438 L×0.0985 mol/L=4.314×10-3 mol The NaOH and CH3COOH reacts in 1:1 ratio. So, moles of CH3COOH=4.314×10-3 mol.

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Average volume NaOH used (mL) Average volume NaOH used (L) Moles NaOH used (from avg volume) Moles CH3COOH

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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