Data Sheet with Sample Results Moles of acetic acid contained in 50.0 mL of Mass of sodium acetate needed to equalize the a 0.10 M sample solution: _ moles of acetic acid in sample solution: Buffer pH Study Buffer Sample initial pH: 4.75 pH after addition of 0.100 M HC1 pH after addition of 0.100 M NaOH Total volume added pH Total volume added pH 1.00 mL 4.70 1.00 mL 4.80 2.00 mL 4.65 2.00 mL 4.85 3.00 mL 4.60 3.00 mL 4.90 4.00 mL 4.54 4.00 mL 4.97 5.00 mL 4.48 5.00 mL 5.05 Diluted Buffer Sample initial pH: 4.80 pH after addition of 0.100 M HC1 pH after addition of 0.100 M NaOH Total volume added pH Total volume added pH 1.00 mL 4.50 1.00 mL 5.00 2.00 mL 4.20 2.00 mL 5.25 3.00 mL 3.35 3.00 mL 5.70 4.00 mL 2.50 4.00 mL 9.50 5.00 mL 2.20 5.00 mL 11.67 Initial pH of unbuffered HCl: 3.00 pH after addition of 1.0 mL NAOH: 10.50
Data Sheet with Sample Results Moles of acetic acid contained in 50.0 mL of Mass of sodium acetate needed to equalize the a 0.10 M sample solution: _ moles of acetic acid in sample solution: Buffer pH Study Buffer Sample initial pH: 4.75 pH after addition of 0.100 M HC1 pH after addition of 0.100 M NaOH Total volume added pH Total volume added pH 1.00 mL 4.70 1.00 mL 4.80 2.00 mL 4.65 2.00 mL 4.85 3.00 mL 4.60 3.00 mL 4.90 4.00 mL 4.54 4.00 mL 4.97 5.00 mL 4.48 5.00 mL 5.05 Diluted Buffer Sample initial pH: 4.80 pH after addition of 0.100 M HC1 pH after addition of 0.100 M NaOH Total volume added pH Total volume added pH 1.00 mL 4.50 1.00 mL 5.00 2.00 mL 4.20 2.00 mL 5.25 3.00 mL 3.35 3.00 mL 5.70 4.00 mL 2.50 4.00 mL 9.50 5.00 mL 2.20 5.00 mL 11.67 Initial pH of unbuffered HCl: 3.00 pH after addition of 1.0 mL NAOH: 10.50
Chemical Principles in the Laboratory
11th Edition
ISBN:9781305264434
Author:Emil Slowinski, Wayne C. Wolsey, Robert Rossi
Publisher:Emil Slowinski, Wayne C. Wolsey, Robert Rossi
Chapter24: The Standardization Of A Basic Solution And The Determination Of The Molar Mass Of An Acid
Section: Chapter Questions
Problem 3ASA: A 0.3012g sample of an unknown monoprotic acid requires 24.13mL of 0.0944MNaOH for neutralization to...
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I need help to calculate this data please
thanks a lot
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